url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://definitionwiki.com/what-is-diameter/ | 1,656,531,717,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00511.warc.gz | 247,354,068 | 17,934 | # What Is Diameter
338
What is the diameter? It’s a question asked of many people, and their answers are not always clear. The definition of a circle is a circle that has no holes in it. Diameter comes from the Latin word “dixi” meaning divided or subdivided. A circle’s diameter can be compared to the distance between any two points on the circumference of the circle.
Let’s examine the definition of what is the diameter more closely. Each circle has a circumference that is equal to its circumference as measured by the surface on which it is drawn. However, a circle with no holes in it is not a circle, but an ellipse. An ellipse is a flattened plane. To measure the area inside of an ellipse, you need to add the inner and outer radii together. The result is the area of the circle.
If you know what is the diameter, you can determine the areas of various shapes and sizes. For example, if you measure the diameter of a circle, you can find out the area of the circular object. This is because a circular object has a definite perimeter. You can also determine what is the diameter from the formula that was described above. Remember that radii are measured in degrees, and the formula uses radii of one degree to measure radii of all possible degrees.
There are actually four different formulas for measuring a circle’s circumference. The most commonly used formula is the mathematical one, which is very similar to the English definition of the measurement. The circumference of a circle is found by adding the inner or outer radii together. Another way to find out what is the diameter of a circle is to make a line connecting the points along the circumference with the starting point of the line and the outer or starting point of the radius.
To use the formula for finding out what is the diameter of a circle, you will need to know how to do a radius calculation between the points on the circle. You can do this with a radius access server. A radius access server is a computer program that allows you to connect to a remote computer through the Internet so that you can use a computer program to do calculations such as these. You can put any destination you want to calculate your radius on a radial access server and then put in the inner or outer radii of your circle.
For your calculation, you will need to determine the inner and outer radii. The inner radii refers to the radius of your circle. It’s called the inner radius because it goes all the way from the surface of the circle to its center. The outer radii, on the other hand, is what represents the circumference of the outer circle and is equal to half the inner radii.
In order to know what is the diameter, it’s a good idea to know the square of the inner rims. This tells you how many times the inner rims are equal to the outer rims. This comes from how the inner rim is the outer rim over a complete cycle of the revolution of the earth. When you divide this by the number of sides, you get the ratio of the inner to outer rims.
So, what is the diameter? Determining what is the diameter involves knowing what is the outermost point of a circle and what is the innermost point of a circle. Those numbers will help you figure out what is the diameter. This helps you make sure that you are getting the right measurements when you are working with circular objects. It also helps you see what is diameter when you’re doing measurements of other objects. | 714 | 3,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-27 | latest | en | 0.954242 |
https://similarworlds.com/7494145-I-Like-Mathematics/3869945-Predict-lottery-through-Probability-Theory | 1,638,537,225,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362879.45/warc/CC-MAIN-20211203121459-20211203151459-00368.warc.gz | 588,693,712 | 12,377 | Only logged in members can reply and interact with the post.
Join Similar Worlds today 禄
# Predict lottery through Probability Theory?
Here is a curious question for you mathematicians: Is it possible to predict a significant amount of lottery numbers (let's say 5 out of 6 so you win hundreds of thousands) using Probability Theory - or are the numbers truly so random they are really hard to predict? I would figure that perhaps using a combination of Probability Theory and Forecasting, a modest amount of success could be reached, if not a great amount.
I don't know if it's either been done already or if it's possible.
Repete61-69, M
Probability is just that probably not a definite and getting 5 out of six all at the same time , ina row is probably not going to happen often enough to say ( hey this system works , I鈥檓 going to be a billionaire, 馃帀 馃拑 )
There is one issue with your theory. (Aside from the obvious one that randomly generated numbers have no predictability and there is no reduction of odds as one is drawn. It can easily be drawn again, or not)
The problem no one thinks about is that lotteries have a percentage of the pool removed at the start. The whole pool is never going to be returned and that tilts the odds firmly against the player.
In short.. The House always wins..
CoralineF
[@621893,whowasthatmaskedman] if you have a bigger money pool then the casino then the odds are actually in your favor
ExistentiorM
[@16438,Coraline] Yep. : )
[@16438,Coraline] If you cant see the difference between only returning a percentage of the overall take to the winners, and one winner winning big, but the rest losing all, I cant help you. With any kind of lottery ticket you buy, you are not actually beying w real chance of wunning. You are actually buying a period of hope, where you might win.
CoralineF
a true random system is very hard to make in reality... but for all practical purposes in the universe it can be assumed to be truely random
CoralineF
[@469824,Existentior] its really great.. just maybe dont watch it with kids even though its 6+ rated
ExistentiorM
[@16438,Coraline] Thank goodness I don't have kids, lmao. I'll watch it tonight! : )
ExistentiorM
[@16438,Coraline] Mm, alright, so the movie Coraline is done. I think it was fun to watch and I think it has a few metaphors in it, such as: People on the outside pretend to be happy but they need someone authentic to fill their void facade. etc. That which is real is best.
And to always look out for yourself and others. And other metaphors.
59 people following
I Like Mathematics
Hope you add questions you enjoyed..
Hobbies
Group Members | 632 | 2,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-49 | latest | en | 0.952338 |
https://www.transtutors.com/questions/let-the-internal-dimensions-of-a-coaxial-capacitor-be-a-1-2-cm-b-4-cm-and-l-40-cm-th-2007562.htm | 1,619,035,804,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039550330.88/warc/CC-MAIN-20210421191857-20210421221857-00205.warc.gz | 1,143,187,698 | 13,827 | # Let the internal dimensions of a coaxial capacitor be a = 1.2 cm, b = 4 cm, and l = 40 cm. The... 1 answer below »
Let the internal dimensions of a coaxial capacitor be a = 1.2 cm, b = 4 cm, and l = 40 cm. The homogeneous material inside the capacitor has the parameters = 10−11 F/m, µ = 10−5 H/m, and σ = 10−5 S/m. If the electric field intensity is E = (106/ρ) cos 105taρ V/m, find (a) J; (b) the total conduction current Ic through the capacitor; (c) the total displacement current Id through the capacitor; (d) the ratio of the amplitude of Id to that of Ic, the quality factor of the capacitor.
VENKATALAKSHMI S
(a)
Current Density
(b)
EFI
(b)
Conduction Current
Where
from eqn1 and 2
(c)
Displacement Current
(d)
## Plagiarism Checker
Submit your documents and get free Plagiarism report
Free Plagiarism Checker
## Recent Questions in Mechanical Engineering
Looking for Something Else? Ask a Similar Question | 266 | 934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-17 | latest | en | 0.769949 |
https://greenemath.com/College_Algebra/165/Exponential-Logarithmic-InequalitiesLesson.html | 1,695,912,470,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00030.warc.gz | 308,445,561 | 7,180 | Lesson Objectives
• Demonstrate an understanding of how to solve exponential equations
• Demonstrate an understanding of how to solve logarithmic equations
• Learn how to solve exponential and logarithmic inequalities
## How to Solve Exponential and Logarithmic Inequalities
In this lesson, we will learn how to solve exponential and logarithmic inequalities. Throughout our course, we have encountered various types of inequalities and explored different strategies to solve them. Here, we will provide a general method that works for all cases. Note, the method given here is usually not the fastest but is easy to follow and removes some complexity from other methods.
### General Method for Solving an Exponential or Logarithmic Inequality
1. Identify the domain
• Remember that the argument for a logarithm is a positive real number
2. Use the addition property of inequality to change the right side into 0 and then simplify the left side of the inequality as much as possible
3. Find the critical values
• Replace the inequality symbol with an equality symbol and solve the resulting equation
• These critical values are where the function changes sign
• Critical values also occur where the function is undefined, found in step 1
4. Split the number line up into intervals based on the critical values
• Solution(s) to our related equation
• Any undefined points
5. Test a point in each interval
• Use a sign chart to keep track of the sign of the function
• Consider the critical values separately
6. The solution is the union of all intervals where the function is positive or negative, depending on the inequality sign given in the problem
Example #1: Solve each inequality. $$32^{3x}\cdot \left(\frac{1}{16}\right)^{2x - 3}> 64$$ 1) What is the domain? The set of real numbers. $$\text{domain}: \left\{x | x ∈ \mathbb{R}\right\}$$ 2.1) Change the right side into 0. Subtract 64 away from each side: $$32^{3x}\cdot \left(\frac{1}{16}\right)^{2x - 3}- 64 > 0$$ 2.2) Simplify the left side. To do this, let's write everything with a common base of 2: $$(2^5)^{3x}\cdot (2^{-4})^{2x - 3}- 2^6 > 0$$ $$2^{15x}\cdot 2^{-8x + 12}- 2^6 > 0$$ $$2^{7x + 12}- 2^6 > 0$$ 3) Find the critical values. Since the domain is the set of real numbers, we just need to solve the resulting equation (replace the inequality symbol with an equality symbol). $$2^{7x + 12}- 2^{6}=0$$ $$2^{7x + 12}=2^{6}$$ Since the base is 2 on each side, we can just set the exponents equal and solve. $$7x + 12=6$$ Subtract 12 away from each side: $$7x=-6$$ Divide both sides by 7: $$x=-\frac{6}{7}$$ 4) Split the number line up into intervals based on the critical values.
$$\left(-\infty, -\frac{6}{7}\right)$$ $$\left(-\frac{6}{7}, \infty\right)$$
5) Test a point in each interval. We can use the inequality from 2.2: $$2^{7x + 12}- 2^6 > 0$$ Pick anything less than -6/7 and plug in for x. We just want to see if the left side is greater than 0. Then pick anything greater than -6/7 and plug in for x. Again, we just want to see if the left side is greater than 0.
$$\left(-\infty, -\frac{6}{7}\right)$$ $$\left(-\frac{6}{7}, \infty\right)$$
(-)(+)
6) Our solution for x will be anything that is greater than -6/7. $$x > -\frac{6}{7}$$ Interval Notation: $$\left(-\frac{6}{7}, \infty\right)$$ Graphing the Interval:
### A faster approach for exponential inequalities with like bases
Seems like quite a bit of work for such a simple problem. You may be asking, is there a faster method? Yes, but it involves a bit of a rule that must be memorized. $$a > 1 \hspace{.1em}\text{and}\hspace{.1em}x > y$$ then: $$a^x > a^y$$ Notice that this is true for a is greater than 1. We will see in a moment why this may cause confusion and lead to a wrong answer when the base is between 0 and 1 (not including either). Let's rework the problem using this strategy. $$32^{3x}\cdot \left(\frac{1}{16}\right)^{2x - 3}> 64$$ Simplify and write each side with a common base. $$2^{7x + 12}> 2^{6}$$ You can just set 7x + 12 > 6 and solve the simple linear inequality. $$7x + 12 > 6$$ Subtract 12 away from each side: $$7x > -6$$ Divide both sides by 7: $$x > -\frac{6}{7}$$ Of course this is much quicker but let's see the issue that will come up on the next problem.
Example #2: Solve each inequality. $$\left(\frac{1}{2}\right)^x \cdot \left(\frac{1}{2}\right)^{2x + 3}> \left(\frac{1}{2}\right)^{12}$$ Let's begin with our given procedure.
1) What is the domain? The set of real numbers. $$\text{domain}: \left\{x | x ∈ \mathbb{R}\right\}$$ 2.1) Change the right side into 0. Subtract (1/2)12 away from each side: $$\left(\frac{1}{2}\right)^x \cdot \left(\frac{1}{2}\right)^{2x + 3}- \left(\frac{1}{2}\right)^{12}> 0$$ 2.2) Simplify the left side. $$\left(\frac{1}{2}\right)^{3x + 3}- \left(\frac{1}{2}\right)^{12}> 0$$ 3) Find the critical values. Since the domain is the set of real numbers, we just need to solve the resulting equation (replace the inequality symbol with an equality symbol). $$\left(\frac{1}{2}\right)^{3x + 3}- \left(\frac{1}{2}\right)^{12}=0$$ $$\left(\frac{1}{2}\right)^{3x + 3}=\left(\frac{1}{2}\right)^{12}$$ Since the base is 1/2 on each side, we can just set the exponents equal and solve. $$3x + 3=12$$ Subtract 3 away from each side: $$3x=9$$ Divide both sides by 3: $$x=3$$ 4) Split the number line up into intervals based on the critical values.
$$\left(-\infty, 3\right)$$ $$\left(3, \infty\right)$$
5) Test a point in each interval. We can use the inequality from 2.2: $$\left(\frac{1}{2}\right)^{3x + 3}- \left(\frac{1}{2}\right)^{12}> 0$$ Pick anything less than 3 and plug in for x. We just want to see if the left side is greater than 0. Then pick anything greater than 3 and plug in for x. Again, we just want to see if the left side is greater than 0.
$$\left(-\infty, 3\right)$$ $$\left(3, \infty\right)$$
(+)(-)
6) Our solution for x will be anything that is less than 3. $$x < 3$$ Interval Notation: $$\left(-\infty, 3\right)$$ Graphing the Interval: Using the faster approach we must modify our formula, otherwise, we will get the wrong answer. $$0 < a < 1 \hspace{.1em}\text{and}\hspace{.1em}x > y$$ then: $$a^x < a^y$$ Notice the flipping of the sign here. If you don't do this, you will get the wrong answer which is why the general strategy may be preferred. There is just less to remember and it always works.
The reasoning behind flipping the sign is very straightforward. Let's just think about something like 1/2. $$\left(\frac{1}{2}\right)^2=\frac{1}{4}$$ $$\left(\frac{1}{2}\right)^4=\frac{1}{16}$$ $$\left(\frac{1}{2}\right)^6=\frac{1}{64}$$ As the exponent is increasing, the result is decreasing. So if x > y, meaning the exponent is larger and we have some base that is between 0 and 1 (not including either), then a larger exponent results in a smaller number and so the sign must be flipped. Let's rework the problem using this strategy. $$\left(\frac{1}{2}\right)^x \cdot \left(\frac{1}{2}\right)^{2x + 3}> \left(\frac{1}{2}\right)^{12}$$ Simplify and write each side with a common base. $$\left(\frac{1}{2}\right)^{3x + 3}> \left(\frac{1}{2}\right)^{12}$$ You can't just set 3x + 3 > 12 and solve. You must flip the direction of the inequality symbol. $$3x + 3 > 12{\color{red}✗}\text{wrong!}$$ $$3x + 3 < 12{\color{green}✓}\text{correct!}$$ Let's solve the inequality: $$3x + 3 < 12$$ Subtract 3 from each side: $$3x < 9$$ Divide both sides by 3: $$x < 3$$ Now, we see the problems that can arise when using the shorter method. From here on out, we will just stick with the general strategy given.
### Solving Exponential Inequalities with Unlike Bases
Example #3: Solve each inequality. $$3 \cdot 7^{4x + 2}+ 9 ≤ 51$$ 1) What is the domain? The set of real numbers. $$\text{domain}: \left\{x | x ∈ \mathbb{R}\right\}$$ 2.1) Change the right side into 0. Subtract 51 away from each side: $$3 \cdot 7^{4x + 2}- 42 ≤ 0$$ 2.2) Simplify the left side. To do this, let's divide each side by 3: $$7^{4x + 2}- 14 ≤ 0$$ 3) Find the critical values. Since the domain is the set of real numbers, we just need to solve the resulting equation (replace the inequality symbol with an equality symbol). $$7^{4x + 2}- 14=0$$ Add 14 to each side: $$7^{4x + 2}=14$$ Take log base 7 of each side: $$\log_7(7^{4x + 2})=\log_7(14)$$ $$4x + 2=\log_7(14)$$ Subtract 2 away from each side: $$4x=\log_7(14) - 2$$ Divide both sides by 4: $$x=\frac{\log_7(14) - 2}{4}$$ 4) Split the number line up into intervals based on the critical values.
$$\left(-\infty, \frac{\log_7(14) - 2}{4}\right)$$ $$\left(\frac{\log_7(14) - 2}{4}, \infty\right)$$
5) Test a point in each interval. We can use the inequality from 2.2: $$7^{4x + 2}- 14 ≤ 0$$ We can just use a simple approximation of -0.16. We just need to check values on each side, so we can pick something for x such as -1, which is to the left of -0.16, and 0, which is to the right of -0.16. Again, we just want to see if the left side is less than 0.
$$\left(-\infty, \frac{\log_7(14) - 2}{4}\right)$$ $$\left(\frac{\log_7(14) - 2}{4}, \infty\right)$$
(-)(+)
6) Our solution for x will be anything that is less than or equal to about -0.16. $$x ≤ \frac{\log_7(14) - 2}{4}$$ Interval Notation: $$\left(-\infty, \frac{\log_7(14) - 2}{4}\right]$$ Graphing the Interval:
### Solving Logarithmic Inequalities
Let's look at a few problems with logarithms.
Example #4: Solve each inequality. $$\log_6(x + 1) ≥ \log_6(4 - 2x)$$ 1) What is the domain? Here, the process is more involved. Remember that the argument of a logarithm is a positive real number. $$x + 1 > 0$$ $$\text{and}$$ $$4 - 2x > 0$$ Both arguments must be positive, so we want the intersection of the two inequalities.
First inequality: $$x + 1 > 0$$ Subtract 1 away from each side: $$x > -1$$ Second inequality: $$4 - 2x > 0$$ Subtract 4 away from each side: $$-2x > -4$$ Divide both sides by -2, don't forget to flip the direction of the inequality symbol: $$x < 2$$ Now we have the following: $$x > -1$$ $$\text{and}$$ $$x < 2$$ We can use a three-part inequality to write the domain: $$\text{domain}: \{x |{-}1 < x < 2\}$$ 2.1) Change the right side into 0. Subtract log6(4 - 2x) away from each side: $$\log_6(x + 1) - \log_6(4 - 2x) ≥ 0$$ 2.2) Simplify the left side. To do this, let's use our quotient rule for logarithms. $$\log_6\left(\frac{x + 1}{-2x + 4}\right) ≥ 0$$ 3) Find the critical values. Let's solve the following equation. $$\log_6(x + 1)=\log_6(4 - 2x)$$ $$x + 1=4 - 2x$$ Add 2x to each side: $$3x + 1=4$$ Subtract 1 away from each side: $$3x=3$$ Divide both sides by 3: $$x=1$$ 4) Split the number line up into intervals based on the critical values. We will use our domain from part 1 along with our solution from the equation in part 3.
$$\left(-\infty, -1\right)$$ $$\left(-1, 1\right)$$ $$\left(1, 2\right)$$ $$\left(2, \infty\right)$$
5) Test a point in each interval. We can use the inequality from 2.2: $$\log_6\left(\frac{x + 1}{-2x + 4}\right) ≥ 0$$
$$\left(-\infty, -1\right)$$ $$\left(-1, 1\right)$$ $$\left(1, 2\right)$$ $$\left(2, \infty\right)$$
undefined(-)(+)undefined
6) Our solution for x will be anything that is between 1 and 2. We also need to include 1 here because we have a non-strict inequality. $$1 ≤ x < 2$$ Interval Notation: $$[1, 2)$$ Graphing the Interval: For this type of problem, it can be helpful to look at the graph of the original inequality. $$\log_6(x + 1) ≥ \log_6(4 - 2x)$$ $$\log_6(x + 1) - \log_6(4 - 2x)≥ 0$$ Let's graph this as: $$y=\log_6(x + 1) - \log_6(4 - 2x)$$ Here we are looking for the x-values for which the y-values are greater than or equal to 0. We can see from the graph below, this occurs when x is between 1 and 2, where 1 is included and 2 is excluded. Example #5: Solve each inequality. $$\log_4(x^2 + 8) - \log_4(6) < \log_4(2)$$ 1) What is the domain? Here, the process is more involved. Remember that the argument of a logarithm is a positive real number. $$x^2 + 8 > 0$$ The solution to this inequality is all real numbers, therefore, our domain will include all real numbers. $$\text{domain}: \left\{x | x ∈ \mathbb{R}\right\}$$ 2.1) Change the right side into 0. Subtract log4(2) away from each side: $$\log_4(x^2 + 8) - \log_4(6) - \log_4(2) < 0$$ 2.2) Simplify the left side. To do this, let's use our product and quotient rules for logarithms. $$\log_4(x^2 + 8) - (\log_4(6) + \log_4(2)) < 0$$ $$\log_4(x^2 + 8) - \log_4(12) < 0$$ $$\log_4\left(\frac{x^2 + 8}{12}\right) < 0$$ 3) Find the critical values. Let's solve the following equation. $$\log_4\left(\frac{x^2 + 8}{12}\right)=0$$ $$4^{0}=\frac{x^2 + 8}{12}$$ Switch sides, write 40 as 1: $$\frac{x^2 + 8}{12}=1$$ Multiply both sides by 12: $$x^2 + 8=12$$ Subtract 8 away from each side: $$x^2=4$$ Square root property: $$x=\pm 2$$ 4) Split the number line up into intervals based on the critical values. We will use our domain from part 1 along with our solution from the equation in part 3.
$$\left(-\infty, -2\right)$$ $$\left(-2, 2\right)$$ $$\left(2, \infty\right)$$
5) Test a point in each interval. We can use the inequality from 2.2: $$\log_4\left(\frac{x^2 + 8}{12}\right) < 0$$
$$\left(-\infty, -2\right)$$ $$\left(-2, 2\right)$$ $$\left(2, \infty\right)$$
(+)(-)(+)
6) Our solution for x will be anything that is between -2 and 2 with both being excluded. $$-2 < x < 2$$ Interval Notation: $$(-2, 2)$$ Graphing the Interval: Again, just like in the last example, a graph can be helpful. $$\log_4(x^2 + 8) - \log_4(6) < \log_4(2)$$ $$\log_4(x^2 + 8) - \log_4(6) - \log_4(2) < 0$$ Let's graph this as: $$y=\log_4(x^2 + 8) - \log_4(6) - \log_4(2)$$ Here we are looking for the x-values for which the y-values are less than 0. We can see from the graph below, this occurs when x is between -2 and 2, where both are excluded.
#### Skills Check:
Example #1
Solve each inequality.
$$\frac{27^{x - 1}}{243^{3 - 3x}}> 27^x$$
A
$$x > \frac{6}{5}$$
B
$$x < \frac{6}{5}$$
C
$$\text{No Solution}$$
D
$$-\frac{2}{3}< x < \frac{2}{3}$$
E
$$x > -1$$
Example #2
Solve each inequality.
$$\log_2(x^2 - 8x) < \log_2(3x - 18)$$
A
$$x < 9$$
B
$$x > 9$$
C
$$8 < x < 9$$
D
$$-\frac{1}{3}< x < \frac{1}{3}$$
E
$$\text{No Solution}$$
Example #3
Solve each inequality.
$$\log_3(x - 4) - \log_3(x + 3) > 2$$
A
$$x > -\frac{2}{5}$$
B
$$\text{No Solution}$$
C
$$x > \frac{1}{27}$$
D
$$-3 < x < 3$$
E
$$x > -2$$
Congrats, Your Score is 100%
Better Luck Next Time, Your Score is %
Try again? | 4,932 | 14,353 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 5.03125 | 5 | CC-MAIN-2023-40 | latest | en | 0.834197 |
http://mathhelpforum.com/math-topics/15999-quadratic-functions-maximum-minimum-problems.html | 1,481,330,707,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542851.96/warc/CC-MAIN-20161202170902-00159-ip-10-31-129-80.ec2.internal.warc.gz | 181,631,198 | 11,271 | 1. ## Quadratic Functions- Maximum and Minimum Problems
This problem is from the Addison-Wesley Mathematics 11 Text Book (Western Canada Edition)
Section 2.5 - Maximum and Minimum problems
pg. 131 Problem #16
A 30-cm piece of wire is cut in two. One piece is bent into the shape of a square. The other piece is bent into the shape of a rectangle with a length-to-width ratio of 2:1. The sum of the areas of the square and rectangle is a minimum. What are the lengths of the two pieces?
I have tried to solve most of the problem already; I just don't know how to end it. The back of the text says that the answer is 14.1cm, 15.9 cm but it could always be wrong. I would really appreciate if someone could help me Here is my attempt at it: (I’m sorry if it’s messy, I tried to do the best I could on the computer.)
Piece #1 + Piece #2 = 30cm
Piece #1 = x
Piece #2 = 30 - x
Area of the square = 1/4x2
Area of the rectangle = 2w2
6w = 30-x
w = (30-x)
6
w = 5 - (x/6)
Area of the rectangle = 2w2
= 2(5 - (x/6))2
= (10 – (x/3))2
Sum of the areas = 1/4x2 + (10 – (x/3))2
I tired this equation in my graphing calculator but came up with a minimum vertex of x=9.23 and y=69.23, which is impossible because y must be less than 30.
Thank you!
-Olivia
2. Hello, Olivia!
You made a vailiant effort, but your set-up is off . . .
A 30-cm piece of wire is cut in two.
One piece is bent into the shape of a square.
The other piece is bent into the shape of a rectangle with a length-to-width ratio of 2:1.
The sum of the areas of the square and rectangle is a minimum.
What are the lengths of the two pieces?
Let $x$ = length of wire for the rectangle.
Then $30-x$ = length of wire for the square.
The rectangle looks like this:
Code:
2k
* - - - - - - - *
| |
k | | k
| |
* - - - - - - - *
2k
The perimeter of the rectangle is: . $2k + k + 2k + k \:=\:6k$
Its wire is $x$ cm long: . $6k = x\quad\Rightarrow\quad k = \frac{x}{6}$ . [1]
The area of the rectangle is: . $A_r \:=\:(2k)(k) \:=\:2k^2$ . [2]
Substitute [1] into [2]: . $A_r \;=\;2\left(\frac{x}{6}\right)^2\quad\Rightarrow\q uad\boxed{A_r\:=\:\frac{x^2}{18}}$
The square looks like this:
Code:
* - - - - - - *
| |
| |
| | ¼(30-x)
| |
| |
* - - - - - - *
¼(30-x)
The perimeter of the square is: $30 - x$ cm,
. . so its side is: $\frac{1}{4}(30 - x)$
The area of the square is: . $A_s \;=\;\left(\frac{30-x}{4}\right)^2\quad\Rightarrow\quad\boxed{A_s\;=\; \frac{x^2 - 60x + 900}{16}}$
Hence, the total area is: . $A \;=\;\frac{x^2}{18} + \frac{x^2 - 60x + 900}{16} \;=\;\frac{1}{144}(17x^2 - 540x + 8100)$
And that is the function we must minimize . . .
3. ## Thank you
Thank you very much, I really do appreciate it | 924 | 2,804 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2016-50 | longest | en | 0.922645 |
https://conceptdraw.com/a1942c3/p4/preview/640 | 1,714,054,912,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00252.warc.gz | 150,831,108 | 2,308 | www.ConceptDraw.com
# Cone
A cone is a three-dimensional geometric shape that tapers smoothly from a flat base (frequently, though not necessarily, circular) to a point called the apex or vertex. More precisely, it is the solid figure bounded by a base in a plane and by a surface (called the lateral surface) formed by the locus of all straight line segments joining the apex to the perimeter of the base. The term "cone" sometimes refers just to the surface of this solid figure, or just to the lateral surface. The axis of a cone is the straight line (if any), passing through the apex, about which the base has a rotational symmetry. In common usage in elementary geometry, cones are assumed to be right circular, where circular means that the base is a circle and right means that the axis passes through the centre of the base at right angles to its plane. Contrasted with right cones are oblique cones, in which the axis does not pass perpendicularly through the centre of the base. In general, however, the base may be any shape and the apex may lie anywhere (though it is usually assumed that the base is bounded and therefore has finite area, and that the apex lies outside the plane of the base). A cone with a polygonal base is called a pyramid. [Cone. Wikipedia]
Cone | 276 | 1,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-18 | latest | en | 0.955747 |
https://math.answers.com/basic-math/What_is_the_greatest_common_factor_of_12_and_78 | 1,709,314,999,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475422.71/warc/CC-MAIN-20240301161412-20240301191412-00674.warc.gz | 373,726,839 | 44,926 | 0
# What is the greatest common factor of 12 and 78?
Updated: 4/28/2022
Wiki User
8y ago
The greatest common factor of 12 and 78 is 6.
Wiki User
8y ago
Wiki User
13y ago
GCF of 12 and 78 is 6.
Earn +20 pts
Q: What is the greatest common factor of 12 and 78?
Submit
Still have questions?
Related questions
### Greatest common factor of 12 and 78?
The greatest common factor of 12 and 78 is 6.
### What is the greatest common factor of 99 78 and 12?
The greatest common factor of 99, 78, and 12 is 3.
### What is the greatest common factor of 78 and 546?
The greatest common factor of 78 and 546 is 78.
The GCF is 6.
### What is the greatest common factor of 78 and 8?
The greatest common factor of 78 , 8 = 2
### What is the greatest common factor of 78 and 41?
The greatest common factor of 78 , 41 = 1
### What is the greatest common factor of 78 and 120?
The greatest common factor of 78 and 120 is 6.
The GCF is 6.
The GCF is 6.
### What is the greatest common factor of 78 and 156?
Greatest common factor of 78 and 286 is 26.
### What is the greatest common factor of 78 91?
The GCF is 13.The Greatest Common Factor (GCF) is: 13
### What is the greatest common factor of 6 78 30?
The greatest common factor of 6, 78 and 30 is 6. | 387 | 1,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-10 | latest | en | 0.909176 |
https://physics.stackexchange.com/questions/758306/imaging-a-diffraction-grating-as-it-passes-through-a-lens-focal-point | 1,701,276,286,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100112.41/warc/CC-MAIN-20231129141108-20231129171108-00314.warc.gz | 526,666,620 | 41,589 | # Imaging a diffraction grating as it passes through a lens' focal point
I am trying to understand the image created when a coherent light source is incident on a diffraction grating as it is swept through the focal point of a lens. The situation is illustrated in the figure below,
where $$P_1$$ is the object (diffraction grating) plane, located at the lens focal point. $$P_2$$ is the image plane. $$s$$ and $$s'$$ are the distances between $$P_1$$ and the (convex) lens, and $$P_2$$ and the lens, respectively. $$F$$ is the back focal plane of the lens at distance $$f$$ from the lens. $$\theta$$ is the angle of diffraction due to the grating. $$D$$ is the distance between diffraction orders, which have been focused to form bright spots in the back focal plane $$F$$.
In the case where $$P_1$$ is located at the focal point of the lens, and the lens' numerical aperture is large enough to capture sufficient diffraction orders, the image at $$P_2$$ will be a real, magnified, inverted copy of the grating. This is explained nicely at the following link: https://users.physics.ox.ac.uk/~lvovsky/471/labs/abbe.pdf
My question is, assuming that everything else remains the constant, how will the image change as $$s$$ is increased or decreased such that $$P_1$$ no longer lays at the focal point?
I understand that as $$s$$ is increased, higher diffraction orders that lay outside of the lens' numerical aperture will be filtered out, resulting in interference fringes appearing in the grating image. I also believe that there is the effect of point-spread function (PSF) blur to consider? Although I am slightly confused as to how this differs from the aforementioned filtering. Finally, what about the case where $$s$$ is decreased?
Ultimately I am trying to understand whether or not the image at $$P_2$$ can be used to determine how far $$P_1$$ is from the lens focal point.
The general result can be found in Chapter 5.2.2 of Goodman: Introduction to Fourier Optics. Goodman derives Eq.5-19 according to which, aside from a quadratic phase modulation, the focal plane distribution is still the Fourier transform of the input transparency (grating) for any position preceding the lens (In Goodman's notation $$d$$ is what you write for $$s$$). This result is modified in Eq.5-20 to include the vignetting effect of the aperture. (Goodman's book is excellent and it deserves a careful reading.) He makes the point that if you just want to measure the Fourier Transform in the back focal plane then you might as well place the transparency (gratint) right against the lens to minimize vignetting. | 603 | 2,609 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.951 |
https://mathoverflow.net/questions/319715/boundary-condition-for-elliptic-problems-and-domain-decomposition | 1,548,112,036,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583814455.32/warc/CC-MAIN-20190121213506-20190121235506-00182.warc.gz | 564,489,300 | 30,389 | # Boundary condition for elliptic problems and domain decomposition
This question is motivated by one that has been previously asked on this website: Elliptic problem on a domain split in two subdomains
Consider an open domain $$U$$ split in two non-overlapping subdomains: $$U = U_1 \cup U_2$$.
For a model case, consider a ball split in a smaller ball and an anulus.
Consider the following elliptic problem:
\begin{align*} -&\Delta u=f_1 \ &\text{ in } U_1\\ -&\Delta u =f_2 & \text{ in } U_2\\ & u=g & \text{ on } \partial U \end{align*}
• To obtain existence and uniqueness results for this problem, do we need to impose compatibility conditions at the interface between $$U_1$$ and $$U_2$$?
• What is a reference on this kind of problems?
The place where $$U_1$$ and $$U_2$$ meet is known as an interface, and so this is a Poisson interface problem which can be read about in the paper On the Existence and Uniqueness of Solutions of the Poisson Interface Problem by D. P. Squier found here.
• Yes. The question of existence is already answered because of the existence of a solution with added conditions. One is free to change the constant $K$ that appears in the added constraints so there are in fact multiple solutions and so some additional constraints are needed generally to ensure uniqueness. As for general divergence form operators, on the first page of the paper there is mention of work by Olenik in this direction. – Josiah Park Dec 30 '18 at 0:56 | 363 | 1,473 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-04 | longest | en | 0.925474 |
https://wikimili.com/en/Variational_Monte_Carlo | 1,685,304,754,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644506.21/warc/CC-MAIN-20230528182446-20230528212446-00305.warc.gz | 682,640,597 | 20,866 | # Variational Monte Carlo
Last updated
In computational physics, variational Monte Carlo (VMC) is a quantum Monte Carlo method that applies the variational method to approximate the ground state of a quantum system. [1]
## Contents
The basic building block is a generic wave function ${\displaystyle |\Psi (a)\rangle }$ depending on some parameters ${\displaystyle a}$. The optimal values of the parameters ${\displaystyle a}$ is then found upon minimizing the total energy of the system.
In particular, given the Hamiltonian ${\displaystyle {\mathcal {H}}}$, and denoting with ${\displaystyle X}$ a many-body configuration, the expectation value of the energy can be written as: [2]
${\displaystyle E(a)={\frac {\langle \Psi (a)|{\mathcal {H}}|\Psi (a)\rangle }{\langle \Psi (a)|\Psi (a)\rangle }}={\frac {\int |\Psi (X,a)|^{2}{\frac {{\mathcal {H}}\Psi (X,a)}{\Psi (X,a)}}\,dX}{\int |\Psi (X,a)|^{2}\,dX}}.}$
Following the Monte Carlo method for evaluating integrals, we can interpret ${\displaystyle {\frac {|\Psi (X,a)|^{2}}{\int |\Psi (X,a)|^{2}\,dX}}}$ as a probability distribution function, sample it, and evaluate the energy expectation value ${\displaystyle E(a)}$ as the average of the so-called local energy ${\displaystyle E_{\textrm {loc}}(X)={\frac {{\mathcal {H}}\Psi (X,a)}{\Psi (X,a)}}}$. Once ${\displaystyle E(a)}$ is known for a given set of variational parameters ${\displaystyle a}$, then optimization is performed in order to minimize the energy and obtain the best possible representation of the ground-state wave-function.
VMC is no different from any other variational method, except that the many-dimensional integrals are evaluated numerically. Monte Carlo integration is particularly crucial in this problem since the dimension of the many-body Hilbert space, comprising all the possible values of the configurations ${\displaystyle X}$, typically grows exponentially with the size of the physical system. Other approaches to the numerical evaluation of the energy expectation values would therefore, in general, limit applications to much smaller systems than those analyzable thanks to the Monte Carlo approach.
The accuracy of the method then largely depends on the choice of the variational state. The simplest choice typically corresponds to a mean-field form, where the state ${\displaystyle \Psi }$ is written as a factorization over the Hilbert space. This particularly simple form is typically not very accurate since it neglects many-body effects. One of the largest gains in accuracy over writing the wave function separably comes from the introduction of the so-called Jastrow factor. In this case the wave function is written as ${\textstyle \Psi (X)=\exp(\sum {u(r_{ij})})}$, where ${\displaystyle r_{ij}}$ is the distance between a pair of quantum particles and ${\displaystyle u(r)}$ is a variational function to be determined. With this factor, we can explicitly account for particle-particle correlation, but the many-body integral becomes unseparable, so Monte Carlo is the only way to evaluate it efficiently. In chemical systems, slightly more sophisticated versions of this factor can obtain 8090% of the correlation energy (see electronic correlation) with less than 30 parameters. In comparison, a configuration interaction calculation may require around 50,000 parameters to reach that accuracy, although it depends greatly on the particular case being considered. In addition, VMC usually scales as a small power of the number of particles in the simulation, usually something like N24 for calculation of the energy expectation value, depending on the form of the wave function.
## Wave function optimization in VMC
QMC calculations crucially depend on the quality of the trial-function, and so it is essential to have an optimized wave-function as close as possible to the ground state. The problem of function optimization is a very important research topic in numerical simulation. In QMC, in addition to the usual difficulties to find the minimum of multidimensional parametric function, the statistical noise is present in the estimate of the cost function (usually the energy), and its derivatives, required for an efficient optimization.
Different cost functions and different strategies were used to optimize a many-body trial-function. Usually three cost functions were used in QMC optimization energy, variance or a linear combination of them. The variance optimization method has the advantage that the exact wavefunction's variance is known. (Because the exact wavefunction is an eigenfunction of the Hamiltonian, the variance of the local energy is zero). This means that variance optimization is ideal in that it is bounded by below, it is positive defined and its minimum is known. Energy minimization may ultimately prove more effective, however, as different authors recently showed that the energy optimization is more effective than the variance one.
There are different motivations for this: first, usually one is interested in the lowest energy rather than in the lowest variance in both variational and diffusion Monte Carlo; second, variance optimization takes many iterations to optimize determinant parameters and often the optimization can get stuck in multiple local minimum and it suffers of the "false convergence" problem; third energy-minimized wave functions on average yield more accurate values of other expectation values than variance minimized wave functions do.
The optimization strategies can be divided into three categories. The first strategy is based on correlated sampling together with deterministic optimization methods. Even if this idea yielded very accurate results for the first-row atoms, this procedure can have problems if parameters affect the nodes, and moreover density ratio of the current and initial trial-function increases exponentially with the size of the system. In the second strategy one use a large bin to evaluate the cost function and its derivatives in such way that the noise can be neglected and deterministic methods can be used.
The third approach, is based on an iterative technique to handle directly with noise functions. The first example of these methods is the so-called Stochastic Gradient Approximation (SGA), that was used also for structure optimization. Recently an improved and faster approach of this kind was proposed the so-called Stochastic Reconfiguration (SR) method.
## VMC and deep learning
In 2017, Giuseppe Carleo and Matthias Troyer [3] used a VMC objective function to train an artificial neural network to find the ground state of a quantum mechanical system. More generally, artificial neural networks are being used as a wave function ansatz (known as neural network quantum states) in VMC frameworks for finding ground states of quantum mechanical systems. The use of neural network ansatzes for VMC has been extended to fermions, enabling electronic structure calculations that are significantly more accurate than VMC calculations which do not use neural networks. [4] [5] [6]
### General
• McMillan, W. L. (19 April 1965). "Ground State of Liquid He4". Physical Review. American Physical Society (APS). 138 (2A): A442–A451. Bibcode:1965PhRv..138..442M. doi:10.1103/physrev.138.a442. ISSN 0031-899X.
• Ceperley, D.; Chester, G. V.; Kalos, M. H. (1 September 1977). "Monte Carlo simulation of a many-fermion study". Physical Review B. American Physical Society (APS). 16 (7): 3081–3099. Bibcode:1977PhRvB..16.3081C. doi:10.1103/physrevb.16.3081. ISSN 0556-2805.
## Related Research Articles
Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It is the foundation of all quantum physics including quantum chemistry, quantum field theory, quantum technology, and quantum information science.
In quantum mechanics, the uncertainty principle is any of a variety of mathematical inequalities asserting a fundamental limit to the accuracy with which the values for certain pairs of physical quantities of a particle, such as position, x, and momentum, p, can be predicted from initial conditions.
The de Broglie–Bohm theory, also known as the pilot wave theory, Bohmian mechanics, Bohm's interpretation, and the causal interpretation, is an interpretation of quantum mechanics. In addition to the wavefunction, it also postulates an actual configuration of particles exists even when unobserved. The evolution over time of the configuration of all particles is defined by a guiding equation. The evolution of the wave function over time is given by the Schrödinger equation. The theory is named after Louis de Broglie (1892–1987) and David Bohm (1917–1992).
Loop quantum gravity (LQG) is a theory of quantum gravity, which aims to merge quantum mechanics and general relativity, incorporating matter of the Standard Model into the framework established for the pure quantum gravity case. It is an attempt to develop a quantum theory of gravity based directly on Einstein's geometric formulation rather than the treatment of gravity as a force. As a theory LQG postulates that the structure of space and time is composed of finite loops woven into an extremely fine fabric or network. These networks of loops are called spin networks. The evolution of a spin network, or spin foam, has a scale above the order of a Planck length, approximately 10−35 meters, and smaller scales are meaningless. Consequently, not just matter, but space itself, prefers an atomic structure.
In quantum mechanics, einselections, short for "environment-induced superselection", is a name coined by Wojciech H. Zurek for a process which is claimed to explain the appearance of wavefunction collapse and the emergence of classical descriptions of reality from quantum descriptions. In this approach, classicality is described as an emergent property induced in open quantum systems by their environments. Due to the interaction with the environment, the vast majority of states in the Hilbert space of a quantum open system become highly unstable due to entangling interaction with the environment, which in effect monitors selected observables of the system. After a decoherence time, which for macroscopic objects is typically many orders of magnitude shorter than any other dynamical timescale, a generic quantum state decays into an uncertain state which can be expressed as a mixture of simple pointer states. In this way the environment induces effective superselection rules. Thus, einselection precludes stable existence of pure superpositions of pointer states. These 'pointer states' are stable despite environmental interaction. The einselected states lack coherence, and therefore do not exhibit the quantum behaviours of entanglement and superposition.
In theoretical physics, the pilot wave theory, also known as Bohmian mechanics, was the first known example of a hidden-variable theory, presented by Louis de Broglie in 1927. Its more modern version, the de Broglie–Bohm theory, interprets quantum mechanics as a deterministic theory, avoiding troublesome notions such as wave–particle duality, instantaneous wave function collapse, and the paradox of Schrödinger's cat. To solve these problems, the theory is inherently nonlocal.
The Wheeler–DeWitt equation for theoretical physics and applied mathematics, is a field equation attributed to John Archibald Wheeler and Bryce DeWitt. The equation attempts to mathematically combine the ideas of quantum mechanics and general relativity, a step towards a theory of quantum gravity.
Quantum Monte Carlo encompasses a large family of computational methods whose common aim is the study of complex quantum systems. One of the major goals of these approaches is to provide a reliable solution of the quantum many-body problem. The diverse flavors of quantum Monte Carlo approaches all share the common use of the Monte Carlo method to handle the multi-dimensional integrals that arise in the different formulations of the many-body problem.
Car–Parrinello molecular dynamics or CPMD refers to either a method used in molecular dynamics or the computational chemistry software package used to implement this method.
Diffusion Monte Carlo (DMC) or diffusion quantum Monte Carlo is a quantum Monte Carlo method that uses a Green's function to solve the Schrödinger equation. DMC is potentially numerically exact, meaning that it can find the exact ground state energy within a given error for any quantum system. When actually attempting the calculation, one finds that for bosons, the algorithm scales as a polynomial with the system size, but for fermions, DMC scales exponentially with the system size. This makes exact large-scale DMC simulations for fermions impossible; however, DMC employing a clever approximation known as the fixed-node approximation can still yield very accurate results.
The Ghirardi–Rimini–Weber theory (GRW) is a spontaneous collapse theory in quantum mechanics, proposed in 1986 by Giancarlo Ghirardi, Alberto Rimini, and Tullio Weber.
In applied mathematics, the numerical sign problem is the problem of numerically evaluating the integral of a highly oscillatory function of a large number of variables. Numerical methods fail because of the near-cancellation of the positive and negative contributions to the integral. Each has to be integrated to very high precision in order for their difference to be obtained with useful accuracy.
David Matthew Ceperley is a theoretical physicist in the physics department at the University of Illinois Urbana-Champaign or UIUC. He is a world expert in the area of Quantum Monte Carlo computations, a method of calculation that is generally recognised to provide accurate quantitative results for many-body problems described by quantum mechanics.
In condensed matter physics, biexcitons are created from two free excitons.
In theoretical physics, the logarithmic Schrödinger equation is one of the nonlinear modifications of Schrödinger's equation. It is a classical wave equation with applications to extensions of quantum mechanics, quantum optics, nuclear physics, transport and diffusion phenomena, open quantum systems and information theory, effective quantum gravity and physical vacuum models and theory of superfluidity and Bose–Einstein condensation. Its relativistic version was first proposed by Gerald Rosen. It is an example of an integrable model.
The Koopman–von Neumann mechanics is a description of classical mechanics in terms of Hilbert space, introduced by Bernard Koopman and John von Neumann in 1931 and 1932, respectively.
The light front quantization of quantum field theories provides a useful alternative to ordinary equal-time quantization. In particular, it can lead to a relativistic description of bound systems in terms of quantum-mechanical wave functions. The quantization is based on the choice of light-front coordinates, where plays the role of time and the corresponding spatial coordinate is . Here, is the ordinary time, is one Cartesian coordinate, and is the speed of light. The other two Cartesian coordinates, and , are untouched and often called transverse or perpendicular, denoted by symbols of the type . The choice of the frame of reference where the time and -axis are defined can be left unspecified in an exactly soluble relativistic theory, but in practical calculations some choices may be more suitable than others.
The time-dependent variational Monte Carlo (t-VMC) method is a quantum Monte Carlo approach to study the dynamics of closed, non-relativistic quantum systems in the context of the quantum many-body problem. It is an extension of the variational Monte Carlo method, in which a time-dependent pure quantum state is encoded by some variational wave function, generally parametrized as
In nuclear physics, ab initio methods seek to describe the atomic nucleus from the bottom up by solving the non-relativistic Schrödinger equation for all constituent nucleons and the forces between them. This is done either exactly for very light nuclei or by employing certain well-controlled approximations for heavier nuclei. Ab initio methods constitute a more fundamental approach compared to e.g. the nuclear shell model. Recent progress has enabled ab initio treatment of heavier nuclei such as nickel.
Neural Network Quantum States is a general class of variational quantum states parameterized in terms of an artificial neural network. It was first introduced in 2017 by the physicists Giuseppe Carleo and Matthias Troyer to approximate wave functions of many-body quantum systems.
## References
1. Scherer, Philipp O.J. (2017). Computational Physics. Graduate Texts in Physics. Cham: Springer International Publishing. doi:10.1007/978-3-319-61088-7. ISBN 978-3-319-61087-0.
2. Kalos, Malvin H., ed. (1984). Monte Carlo Methods in Quantum Problems. Dordrecht: Springer Netherlands. doi:10.1007/978-94-009-6384-9. ISBN 978-94-009-6386-3.
3. Carleo, Giuseppe; Troyer, Matthias (2017). "Solving the Quantum Many-Body Problem with Artificial Neural Networks". Science. 355 (6325). arXiv:. doi:10.1126/science.aag2302.
4. Pfau, David; Spencer, James; Matthews, Alexander G. de G.; Foulkes, W. M. C. (2020). "Ab-initio Solution of the Many-Electron Schrödinger Equation with Deep Neural Networks". Physical Review Research. 2 (3). arXiv:. doi:10.1103/PhysRevResearch.2.033429.
5. Hermann, Jan; Schätzle, Zeno; Noé, Frank (2020). "Deep Neural Network Solution of the Electronic Schrödinger Equation". Nature Chemistry. 12. arXiv:. doi:10.1038/s41557-020-0544-y.
6. Choo, Kenny; Mezzacapo, Antonio; Carleo, Giuseppe (2020). "Fermionic Neural-Network States for Ab-initio Electronic Structure". Nature Communications. 11. arXiv:. doi:10.1038/s41467-020-15724-9. PMC . | 3,802 | 17,844 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-23 | latest | en | 0.774474 |
https://pvk.ca/posts/2/ | 1,544,682,641,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824525.29/warc/CC-MAIN-20181213054204-20181213075704-00497.warc.gz | 705,910,538 | 27,072 | # The Confidence Sequence Method: A Computer-age Test for Statistical SLOs
This post goes over some code that I pushed to github today. All the snippets below should be in the repo, which also includes C and Python code with the same structure.
I recently resumed thinking about balls and bins for hash tables. This time, I’m looking at large bins (on the order of one 2MB huge page). There are many hashing methods with solid worst-case guarantees that unfortunately query multiple uncorrelated locations; I feel like we could automatically adapt them to modern hierarchical storage (or address translation) to make them more efficient, for a small loss in density.
In theory, large enough bins can be allocated statically with a minimal waste of space. I wanted some actual non-asymptotic numbers, so I ran numerical experiments and got the following distribution of global utilisation (fill rate) when the first bin fills up.
It looks like, even with one thousand bins of thirty thousand values, we can expect almost 98% space utilisation until the first bin saturates. I want something more formal.
Could I establish something like a service level objective, “When distributing balls randomly between one thousand bins with individual capacity of thirty thousand balls, we can utilise at least 98% of the total space before a bin fills up, x% of the time?”
The natural way to compute the “x%” that makes the proposition true is to first fit a distribution on the observed data, then find out the probability mass for that distribution that lies above 98% fill rate. Fitting distributions takes a lot of judgment, and I’m not sure I trust myself that much.
Alternatively, we can observe independent identically distributed fill rates, check if they achieve 98% space utilisation, and bound the success rate for this Bernoulli process.
There are some non-trivial questions associated with this approach.
1. How do we know when to stop generating more observations… without fooling ourselves with $$p$$-hacking?
2. How can we generate something like a confidence interval for the success rate?
Thankfully, I have been sitting on a software package to compute satisfaction rate for exactly this kind of SLO-type properties, properties of the form “this indicator satisfies $PREDICATE x% of the time,” with arbitrarily bounded false positive rates. The code takes care of adaptive stopping, generates a credible interval, and spits out a report like this : we see the threshold (0.98), the empirical success rate estimate (0.993 ≫ 0.98), a credible interval for the success rate, and the shape of the probability mass for success rates. This post shows how to compute credible intervals for the Bernoulli’s success rate, how to implement a dynamic stopping criterion, and how to combine the two while compensating for multiple hypothesis testing. It also gives two examples of converting more general questions to SLO form, and answers them with the same code. # Credible intervals for the Binomial If we run the same experiment $$n$$ times, and observe $$a$$ successes ($$b = n - a$$ failures), it’s natural to ask for an estimate of the success rate $$p$$ for the underlying Bernoulli process, assuming the observations are independent and identically distributed. Intuitively, that estimate should be close to $$a / n$$, the empirical success rate, but that’s not enough. I also want something that reflects the uncertainty associated with small $$n$$, much like in the following ridge line plot, where different phrases are assigned not only a different average probability, but also a different spread. I’m looking for an interval of plausible success rates $$p$$ that responds to both the empirical success rate $$a / n$$ and the sample size $$n$$; that interval should be centered around $$a / n$$, be wide when $$n$$ is small, and become gradually tighter as $$n$$ increases. The Bayesian approach is straightforward, if we’re willing to shut up and calculate. Once we fix the underlying success rate $$p = \hat{p}$$, the conditional probability of observing $$a$$ successes and $$b$$ failures is $P((a, b) | p = \hat{p}) \sim \hat{p}\sp{a} \cdot (1 - \hat{p})\sp{b},$ where the right-hand side is a proportion1, rather than a probability. We can now apply Bayes’s theorem to invert the condition and the event. The inversion will give us the conditional probability that $$p = \hat{p}$$, given that we observed $$a$$ successes and $$b$$ successes. We only need to impose a prior distribution on the underlying rate $$p$$. For simplicity, I’ll go with the uniform $$U[0, 1]$$, i.e., every success rate is equally plausible, at first. We find $P(p = \hat{p} | (a, b)) = \frac{P((a, b) | p = \hat{p}) P(p = \hat{p})}{P(a, b)}.$ We already picked the uniform prior, $$P(p = \hat{p}) = 1\,\forall \hat{p}\in [0,1],$$ and the denominator is a constant with respect to $$\hat{p}$$. The expression simplifies to $P(p = \hat{p} | (a, b)) \sim \hat{p}\sp{a} \cdot (1 - \hat{p})\sp{b},$ or, if we normalise to obtain a probability, $P(p = \hat{p} | (a, b)) = \frac{\hat{p}\sp{a} \cdot (1 - \hat{p})\sp{b}}{\int\sb{0}\sp{1} \hat{p}\sp{a} \cdot (1 - \hat{p})\sp{b}\, d\hat{p}} = \textrm{Beta}(a+1, b+1).$ A bit of calculation, and we find that our credibility estimate for the underlying success rate follows a Beta distribution. If one is really into statistics, they can observe that the uniform prior distribution is just the $$\textrm{Beta}(1, 1)$$ distribution, and rederive that the Beta is the conjugate distribution for the Binomial distribution. For me, it suffices to observe that the distribution $$\textrm{Beta}(a+1, b+1)$$ is unimodal, does peak around $$a / (a + b)$$, and becomes tighter as the number of observations grows. In the following image, I plotted three Beta distributions, all with empirical success rate 0.9; red corresponds to $$n = 10$$ ($$a = 9$$, $$b = 1$$, $$\textrm{Beta}(10, 2)$$), black to $$n = 100$$ ($$\textrm{Beta}(91, 11)$$), and blue to $$n = 1000$$ ($$\textrm{Beta}(901, 101)$$). We calculated, and we got something that matches my intuition. Before trying to understand what it means, let’s take a detour to simply plot points from that un-normalised proportion function $$\hat{p}\sp{a} \cdot (1 - \hat{p})\sp{b}$$, on an arbitrary $$y$$ axis. Let $$\hat{p} = 0.4$$, $$a = 901$$, $$b = 101$$. Naïvely entering the expression at the REPL yields nothing useful. CL-USER> (* (expt 0.4d0 901) (expt (- 1 0.4d0) 101)) 0.0d0 The issue here is that the un-normalised proportion is so small that it underflows double floats and becomes a round zero. We can guess that the normalisation factor $$\frac{1}{\mathrm{Beta}(\cdot,\cdot)}$$ quickly grows very large, which will bring its own set of issues when we do care about the normalised probability. How can we renormalise a set of points without underflow? The usual trick to handle extremely small or large magnitudes is to work in the log domain. Rather than computing $$\hat{p}\sp{a} \cdot (1 - \hat{p})\sp{b}$$, we shall compute $\log\left[\hat{p}\sp{a} \cdot (1 - \hat{p})\sp{b}\right] = a \log\hat{p} + b \log (1 - \hat{p}).$ CL-USER> (+ (* 901 (log 0.4d0)) (* 101 (log (- 1 0.4d0)))) -877.1713374189787d0 CL-USER> (exp *) 0.0d0 That’s somewhat better: the log-domain value is not $$-\infty$$, but converting it back to a regular value still gives us 0. The $$\log$$ function is monotonic, so we can find the maximum proportion value for a set of points, and divide everything by that maximum value to get plottable points. There’s one last thing that should change: when $$x$$ is small, $$1 - x$$ will round most of $$x$$ away. Instead of (log (- 1 x)), we should use (log1p (- x)) to compute $$\log (1 + -x) = \log (1 - x)$$. Common Lisp did not standardise log1p, but SBCL does have it in internals, as a wrapper around libm. We’ll just abuse that for now. CL-USER> (defun proportion (x) (+ (* 901 (log x)) (* 101 (sb-kernel:%log1p (- x))))) PROPORTION CL-USER> (defparameter *points* (loop for i from 1 upto 19 collect (/ i 20d0))) *POINTS* CL-USER> (reduce #'max *points* :key #'proportion) -327.4909190001001d0 We have to normalise in the log domain, which is simply a subtraction: $$\log(x / y) = \log x - \log y$$. In the case above, we will subtract $$-327.49\ldots$$, or add a massive $$327.49\ldots$$ to each log proportion (i.e., multiply by $$10\sp{142}$$). The resulting values should have a reasonably non-zero range. CL-USER> (mapcar (lambda (x) (cons x (exp (- (proportion x) *)))) *points*) ((0.05d0 . 0.0d0) (0.1d0 . 0.0d0) [...] (0.35d0 . 3.443943164733533d-288) [...] (0.8d0 . 2.0682681158181894d-16) (0.85d0 . 2.6252352579425913d-5) (0.9d0 . 1.0d0) (0.95d0 . 5.65506756824607d-10)) There’s finally some signal in there. This is still just an un-normalised proportion function, not a probability density function, but that’s already useful to show the general shape of the density function, something like the following, for $$\mathrm{Beta}(901, 101)$$. Finally, we have a probability density function for the Bayesian update of our belief about the success rate after $$n$$ observations of a Bernoulli process, and we know how to compute its proportion function. Until now, I’ve carefully avoided the question of what all these computations even mean. No more (: The Bayesian view assumes that the underlying success rate (the value we’re trying to estimate) is unknown, but sampled from some distribution. In our case, we assumed a uniform distribution, i.e., that every success rate is a priori equally likely. We then observe $$n$$ outcomes (successes or failures), and assign an updated probability to each success rate. It’s like a many-world interpretation in which we assume we live in one of a set of worlds, each with a success rate sampled from the uniform distribution; after observing 900 successes and 100 failures, we’re more likely to be in a world where the success rate is 0.9 than in one where it’s 0.2. With Bayes’s theorem to formalise the update, we assign posterior probabilities to each potential success rate value. We can compute an equal-tailed credible interval from that $$\mathrm{Beta}(a+1,b+1)$$ posterior distribution by excluding the left-most values, $$[0, l)$$, such that the Beta CDF (cumulative distribution function) at $$l$$ is $$\varepsilon / 2$$, and doing the same with the right most values to cut away $$\varepsilon / 2$$ of the probability density. The CDF for $$\mathrm{Beta}(a+1,b+1)$$ at $$x$$ is the incomplete beta function, $$I\sb{x}(a+1,b+1)$$. That function is really hard to compute (this technical report detailing Algorithm 708 deploys five different evaluation strategies), so I’ll address that later. The more orthodox “frequentist” approach to confidence intervals treats the whole experiment, from data colleaction to analysis (to publication, independent of the observations 😉) as an Atlantic City algorithm: if we allow a false positive rate of $$\varepsilon$$ (e.g., $$\varepsilon=5\%$$), the experiment must return a confidence interval that includes the actual success rate (population statistic or parameter, in general) with probability $$1 - \varepsilon$$, for any actual success rate (or underlying population statistic / parameter). When the procedure fails, with probability at most $$\varepsilon$$, it is allowed to fail in an arbitrary manner. The same Atlantic City logic applies to $$p$$-values. An experiment (data collection and analysis) that accepts when the $$p$$-value is at most $$0.05$$ is an Atlantic City algorithm that returns a correct result (including “don’t know”) with probability at least $$0.95$$, and is otherwise allowed to yield any result with probability at most $$0.05$$. The $$p$$-value associated with a conclusion, e.g., “success rate is more than 0.8” (the confidence level associated with an interval) means something like “I’m pretty sure that the success rate is more than 0.8, because the odds of observing our data if that were false are small (less than 0.05).” If we set that threshold (of 0.05, in the example) ahead of time, we get an Atlantic City algorithm to determine if “the success rate is more than 0.8” with failure probability 0.05. (In practice, reporting is censored in all sorts of ways, so…) There are ways to recover a classical confidence interval, given $$n$$ observations from a Bernoulli. However, they’re pretty convoluted, and, as Jaynes argues in his note on confidence intervals, the classical approach gives values that are roughly the same2 as the Bayesian approach… so I’ll just use the Bayesian credibility interval instead. See this stackexchange post for a lot more details. # Dynamic stopping for Binomial testing The way statistics are usually deployed is that someone collects a data set, as rich as is practical, and squeezes that static data set dry for significant results. That’s exactly the setting for the credible interval computation I sketched in the previous section. When studying the properties of computer programs or systems, we can usually generate additional data on demand, given more time. The problem is knowing when it’s ok to stop wasting computer time, because we have enough data… and how to determine that without running into multiple hypothesis testing issues (ask anyone who’s run A/B tests). Here’s an example of an intuitive but completely broken dynamic stopping criterion. Let’s say we’re trying to find out if the success rate is less than or greater than 90%, and are willing to be wrong 5% of the time. We could get $$k$$ data points, run a statistical test on those data points, and stop if the data let us conclude with 95% confidence that the underlying success rate differs from 90%. Otherwise, collect $$2k$$ fresh points, run the same test; collect $$4k, \ldots, 2\sp{i}k$$ points. Eventually, we’ll have enough data. The issue is that each time we execute the statistical test that determines if we should stop, we run a 5% risk of being totally wrong. For an extreme example, if the success rate is exactly 90%, we will eventually stop, with probability 1. When we do stop, we’ll inevitably conclude that the success rate differs from 90%, and we will be wrong. The worst-case (over all underlying success rates) false positive rate is 100%, not 5%! In my experience, programmers tend to sidestep the question by wasting CPU time with a large, fixed, number of iterations… people are then less likely to run our statistical tests, since they’re so slow, and everyone loses (the other popular option is to impose a reasonable CPU budget, with error thresholds so lax we end up with a smoke test). Robbins, in Statistical Methods Related to the Law of the Iterated Logarithm, introduces a criterion that, given a threshold success rate $$p$$ and a sequence of (infinitely many!) observations from the same Bernoulli with unknown success rate parameter, will be satisfied infinitely often when $$p$$ differs from the Bernoulli’s success rate. Crucially, Robbins also bounds the false positive rate, the probability that the criterion be satisfied even once in the infinite sequence of observations if the Bernoulli’s unknown success rate is exactly equal to $$p$$. That criterion is ${n \choose a} p\sp{a} (1-p)\sp{n-a} \leq \frac{\varepsilon}{n+1},$ where $$n$$ is the number of observations, $$a$$ the number of successes, $$p$$ the threshold success rate, and $$\varepsilon$$ the error (false positive) rate. As the number of observation grows, the criterion becomes more and more stringent to maintain a bounded false positive rate over the whole infinite sequence of observations. There are similar “Confidence Sequence” results for other distributions (see, for example, this paper of Lai), but we only care about the Binomial here. More recently, Ding, Gandy, and Hahn showed that Robbins’s criterion also guarantees that, when it is satisfied, the empirical success rate ($$a/n$$) lies on the correct side of the threshold $$p$$ (same side as the actual unknown success rate) with probability $$1-\varepsilon$$. This result leads them to propose the use of Robbins’s criterion to stop Monte Carlo statistical tests, which they refer to as the Confidence Sequence Method (CSM). (defun csm-stop-p (successes failures threshold eps) "Pseudocode, this will not work on a real machine." (let ((n (+ successes failures))) (<= (* (choose n successes) (expt threshold successes) (expt (- 1 threshold) failures)) (/ eps (1+ n))))) We may call this predicate at any time with more independent and identically distributed results, and stop as soon as it returns true. The CSM is simple (it’s all in Robbins’s criterion), but still provides good guarantees. The downside is that it is conservative when we have a limit on the number of observations: the method “hedges” against the possibility of having a false positive in the infinite number of observations after the limit, observations we will never make. For computer-generated data sets, I think having a principled limit is pretty good; it’s not ideal to ask for more data than strictly necessary, but not a blocker either. In practice, there are still real obstacles to implementing the CSM on computers with finite precision (floating point) arithmetic, especially since I want to preserve the method’s theoretical guarantees (i.e., make sure rounding is one-sided to overestimate the left-hand side of the inequality). If we implement the expression well, the effect of rounding on correctness should be less than marginal. However, I don’t want to be stuck wondering if my bad results are due to known approximation errors in the method, rather than errors in the code. Moreover, if we do have a tight expression with little rounding errors, adjusting it to make the errors one-sided should have almost no impact. That seems like a good trade-off to me, especially if I’m going to use the CSM semi-automatically, in continuous integration scripts, for example. One look at csm-stop-p shows we’ll have the same problem we had with the proportion function for the Beta distribution: we’re multiplying very small and very large values. We’ll apply the same fix: work in the log domain and exploit $$\log$$’s monotonicity. ${n \choose a} p\sp{a} (1-p)\sp{n-a} \leq \frac{\varepsilon}{n+1}$ becomes $\log {n \choose a} + a \log p + (n-a)\log (1-p) \leq \log\varepsilon -\log(n+1),$ or, after some more expansions, and with $$b = n - a$$, $\log n! - \log a! - \log b! + a \log p + b \log(1 - p) + \log(n+1) \leq \log\varepsilon.$ The new obstacle is computing the factorial $$x!$$, or the log-factorial $$\log x!$$. We shouldn’t compute the factorial iteratively: otherwise, we could spend more time in the stopping criterion than in the data generation subroutine. Robbins has another useful result for us: $\sqrt{2\pi} n\sp{n + ½} \exp(-n) \exp\left(\frac{1}{12n+1}\right) < n! < \sqrt{2\pi} n\sp{n + ½} \exp(-n) \exp\left(\frac{1}{12n}\right),$ or, in the log domain, $\log\sqrt{2\pi} + \left(n + \frac{1}{2}\right)\log n -n + \frac{1}{12n+1} < \log n! < \log\sqrt{2\pi} + \left(n + \frac{1}{2}\right)\log n -n +\frac{1}{12n}.$ This double inequality gives us a way to over-approximate $$\log {n \choose a} = \log \frac{n!}{a! b!} = \log n! - \log a! - \log b!,$$ where $$b = n - a$$: $\log {n \choose a} < -\log\sqrt{2\pi} + \left(n + \frac{1}{2}\right)\log n -n +\frac{1}{12n} - \left(a + \frac{1}{2}\right)\log a +a - \frac{1}{12a+1} - \left(b + \frac{1}{2}\right)\log b +b - \frac{1}{12b+1},$ where the right-most expression in Robbins’s double inequality replaces $$\log n!$$, which must be over-approximated, and the left-most $$\log a!$$ and $$\log b!$$, which must be under-approximated. Robbins’s approximation works well for us because, it is one-sided, and guarantees that the (relative) error in $$n!$$, $$\frac{\exp\left(\frac{1}{12n}\right) - \exp\left(\frac{1}{12n+1}\right)}{n!},$$ is small, even for small values like $$n = 5$$ (error $$< 0.0023\%$$), and decreases with $$n$$: as we perform more trials, the approximation is increasingly accurate, thus less likely to spuriously prevent us from stopping. Now that we have a conservative approximation of Robbins’s criterion that only needs the four arithmetic operations and logarithms (and log1p), we can implement it on a real computer. The only challenge left is regular floating point arithmetic stuff: if rounding must occur, we must make sure it is in a safe (conservative) direction for our predicate. Hardware usually lets us manipulate the rounding mode to force floating point arithmetic operations to round up or down, instead of the usual round to even. However, that tends to be slow, so most language (implementations) don’t support changing the rounding mode, or do so badly… which leaves us in a multi-decade hardware/software co-evolution Catch-22. I could think hard and derive tight bounds on the round-off error, but I’d rather apply a bit of brute force. IEEE-754 compliant implementations must round the four basic operations correctly. This means that $$z = x \oplus y$$ is at most half a ULP away from $$x + y,$$ and thus either $$z = x \oplus y \geq x + y,$$ or the next floating point value after $$z,$$ $$z^\prime \geq x + y$$. We can find this “next value” portably in Common Lisp, with decode-float/scale-float, and some hand-waving for denormals. (defun next (x &optional (delta 1)) "Increment x by delta ULPs. Very conservative for small (0/denormalised) values." (declare (type double-float x) (type unsigned-byte delta)) (let* ((exponent (nth-value 1 (decode-float x))) (ulp (max (scale-float double-float-epsilon exponent) least-positive-normalized-double-float))) (+ x (* delta ulp)))) I prefer to manipulate IEEE-754 bits directly. That’s theoretically not portable, but the platforms I care about make sure we can treat floats as sign-magnitude integers. CL-USER> (double-float-bits pi) 4614256656552045848 CL-USER> (double-float-bits (- pi)) -4614256656552045849 The two’s complement value for pi is one less than (- (double-float-bits pi)) because two’s complement does not support signed zeros. CL-USER> (eql 0 (- 0)) T CL-USER> (eql 0d0 (- 0d0)) NIL CL-USER> (double-float-bits 0d0) 0 CL-USER> (double-float-bits -0d0) -1 We can quickly check that the round trip from float to integer and back is an identity. CL-USER> (eql pi (bits-double-float (double-float-bits pi))) T CL-USER> (eql (- pi) (bits-double-float (double-float-bits (- pi)))) T CL-USER> (eql 0d0 (bits-double-float (double-float-bits 0d0))) T CL-USER> (eql -0d0 (bits-double-float (double-float-bits -0d0))) T We can also check that incrementing or decrementing the integer representation does increase or decrease the floating point value. CL-USER> (< (bits-double-float (1- (double-float-bits pi))) pi) T CL-USER> (< (bits-double-float (1- (double-float-bits (- pi)))) (- pi)) T CL-USER> (bits-double-float (1- (double-float-bits 0d0))) -0.0d0 CL-USER> (bits-double-float (1+ (double-float-bits -0d0))) 0.0d0 CL-USER> (bits-double-float (1+ (double-float-bits 0d0))) 4.9406564584124654d-324 CL-USER> (bits-double-float (1- (double-float-bits -0d0))) -4.9406564584124654d-324 The code doesn’t handle special values like infinities or NaNs, but that’s out of scope for the CSM criterion anyway. That’s all we need to nudge the result of the four operations to guarantee an over- or under- approximation of the real value. We can also look at the documentation for our libm (e.g., for GNU libm) to find error bounds on functions like log; GNU claims their log is never off by more than 3 ULP. We can round up to the fourth next floating point value to obtain a conservative upper bound on $$\log x$$. I could go ahead and use the building blocks above (ULP nudging for directed rounding) to directly implement Robbins’s criterion, $\log {n \choose a} + a \log p + b\log (1-p) + \log(n+1) \leq \log\varepsilon,$ with Robbins’s factorial approximation, $\log {n \choose a} < -\log\sqrt{2\pi} + \left(n + \frac{1}{2}\right)\log n -n +\frac{1}{12n} - \left(a + \frac{1}{2}\right)\log a +a - \frac{1}{12a+1} - \left(b + \frac{1}{2}\right)\log b +b - \frac{1}{12b+1}.$ However, even in the log domain, there’s a lot of cancellation: we’re taking the difference of relatively large numbers to find a small result. It’s possible to avoid that by re-associating some of the terms above, e.g., for $$a$$: $-\left(a + \frac{1}{2}\right) \log a + a - a \log p = -\frac{\log a}{2} + a (-\log a + 1 - \log p).$ Instead, I’ll just brute force things (again) with Kahan summation. Shewchuk’s presentation in Adaptive Precision Floating-Point Arithmetic and Fast Robust Geometric Predicates highlights how the only step where we may lose precision to rounding is when we add the current compensation term to the new summand. We can implement Kahan summation with directed rounding in only that one place: all the other operations are exact! We need one last thing to implement $$\log {n \choose a}$$, and then Robbins’s confidence sequence: a safely rounded floating-point value approximation of $$-\log \sqrt{2 \pi}$$. I precomputed one with computable-reals: CL-USER> (computable-reals:-r (computable-reals:log-r (computable-reals:sqrt-r computable-reals:+2pi-r+))) -0.91893853320467274178... CL-USER> (computable-reals:ceiling-r (computable-reals:*r * (ash 1 53))) -8277062471433908 -0.65067431749790398594... CL-USER> (* -8277062471433908 (expt 2d0 -53)) -0.9189385332046727d0 CL-USER> (computable-reals:-r (rational *) ***) +0.00000000000000007224... We can safely replace $$-\log\sqrt{2\pi}$$ with -0.9189385332046727d0, or, equivalently, (scale-float -8277062471433908.0d0 -53), for an upper bound. If we wanted a lower bound, we could decrement the integer significand by one. We can quickly check against an exact implementation with computable-reals and a brute force factorial. CL-USER> (defun cr-log-choose (n s) (computable-reals:-r (computable-reals:log-r (alexandria:factorial n)) (computable-reals:log-r (alexandria:factorial s)) (computable-reals:log-r (alexandria:factorial (- n s))))) CR-LOG-CHOOSE CL-USER> (computable-reals:-r (rational (robbins-log-choose 10 5)) (cr-log-choose 10 5)) +0.00050526703375914436... CL-USER> (computable-reals:-r (rational (robbins-log-choose 1000 500)) (cr-log-choose 1000 500)) +0.00000005551513197557... CL-USER> (computable-reals:-r (rational (robbins-log-choose 1000 5)) (cr-log-choose 1000 5)) +0.00025125559085509706... That’s not obviously broken: the error is pretty small, and always positive. Given a function to over-approximate log-choose, the Confidence Sequence Method’s stopping criterion is straightforward. The other, much harder, part is computing credible (Bayesian) intervals for the Beta distribution. I won’t go over the code, but the basic strategy is to invert the CDF, a monotonic function, by bisection3, and to assume we’re looking for improbable ($$\mathrm{cdf} < 0.5$$) thresholds. This assumption lets us pick a simple hypergeometric series that is normally useless, but converges well for $$x$$ that correspond to such small cumulative probabilities; when the series converges too slowly, it’s always conservative to assume that $$x$$ is too central (not extreme enough). That’s all we need to demo the code. Looking at the distribution of fill rates for the 1000 bins @ 30K ball/bin facet in it looks like we almost always hit at least 97.5% global density, let’s say with probability at least 98%. We can ask the CSM to tell us when we have enough data to confirm or disprove that hypothesis, with a 0.1% false positive rate. Instead of generating more data on demand, I’ll keep things simple and prepopulate a list with new independently observed fill rates. CL-USER> (defparameter *observations* '(0.978518900 0.984687300 0.983160833 [...])) CL-USER> (defun test (n) (let ((count (count-if (lambda (x) (>= x 0.975)) *observations* :end n))) (csm:csm n 0.98d0 count (log 0.001d0)))) CL-USER> (test 10) NIL 2.1958681996231784d0 CL-USER> (test 100) NIL 2.5948497850893184d0 CL-USER> (test 1000) NIL -3.0115331544604658d0 CL-USER> (test 2000) NIL -4.190687115879456d0 CL-USER> (test 4000) T -17.238559826956475d0 We can also use the inverse Beta CDF to get a 99.9% credible interval. After 4000 trials, we found 3972 successes. CL-USER> (count-if (lambda (x) (>= x 0.975)) *observations* :end 4000) 3972 These values give us the following lower and upper bounds on the 99.9% CI. CL-USER> (csm:beta-icdf 3972 (- 4000 3972) 0.001d0) 0.9882119750976562d0 1.515197753898523d-5 CL-USER> (csm:beta-icdf 3972 (- 4000 3972) 0.001d0 t) 0.9963832682169742d0 2.0372679238045424d-13 And we can even re-use and extend the Beta proportion code from earlier to generate this embeddable SVG report. There’s one small problem with the sample usage above: if we compute the stopping criterion with a false positive rate of 0.1%, and do the same for each end of the credible interval, our total false positive (error) rate might actually be 0.3%! The next section will address that, and the equally important problem of estimating power. # Monte Carlo power estimation It’s not always practical to generate data forever. For example, we might want to bound the number of iterations we’re willing to waste in an automated testing script. When there is a bound on the sample size, the CSM is still correct, just conservative. We would then like to know the probability that the CSM will stop successfully when the underlying success rate differs from the threshold rate $$p$$ (alpha in the code). The problem here is that, for any bounded number of iterations, we can come up with an underlying success rate so close to $$p$$ (but still different) that the CSM can’t reliably distinguish between the two. If we want to be able to guarantee any termination rate, we need two thresholds: the CSM will stop whenever it’s likely that the underlying success rate differs from either of them. The hardest probability to distinguish from both thresholds is close to the midpoint between them. With two thresholds and the credible interval, we’re running three tests in parallel. I’ll apply a Bonferroni correction, and use $$\varepsilon / 3$$ for each of the two CSM tests, and $$\varepsilon / 6$$ for each end of the CI. That logic is encapsulated in csm-driver. We only have to pass a success value generator function to the driver. In our case, the generator is itself a call to csm-driver, with fixed thresholds (e.g., 96% and 98%), and a Bernoulli sampler (e.g., return T with probability 97%). We can see if the driver returns successfully and correctly at each invocation of the generator function, with the parameters we would use in production, and recursively compute an estimate for that procedure’s success rate with CSM. The following expression simulates a CSM procedure with thresholds at 96% and 98%, the (usually unknown) underlying success rate in the middle, at 97%, a false positive rate of at most 0.1%, and an iteration limit of ten thousand trials. We pass that simulation’s result to csm-driver, and ask whether the simulation’s success rate differs from 99%, while allowing one in a million false positives. We find that yes, we can expect the 96%/98%/0.1% false positive/10K iterations setup to succeed more than 99% of the time. The code above is available as csm-power, with a tighter outer false positive rate of 1e-9. If we only allow 1000 iterations, csm-power quickly tells us that, with one CSM success in 100 attempts, we can expect the CSM success rate to be less than 99%. CL-USER> (csm:csm-power 0.97d0 0.96d0 1000 :alpha-hi 0.98d0 :eps 1d-3 :stream *standard-output*) 1 0.000e+0 1.250e-10 10.000e-1 1.699e+0 10 0.000e+0 0.000e+0 8.660e-1 1.896e+1 20 0.000e+0 0.000e+0 6.511e-1 3.868e+1 30 0.000e+0 0.000e+0 5.099e-1 5.851e+1 40 2.500e-2 5.518e-7 4.659e-1 7.479e+1 50 2.000e-2 4.425e-7 3.952e-1 9.460e+1 60 1.667e-2 3.694e-7 3.427e-1 1.144e+2 70 1.429e-2 3.170e-7 3.024e-1 1.343e+2 80 1.250e-2 2.776e-7 2.705e-1 1.542e+2 90 1.111e-2 2.469e-7 2.446e-1 1.741e+2 100 1.000e-2 2.223e-7 2.232e-1 1.940e+2 100 iterations, 1 successes (false positive rate < 1.000000e-9) success rate p ~ 1.000000e-2 confidence interval [2.223495e-7, 0.223213 ] p < 0.990000 max inner iteration count: 816 T T 0.01d0 100 1 2.2234953205868331d-7 0.22321314110840665d0 # SLO-ify all the things with this Exact test Until now, I’ve only used the Confidence Sequence Method (CSM) for Monte Carlo simulation of phenomena that are naturally seen as boolean success / failures processes. We can apply the same CSM to implement an exact test for null hypothesis testing, with a bit of resampling magic. Looking back at the balls and bins grid, the average fill rate seems to be slightly worse for 100 bins @ 60K ball/bin, than for 1000 bins @ 128K ball/bin. How can we test that with the CSM? First, we should get a fresh dataset for the two setups we wish to compare. CL-USER> (defparameter *100-60k* #(0.988110167 0.990352500 0.989940667 0.991670667 [...])) CL-USER> (defparameter *1000-128k* #(0.991456281 0.991559578 0.990970109 0.990425805 [...])) CL-USER> (alexandria:mean *100-60k*) 0.9897938 CL-USER> (alexandria:mean *1000-128k*) 0.9909645 CL-USER> (- * **) 0.0011706948 The mean for 1000 bins @ 128K ball/bin is slightly higher than that for 100 bins @ 60k ball/bin. We will now simulate the null hypothesis (in our case, that the distributions for the two setups are identical), and determine how rarely we observe a difference of 0.00117 in means. I only use a null hypothesis where the distributions are identical for simplicity; we could use the same resampling procedure to simulate distributions that, e.g., have identical shapes, but one is shifted right of the other. In order to simulate our null hypothesis, we want to be as close to the test we performed as possible, with the only difference being that we generate data by reshuffling from our observations. CL-USER> (defparameter *resampling-data* (concatenate 'simple-vector *100-60k* *1000-128k*)) *RESAMPLING-DATA* CL-USER> (length *100-60k*) 10000 CL-USER> (length *1000-128k*) 10000 The two observation vectors have the same size, 10000 values; in general, that’s not always the case, and we must make sure to replicate the sample sizes in the simulation. We’ll generate our simulated observations by shuffling the *resampling-data* vector, and splitting it in two subvectors of ten thousand elements. CL-USER> (let* ((shuffled (alexandria:shuffle *resampling-data*)) (60k (subseq shuffled 0 10000)) (128k (subseq shuffled 10000))) (- (alexandria:mean 128k) (alexandria:mean 60k))) 6.2584877e-6 We’ll convert that to a truth value by comparing the difference of simulated means with the difference we observed in our real data, $$0.00117\ldots$$, and declare success when the simulated difference is at least as large as the actual one. This approach gives us a one-sided test; a two-sided test would compare the absolute values of the differences. CL-USER> (csm:csm-driver (lambda (_) (declare (ignore _)) (let* ((shuffled (alexandria:shuffle *resampling-data*)) (60k (subseq shuffled 0 10000)) (128k (subseq shuffled 10000))) (>= (- (alexandria:mean 128k) (alexandria:mean 60k)) 0.0011706948))) 0.005 1d-9 :alpha-hi 0.01 :stream *standard-output*) 1 0.000e+0 7.761e-11 10.000e-1 -2.967e-1 10 0.000e+0 0.000e+0 8.709e-1 -9.977e-1 20 0.000e+0 0.000e+0 6.577e-1 -1.235e+0 30 0.000e+0 0.000e+0 5.163e-1 -1.360e+0 40 0.000e+0 0.000e+0 4.226e-1 -1.438e+0 50 0.000e+0 0.000e+0 3.569e-1 -1.489e+0 60 0.000e+0 0.000e+0 3.086e-1 -1.523e+0 70 0.000e+0 0.000e+0 2.718e-1 -1.546e+0 80 0.000e+0 0.000e+0 2.427e-1 -1.559e+0 90 0.000e+0 0.000e+0 2.192e-1 -1.566e+0 100 0.000e+0 0.000e+0 1.998e-1 -1.568e+0 200 0.000e+0 0.000e+0 1.060e-1 -1.430e+0 300 0.000e+0 0.000e+0 7.207e-2 -1.169e+0 400 0.000e+0 0.000e+0 5.460e-2 -8.572e-1 500 0.000e+0 0.000e+0 4.395e-2 -5.174e-1 600 0.000e+0 0.000e+0 3.677e-2 -1.600e-1 700 0.000e+0 0.000e+0 3.161e-2 2.096e-1 800 0.000e+0 0.000e+0 2.772e-2 5.882e-1 900 0.000e+0 0.000e+0 2.468e-2 9.736e-1 1000 0.000e+0 0.000e+0 2.224e-2 1.364e+0 2000 0.000e+0 0.000e+0 1.119e-2 5.428e+0 NIL T 0.0d0 2967 0 0.0d0 0.007557510165262294d0 We tried to replicate the difference 2967 times, and did not succeed even once. The CSM stopped us there, and we find a CI for the probability of observing our difference, under the null hypothesis, of [0, 0.007557] (i.e., $$p < 0.01$$). Or, for a graphical summary, . We can also test for a lower $$p$$-value by changing the thresholds and running the simulation more times (around thirty thousand iterations for $$p < 0.001$$). This experiment lets us conclude that the difference in mean fill rate between 100 bins @ 60K ball/bin and 1000 @ 128K is probably not due to chance: it’s unlikely that we observed an expected difference between data sampled from the same distribution. In other words, “I’m confident that the fill rate for 1000 bins @ 128K ball/bin is greater than for 100 bins @ 60K ball/bins, because it would be highly unlikely to observe a difference in means that extreme if they had the same distribution ($$p < 0.01$$)”. In general, we can use this exact test when we have two sets of observations, $$X\sb{0}$$ and $$Y\sb{0}$$, and a statistic $$f\sb{0} = f(X\sb{0}, Y\sb{0})$$, where $$f$$ is a pure function (the extension to three or more sets of observations is straightforward). The test lets us determine the likelihood of observing $$f(X, Y) \geq f\sb{0}$$ (we could also test for $$f(X, Y) \leq f\sb{0}$$), if $$X$$ and $$Y$$ were taken from similar distributions, modulo simple transformations (e.g., $$X$$’s mean is shifted compared to $$Y$$’s, or the latter’s variance is double the former’s). We answer that question by repeatedly sampling without replacement from $$X\sb{0} \cup Y\sb{0}$$ to generate $$X\sb{i}$$ and $$Y\sb{i}$$, such that $$|X\sb{i}| = |X\sb{0}|$$ and $$|Y\sb{i}| = |Y\sb{0}|$$ (e.g., by shuffling a vector and splitting it in two). We can apply any simple transformation here (e.g., increment every value in $$Y\sb{i}$$ by $$\Delta$$ to shift its mean by $$\Delta$$). Finally, we check if $$f(X\sb{i}, Y\sb{i}) \geq f\sb{0} = f(X\sb{0}, Y\sb{0})$$; if so, we return success for this iteration, otherwise failure. The loop above is a Bernoulli process that generates independent, identically distributed (assuming the random sampling is correct) truth values, and its success rate is equal to the probability of observing a value for $$f$$ “as extreme” as $$f\sb{0}$$ under the null hypothesis. We use the CSM with false positive rate $$\varepsilon$$ to know when to stop generating more values and compute a credible interval for the probability under the null hypothesis. If that probability is low (less than some predetermined threshold, like $$\alpha = 0.001$$), we infer that the null hypothesis does not hold, and declare that the difference in our sample data points at a real difference in distributions. If we do everything correctly (cough), we will have implemented an Atlantic City procedure that fails with probability $$\alpha + \varepsilon$$. Personally, I often just set the threshold and the false positive rate unreasonably low and handwave some Bayes. # That’s all! I pushed the code above, and much more, to github, in Common Lisp, C, and Python (probably Py3, although 2.7 might work). Hopefully anyone can run with the code and use it to test, not only SLO-type properties, but also answer more general questions, with an exact test. I’d love to have ideas or contributions on the usability front. I have some throw-away code in attic/, which I used to generate the SVG in this post, but it’s not great. I also feel like I can do something to make it easier to stick the logic in shell scripts and continuous testing pipelines. When I passed around a first draft for this post, many readers that could have used the CSM got stuck on the process of moving from mathematical expressions to computer code; not just how to do it, but, more fundamentally, why we can’t just transliterate Greek to C or CL. I hope this revised post is clearer. Also, I hope it’s clear that the reason I care so much about not introducing false positive via rounding isn’t that I believe they’re likely to make a difference, but simply that I want peace of mind with respect to numerical issues; I really don’t want to be debugging some issue in my tests and have to wonder if it’s all just caused by numerical errors. The reason I care so much about making sure users can understand what the CSM codes does (and why it does what it does) is that I strongly believe we should minimise dependencies whose inner working we’re unable to (legally) explore. Every abstraction leaks, and leakage is particularly frequent in failure situations. We may not need to understand magic if everything works fine, but, everything breaks eventually, and that’s when expertise is most useful. When shit’s on fire, we must be able to break the abstraction and understand how the magic works, and how it fails. This post only tests ideal SLO-type properties (and regular null hypothesis tests translated to SLO properties), properties of the form “I claim that this indicator satisfies$PREDICATE x% of the time, with false positive rate y%” where the indicator’s values are independent and identically distributed.
The last assumption is rarely truly satisfied in practice. I’ve seen an interesting choice, where the service level objective is defined in terms of a sample of production requests, which can replayed, shuffled, etc. to ensure i.i.d.-ness. If the nature of the traffic changes abruptly, the SLO may not be representative of behaviour in production; but, then again, how could the service provider have guessed the change was about to happen? I like this approach because it is amenable to predictive statistical analysis, and incentivises communication between service users and providers, rather than users assuming the service will gracefully handle radically new crap being thrown at it.
Even if we have a representative sample of production, it’s not true that the service level indicators for individual requests are distributed identically. There’s an easy fix for the CSM and our credible intervals: generate i.i.d. sets of requests by resampling (e.g., shuffle the requests sample) and count successes and failures for individual requests, but only test for CSM termination after each resampled set.
On a more general note, I see the Binomial and Exact tests as instances of a general pattern that avoids intuitive functional decompositions that create subproblems that are harder to solve than the original problem. For example, instead of trying to directly determine how frequently the SLI satisfies some threshold, it’s natural to first fit a distribution on the SLI, and then compute percentiles on that distribution. Automatically fitting an arbitrary distribution is hard, especially with the weird outliers computer systems spit out. Reducing to a Bernoulli process before applying statistics is much simpler. Similarly, rather than coming up with analytical distributions in the Exact test, we brute-force the problem by resampling from the empirical data. I have more examples from online control systems… I guess the moral is to be wary of decompositions where internal subcomponents generate intermediate values that are richer in information than the final output.
Thank you Jacob, Ruchir, Barkley, and Joonas for all the editing and restructuring comments.
1. Proportions are unscaled probabilities that don’t have to sum or integrate to 1. Using proportions instead of probabilities tends to make calculations simpler, and we can always get a probability back by rescaling a proportion by the inverse of its integral.
2. Instead of a $$\mathrm{Beta}(a+1, b+1)$$, they tend to bound with a $$\mathrm{Beta}(a, b)$$. The difference is marginal for double-digit $$n$$.
3. I used the bisection method instead of more sophisticated ones with better convergence, like Newton’s method or the derivative-free Secant method, because bisection already adds one bit of precision per iteration, only needs a predicate that returns “too high” or “too low,” and is easily tweaked to be conservative when the predicate declines to return an answer. | 12,249 | 44,226 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-51 | longest | en | 0.905929 |
https://www.coursehero.com/file/1656995/HW-8/ | 1,524,277,931,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944851.23/warc/CC-MAIN-20180421012725-20180421032725-00322.warc.gz | 769,336,121 | 27,495 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
# HW 8 - _Rkfi flgngLEM 5-3.1 The components of plane...
This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: __Rkfi flgngLEM 5-3.1; The components of plane stress acting on an element of a bar , SLbjiected to torsion are shown in Fig. 5-7. Assuming the stress 1,, to be known. use Mohr’s circle to determine the principal ‘ st: asses and the maximum in-ptane shear scess and show them ' acting on properly oriented elements. i flLQTION: The center of the circle is at : i i Oat-0', 0',+0 a; i a": = =— i 2 2 2 Tc=0 MPa Th 3 radius of the circle is: r=V(ax_dc)2+(fxy)2 =V(_0'x“0)2+02=0'1 T We see from Mohr’s circle that: ANS 01:55: 0'2 =0 Irwl=crx i E; V i ‘3‘ The components of plane stress acting on an element of a bar subjected to torsion are shown in Fig. 5-8. Assuming the stress :1, to be known. use Mohr’s circle to determine the principal stresses and the maximum in-plane shear stress and show them ‘ i acti '19 on properly oriented elements. i i ' seems: . 1 The center of the circle is at : Q at =JX+Uy =_0_+_Q=0 ‘ 2 2 " rc = 1,, O I; Fhe radius of the circle is: ll | H r=‘i(crJr ~a-c)2 +(r‘1tv)2 =1i(0-—0)2 +33? =er I lie see from Mohr’s circle that: P Ii {NS J] = T”, 0'2 2—“), [TMAXI = 1-") —-—- ’i‘ “‘ 271 III ___________—__—____.___—] EBQW - I F or tr e state of plane stress c. = 8 ksi, 0,, = 6 ksi, and 11, = - 6 ksi, 3e Mohr’s circle to determine the absolute maximum shear stress. 3 ;TRATEGY: Use Mohr‘s circle to determine the principal stresses 31d then determine the absolute maximum shear stress from the a <pressions (5-24)} O (ksi) § fllTlON: . o- + o- ' - T we center of the circle is located at: ' 0-6 = ’_2.L = w = 71m O (6.6) 1,.y = 0 he radius of the circle is: r = (8161' — 7131'): + (* 61602 = 6.081(51' 17-06;) T 13 principal stresses are: 01 = etc + r = 7 ksi + 6.08 ksi = 13.08 ksi 52 = cc - r = T ksi - 6.08 ksi = 0.92 ksi L sing equations (5.24) to determine the absolute maximum shear stress: 13.08ks‘i 0.92k51‘ 2 6.54ksi — (0.46131) 2 0'1 0'2 2 a, —o'. 2 = 6.54161? = 3.04ksi = 0.46131' ANS WSW=654 k§i ER_0I3LEM 5-3.;1 For the state of plane stress c,= 240 MPa, cr, = -120 MPa. and 1x, = 240 MPa, use Mohr’s circle to determine the a )solJte maximum shear stress. ggigy'nou T 1 c enter of the circle is located at: JC ._. 3% = W = 60MB: xx, = 240 MPa he radius of the circle is: r = 1[moms - son/119.2)2 + (240MHz)2 = 3OOMPa 1' 1e principal stresses are: c1=cc+r =60MPa+300MPa=360MPa 0’2 = O"; - r = 60 MPa — 300 MPa = -240 MPa 1’ ie ansolute maximum shear stress is: a. vs W = 39g MPa (240,240) TWPO) 275 ...
View Full Document
{[ snackBarMessage ]} | 1,080 | 2,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-17 | latest | en | 0.76935 |
https://mathematica.stackexchange.com/questions/264626/finding-the-curvature-of-the-path-of-a-particle | 1,686,079,212,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653071.58/warc/CC-MAIN-20230606182640-20230606212640-00012.warc.gz | 427,651,176 | 38,055 | # Finding the Curvature of the path of a particle
How do I find the curvature of the vector when t=5?
The function is:
v[t_] := {2 t^4 + 2, 6 t^3 + 1, 3 t^2 + 4}
• Please read my comment! mathematica.stackexchange.com/questions/264624/… Mar 4, 2022 at 4:51
• FrenetSerretSystem[v[t], t][[1,1]]/.t->5 Mar 4, 2022 at 4:52
• @cvgmt as you can see from my answer, I kind of guessed the FrenetSerretSystem and was hoping that you'd write an answer :-) Mar 4, 2022 at 4:55
v[t_] := {2 t^4 + 2, 6 t^3 + 1, 3 t^2 + 4}
You sent t to 5 like so:
N[curv /. t -> 5] | 223 | 562 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | latest | en | 0.809601 |
https://www.scribd.com/document/332518009/New-Text-Document-2 | 1,568,898,417,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573519.72/warc/CC-MAIN-20190919122032-20190919144032-00475.warc.gz | 1,043,434,953 | 49,055 | You are on page 1of 1
# Now take the battery and connect its terminals across
## the terminals of the capacitor and start the stop watch.
Note the readings at 20sec intervals and write them
down. [NOTE:- Reading the previous statement could be
astonishing as it says that measure voltage at 20sec
interval but one could question that current move at
very high speed so how could one measure the
changing readings! But believe me it wasn t an easy
13
task but since the voltage depends on reciprocal of
exponential function and as time passes by the
changing readings will get slowed down and even after
infinite time the capacitor could not be charged up to
max voltage. Also since its time constant is 100sec
which is quite practical to measure at and hence this
experiment is very much justified.]. Take 10 readings
and if required the 20sec gap could be increased
because as the time passes by the change in voltage
becomes smaller and smaller.
(v) Now let the capacitor be charged up to 460 sec
because then it will become 99.99% charged [since we
have a limited time and we can t wait for infinite time
for it to charge completely!]. Now remove the battery
and now attach a wire in place of the battery terminals
and again note the multimeter readings changing and
record them.
(vi) Plot a graph between voltage and time for charging as
well as discharging. | 310 | 1,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-39 | latest | en | 0.966328 |
http://mathhelpforum.com/calculus/203333-differentiation-i-thought-i-had-finally-got-but-im-stuck-again.html | 1,527,295,273,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00343.warc.gz | 175,438,513 | 9,527 | # Thread: Differentiation - I thought I had finally got it, but im stuck again!
1. ## Differentiation - I thought I had finally got it, but im stuck again!
Ok, the question is:
Differentiate: five over the square root of four + x
2. ## Re: Differentiation - I thought I had finally got it, but im stuck again!
$\displaystyle \frac{5}{\sqrt{4+x}}=5(4+x)^{-1/2}$
Use the chain rule.
3. ## Re: Differentiation - I thought I had finally got it, but im stuck again!
thanks, i got that far. But its after that that I get lost.
The answer says -5 over 2 times the square root of 4 + x cubed
thankyou! | 170 | 602 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-22 | latest | en | 0.944841 |
https://www.studyclix.ie/Discuss/Leaving-Cert-Mathematics/help-explain-geometric-seq-ans | 1,490,582,651,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189377.63/warc/CC-MAIN-20170322212949-00590-ip-10-233-31-227.ec2.internal.warc.gz | 974,805,868 | 4,026 | All Leaving Cert Mathematics posts
• #### Help explain geometric seq. ans c_gall
HL 2011 Q 4(c)
Have solution here showing
Q(t-1)=Ae^-b(e^-b)
then turns to
Ae^-b(t+k)=0.5(Ae^bt)
Can someone please explain as to where k and 0.5 come from?
Much appreciated.
1. #### Dazzla16
The part where you mentioned Ae^(-b(t + k)) = 0.5(Ae^-bt) refers to Q 4(d). Maybe that was the thing that was confusing you. Q 4(d) is not part of Q 4(c) (i.e. you're not using the same equation for the two questions). | 162 | 500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-13 | latest | en | 0.897256 |
https://chess.stackexchange.com/questions/6081/which-piece-is-least-likely-to-be-taken | 1,716,953,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059167.30/warc/CC-MAIN-20240529010927-20240529040927-00515.warc.gz | 152,508,841 | 44,154 | # Which piece is least likely to be taken?
If one were to analyse a large number of chess games, which specific piece (not type of piece) would be found to be taken the least number of times? And yes, in light of SomePatzer's comment, besides the kings of course.
• Probably the king. ;-)
– user3598
Aug 14, 2014 at 15:49
• Lyndsey, I imagine that your question is not asking which type of piece is captured least often, but rather which of the 30 non-king pieces is captured least often (so that, for example, someone answering with "a2 pawn" would be giving a different answer than "f7 pawn"). But it isn't clear from your question as written whether my interpretation is right or wrong; could you please clarify?
– ETD
Aug 14, 2014 at 22:28
• Yes, your interpretation is correct, I would like to know the specific piece, not just the type. Aug 14, 2014 at 23:01
• So suppose somebody does the analysis and finds that the queen's rook is the least-often captured piece. What does that teach us about chess? Aug 15, 2014 at 13:47
• @DavidRicherby: Before Einstein's use of it in special relativity, Minkowski space was merely an abstract setting. What does that teach us about pragmatism? ;-) Jun 22, 2016 at 13:25
I recently looked into that very thing a couple of weeks ago, and found an excellent breakdown on it (just this one, though):
The short answer is that aside from Kings, h2 & h7 are the least likely to be captured pieces (with survival rates claimed to be 73.92% and 72.29%, respectively).
Regarding individual pieces' survival (from a 2,196,968 game database), http://www.quora.com/What-are-the-chances-of-survival-of-individual-chess-pieces-in-average-games/answer/Oliver-Brennan gives a table (and an accompanying graphic) of each piece's respective chance of survival.
While I was at it, I found some data on square usage. I've included the sources below not because square usage relates to your question directly, but because the nature of the two topics is so close that I figured that they might be of interest to you (or future readers).
Using 509 games selected randomly [and analyzed] out of a 567,000 game database, these two URLs give tables of raw events and normalized through-traffic usage of squares, and heat maps of same:
This URL analyzes the moves of 12 Grandmasters' entire careers (separated by white and black, all normalized [no raw counts]), along with some graphs of average-data counts:
There are also some excellent analyses (per side) of each type-of-piece's square usage at http://philanalytics.blogspot.com/2014/11/chess-analytics-analyzing-championship.html .
• It's an interesting argument, and not surprising. Those pawns are often at the base of a pawn chain. Still - in that analysis promotion counts as 'survival' so that also increases their chances (as Queens survive well too!). I wonder how the data is changed if promotion isn't considered 'survival'. Feb 8, 2017 at 14:06
Probably the rooks. That's why the greatest number of endings are the rook endings.
The knights and bishops fight it out early on and get "killed." Later, the queens are targets because they're so powerful, and they get traded. The rooks are left on the board. Sometimes one or both sets are traded, but often, they remain until one player resigns.
Between the rooks, the queen rook is less exposed, and therefore less likely to be captured.
The obvious answer is the king. It will be taken 0 times. The most frequent one would be a pawn: there are 16 pawns and you can hardly find a game, where a pawn is not taken. I would bet that the least frequently taken (apart from king) would be a queen.
I do not have statistics to back up my assertion, but you have only 2 queens and 4 rooks, bishops, knights. Also pro players prefer not to exchange queens early, so they will go only after some of the other pieces would be exchanged.
• I should have been more specific in my question. I did not mean the type of piece, e.g. pawn, but the actual piece, e.g. a2 pawn. Aug 14, 2014 at 23:08
• @Lyndsey so please be more specific and edit your question, because I highly doubt someone would read my answer to check for your comment. Aug 14, 2014 at 23:12
The h2/h7 pawns as confirmed by @Charles Rockafellor seems to be the most likely to me; at least for normal games above Patzer level.
If you look at games there are very few pawnless end positions. And since more people castle short than long, it makes sense to keep those protective f-g-h pawns a bit longer. Also the h pawns being where they are at the rim, it is very hard to get at them and if they'd ever be involved in some exchange action it could only be to one side, towards the g file.
If by taken, you mean captured, then the least likely to be taken might be the queen, although it's impossible to statistically prove it as there are as many possible chess games as atoms in the universe.
By taken, if you consider an exchange, I would bet on the rooks.
• The power of statistics is that you don't have to consider all possible cases (games in this case), just a large random selection. If you were to log all the pieces left on the board at the end of, say, ten thousand games you would be able to identify the piece most likely to avoid capture - probably a rook, as you say. Aug 16, 2014 at 8:42
• I don't see why you've interpreted the question as being "If we count up all the theoretically possible games of chess, ..." rather than "In games of chess as typically played by humans, ..." And you seem to think that "capture" has some sort of special meaning, more or less equivalent to "I get your piece but you don't get one of mine". That is not what the word means: a capture is any removal of an enemy piece from the board by placing one of your own pieces on the same square. Most captures are part of exchanges; some are not. But they're all captures. Aug 17, 2014 at 9:42
• we clearly have different philosophies of 'capture' in chess. I will stick with mine and I appreciate your input :) "in games of chess as typically played by humans" you mentioned. Well sure, if that was mentioned in the question specifically, I would have considered it. There are engine vs engine games too, to consider and engine vs human games. Aug 17, 2014 at 18:18 | 1,519 | 6,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.965366 |
http://www.newgre.org/tag/geometry/ | 1,603,779,054,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893402.83/warc/CC-MAIN-20201027052750-20201027082750-00593.warc.gz | 156,659,865 | 10,246 | # Tag Archives: geometry
## Quantitative Comparison Example Problem
The Quantitative Comparison question type on the GRE can be very challenging. Essentially, you are given some information (usually in the form of a sentence, equation, or picture) and then two quantities. You are then asked to determine: A if … Continue reading
Posted in Example Problems, Revised GRE | | Leave a comment
## GRE Math: Two More Example Geometry Problems and Solutions
Example Geometry Problem #3 The problem states that the triangle depicted is an isosceles triangle. Remember, there are two qualities that define an isosceles triangle: 1) two identical sides 2) two identical angles If you have one, you will automatically … Continue reading
Posted in Example Problems | | 4 Comments
## GRE Math: Two Difficult Example Problems and Solutions
Example Difficult Arithmetic Problem These problems aren’t difficult because they use particularly complicated mathematical principles — they’re difficult because they’re a little tricky. They require you to think a little beyond what’s given directly to you. This problem looks a … Continue reading
Posted in Example Problems | | 13 Comments
## GRE Math: Two Example Math Problems
Example Problem #1 This is a fun problem. The problem tells us that ∆ABC and ∆DEF have the same area. It also tells us that AD > CF. The problem is asking us about altitude, which is height. From this … Continue reading
Posted in Example Problems | | 3 Comments
## GRE Math: Two Example Geometry Problems and Solutions
I’m still trying to hammer out a coherent and apposite answer to the nightmare prompt from yesterday’s post, so I haven’t made much other progress on my applications. Since there’s nothing to report, today we’ll just do a couple geometry … Continue reading
Posted in Example Problems | | 3 Comments | 388 | 1,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-45 | latest | en | 0.859062 |
https://www.speedsolving.com/wiki/index.php?title=4x4x4_parity_algorithms&diff=cur&oldid=37149 | 1,624,369,984,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00579.warc.gz | 920,649,137 | 75,153 | # Difference between revisions of "4x4x4 parity algorithms"
For those who are just looking for lists of commonly used speedsolving (speedcubing) 4x4x4 parity algorithms and cannot find what they are looking for just based on looking at the Table of Contents, this section is for you,! However, it's important to read the Introduction to answer any questions you may have regarding the formatting of this page. (Note: Some of these algorithm lists are LONG. If you don't find the type of algorithms that you are looking for, keep scrolling down and chances are you will find them!)
Last two dedge pairing algorithms
PLL Parity
OLL Parity (Speedcubing Form)
Single Parity ("Pure" Form) - See this section of the Introduction to see why "pure" is an informal way of describing these type of algorithms.
Double Parity ("Pure" Form)
Algorithms for other 3x3x3 methods besides the CFOP method (or any 3x3x3 method which solves the cube layer-by-layer, in general)
# Introduction
This page attempts to list move optimal algorithms for every common form of parity encountered in popular 4x4x4 (Rubik's Revenge) solving methods.
Solutions listed under a case image which are not move optimal (in the move metric in which algorithms are sorted by):
• are faster to execute,
• demonstrate a notable alternative way to solve the case (including, but not limited to using different types of moves),
• have a different effect on the 4x4x4 supercube, or
## Reduction parity
The term "parity" can be used to describe a number of situations that occur during a 4x4x4 solve which cannot manifest during a 3x3x3 (standard size Rubik's cube) solve. In fact, there has been debate about what situations are considered to be a parity case, but there is one situation of which any cuber who uses the term "parity" for the 4x4x4 identifies as parity: the single dedge flip.
The most popular 2-cycle (a swap of two pieces) besides the single dedge flip case is the following. This 2-cycle of wings is as common during a K4 Method solve as the single dedge flip is, but it should never arise during a solve using the Reduction Method because two dedges are not paired up. However, many who solve the 5x5x5 Rubik's cube using some variant of the Reduction Method will come across this case; and thus several (but not all) of the algorithms listed on this page which solve this case directly can be used for completing the tredge-tripling stage of a 5x5x5 Reduction solve.
An equally well-known form of reduction parity (this term will be defined formally soon) besides the single dedge flip is switching two opposite dedges in the same face.
This parity situation can be transformed into 21 other last layer forms of what is commonly called PLL parity by performing a 3x3x3 PLL and adjusting the upper face (AUF) as needed. That is, there is a total of 22 PLL parity cases. (See the PLL Parity section for details.) The remaining PLL parity cases which involve the fewest number of pieces (besides the most popular case above) are the following.
Despite that one can technically solve all 22 PLL parity cases by executing an algorithm meant to solve any one of them (to any face) and then finish solving the 4x4x4 as if it was a 3x3x3, special algorithms have been developed for every case. This allows one to use fewer moves to solve any given case and gives one more options.
Combining some form of PLL parity and a single dedge flip creates one of the many cases of what's commonly called double parity. For example, performing a swap of dedges to a fully solved 4x4x4 and then flipping the front dedge resulting from that swap gives us the following.
Since the double parity case above and the single dedge flip case both have a single dedge flipped, and since OLL algorithms do not necessarily aim to permute (move) the pieces that they correctly orient in any particular fashion, any 4x4x4 algorithm which solves:
• a case containing an odd number of flipped dedges (which will be called "single parity" on this page)
or
• a case which additionally has an odd permutation of dedges and an even permutation of corners (or vice versa) (which will be called "double parity" on this page)
is called an OLL parity algorithm.
It is common convention among the speedcubing community to use algorithms which contain wide (double layer) turns to solve OLL parity instead of single inner layer slices. The notation used in Chris Hardwick's Rubik's Revenge Solution signifies a wide right turn as (Rr), for example. In old WCA notation (the notation used on this page), (Rr) is expressed as Rw. (The "w" is short for "wide".) In fact, the most popular speedcubing single parity algorithms perform additional swaps besides flipping a single dedge due to the use of wide turns.
Such an algorithm is called a non-pure algorithm when compared to algorithms which just flip a single dedge, which are often called pure flips. However, the term pure is more formally associated with an algorithm being supercube safe--algorithms which do not permute (move) any centers in the supercube version of a given order.
Most of the algorithms on this page affect some centers of the 4x4x4 supercube: not all algorithms affect the supercube centers in the same manner. Algorithms followed by a "//Safe" are supercube safe.
There are many types of parity cases which can occur during a 4x4x4 solve (when considering the parity cases which can arise in every 4x4x4 method), but the most widely known cases are the cases which can manifest after reducing a fully scrambled 4x4x4 into a pseudo 3x3x3 -- an nxnxn cube in which all of its composite edges are complete and all of its centers are complete and are in the correct locations with respect to each other with regards to the cube's Color Scheme -- are called reduction parity cases. (Because this is the objective of the Reduction Method and its variants.)
In May 2013, Michael Gottlieb defined reduction parity in detail.
Reduction parity occurs when you try to reduce the puzzle so it can be solved by a constrained set of moves, putting it into some subset of the positions. However, you can often reach a position which seems like it is in your subset, but which is actually not, and to solve the puzzle you have to briefly go outside your constrained set of moves to bring the puzzle back into the subset you want. Typically the number of positions you can encounter is some small multiple of the number of positions you expect. The obvious example is PLL parity in 4x4x4: all the centers and edges are properly paired, so you expect to be able to finish the puzzle with only outer layer turns, but this isn't quite possible. OLL parity falls under this definition too (so the reduced 4x4x4 has four times as many positions as you would expect).[1]
There is a 50/50 chance that PLL parity will be present (assuming that one is doing a solve using reduction). Independently, there is a 50/50 chance to get a single dedge flip. Therefore there is a 1/2×1/2 = (chance of PLL parity)×(chance of single parity) = a 1/4 chance of getting all possibilities, separately. That is, for the reduction method (and its variants), PLL parity occurs by itself 25% of the time, single parity occurs 25% of the time, double parity occurs 25% of the time, and there is no reduction parity 25% of the time.
This page will keep strong focus on reduction parity (OLL parity and PLL parity) cases, but it will also include a limited number of other parity situations which are also common in other solving methods, as well as cases which share some characteristics with reduction parity algorithms.
The key characteristics of 4x4x4 reduction parity algorithms are:
• They preserve the colors of the centers. (Note that most reduction parity algorithms do not technically preserve the centers themselves, because if they are applied to a fully solved 4x4x4 supercube, one can see that same color centers are swapped with each other.)
• They do not break up (and therefore do not pair up) any dedges. (They preserve the coupling of the dedges, but they may move entire dedges.)
• All OLL parity algorithms contain an odd number of inner slice quarter turns. (They are called odd parity algorithms.)
• All PLL parity algorithms contain an even number of inner slice quarter turns. (They are called even parity algorithms.)
## Other parity cases
It turns out that we are not limited to using well-known dedge-pairing (even parity) algorithms to pair dedges.
Websites such as bigcubes.com present algorithms with an even number of inner slice quarter turns for the dedge-pairing stage of the Reduction Method.
Algorithms which solve/generate any case in which dedges are broken up (including odd parity algorithms) can be used. For example, algorithms for this parity case (mentioned previously) can be used. This page contains quite a few algorithms which solve that case and other 2-cycle cases like
as well as 4-cycle cases which are contained within two dedges (which also can be used to pair the last two dedges).
There is actually a total of 110 last layer 4-cycles, but since 4-cycles in two dedges are the only ones encountered using the most popular 4x4x4 solving methods, they are the only ones shown on this page. However, this PDF includes all 110 cases and relatively short algorithms to solve each one directly.
(All of these cases often occur in the ELL stage of the K4 Method.)
Algorithms for the Cage Method, as well as algorithms for theoretical purposes and general 4x4x4 exploration are present as well.
## Some short/easy parity fixes
Since all OLL parity algorithms contain an odd number of inner slice quarter turns, one can technically fix any 4x4x4 wing edge odd parity case by executing a single slice quarter turn and then resolve the cube using an even number of inner slice quarter turns. Here's one video tutorial that illustrates the typical process.
Similar to doing an inner slice quarter turn like r to technically fix the single dedge flip parity, an inner slice half turn such as r2 is technically all that is needed to fix PLL parity. In fact, only two inner slice quarter turns + 3x3x3 turns is all that is needed to create/solve PLL parity on the 4x4x4. One can split up r2 as r r or as r' r' and insert 3x3x3 moves to obtain the pure form of PLL parity. Below is an example algorithm found in December of 2013. (r' U' D' R2 U' D' S2 r')(F' U' D' F2 R F2 U2 R E' R F2 L' U2 R D2 R B2 R' x2 y') Animation However, we can also just use the inner slice turns r and r' as well. (r' U D S2 U' D' S2 r)(U2 B D2 U2 F' U2 F2 L' R D' B2 L' R D2 F') Animation
Should one wish to induce an odd permutation in the wing edges of the 4x4x4 with a short algorithm without having to restore the cube as much as applying an inner slice quarter turn requires, below are fairly short (and simple) algorithms one can use.
• The shortest quarter turn 4x4x4 cube odd parity fix which preserves the colors of the centers was (found by Tom Rokicki on April 27, 2021) is u2 r u M u M' u r' (9,8).
Although these algorithms are not listed under a case image on this page, they would appear in the following format (in an "algorithm bar") if they were.
f2 r E2 r E2 r f2 (11,7) N Tom Rokicki and Ed Trice [2]
u2 r u M u M' u r' (9,8) N Tom Rokicki [3]
There are links to either forum posts or video URLs in the right-most column of many "algorithm bars". These links are intended to either show one of the (if not the) first place the algorithm (or a simple conjugation, transformation, and/or a directly related version of it) was first published, and/or show one of the (if not the) earliest date of publish. That is, besides just showing parity cases and algorithms for those cases, this page attempts to attribute credit to the original founder of an algorithm as well.
(More will be explained about what other pieces of information in the algorithm bars above mean later.)
Since the shortest 4x4x4 (half turn) odd parity fix contains move repetition, it can be written more compactly as f2 (r E2)2 r f2.
Clearly this algorithm does not preserve the pairing of dedges, but it does preserve the colors of the centers; and it contains 7 inner slice quarter turns, an odd number.
We can break up this algorithm as (f f r E E r E E r f f) to count 4 f's and 3 r's. At the same time, we can count a total of 11 block quarter turn moves (BQTM). We can count that this algorithm has 7 block half turn moves (BHTM) without breaking it up. In addition to the fact that all parity cases on this page are each represented by a case image, The number of moves an algorithm contains in these two big cube move metrics is written next to them in the form of the ordered pair, (BQTM, BHTM). (We can clearly see this in the above algorithm bar.) Algorithms with fewer BHTM are listed first in each category. Algorithms with fewer BQTM are listed before other algorithms which have the same number of BHTM as they.
The shortest (and well-known) nxnxn cube odd parity fix which preserves the colors of the centers is (r U2)4 r (13,9).
Although this algorithm affects the same number of pieces as the (11,7) move algorithm above on the 4x4x4, it also "works"/does the same for the nxnxn cube. The (11,7) above discolors centers on, say, the 5x5x5 cube. Many of the algorithms on this page need to be "adjusted" to work for the nxnxn Rubik's cube. Some algorithms may only be translatable to higher order even cubes (6x6x6, and larger). All algorithms can be applied to the 6x6x6 if instead of turning the outer 2 layers, turn the outer 3 layers; instead of turning 1 inner layer slice, turn 2 inner layer slices.
Finally, one of the simplest OLL parity (more specifically, a double parity) algorithms (found in December of 2017) to remember also consists of a short repeated sequence:
Rw' (F2 U' Lw' U)5 Rw (27,22).
This was deduced from the same idea that Floyd Newberry came up with for using a short repeated sequence to directly solve a 2-cycle. Below are two single dedge flip (2-cycle) algorithms illustrating the idea.
(Rw B' z')(r' F U2 F')4 r' (z B Rw') (25,21) N Floyd Newberry [4]
(Rw B' z')(r F U2 F')4 r (z B Rw') (25,21) N Floyd Newberry [5]
Besides the notes mentioned already about what types of algorithms are contained within this page, including some of the specific common characteristics they share, this section touches on how they "look" and "feel" when they are displayed in notation and executed on a cube, respectively.
### Notation and animations
• All algorithms on this page are written in old WCA notation, where lowercase letters represent inner slice turns, uppercase letters and xyz have the same meaning as standard 3x3x3 notation, and a move like Rw, for example, means to turn both the face and inner slice parallel and adjacent to it simultaneously in the same direction. (Using the same example, Rw = (Rr).)
• Click the move length ordered pairs of the form, (number of block quarter turns, number of block half turns) (which are to the right of the algorithm in the algorithm bar), to see the algorithm animated in SiGN notation at alg.cubing.net.
### Move sets
Two algorithms of similar length (the number of moves an algorithm contains) can look (and feel, when executing) very different. This is especially common if two algorithms are in a different move set (consist only of certain types of turns).
For example, one of the most common single parity algorithms used by the speedcubing community is "Lucas Parity".
Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw' (25,17)
Notice that it only contains the (and inverses of the) moves Rw, U2, x and Lw.
Interesting Note: On January 24, 2010, speedsolving.com forum member reThinking the Cube showed that Lucas Garron's "Lucas Parity" (published in mid 2008) is equivalent to Stefan Pochmann's "New Dedge Flip" from his webpage (published in or before early 2007).
An alternate algorithm of the same move length in BHTM (but contains 5 fewer BQTM) is the following.
Lw E Rw2 Uw r' Uw' Rw2 Dw R2 Uw r' Uw' Rw R Dw' E' Lw' (20,17)
Clearly this algorithm has much more of a variety of moves than "Lucas Parity". It is also clearly not a speedsolving algorithm as "Lucas Parity" is.
This page not only contains commonly practiced speedsolving algorithms: it also contains algorithms which illustrate the veracity of the 4x4x4 cube parity algorithm domain.
### Single slices vs. wide turns
Two of the most popular 15 BHTM move algorithms which flip a single dedge on the 4x4x4 are the following.
r' U2 l F2 l' F2 r2 U2 r U2 r' U2 F2 r2 F2 (25,15)
r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2 (25,15)
Their inner slice turns may all be replaced with wide turns and still preserve the first three layers (F3L) of the 4x4x4 and flip one dedge.
Rw' U2 Lw F2 Lw' F2 Rw2 U2 Rw U2 Rw' U2 F2 Rw2 F2 (25,15)
Rw2 B2 U2 Lw U2 Rw' U2 Rw U2 F2 Rw F2 Lw' B2 Rw2 (25,15)
The fourth column of algorithm bars is either labelled:
• Y (short for "Yes"), if all of an algorithm's inner slice turns can be substituted with their corresponding wide turns to still preserve F3L (or if an algorithm does not contain any inner layer slices)
or
• N (short for "No"), if their inner slices cannot be substituted with wide turns and still preserve F3L.
Note that with many algorithms, it's not "all or nothing". A few of the slice turns can be wide to still just flip a single dedge, for example. For convenience, an algorithm is written with the maximum number of wide turns, should that version of it still preserve as much as the version of it without any wide turns.
For example, the second 15 BHTM algorithm mentioned above could be expressed later on this page with the following algorithm bar, since all of its inner slice turns can be made wide (hence the "Y" instead of an "N") and its first and last moves can be wide and still solve the pure dedge flip case (hence why the algorithm begins and ends with Rw2 instead of r2).
For illustration of how algorithm bars are going to be labelled, let us temporarily name it "Old Standard Alg" and called the author "anonymous". (Algorithm names will be explained next.)
Rw2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 Rw2 (25,15) Old Standard Alg Y anonymous [6]
However, despite that all (25,15) single dedge flip solutions which begin and end with an l2 or r2 move can instead be Lw2 and Rw2, respectively, all slices will be expressed as single slice (lowercase) turns for simplicity for all (25,15) solutions.
### Names/labels
Although the third column in the majority of the algorithm bars on this page is blank, when it is not blank, it is either an algorithm name (given by the algorithm author) or an algorithm label (for organizational or classification purposes).
• Some algorithms like "Lucas Parity" have been named.
• Some algorithms are labelled (named) "Alg1(v1)", "v3", etc., (short for "algorithm 1, version 1" and "version 3", respectively) to:
• represent consecutive ordered version algorithms which are different versions (transformations) of each other
OR
• prove things like "there are actually fewer distinct paths than it appears for an algorithm in this move set and of this length" and also to show that "one algorithm is just another in disguise".
• Algorithms that have been named "5x5x5 Alg" are significantly longer algorithms than their related 4x4x4 algorithm counterparts, but they are algorithms which follow a lot of constraints and manage to work on the 4x4x4 and the 5x5x5.
# PLL Parity
• This section shows all 22 possible PLL parity cases. In addition, algorithms for the two cases when two dedges are unoriented (and corners are solved) and the three cases when all four last layer dedges are unoriented (and the corners are solved) are also included.
• Of course, all PLL parity cases can be handled by using the first case shown, commonly called "opposite parity" (OP).
## Dedges
### Two dedges (oriented)
#### I permutation
r2 U2 r2 Uw2 r2 u2 (12,6) N Chris Hardwick [7]
r2 U2 r2 U2 d2 r2 d2 (14,7) N []
U2 r2 U2 r2 Uw2 r2 Uw2 (14,7) N Chris Hardwick [8]
(Uw2 Rw2 U2) r2 (U2 Rw2 Uw2) (14,7) SP01 N Stefan Pochmann [9]
(Dw2 Rw2 U2) r2 (U2 Rw2 Dw2) (14,7) N []
(Rw2 F2 U2) r2 (U2 F2 Rw2) (14,7) N Stefan Pochmann [10]
(Rw2 B2 U2) r2 (U2 B2 Rw2) (14,7) N Stefan Pochmann []
Rw2 (U2 r U2 S2)2 Rw2//Safe (18,10) Y Walter Randelshofer [11]
r2 U2 r E2 r2 E2 r' D2 l2 E2 y2//Safe (18,10) Y []
r2 D2 l E2 l2 E2 l' U2 l2 E2 y2 //Safe (18,10) Y []
r2 U2 r' E2 r2 E2 r D2 l2 E2 y2//Safe (18,10) Y []
r2 D2 l' E2 l2 E2 l U2 l2 E2 y2 //Safe (18,10) Y []
r2 U2 l S2 r2 S2 l' D2 l2 E2 y2//Safe (18,10) Y []
r2 D2 r S2 l2 S2 r' U2 l2 E2 y2 //Safe (18,10) Y []
r2 U2 l' S2 r2 S2 l D2 l2 E2 y2//Safe (18,10) Y []
r2 D2 r' S2 l2 S2 r U2 l2 E2 y2 //Safe (18,10) Y []
r2 U2 B2 l r U2 M' U2 r2 B2 U2//Safe (19,11) N Stefan Pochmann []
M U2 l2 B2 D2 r2 D2 B2 r l U2//Safe (19,11) Y []
M' U2 r2 B2 U2 r2 U2 B2 r l U2//Safe (19,11) Y []
M' U2 l' r' D2 B2 r2 B2 D2 l2 U2//Safe (19,11) Y []
M U2 l' r' U2 B2 r2 B2 U2 r2 U2//Safe (19,11) Y []
M U2 l r D2 F2 r2 F2 D2 l2 U2//Safe (19,11) Y []
M' U2 l r U2 F2 r2 F2 U2 r2 U2//Safe (19,11) Y []
(y Rw2 U2) Rw U2 (Rw2 U2)2 Rw (U2 Rw2 y') (20,11) v1 Y []
(y Rw2 U2) Rw' U2 (Rw2 U2)2 Rw' (U2 Rw2 y') (20,11) v2 Y []
(Rw2 U2)(E r2 E' r)2 (U2 Rw2)//Safe (15,12) Y Werner Randelshofer, Walter Randelshofer, and André Boulouard []
(Rw2 U2)(E' r2 E r)2 (U2 Rw2)//Safe (15,12) Y Werner Randelshofer, Walter Randelshofer, and André Boulouard []
(Rw2 U2)(E r2 E' r')2 (U2 Rw2)//Safe (15,12) Y Werner Randelshofer, Walter Randelshofer, and André Boulouard []
(Rw2 U2)(E' r2 E r')2 (U2 Rw2)//Safe (15,12) Y Werner Randelshofer, Walter Randelshofer, and André Boulouard []
r2 U2 r U2 r2 U2 r2 U2 r U2 r2 U2//Safe (22,12) Y []
r' F U' R F' U l r U' F R' U F' l'//Safe (14,14) N []
(Rw2 U2)(r E2)6 (U2 Rw2)//Safe (24,15) N Christopher Mowla []
(Rw2 U2)(r' E2)6 (U2 Rw2)//Safe (24,15) N Christopher Mowla []
y' Rw2 3Uw2 Rw 3Uw Rw2 3Uw2 Rw2 3Uw' Rw2 3Uw Rw2 3Uw' Rw2 3Uw' Rw2 3Uw Rw2 3Uw Rw2 3Uw' Rw 3Uw2 Rw2 y (36,23) Y Ben Whitmore [12]
M2 Uw M Uw M' Uw' M' Uw M Uw M Uw2 M' Uw2 M' Uw' M' Uw M Uw2 M Uw' M' Uw2 M' Uw' M Uw' M' Uw' M Uw M2 (38,33) N Ben Whitmore [13]
#### Double slash (//) permutation
(R2 D' x) r2 U2 r2 Uw2 r2 u2 (x' D R2) (18,10) N Chris Hardwick [14]
(R2 D' x Uw2 Rw2 U2) r2 (U2 Rw2 Uw2 x' D R2) (20,11) SP02 N Stefan Pochmann [15]
(R2 D' Rw2 U2 F2) r2 (F2 U2 Rw2 D R2) (20,11) FB02 N Frédérick Badie [16]
(F2 U Rw2 U2 F2) r2 (F2 U2 Rw2 U' F2) (20,11) N Stefan Pochmann []
(R2 D' x Rw2 F2 U2) r2 (U2 F2 Rw2 x' D R2) (20,11) N Stefan Pochmann []
(R2 D' x Rw2 B2 U2) r2 (U2 B2 Rw2 x' D R2) (20,11) N Stefan Pochmann []
(F U' F') r2 U2 r2 Uw2 r2 u2 (F U F') (18,12) N Chris Hardwick []
(F U' F' Rw2 F2 U2) r2 (U2 F2 Rw2 F U F') (20,13) N Stefan Pochmann []
(y' R' F) l E F2 E' l' r' E F2 E' r (F' R y)//Safe (16,14) Alg.1 N Christopher Mowla [17]
(y' R' F) r' E F2 E' l r E F2 E' l' (F' R y)//Safe (16,14) Inverse[Alg.1] N Christopher Mowla []
(y' R' F) l E' F2 E l' r' E' F2 E r (F' R y)//Safe (16,14) Alg.2 N Christopher Mowla []
(y' R' F) r' E' F2 E l r E' F2 E l' (F' R y)//Safe (16,14) Inverse[Alg.2] N Christopher Mowla []
(R B') l S U2 S' l' r' S U2 S' r (B R')//Safe (16,14) Alg.3 N Walter Randelshofer []
(R B') r' S U2 S' l r S U2 S' l' (B R')//Safe (16,14) Inverse[Alg.3] N Walter Randelshofer []
(R B') l S' U2 S l' r' S' U2 S r (B R')//Safe (16,14) Alg.4 N Walter Randelshofer []
(R B') r' S' U2 S l r S' U2 S l' (B R')//Safe (16,14) Inverse[Alg.4] N Walter Randelshofer []
(R U R' U') r2 U2 r2 Uw2 r2 Uw2 (U' R U' R') (19,14) N Chris Hardwick [18]
(R U R') U r2 U2 r2 Uw2 r2 Uw2 U (R U' R') (19,14) N Chris Hardwick []
(Uw' R U' R' Uw' Rw2 U2) r2 (U2 Rw2 Uw R U R' Uw) (20,15) N xyzzy [19]
(y Uw' R U R' Uw' Rw2 U2) r2 (U2 Rw2 Uw R U' R' Uw y') (20,15) N xyzzy [20]
(y2 Uw R U' R' Uw Rw2 U2) r2 (U2 Rw2 Uw' R U R' Uw' y2) (20,15) N xyzzy [21]
(y' Uw R U R' Uw Rw2 U2) r2 (U2 Rw2 Uw' R U' R' Uw' y) (20,15) N xyzzy [22]
(y2 Uw2 R U R' U' Rw2 U2) r2 (U2 Rw2 U R U' R' Uw2 y2) (22,15) N xyzzy [23]
(R2 D' x) r2 U2 B2 l r U2 M' U2 r2 B2 U2 (x' D R2)//Safe (24,15) N []
(Rw' U R U Lw' U2 Rw' U2) r2 (U2 Rw U2 Lw U' R' U' Rw) (22,17) v1 N Christopher Mowla [24]
y' (Rw U' R' U' Rw B2) (Rw B2 r2 B2 Rw') (B2 Rw' U R U Rw') y (22,17) v2 N Christopher Mowla &Robert Yau [25]
y2 Rw U Rw' R U' Rw' U' Rw U Rw U' Rw' U' Rw' R U Rw U R' U' R' U y2 (20,20) Y António Gomes [26]
y2 R' U' R' U Rw U R Rw' U' Rw' U' Rw U Rw U' Rw' U' R Rw' U Rw U y2 (20,20) Y António Gomes []
y R' U Rw U R Rw' U' Rw' U' Rw U Rw U' Rw' U' R Rw' U Rw U R' U' y' (20,20) Y António Gomes []
(y2 Bw2 R2 Bw2) Rw U Rw' U' Lw' U' Lw U2 Rw U' Rw' U' Lw' U Lw (Bw2 R2 Bw2 y2) (28,21) Y Moritz Karl [27]
y Rw' U2 Rw U2 Rw' U2 Rw' U' Rw U' Rw U Rw' U' Rw U Rw2 U Rw U' Rw U Rw' U Rw U y' (30,26) Y Ben Whitmore []
### Two dedges (unoriented)
#### I permutation
F2 l E F2 E' l' r' E F2 E' r F2//Safe (16,12) Alg.1 N Christopher Mowla [28]
F2 r' E' F2 E l r E' F2 E l' F2//Safe (16,12) Mirror[Alg.1] N Christopher Mowla []
F2 r' E F2 E' l r E F2 E' l' F2//Safe (16,12) Inverse[Alg.1] N Christopher Mowla []
F2 l E' F2 E l' r' E' F2 E r F2//Safe (16,12) Mirror[Inverse[Alg.1] N Christopher Mowla []
F2 l' S U2 S' l r S U2 S' r' F2//Safe (16,12) Alg.2 N Walter Randelshofer []
F2 r S' U2 S l' r' S' U2 S l F2//Safe (16,12) Mirror[Alg.2] N Walter Randelshofer []
F2 r S U2 S' l' r' S U2 S' l F2//Safe (16,12) Inverse[Alg.2] N Walter Randelshofer []
F2 l' S' U2 S l r S' U2 S r' F2//Safe (16,12) Mirror[Inverse[Alg.2] N Walter Randelshofer [29]
(y R' U F') r2 U2 r2 Uw2 r2 u2 (F U' R y') (18,12) N Chris Hardwick [30]
r U2 l D2 l' U2 M U2 r D2 r' U2 l'//Safe (19,13) N Kenneth Gustavsson [31]
(y' R' F U' Rw2 U2 F2) r2 (F2 U2 Rw2 U F' R y) (20,13) N Stefan Pochmann [32]
(y R' U F' Rw2 F2 U2) r2 (U2 F2 Rw2 F U' R y') (20,13) N []
(F E' F Rw2 F2 U2) r2 (U2 F2 Rw2 F' E F') (20,13) N Per Kristen Fredlund [33]
F' U R d2 R' U' F U2 F' U R d2 R' U' F U2//Safe (20,16) N Per Kristen Fredlund [34]
l U2 M l U2 M' l' U2 M' U2 Rw M' U2 M r' U2 Rw'//Safe (23,17) N Kenneth Gustavsson []
(F r U' R U' Lw U2 Rw U2) r2 (U2 Rw' U2 Lw' U R' U r' F') (24,19) N Christopher Mowla []
(Rw U2 Rw' U l' U' Lw' U2 Rw' U' l2 U') r2 (U l2 U Rw U2 Lw U l U' Rw U2 Rw') (32,25) N Christopher Mowla []
#### Double slash (//) permutation
(R B) r2 U2 r2 Uw2 r2 u2 (B' R') (16,10) N Chris Hardwick [35]
(R B Rw2 F2 U2) r2 (U2 F2 Rw2 B' R') (18,11) N Stefan Pochmann [36]
(3Lw U Rw2 U2 F2) r2 (F2 U2 Rw2 U' 3Lw') (18,11) N Stefan Pochmann [37]
(R2 D) l' E F2 E' l r E F2 E' r' (D' R2)//Safe (16,14) N Per Kristen Fredlund [38]
F' R d2 R' F U F' U2 R d2 R' U2 F U'//Safe (16,14) N Per Kristen Fredlund [39]
R' U2 B u2 B' U2 R U' R' B u2 B' R U//Safe (16,14) N Walter Randelshofer [40]
R B U2 r2 U2 B2 l r U2 M' U2 r2 B R'//Safe (21,14) N Stefan Pochmann []
The four cases above clearly switch two dedges, but they can also be interpreted as doing two separate swaps of wing edges. Recalling that the term "2-cycle" is interchangeable with the common term "swap", these cases perform 2 2-cycles of wing edges. (They are called "2 2-cycles" for short.) There are actually 58 of these cases in the last layer, in general. However, the other 54 will only be encountered during a K4 Method solve. This PDF includes all 58 cases and short algorithms to solve each one.
### Four dedges (oriented)
#### O permutation
##### Oa
r2 Uw2 b2 U' b2 r2 U b2 r2 Uw2 r2 (20,11) CG03 N Clément Gallet [41]
f2 Uw2 r2 U' r2 f2 U r2 f2 Uw2 f2 (20,11) N Clément Gallet []
y2 M2 U' M2 u Uw f2 M2 Dw2 b2 U2 b2 (20,11) N Clément Gallet []
M' U2 Uw2 r2 Uw2 r2 U2 l2 U2 M' U M2 U M2 (22,13) N []
M U2 Uw2 l2 Uw2 l2 U2 r2 U2 M U M2 U M2 (22,13) N []
y2 D2 M E2 M U M2 U' l2 U2 r2 Uw2 r2 Uw2 (22,13) PKF03 N Per Kristen Fredlund [42]
(E2 M)2 U M2 U' l2 U2 r2 Uw2 r2 u2 (22,13) PKF03 N Per Kristen Fredlund [43]
l2 U2 f2 U' l2 f2 U u2 f2 u2 l2 U2 l2 (23,13) N Clément Gallet []
l2 U2 f2 U' l2 f2 U u2 f2 l2 U2 l2 u2 (23,13) N Clément Gallet []
r2 Uw2 b2 U' b2 r2 U b2 R2 Uw2 Rw2 Uw2 R2 (24,13) N Clément Gallet [44]
M2 U M U2 r' l' U2 F2 r2 F2 U2 r2 U M2 (23,14) N []
M2 U S2 l2 S2 U2 r S2 l2 S2 r' U2 r2 U' M2 //Safe (26,15) N Christopher Mowla [45]
M2 U S2 l2 S2 U2 r' S2 l2 S2 r U2 r2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 l2 S2 U2 l E2 l2 E2 l' U2 r2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 l2 S2 U2 l' E2 l2 E2 l U2 r2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 r2 S2 D2 r S2 l2 S2 r' D2 l2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 r2 S2 D2 r' S2 l2 S2 r D2 l2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 r2 S2 D2 l E2 l2 E2 l' D2 l2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 r2 S2 D2 l' E2 l2 E2 l D2 l2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 l2 D2 l S2 r2 S2 l' D2 S2 r2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 l2 D2 l' S2 r2 S2 l D2 S2 r2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 l2 D2 r E2 r2 E2 r' D2 S2 r2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 l2 D2 r' E2 r2 E2 r D2 S2 r2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 r2 U2 l S2 r2 S2 l' U2 S2 l2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 r2 U2 l' S2 r2 S2 l U2 S2 l2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 r2 U2 r E2 r2 E2 r' U2 S2 l2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U S2 r2 U2 r' E2 r2 E2 r U2 S2 l2 U' M2 //Safe (26,15) N Christopher Mowla []
M2 U' F2 M' F2 M U2 r2 D2 F2 r2 F2 D2 r2 U' M2//Safe (28,16) N Christopher Mowla [46]
F' L B D2 (R d L R' E2 M' f' M E2 L') D2 B' L' F//Safe (21,17) N Walter Randelshofer [47]
l' r' 3Dw L' R' u L R 3Dw' M2 3Dw' L R u L' R' 3Dw l' r'//Safe (20,19) N Christopher Mowla []
l r 3Dw L R u' L' R' 3Dw' M2 3Dw' L' R' u' L R 3Dw l r//Safe (20,19) N Christopher Mowla []
l' r' 3Dw' L' R' u' L R 3Dw M2 3Dw L R u' L' R' 3Dw' l' r' //Safe (20,19) N Christopher Mowla []
l r y' U' L R u L' R' 3Dw M2 3Dw L' R' u L R y' U' l r//Safe (20,19) N Christopher Mowla []
U2 (R' U' R' F R F' U R) (Uw2 r2 Uw2 r2 U2 r2) (F U R U' R' F') (28,21) N Antoine Cantin [48]
(F2 U2 M U Fw2 r2 u S' Rw2 u2) M (u2 Rw2 S u' r2 Fw2 U' M' U2 F2) (33,21) N Christopher Mowla []
Rw2 U R2 U R U' R' U R Rw' U2 Rw2 U2 Rw2 U2 Rw' U' R U R' U2 Rw2 U' (31,22) Y António Gomes [49]
Rw2 U2 R' U R U' Rw' U2 Rw2 U2 Rw2 U2 R Rw' U R' U' R U R2 U Rw2 U' (31,22) Y António Gomes []
Rw R U' Rw' R' U' Rw' R U Rw U' R2 U R U R U' Rw U Rw' R U' Rw' U' Rw (24,23) Y António Gomes []
Rw2 U R2 U R U' R U Rw U R' Rw2 U2 R Rw2 U2 R' Rw2 U Rw' U2 Rw2 U' (29,23) Y António Gomes []
Rw2 U2 Rw' U R' Rw2 U2 R Rw2 U2 R' Rw2 U Rw U R U' R U R2 U Rw2 U' (29,23) Y António Gomes []
Rw2 U' R' U' R U R U' Rw U2 Rw2 U2 Rw2 U2 Rw U2 R U' R U R2 Rw2 U' (30,23) Y António Gomes []
Rw2 U' R' U' R U R U' Rw' U2 Rw2 U2 Rw2 U2 Rw' U2 R U' R U R2 Rw2 U' (30,23) Y António Gomes []
Rw2 R2 U R U' R U2 Rw U2 Rw2 U2 Rw2 U2 Rw U' R U R U' R' U' Rw2 U' (30,23) Y António Gomes []
Rw2 R2 U R U' R U2 Rw' U2 Rw2 U2 Rw2 U2 Rw' U' R U R U' R' U' Rw2 U' (30,23) Y António Gomes []
Rw2 R2 U R2 Rw2 U R2 U' Rw2 U' Rw2 U' Rw U' Rw2 U' R2 U Rw2 U Rw' U' R2 (30,23) Y António Gomes []
Rw2 R2 U R2 Rw2 U R2 U' Rw2 U' Rw2 U' Rw' U' Rw2 U' R2 U Rw2 U Rw U' R2 (30,23) Y António Gomes []
R U' R' Rw2 U2 Rw U2 Rw2 U2 Rw2 U2 Rw U2 R2 Rw2 U R U R U' R' U' R2 (31,23) Y António Gomes []
R U' R' Rw2 U2 Rw' U2 Rw2 U2 Rw2 U2 Rw' U2 R2 Rw2 U R U R U' R' U' R2 (31,23) Y António Gomes []
Rw2 U R2 U R U' R' U R Rw U2 Rw2 U2 Rw2 U2 Rw U' R U R' U2 Rw2 U' (32,33) Y António Gomes []
Rw2 U2 R' U R U' Rw U2 Rw2 U2 Rw2 U2 R Rw U R' U' R U R2 U Rw2 U' (32,33) Y António Gomes []
R U2 R U R' Rw2 U2 Rw U2 Rw2 U2 Rw2 U2 Rw U2 R2 Rw2 U R2 U' R' U' R2//Safe (33,23) Y António Gomes []
R U2 R U R' Rw2 U2 Rw' U2 Rw2 U2 Rw2 U2 Rw' U2 R2 Rw2 U R2 U' R' U' R2//Safe (33,23) Y António Gomes []
##### Ob
r2 Uw2 r2 b2 U' r2 b2 U b2 Uw2 r2 (20,11) CG03 N Clément Gallet [50]
f2 Uw2 f2 r2 U' f2 r2 U r2 Uw2 f2 (20,11) N Clément Gallet []
b2 U2 b2 Dw2 M2 f2 Uw' u' M2 U M2 y2 (20,11) N Clément Gallet []
M2 U' M2 U' M U2 l2 U2 r2 Uw2 r2 Uw2 U2 M (22,13) N []
M2 U' M2 U' M' U2 r2 U2 l2 Uw2 l2 Uw2 U2 M' (22,13) N []
Uw2 r2 Uw2 r2 U2 l2 U M2 U' M' E2 M' D2 y2 (22,13) PKF03 N Per Kristen Fredlund [51]
u2 r2 Uw2 r2 U2 l2 U M2 U' (M' E2)2 (22,13) PKF03 N Per Kristen Fredlund [52]
l2 U2 l2 u2 f2 u2 U' f2 l2 U f2 U2 l2 (23,13) N Clément Gallet []
u2 l2 U2 l2 f2 u2 U' f2 l2 U f2 U2 l2 (23,13) N Clément Gallet []
R2 Uw2 Rw2 Uw2 R2 b2 U' r2 b2 U b2 Uw2 r2 (24,13) N Clément Gallet [53]
M2 U' r2 U2 F2 r2 F2 U2 l r U2 M' U' M2 (23,14) N []
M2 U r2 U2 r S2 l2 S2 r' U2 S2 l2 S2 U' M2//Safe (26,15) N Christopher Mowla [54]
M2 U r2 U2 r' S2 l2 S2 r U2 S2 l2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U r2 U2 l E2 l2 E2 l' U2 S2 l2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U r2 U2 l' E2 l2 E2 l U2 S2 l2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U l2 D2 r S2 l2 S2 r' D2 S2 r2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U l2 D2 r' S2 l2 S2 r D2 S2 r2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U l2 D2 l E2 l2 E2 l' D2 S2 r2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U l2 D2 l' E2 l2 E2 l D2 S2 r2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U r2 S2 D2 l S2 r2 S2 l' D2 l2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U r2 S2 D2 l' S2 r2 S2 l D2 l2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U r2 S2 D2 r E2 r2 E2 r' D2 l2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U r2 S2 D2 r' E2 r2 E2 r D2 l2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U l2 S2 U2 l S2 r2 S2 l' U2 r2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U l2 S2 U2 l' S2 r2 S2 l U2 r2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U l2 S2 U2 r E2 r2 E2 r' U2 r2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U l2 S2 U2 r' E2 r2 E2 r U2 r2 S2 U' M2//Safe (26,15) N Christopher Mowla []
M2 U r2 D2 F2 r2 F2 D2 r2 U2 M' F2 M F2 U M2//Safe (28,16) N Christopher Mowla [55]
F' L B D2 (L E2 M' f M E2 L' R d' R') D2 B' L' F//Safe (21,17) N Walter Randelshofer [56]
l r 3Dw' L R u' L' R' 3Dw M2 3Dw L' R' u' L R 3Dw' l r//Safe (20,19) N Christopher Mowla []
l' r' 3Dw' L' R' u L R 3Dw M2 3Dw L R u L' R' 3Dw' l' r'//Safe (20,19) N Christopher Mowla []
l r 3Dw L R u L' R' 3Dw' M2 3Dw' L' R' u L R 3Dw l r//Safe (20,19) N Christopher Mowla []
l' r' U y L' R' u' L R 3Dw' M2 3Dw' L R u' L' R' U y l' r'//Safe (20,19) N Christopher Mowla []
(F R U R' U' F') (r2 U2 r2 Uw2 r2 Uw2) (R' U' F R' F' R U R) U2 (28,21) N Antoine Cantin [57]
(F2 U2 M U Fw2 r2 u S' Rw2 u2) M' (u2 Rw2 S u' r2 Fw2 U' M' U2 F2) (33,21) N Christopher Mowla []
U Rw2 U2 R U' R' U Rw U2 Rw2 U2 Rw2 U2 Rw R' U' R U R' U' R2 U' Rw2 (31,22) Y António Gomes [58]
U Rw2 U' R2 U' R' U R U' Rw R' U2 Rw2 U2 Rw2 U2 Rw U R' U' R U2 Rw2 (31,22) Y António Gomes []
Rw' U Rw U R' Rw U' Rw' U R' U' R' U' R2 U Rw' U' R' Rw U R Rw U R' Rw' (24,23) Y António Gomes []
U Rw2 U2 Rw U' Rw2 R U2 Rw2 R' U2 Rw2 R U' Rw' U' R' U R' U' R2 U' Rw2 (29,23) Y António Gomes []
U Rw2 U' R2 U' R' U R' U' Rw' U' Rw2 R U2 Rw2 R' U2 Rw2 R U' Rw U2 Rw2 (29,23) Y António Gomes []
U Rw2 R2 U' R' U R' U2 Rw' U2 Rw2 U2 Rw2 U2 Rw' U R' U' R' U R U Rw2 (30,23) Y António Gomes []
U Rw2 R2 U' R' U R' U2 Rw U2 Rw2 U2 Rw2 U2 Rw U R' U' R' U R U Rw2 (30,23) Y António Gomes []
U Rw2 U R U R' U' R' U Rw' U2 Rw2 U2 Rw2 U2 Rw' U2 R' U R' U' Rw2 R2 (30,23) Y António Gomes []
U Rw2 U R U R' U' R' U Rw U2 Rw2 U2 Rw2 U2 Rw U2 R' U R' U' Rw2 R2 (30,23) Y António Gomes []
R2 U Rw U' Rw2 U' R2 U Rw2 U Rw' U Rw2 U Rw2 U R2 U' Rw2 R2 U' Rw2 R2 (30,23) Y António Gomes []
R2 U Rw' U' Rw2 U' R2 U Rw2 U Rw U Rw2 U Rw2 U R2 U' Rw2 R2 U' Rw2 R2 (30,23) Y António Gomes []
R2 U R U R' U' R' U' Rw2 R2 U2 Rw' U2 Rw2 U2 Rw2 U2 Rw' U2 Rw2 R U R' (31,23) Y António Gomes []
R2 U R U R' U' R' U' Rw2 R2 U2 Rw U2 Rw2 U2 Rw2 U2 Rw U2 Rw2 R U R' (31,23) Y António Gomes []
U Rw2 U2 R U' R' U Rw' U2 Rw2 U2 Rw2 U2 Rw' R' U' R U R' U' R2 U' Rw2 (32,33) Y António Gomes []
U Rw2 U' R2 U' R' U R U' Rw' R' U2 Rw2 U2 Rw2 U2 Rw' U R' U' R U2 Rw2 (32,33) Y António Gomes []
R2 U R U R2 U' Rw2 R2 U2 Rw' U2 Rw2 U2 Rw2 U2 Rw' U2 Rw2 R U' R' U2 R'//Safe (33,23) Y António Gomes []
R2 U R U R2 U' Rw2 R2 U2 Rw U2 Rw2 U2 Rw2 U2 Rw U2 Rw2 R U' R' U2 R'//Safe (33,23) Y António Gomes []
#### B permutation
• Note: This is commonly called the "W" permutation.
• A list of algorithms for the inverse case are unnecessary. Just do U2 (alg) U2 or y2 (alg) y2.
M2 U M' U2 M U l2 U2 r2 Uw2 r2 Uw2 U2 (20,12) N []
u2 S2 Uw' u' b2 R2 b2 Rw2 f2 r2 U S2 (21,12) N Clément Gallet []
F2 D' f2 d2 b2 D' B2 Dw2 3Uw' M2 3Uw F2 Uw2 (21,13) N Clément Gallet []
(r2 U2 r2 Uw2 r2 u2) (F2 U M' U2 M U F2) (22,13) PKF04 N Per Kristen Fredlund [59]
R2 Uw2 B2 R2 Uw2 B2 R2 U R2 B2 R2 U B2 Uw2 (26,14) CG04 Y Clément Gallet []
F2 Uw' u' M2 U' F2 D S2 Dw d Bw2 u2 b2 U2 y2 (22,14) N Clément Gallet []
(Rw2 F2 U2 r2 U2 F2 Rw2) (F2 U M' U2 M U F2)//Safe (24,14) N Christopher Mowla [60]
l2 D2 B2 Rw2 B2 D2 l2 F2 U' F2 L2 F2 U' L2 (26,14) N Clément Gallet []
y' M2 U' r2 b2 Rw2 Bw2 R2 Bw2 L2 U2 M2 U2 F2 D M2 x2 y (28,15) N Doug [61]
R' U R' U' R' U' R' U R U' Uw2 r2 Uw2 r2 U2 Rw2 (21,16) SP04 N Stefan Pochmann []
R' U R' U' R' U' R' U R U' Uw2 r2 Uw2 r2 U2 Rw2 (21,16) SP04 N Stefan Pochmann []
r2 (U2 r U2 S2)2 Rw2 S2 D' F2 D S2 D' F2 D R2//Safe (32,19) N Walter Randelshofer [62]
R2 U R' U' R2 U R U Rw2 U2 R' Rw U2 R' Rw2 U2 Rw2 U2 Rw U2 Rw2//Safe (30,20) Y António Gomes [63]
R2 U R' U' R2 U R U R' Rw2 U2 R' Rw U2 Rw2 U2 Rw2 U2 Rw U2 Rw2//Safe (30,20) Y António Gomes [64]
R2 U R' U' R2 U R U R' Rw2 U2 R' Rw' U2 Rw2 U2 Rw2 U2 Rw' U2 Rw2//Safe (31,21) Y António Gomes [65]
y R2 U' R' U' R U2 Rw U R Rw' U' Rw' U' Rw U Rw U' Rw' U' R Rw' U R Rw y' (24,22) Y António Gomes [66]
y' U' Rw2 U2 Rw' U Rw2 U' Rw2 U Rw' U Rw U Rw' U' Rw' U Rw U Rw' U Rw2 U' Rw2 U Rw U2 Rw2 y (36,28) Y Ben Whitmore [67]
(M2 U Fw2 r2 u r2 Uw2 S' U' B R B' R2 U) M' (U' R2 B R' B' U S Uw2 r2 u' r2 Fw2 U' M2) (41,29) N Christopher Mowla [68]
### Four Dedges (unoriented)
#### O permutation
##### Oa
F2 M F2 U' l2 U2 r2 Uw2 r2 U' Uw2 B2 M B2 (23,14) N Christopher Mowla []
y' U2 Rw2 U2 r2 Uw2 r2 y Uw2 M U' M2 U' F2 U2 M (24,14) N Christopher Mowla []
R2 S' (u2 r2 Uw2 r2 U2) Rw2 U' S2 U L2 S' L2 (24,14) N Per Kristen Fredlund [69]
R2 S (u2 r2 Uw2 r2 U2) Rw2 U S2 U' L2 S L2 (24,14) N Per Kristen Fredlund []
L2 S L2 U S2 U' Rw2 (U2 r2 Uw2 r2 u2) S R2 (24,14) N Per Kristen Fredlund []
L2 S' L2 U' S2 U Rw2 (U2 r2 Uw2 r2 u2) S' R2 (24,14) N Per Kristen Fredlund []
M2 D R' L' u' R L D2 R L y M2 u' M2 y' R' L' D M2 //Safe (22,17) N Christopher Mowla []
M2 D R L u R' L' D2 R' L' y M2 u M2 y' R L D M2 //Safe (22,17) N Christopher Mowla []
M2 D' R' L' y M2 d' M2 y' R L D2 R L d' R' L' D' M2 //Safe (22,17) N Christopher Mowla []
M2 D' R L y M2 d M2 y' R' L' D2 R' L' d R L D' M2 //Safe (22,17) N Christopher Mowla []
M2 U F2 D2 r F2 r2 F2 D2 l2 F2 D2 r2 D2 l U' M2 (30,17) N Christopher Mowla []
(M2 U Fw2 r2 u S' Rw2 u2) M' (u2 Rw2 S u' r2 Fw2 U' M2) (27,17) N Christopher Mowla []
(M2 U Fw2 r2 u' S' Rw2 u2) M' (u2 Rw2 S u r2 Fw2 U' M2) (27,17) N Christopher Mowla []
Rw U' Rw2 U' r2 U' Rw' U R' U2 R' U Rw' U' r2 U' Rw2 U' Rw U2 (26,20) N xyzzy [70]
##### Ob
B2 M' B2 Uw2 U r2 Uw2 r2 U2 l2 U F2 M' F2 (23,14) N Christopher Mowla []
M' U2 F2 U M2 U M' Uw2 y' r2 Uw2 r2 U2 Rw2 U2 y (24,14) N Christopher Mowla []
L2 S L2 U' S2 U Rw2 (U2 r2 Uw2 r2 u2) S R2 (24,14) N Per Kristen Fredlund [71]
L2 S' L2 U S2 U' Rw2 (U2 r2 Uw2 r2 u2) S' R2 (24,14) N Per Kristen Fredlund []
R2 S' (u2 r2 Uw2 r2 U2) Rw2 U S2 U' L2 S' L2 (24,14) N Per Kristen Fredlund []
R2 S (u2 r2 Uw2 r2 U2) Rw2 U' S2 U L2 S L2 (24,14) N Per Kristen Fredlund []
M2 D' L R y M2 u M2 y' L' R' D2 L' R' u L R D' M2//Safe (22,17) N Christopher Mowla []
M2 D' L' R' y M2 u' M2 y' L R D2 L R u' L' R' D' M2//Safe (22,17) N Christopher Mowla []
M2 D L R d L' R' D2 L' R' y M2 d M2 y' L R D M2//Safe (22,17) N Christopher Mowla []
M2 D L' R' d' L R D2 L R y M2 d' M2 y' L' R' D M2//Safe (22,17) N Christopher Mowla []
M2 U l' D2 r2 D2 F2 l2 D2 F2 r2 F2 r' D2 F2 U' M2 (30,17) N Christopher Mowla []
(M2 U Fw2 r2 u S' Rw2 u2) M (u2 Rw2 S u' r2 Fw2 U' M2) (27,17) N Christopher Mowla []
(M2 U Fw2 r2 u' S' Rw2 u2) M (u2 Rw2 S u r2 Fw2 U' M2) (27,17) N Christopher Mowla []
Rw' U Rw2 U r2 U Rw U' R U2 R U' Rw U r2 U Rw2 U Rw' U2 (26,20) N xyzzy [72]
#### B permutation
• Note: This is commonly called the "W" permutation.
• A list of algorithms for the inverse case are unnecessary. Just do U2 (alg) U2 or y2 (alg) y2.
y' M2 U' D2 r2 S2 l S2 r2 E2 r' U' M2 y'//Safe (19,12) N Christopher Mowla []
M2 U' r U D L2 U D S2 r S2 D' U' L2 D' M2//Safe (22,16) N Christopher Mowla []
y M2 U r' U' D' L2 U' D' S2 r' S2 D U L2 D M2 y'//Safe (22,16) N Christopher Mowla []
M2 D R2 D U r U' D' R2 U' D' S2 r S2 U M2//Safe (22,16) N Christopher Mowla []
y M2 D' R2 D' U' r' U D R2 U D S2 r' S2 U' M2 y'//Safe (22,16) N Christopher Mowla []
(M2 U')(r' E2 l E2)3 (U M2)//Safe (24,16) N Christopher Mowla []
(M2 U')(E2 r' E2 l)3 (U M2)//Safe (24,16) N Christopher Mowla []
(y x' M2 Fw f r2 u r2 S' u2) M (u2 S r2 u' r2 f' Fw' M2 x y') (25,17) N Christopher Mowla [73]
(y x' M2 Fw f r2 u' r2 S' u2) M (u2 S r2 u r2 f' Fw' M2 x y') (25,17) N Christopher Mowla []
(M2 U Fw2 r2 d r2 Uw2 S') M (S Uw2 r2 d' r2 Fw2 U' M2) (27,17) N Christopher Mowla []
(M2 U Fw2 r2 d' r2 Uw2 S') M (S Uw2 r2 d r2 Fw2 U' M2) (27,17) N Christopher Mowla []
## Two corner swaps
### Underscore ( _ ) permutation
F2 R2 B' D' B R2 F' U Fw2 F L2 f2 Lw2 f2 l2 U' (24,16) N Clément Gallet []
L' U Lw2 L B2 l2 Bw2 l2 b2 U' L2 F2 R' D' R F2 (24,16) N Clément Gallet []
y' B2 L U L' B2 R D' r2 F2 r2 Fw2 r2 f2 R D R2 y (25,16) PKF10 N Per Kristen Fredlund [74]
R U' R B2 L' D L B2 R2 U r2 F2 r2 Fw2 r2 f2 (25,16) N []
F2 L2 B D B' L2 F U' F f2 U2 L2 b2 L2 U2 f2 U (26,17) N Clément Gallet []
R U' 3Lw U2' L' B L U2' 3Lw2' B Lw2 F2 U2' l2' U2' F2 Lw2' x' (27,17) N Clément Gallet &Christopher Mowla []
(F2 D R2 U' R2 F2 U' F2 U F2 D')(r2 U2 r2 Uw2 r2 u2) (29,17) N Per Kristen Fredlund [75]
z r2 U2 R' U2 R' U2 R x U2 Rw2 U2 B2' L U2 L' U2 Rw2 U2' z' y' (29,17) N Christopher Mowla [76]
z' Rw2 x U2 R' U2 x' U2 R' U2 R U2 L' x U2 Rw2 U2 Rw2 U2 Rw2 x' U2 Rw2 z U (33,19) Y Christopher Mowla [77]
r2 U2 r2 Uw2 r2 u2 y' R U R' U' R' F R2 U' R' U' R U R' F' y (27,20) N []
y' (R U R' F') U' r2 U2 r2 Uw2 r2 Uw2 U' (R U R' U') R' F R2 U' R' U' y (28,22) N Stefan Lidström []
z' x Rw2 R' U2 Rw' U2 Rw2 U2 Rw2 U2 L Rw' U2 R' U2 R U2 L' R U2 R U2 Rw2 x' z (35,22) Y Christopher Mowla [78]
z' x Rw2 R' U2 Rw U2 Rw2 U2 Rw2 U2 L Rw U2 R' U2 R U2 L' R U2 R U2 Rw2 x' z (35,22) Y Christopher Mowla []
z' (Rw U L U' Rw' U L' U' F)9 z (81,81) Y Christopher Mowla [79]
### Slash (\) permutation
Rw2 f2 U2 Fw2 D Rw2 U2 Fw2 U' Fw2 L2 U2 B2 Lw2 U (27,15) N Clément Gallet [80]
Rw2 f2 U2 Bw2 D' Rw2 U2 Bw2 U Fw2 R2 D2 B2 Lw2 U (27,15) v1 N Clément Gallet []
Rw2 f2 U2 Fw2 U' Rw2 U2 Fw2 U Fw2 R2 U2 F2 Rw2 U (27,15) v2 N Michael Fung [81]
Rw2 f2 D2 Fw2 U' Rw2 U2 Bw2 U Bw2 R2 U2 F2 Rw2 U' (27,15) N Clément Gallet []
Rw2 f2 D2 Fw2 D' Rw2 U2 Bw2 U Bw2 L2 U2 B2 Lw2 U (27,15) N Clément Gallet []
Rw2 f2 D2 Bw2 U Rw2 U2 Fw2 U' Bw2 L2 D2 F2 Rw2 U' (27,15) N Clément Gallet []
Fw2 r2 D2 Lw2 U Bw2 D2 Lw2 D' Rw2 B2 U2 L2 Bw2 U' (27,15) N Clément Gallet []
Fw2 r2 D2 Rw2 U Bw2 D2 Rw2 D' Rw2 F2 D2 L2 Bw2 U (27,15) N Clément Gallet []
Rw2 F2 U2 y Rw2 U' Rw2 U D Lw2' U' Lw2' y' r2 U2 F2 Rw2 U (27,16) N Christopher Mowla [82]
Rw2 F2 U2 y Lw2' U Lw2' U' D' Rw2 U Rw2 y' r2 U2 F2 Rw2 U (27,16) N Christopher Mowla []
r2 F2 L' U2 Lw2 F2 R' D2 R2 U2 B2 R D2 B2 Rw2 U2 (29,16) Y Clément Gallet []
y' z (R2' Uw2' R2 U R2') y (R2 U2 R2' U' R2) y' (R2' Uw2' R2 3Dw' R2) Uw2 U2 x (29,16) reCornerX(BD) v1 N reThinking the Cube [83]
y' U2 Rw2 U2 R U2 F2 R2 F2 R' F2 U2 Rw2 U2 R' F2 r2 y (29,16) reCornerX(BD) v2 N reThinking the Cube [84]
y' Rw' r' S2 U R U' B2 D L' D Lw l D2 F2 Lw l D2 z2 y (22,17) N Frédérick Badie &Clément Gallet []
y' R2 D Rw2 U2' B2 r2 B2 U2' Lw2 R U' 3Lw U2 3Rw' U 3Rw U2 x' y (26,17) N Mario Laurent [85]
F2 L' U2 L R U2 R' F2 Lw2 F2 U2 Lw2' L' U2 F2 Lw2 U2 (28,17) Y Clément Gallet []
y' r2 F2 R' F2 U2 R2 U2 R' F2 R2 U2 Rw' r' U2 F2 Rw2 U2 y (30,17) N Christopher Mowla []
y' r2 B2 R' U2 Rw2 U2 B2 R' B2 r2 U2 Rw2 B2 U2 R' U2 Rw2 y (31,17) N Christopher Mowla [86]
L2 u2 F2 U2 L2 u2 F2 U' F2 L2 U2 L2 U L2 U2 F2 u2 U (32,18) N Clément Gallet []
y' R Rw2 F2 L U L' B D L D' B2 U B r2 U2 F2 Rw2 R' U2 y (25,19) FB14 N Frédérick Badie [87]
y' r2 U2 r2 Uw2 r2 Uw2 L' U R' U2 L U' L' R U R' U2 L U' R U y (29,21) N []
Rw2 R U2 R U2 R' U2 Rw U2 Rw2 U2 Rw2 U2 L Rw U2 R' U2 R U2 L' Rw2 (35,22) Y Christopher Mowla []
Rw2 R U2 R U2 R' U2 Rw' U2 Rw2 U2 Rw2 U2 L Rw' U2 R' U2 R U2 L' Rw2 (35,22) Y Christopher Mowla [88]
l2 U2 F2 l R U2 R' U2 L2 F2 L F2 Lw F2 U2 Lw2 U2 F2 U2 F2 L' U2 (38,22) N Christopher Mowla [89]
(F R U') (R' U' R U) (R' F') U r2 U2 r2 Uw2 r2 Uw2 U (R U R' U') (R' F R F') (30,25) N Stefan Lidström []
F' (Rw U' L' U Rw' U' L U2)9 F (83,74) N Christopher Mowla [90]
## Two corner swaps (only 2 supercube center piece exchange)
### Underscore ( _ ) permutation
x' Uw l S' D2 B d2 B' D2 S L F l2 F' D' Fw' D' Fw D2 Lw' Uw' Rw F' Lw' (28,23) N Tom Rokicki [91]
U Rw2 Bw2 Rw F Rw' Bw2 Rw F' L2 r U' b2 U L' F' L U' b2 U L' F M x (29,23) N Christopher Mowla [92]
z' Fw' L2 Uw F' U R U' F r2 F' U R' U' F r2 Fw Uw' L Uw Fw' Uw' L Fw L z (27,24) N Christopher Mowla []
z' Fw' L2 Uw r2 F' U R U' F r2 F' U R' U' F Fw Uw' L Uw Fw' Uw' L Fw L z (27,24) N Christopher Mowla []
y' (Rw' R2 U' R' U R' Rw U' R U2' R' U' R' Rw U R Rw' U' Rw' F Rw2 U' Rw' U') (Rw U Rw' F') y (27,25) Y Robert Yau []
y' Rw' U' R U r U' R' U R2 U R' U' r U r' F' Rw U Rw' U' Rw' F Rw2 U' Rw' U' y (28,26) N Robert Yau [93]
y' Rw' U' R U r U' R' U R2 U R' U' r U r' U' Rw' F Rw2 U' Rw' U' Rw U Rw' F' y (28,26) N Robert Yau [94]
y' Rw' U' R U r U' R' U R2 U R' U' r U R U' Lw Uw2 Rw' U' Rw Dw2 Rw' U Rw' F' x y' (29,26) N Robert Yau [95]
U Rw2 Bw2 Rw F Rw' Bw2 Rw F' D' f' D F2 D' f D F2 Rw L F' L' f' L F L' f (31,26) N Chris Hardwick [96]
y' Rw U Rw' U' Rw' F Rw2 U' Rw' U' Rw U Rw' R' F' r' F R F' r2 B' R B r' B' R' B y (29,27) N []
y' Rw U R' U' r' U R U' R2 U' R U r' U' R' Lw U Lw' U' Rw U Lw U' Lw' Rw' U Rw U y (29,28) N Christopher Mowla [97]
### Slash (\) permutation
F r2 F' U R' U' F r2 Fw Uw' L Uw Fw' Uw' L Fw L Fw' L2 Uw F' U R U' (27,24) N Christopher Mowla []
Fw R Fw' U' Fw2 Rw2 Fw' R' Fw Rw2 Fw' U2 R2 U Fw L2 Fw' U' R2 U' R' Fw L2 Fw2 (33,24) N Tom Rokicki [98]
F Rw' F Rw2 U' Rw' U' Rw' U' R U r U' R' U R2 U R' U' r U r' F' Rw U Rw' U' F' (30,28) N Robert Yau [99]
### Two X-center piece swap algorithms
• These algorithms, and other algorithms like them, are used to make two corner swap algorithms which only swap two X-center pieces on the 4x4x4 supercube.
y Rw U' Lw Uw2 Rw' U Rw Uw2 x Rw2 U y' (13,10) Y []
Fw' U' Rw Fw Rw' U' Rw Fw' Rw' U2 Fw U' (13,12) Y Christopher Mowla []
Rw U Rw' U' Rw' F Rw2 U' Rw' U' Rw U Rw' F' (15,14) Y []
Rw' U' Lw U Lw' U' Rw U Lw U' Lw' Rw' U Rw U (15,15) Y Christopher Mowla []
## Two corner swap and a dedge 3-cycle
### D permutation
#### Da
z2 F2 R2 Dw2 F2 U' R2 Uw2 U R2 U F2 Uw2 U M2 F2 U' (25,16) Y Clément Gallet []
F2 U Rw2 U' B2 U' F2 U B2 U' r2 F2 U2 Rw2 U' F2 U' (27,17) FB07 N Frédérick Badie []
y2 U' R2 b2 l2 Fw2 l2 F2 r2 L D L' B2 L D' L' F2 R2 (27,17) N Clément Gallet []
R2 U' f2 D2 L2 b2 D' R2 D L2 D' R2 D' f2 U R2 U' (27,17) N Clément Gallet []
y2 U' Uw' R U R' Uw' Rw2 U2' Rw2 R D' R U R' D R U Rw2 Uw R U' R' Uw y2 (27,23) N xyzzy [100]
#### Db
U F2 M2 Uw2 U' F2 U' R2 Uw2 U' R2 U F2 Dw2 R2 F2 z2 (25,16) Y Clément Gallet []
U F2 U Rw2 U2 F2 r2 U B2 U' F2 U B2 U Rw2 U' F2 (27,17) FB07 N Frédérick Badie []
R2 F2 L D L' B2 L D' L' r2 F2 l2 Fw2 l2 b2 R2 U y2 (27,17) N Clément Gallet []
U R2 U' f2 D R2 D L2 D' R2 D b2 L2 D2 f2 U R2 (27,17) N Clément Gallet []
y2 Uw' R U R' Uw' Rw2 U' R' D' R U' R' D R' Rw2 U2 Rw2 Uw R U' R' Uw U y2 (27,23) N xyzzy [101]
### K permutation
#### Ka
U R U' L U2 R' U L l2 F2 Lw2 Fw2 l2 Fw2 (20,14) N Clément Gallet []
Dw2 R2 D B2 Dw2 D' B2 D' R2 Dw2 D' B2 L2 U' L2 (23,15) Y Clément Gallet []
u2 r2 Uw2 r2 U2 Rw2 D R D' R F2 L' U L F2 (23,15) PKF06 N Per Kristen Fredlund [102]
d2 D' B2 L2 U d2 L2 D' B2 U R2 U R2 Dw2 L2 (24,15) N Clément Gallet []
y (Rw2' D' Rw U2 Rw' D Rw U2 Rw) U (Rw U2 Rw D Rw' U2 Rw D' Rw2') U2 y' (27,20) Y Daniel Sheppard [103]
R U R' F' R U R' U' R' F R2 U' R' U (r2 U2 r2 Uw2 r2 Uw2) (27,20) N Stefan Lidström []
#### Kb
Fw2 l2 Fw2 Lw2 F2 l2 L' U' R U2 L' U R' U' (20,14) N Clément Gallet []
L2 U L2 B2 Dw2 D R2 D B2 Dw2 D B2 D' R2 Dw2 (23,15) Y Clément Gallet []
F2 L' U' L F2 R' D R' D' Rw2 U2 r2 Uw2 r2 u2 (23,15) PKF06 N Per Kristen Fredlund [104]
L2 Dw2 R2 U' R2 U' B2 D L2 d2 U' L2 B2 D d2 (24,15) N Clément Gallet []
y' (Rw2 D Rw' U2 Rw D' Rw' U2 Rw') U' (Rw' U2 Rw' D' Rw U2 Rw' D Rw2') y U2 (27,20) Y Daniel Sheppard [105]
(r2 U2 r2 Uw2 r2 Uw2 U2) R U R' F' R U R' U' R' F R2 U' R' U' (27,20) N Stefan Lidström []
### P permutation
#### Pa
U' z' U2 B2 Lw2 U2 L' B2 Lw2' L B2 L U2 Lw2' L U2 z (23,15) CG05 Y Clément Gallet [106]
y U L2 3Uw' L2 U L2 y u2 l2 Uw2 l2 U2 Lw2 3Uw L2 U' L2 y2 (27,16) N Stefan Pochmann [107]
R2 U R2 U' R2 F2 U' (Fw2 r2)2 F2 r2 D R2 D' (27,16) PKF05 N Per Kristen Fredlund []
y2 L' U' L U L F' L2 U L (r2 U2 r2 Uw2 r2 Uw2) U' L' U' L F y2 (26,20) N Stefan Lidström []
#### Pb
z' U2 Lw2 L' U2 L' B2 Lw2' L' B2 L U2 Lw2' B2 U2 z U (23,15) CG05 Y Clément Gallet [108]
y2 L2 U L2 3Uw' Lw2 U2 l2 Uw2 l2 u2 y' L2 U' L2 3Uw L2 U' y' (27,16) N Stefan Pochmann [109]
R2 D' r2 F2 (r2 Fw2)2 U F2 R2 U R2 U' R2 D (27,16) PKF05 N Per Kristen Fredlund [110]
R U R' U (r2 U2 r2 Uw2 r2 Uw2) R' F R2 U' R' U' R U R' F' (27,20) N Stefan Lidström []
### b permutation
#### Ba
z' U2 Lw2' L U2 L F2 Lw2' L F2 L' U2 Lw2 F2 U2 z U' (23,15) CG05 Y Clément Gallet [111]
R2' U' R2 3Uw Rw2 U2 r2 Uw2 r2 u2 y R2 U R2' 3Uw' R2 U y' (27,16) N Stefan Pochmann [112]
R2 D r2 B2 (r2 Bw2)2 U' B2 R2 U' R2 U R2 D' (27,16) PKF05 N Per Kristen Fredlund []
y2 L' U' L U' (r2 U2 r2 Uw2 r2 Uw2) L F' L2 U L U L' U' L F y2 (27,20) N Stefan Lidström []
#### Bb
U z' U2 F2 Lw2' U2 L F2 Lw2 L' F2 L' U2 Lw2 L' U2 z (23,15) CG05 Y Clément Gallet [113]
y U' R2 3Uw R2 U' R2 y' u2 r2 Uw2 r2 U2 Rw2 3Uw' R2 U R2 (27,16) N Stefan Pochmann [114]
R2 U' R2 U R2 B2 U (Bw2 r2)2 B2 r2 D' R2 D (27,16) PKF05 N Per Kristen Fredlund []
R U R' U' R' F R2 U' R' (r2 U2 r2 Uw2 r2 Uw2) U R U R' F' (25,20) N Stefan Lidström []
### Q permutation
#### Qa
y' z U2 R' U2 Rw2 B2 R U2 Rw2 R' U2 R2 B2 Rw2 R U2 z' y U (25,16) CG11 Y Clément Gallet [115]
B2 L2 U2 L2 d2 D' R2 U2 F2 d2 D B2 U2 R2 U L2 R2 u2 (30,17) N Clément Gallet []
B2 d2 L2 U2 L2 D' R2 U2 F2 d2 D B2 U2 R2 U L2 R2 u2 (30,17) N Clément Gallet []
R' U' R F2 R' U R U F2 U' F2 U' Fw2 U2 f2 Uw2 f2 u2 (27,18) PKF11 N Per Kristen Fredlund []
y F' U' F L2 F' U F U L2 U' L2 U' Lw2 U2 l2 Uw2 l2 u2 y' (27,18) N Per Kristen Fredlund []
#### Qb
y' z U2 R U2 Rw2 F2 R' U2 Rw2 R U2 R2 F2 Rw2 R' U2 z' y U (25,16) CG11 Y Clément Gallet [116]
u2 F2 B2 U B2 U2 R2 D d2 L2 U2 B2 D' d2 F2 U2 F2 R2 (30,17) N Clément Gallet []
u2 F2 B2 U B2 U2 R2 D d2 L2 U2 B2 D' F2 U2 F2 d2 R2 (30,17) N Clément Gallet []
u2 l2 Uw2 l2 U2 Lw2 U' L2 U' L2 3Dw L U L' F2 L U' L' y (27,18) PKF11 N Per Kristen Fredlund []
y2 u2 r2 Uw2 r2 U2 Rw2 U' R2 U' R2 U F U F' R2 F U' F' y2 (27,18) N Per Kristen Fredlund [117]
## Two corner swap and a dedge 2 2-cycle
### C permutation
#### Ca
y' Rw2 F2 U2 r2 U' B2 U' F2 U B2 U' Rw2 U y (21,13) N Clément Gallet []
Lw2 U F2 U' B2 U F2 U l2 U2 B2 Lw2 U' (21,13) AO09 N Alexander Ooms [118]
U Lw2 B2 U2 l2 U' F2 U' B2 U F2 U' Lw2 (21,13) Inv of AO09 N Alexander Ooms [119]
y' Rw r B' R F2 R' B R U2 r2 U2 F2 Rw2 U y (20,14) N Frédérick Badie &Clément Gallet []
r2 F2 R2 D2 l2 D L2 D R2 D' L2 D F2 r2 U' (25,15) N Clément Gallet []
U Dw2 Lw2 B2 U F2 U' B2 U F2 U l2 U2 Lw2 Dw2 (24,15) N Frédérick Badie [120]
l r U2 x U' L U' R2 U L' U' Rw2 U2 x' U2 l' r' U' (21,16) N Christopher Mowla []
#### Cb
y Rw2 B2 U2 r2 U F2 U B2 U' F2 U Rw2 U' y' (21,13) N Clément Gallet []
Lw2 U' B2 U F2 U' B2 U' l2 U2 F2 Lw2 U (21,13) AO09 N Alexander Ooms [121]
U' Lw2 F2 U2 l2 U B2 U F2 U' B2 U Lw2 (21,13) Inv of AO09 N Alexander Ooms [122]
y Rw2 R F R' B2 R F' R' U2 r2 U2 B2 Rw2 U' y' (20,14) N Frédérick Badie &Clément Gallet [123]
r2 B2 R2 D2 l2 D' L2 D' R2 D L2 D' B2 r2 U (25,15) N Clément Gallet []
U Dw2 Rw2 B2 U' F2 U B2 U' F2 U' r2 U2 Rw2 u2 y2 (24,15) N Frédérick Badie [124]
l' r' U2 x' U L' U R2 U' L U Rw2 U2 x U2 l r U (21,16) N Christopher Mowla []
### H permutation
L2 Uw2 L2 U' L2 D' B2 R2 d2 U' R2 D B2 u2 U (24,15) N Clément Gallet [125]
L2 Uw2 L2 U L2 3Uw L2 F2 U z l2 F2 L' U2 Lw2 z' U y (25,15) CG08 N Clément Gallet []
R2 D' F2 D R2 B2 D L2 D' Bw2 U2 b2 Uw2 b2 Uw2 U' (26,16) N Kenneth Gustavsson [126]
Lw2 U F2 L D' L D L' U L U' L' U l2 U2 F2 Lw2 (23,17) AO08 N Alexander Ooms [127]
Rw2 U2 r2 Uw2 r2 u2 D' F2 U F2 R2 U R2 U' R2 D (27,16) N Per Kristen Fredlund [128]
### Theta (θ) permutation
R2 U B2 Uw2 R2 U' B2 Uw2 U B2 U2 R2 Uw2 U' R2 U' (25,16) Y Clément Gallet []
z F2 Rw2 R' F2 R2 U2 Rw2 R U2 R' F2 Rw2 U2 R F2 z' U (25,16) CG12 Y Clément Gallet [129]
u2 D F2 U R2 U2 u2 F2 D' F2 U L2 D2 B2 U' u2 R2 y2 (29,17) N Clément Gallet []
R' B' R F D' R' D B R Fw' f' R2 f2 Rw2 f2 Rw2 U R2 (24,18) N Per Kristen Fredlund &Clément Gallet []
F U F' R2 F U' F' U' R2 U f2 R2 (f2 Rw2)2 U R2 (27,18) PKF12 N Per Kristen Fredlund [130]
F R U' R' U' R U R' F' (r2 U2 r2 Uw2 r2 u2) R U R' U' R' F R F' (29,23) N Stefan Lidström []
### Xi (Ξ) permutation
Uw d B2 D M2 U L2 B2 U B2 Uw2 R2 D F2 Uw2 3Lw2 y' (26,16) N Clément Gallet []
R2 Uw2 U' M2 U R2 F2 Uw2 U' F2 U2 3Rw2 Uw2 U' F2 U2 R2 U (28,18) Y Clément Gallet []
Uw2 Lw2 U' B' U B U2 F U' F' U F' L2 F Lw2 U2 Lw2 Uw2 U' (26,19) FB13 Y Frédérick Badie []
U L2 U2 F2 D B2 u2 D' F2 L2 F2 D B2 D u2 R2 D2 F2 d2 (33,19) N Clément Gallet []
Dw2 l' r' F' L' B' L F L' R' F' R B R' F' l2 F2 Lw2 Bw2 F' x (24,20) N Frédérick Badie &Clément Gallet []
U Uw2 Rw2 U2 Rw2 R' D R' U' R (U2 D') R' D' R U2 R' (U' D) Rw2 Uw2 (25,20) N xyzzy [131]
z Uw2 r2 Uw2 r2 U2 (Lw' Rw' z') U R' U R U' R B2 R' U R' U' R U B2 U y' (29,21) N Matthew Sheerin [132]
z Uw2 r2 Uw2 r2 U2 (Lw' Rw' z') U R' U R U' R B2 R' U R' U' R (z 3Lw) U2 z' U (29,21) N Matthew Sheerin [133]
## The shortest PLL parity fixes in BQTM
• Note that these are the only PLL parity algorithms in this section which only contain two inner layer slice quarter turns.
• This alg set is for theoretical purposes only, as they scramble the pseudo 3x3x3 state of the 4x4x4.
Rw U D L2 U D S2 Rw (10,8) Y Christopher Mowla []
Rw' U D L2 U D S2 Rw' (10,8) Y Christopher Mowla []
Rw U' D' L2 U' D' S2 Rw (10,8) Y Christopher Mowla []
Rw' U' D' L2 U' D' S2 Rw' (10,8) Y Christopher Mowla []
Rw U D R2 U D S2 Rw (10,8) Y Christopher Mowla []
Rw' U D R2 U D S2 Rw' (10,8) Y Christopher Mowla []
Rw U' D' R2 U' D' S2 Rw (10,8) Y Christopher Mowla []
Rw' U' D' R2 U' D' S2 Rw' (10,8) Y Christopher Mowla [134]
Rw U D S2 U' D' S2 Rw' (10,8) Y Christopher Mowla []
Rw' U D S2 U' D' S2 Rw (10,8) Y Christopher Mowla []
Rw U' D' S2 U D S2 Rw' (10,8) Y Christopher Mowla []
Rw' U' D' S2 U D S2 Rw (10,8) Y Christopher Mowla []
• Taking the first algorithm, for example, and solving back the outer layers, we see that these short parity fixes were made from a supercube safe unoriented 4-cycle of dedges in M:
r U D L2 U D S2 r S2 D' U' L2 D' U' (18,14) N Christopher Mowla []
• However, note that this is not the shortest solution to this 4x4x4 position, as the following is a shorter supercube safe algorithm.
(l E2 r' E2)3//Safe (18,12) N Christopher Mowla []
# Pure Flips (Single Parity)
• All algorithms in this "pure flips" section are single parity algorithms.
## One dedge flip
### Shortest half turn move algorithms
• It turns out that the shortest algorithms in the half turn move metric (in both the single slice turn metric and the block turn metric) have a length of 15.
• There are exactly 288 algorithms (with a length of 15 half turns) which flip the upper front dedge and affect the top or front centers of the 4x4x4 supercube.
They all happen to be 25 BQTM as well; and they all happen to be single slice turn-based algorithms (algorithms which predominately consist of single slice turns).
All 288 (25,15) algorithms can be obtained by running the 3x3x3 Classic Setup through a 3x3x3 solver like Cube Explorer and choosing all outputted 15f algorithms which are confined to the move set . Simply convert the turns of the L and R faces into their corresponding inner slice turns and select the resulting algorithms which flip a single dedge on the 4x4x4.
• These 288 (25,15) algorithms have been divided into 36 clusters. There are exactly 8 algorithms in each cluster.
• Every cluster essentially represents a single unique algorithm path, as move transformations (from cube rotations, mirrors, and inverses) can be done to any algorithm in a given cluster to obtain the 7 other algorithms in its cluster. To avoid human error, a custom program was written to perform these move transformations to the first algorithm listed in each cluster and to label the remaining 7 algorithms in its cluster as the transformation they are of the first algorithm.
• Even further, several of these clusters are categorized as being in the same group number. This is because, although transformations are strictly restricted to being in the 36 clusters, some of these clusters of algorithms are directly related to each other by either conjugation and/or by cyclic shifting (a special type of conjugation).
Take for example the first algorithm in the first cluster under Group 1 of non-symmetrical algorithms, r' U2 l F2 l' F2 r2 U2 r U2 r' U2 F2 r2 F2.
Shifting ("rotating the algorithm's moves") 5 times in a clockwise manner gives: F2 r2 U2 r U2 r' U2 F2 r2 F2 r' U2 l F2 l' (animation).
This is a single dedge flip of the upper back dedge.
Rotating about y2 gives B2 l2 U2 l U2 l' U2 B2 l2 B2 l' U2 r B2 r', which is the third algorithm in the first cluster under Group 3 of non-symmetrical algorithms.
• This is why Group 3 is called "Group 3: Cyclic Shift of Group 1", for example.
• Even further, all unique non-symmetrical algorithms are directly related to each other by transformations (resulting from inserting cube rotations in between moves of the algorithms). The same holds true for all unique symmetrical algorithms.
• This implies that, despite that there are 288 15 half turn move algorithms which solve the upper front dedge pure flip case, there are actually only two fundamentally unique (25,15) algorithm paths: the symmetrical approach and the non-symmetrical approach.
• This was shown by Christopher Mowla in 2012.[135].
Additional Terminology An algorithm is called symmetrical if its first and last moves are inverses of each other (or if it can be rewritten as such, should its first and/or last move cancel with the second and/or second to last move in the algorithm, respectively). Symmetrical algorithms are conjugates. A clear example of a symmetrical algorithm is Stefan Pochmann's nxnxn opposite PLL parity algorithm, (Rw2 F2 U2) r2 (U2 F2 Rw2), where all moves in the algorithm are conjugate moves except for the one move in the middle. Although symmetrical algorithms are technically conjugates of non-symmetrical algorithms, non-symmetrical algorithms are algorithms which are solely the result of a composition of one or more separate algorithm pieces, which all together accomplish the desired task. (No "conjugate assistance" is used.) In practice, human creation of symmetrical algorithms requires more trial and error of different paths in both creation of the base (the base is defined as the move sequence B in A B A') and final setup moves, whereas the creation of non-symmetrical algorithms requires having knowledge of forming different pieces individually and knowing how to combine them. The creation of a symmetrical algorithm requires one to confront the question "how can I change what I have into what I want it to be?" (video example), whereas the creation of a non-symmetrical algorithm requires one to confront the question "I have already achieved a portion of the task, but what is the missing piece?" (written example).
#### (Non-symmetrical algorithms)
##### Group 1
r' U2 l F2 l' F2 r2 U2 r U2 r' U2 F2 r2 F2 (25,15) Alg.1(v1) Y Frédérick_Badie [136]
l U2 r' F2 r F2 l2 U2 l' U2 l U2 F2 l2 F2 (25,15) Mirror[Alg.1(v1)] Y Frédérick_Badie []
F2 r2 F2 U2 r U2 r' U2 r2 F2 l F2 l' U2 r (25,15) Inverse[Alg.1(v1)] Y Frédérick_Badie []
l' F2 r U2 r' U2 l2 F2 l F2 l' F2 U2 l2 U2 (25,15) Rotation[Alg.1(v1)] Y Frédérick_Badie []
F2 l2 F2 U2 l' U2 l U2 l2 F2 r' F2 r U2 l' (25,15) Mirror[Inverse[Alg.1(v1)]] Y Frédérick_Badie []
r F2 l' U2 l U2 r2 F2 r' F2 r F2 U2 r2 U2 (25,15) Mirror[Rotation[Alg.1(v1)]] Y Frédérick_Badie []
U2 l2 U2 F2 l F2 l' F2 l2 U2 r U2 r' F2 l (25,15) Inverse[Rotation[Alg.1(v1)]] Y Frédérick_Badie []
U2 r2 U2 F2 r' F2 r F2 r2 U2 l' U2 l F2 r' (25,15) Mirror[Inverse[Rotation[Alg.1(v1)]]] Y Frédérick_Badie []
r' U2 r U2 l' U2 l2 F2 r F2 l' U2 F2 r2 F2 (25,15) Alg.1(v2) Y Frédérick_Badie []
l U2 l' U2 r U2 r2 F2 l' F2 r U2 F2 l2 F2 (25,15) Mirror[Alg.1(v2)] Y Frédérick_Badie []
F2 r2 F2 U2 l F2 r' F2 l2 U2 l U2 r' U2 r (25,15) Inverse[Alg.1(v2)] Y Frédérick_Badie []
l' F2 l F2 r' F2 r2 U2 l U2 r' F2 U2 l2 U2 (25,15) Rotation[Alg.1(v2)] Y Frédérick_Badie []
F2 l2 F2 U2 r' F2 l F2 r2 U2 r' U2 l U2 l' (25,15) Mirror[Inverse[Alg.1(v2)]] Y Frédérick_Badie []
r F2 r' F2 l F2 l2 U2 r' U2 l F2 U2 r2 U2 (25,15) Mirror[Rotation[Alg.1(v2)]] Y Frédérick_Badie []
U2 l2 U2 F2 r U2 l' U2 r2 F2 r F2 l' F2 l (25,15) Inverse[Rotation[Alg.1(v2)]] Y Frédérick_Badie []
U2 r2 U2 F2 l' U2 r U2 l2 F2 l' F2 r F2 r' (25,15) Mirror[Inverse[Rotation[Alg.1(v2)]]] Y Frédérick_Badie []
r B2 r' U2 r U2 r2 F2 r' D2 l D2 B2 r2 F2 (25,15) Alg.1(v3) Y Frédérick_Badie []
l' B2 l U2 l' U2 l2 F2 l D2 r' D2 B2 l2 F2 (25,15) Mirror[Alg.1(v3)] Y Frédérick_Badie []
F2 r2 B2 D2 l' D2 r F2 r2 U2 r' U2 r B2 r' (25,15) Inverse[Alg.1(v3)] Y Frédérick_Badie []
l D2 l' F2 l F2 l2 U2 l' B2 r B2 D2 l2 U2 (25,15) Rotation[Alg.1(v3)] Y Frédérick_Badie []
F2 l2 B2 D2 r D2 l' F2 l2 U2 l U2 l' B2 l (25,15) Mirror[Inverse[Alg.1(v3)]] Y Frédérick_Badie []
r' D2 r F2 r' F2 r2 U2 r B2 l' B2 D2 r2 U2 (25,15) Mirror[Rotation[Alg.1(v3)]] Y Frédérick_Badie []
U2 l2 D2 B2 r' B2 l U2 l2 F2 l' F2 l D2 l' (25,15) Inverse[Rotation[Alg.1(v3)]] Y Frédérick_Badie []
U2 r2 D2 B2 l B2 r' U2 r2 F2 r F2 r' D2 r (25,15) Mirror[Inverse[Rotation[Alg.1(v3)]]] Y Frédérick_Badie []
r B2 r' U2 r U2 l2 B2 r' U2 l U2 F2 l2 F2 (25,15) Alg.1(v4) Y Frédérick_Badie []
l' B2 l U2 l' U2 r2 B2 l U2 r' U2 F2 r2 F2 (25,15) Mirror[Alg.1(v4)] Y Frédérick_Badie []
F2 l2 F2 U2 l' U2 r B2 l2 U2 r' U2 r B2 r' (25,15) Inverse[Alg.1(v4)] Y Frédérick_Badie []
l D2 l' F2 l F2 r2 D2 l' F2 r F2 U2 r2 U2 (25,15) Rotation[Alg.1(v4)] Y Frédérick_Badie []
F2 r2 F2 U2 r U2 l' B2 r2 U2 l U2 l' B2 l (25,15) Mirror[Inverse[Alg.1(v4)]] Y Frédérick_Badie []
r' D2 r F2 r' F2 l2 D2 r F2 l' F2 U2 l2 U2 (25,15) Mirror[Rotation[Alg.1(v4)]] Y Frédérick_Badie []
U2 r2 U2 F2 r' F2 l D2 r2 F2 l' F2 l D2 l' (25,15) Inverse[Rotation[Alg.1(v4)]] Y Frédérick_Badie []
U2 l2 U2 F2 l F2 r' D2 l2 F2 r F2 r' D2 r | 25,985 | 63,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-25 | latest | en | 0.94873 |
http://www.expertsmind.com/library/draw-the-logic-diagram-of-a-2-bit-demultiplexer-5842853.aspx | 1,675,604,841,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00072.warc.gz | 61,864,544 | 16,031 | ### Draw the logic diagram of a 2-bit demultiplexer
Assignment Help Computer Engineering
##### Reference no: EM13842853
1. Using the 3-variable multiplexer chip of figure, implement a function whose output is the parity of the inputs. That is, the output is 1 if and only if an even number of inputs are 1.
2. The 3-variable multiplexer chip, shown in figure, is actually capable of computing an arbitrary function of Jour Boolean variables. Describe how, and as an example, draw the logic diagram for the function that is 0 if the English word for the truth table row has an even number of letters and 1 if it has an odd number of letters.
3. Draw the logic diagram of a 2-bit demultiplexer, a circuit whose single input line is steered to one of the four output lines depending on the state of the two control lines.
4. Draw the logic diagram of a 2-bit encoder, a circuit with four input lines, exactly one of which is high at any instant, and two output lines whose 2-bit binary value tells which input is high.
5. A common chip is a 4-bit adder. Four of these chips can be hooked up to form a 16-bit adder. How many pins would you expect the 4-bit adder chip to have? Why?
### Previous Q& A
#### Calculate the formula weight of sucrose
Calculate the formula weight of sucrose
#### Causes of the downward trend by surveying
Sue Taylor, Director of Global Industrial Sales, is concerned by a deteriorating sales trend. Specifically, the number of customers is stable at 1,500, but they are purchasing less each year. She orders her staff to search for causes of the downwa..
#### What is the molarity?
What is the molarity?
#### How and why did the tokugawas isolate japan?
How and why did the Tokugawas isolate Japan?
#### What is the benefit of using a change model
Which of the following is true about change management. What is the benefit of using a change model
#### Actionable strategic plan for riordan manufacturing
Prepare a convincing and comprehensive presentation of your team's actionable strategic plan for Riordan Manufacturing, one that would take 20-30 minutes to deliver to the board of directors, including intentional and active board involvement
#### What are the needs of the priority population
Who is the priority population? what are the needs of the priority population? which subgroups within the priority population have greatest need? where are the subgroups located geographically
#### Important feature of digital cameras
An important feature of digital cameras is battery life, the number of shots that can be taken before the battery needs to be recharged. The file contains battery life information for 29 subcompact cameras and 16 compact cameras. (Data extracted f..
#### How they are used for measuring quality
Describe how they are used for measuring quality by at least four professionals in a health care organization. You must include at least one administrative health professional, one midlevel provider, one physician, and one auxiliary staff member
#### Develop a leadership action plan
For this last segment to your project, you will develop a leadership action plan. In 1-3 pages, you will describe the activities you will undertake over each of the following: next month, quarter, one year, and three years to improve and maximize ..
### Similar Q& A
#### Listing each prompt that is used in this program.
Write down a new program in pseudo-code, it should input-and calculate the average of 5 numbers entered.
#### Writing report with apa format
Select one of the following languages: Modula, Pascal, Ada, C++, C, C#, Java or other instructor-approved language.
#### Describe a situation where a traditional relational
1.describe a situation where a traditional relational database management system would not be suitable and you would
the password should not be same as employee ID and must contain 4-6 characters
#### Program is internet based and it is expected to boost sale
Sunnyville Inc. is in the business of selling home appliances. Sunnyville Inc. has over 50 sales agents all over the country. The management at Sunnyville Inc. has decided to implement an online sales program. The new sales program is Internet bas..
#### Explain the steps you have taken to maintain and redesign
write a 200- to 300-word short-answer response to the followingdescribe the steps you have taken to maintain and
#### Breaking the version of cipher
Assume an improved version of Vigen ere cipher in which in place of utilizing several shift ciphers, several mono-alphabetic substitution ciphers are utilized.Display how to break this version of the cipher.
#### How to write code for selection sort, insertion sort
How to write code for selection sort, insertion sort. Using your performance of selection, bubble and insertion sort, add a counter in an appropriate place so as to measure the runtime of your code for example this capacity be a counter to track ..
#### Describe the mechanics of Buffer overflows
Prepare a complete tutorial, including an analogy to describe the mechanics and a graphic to support your analogy, on one of the subsequent areas
#### Review sectioncontemporary hardware platform trends and
review sectioncontemporary hardware platform trends and section contemporary software platform trends in of management
#### Simulate the transmssions of information by manually typing
modify a menu that allows the administrator to simulate the transmssions of information by manually typing in the login or logoff data. whenever someone logs in or out the display should be updated
#### Describe disparity between vulnerability, threat and control
Describe disparity between vulnerability, threat and control | 1,170 | 5,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-06 | latest | en | 0.910929 |
https://codesjava.com/algebraic-solutions-of-linear-inequalities | 1,580,125,506,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251700675.78/warc/CC-MAIN-20200127112805-20200127142805-00305.warc.gz | 384,172,854 | 16,457 | # Algebraic solutions of linear inequalities
Algebraic solutions of linear inequalities in one variable and their representation on the number line
Recalling the definition of linear inequalities, “it is an equation that carries a linear function.” It could further be expressed as the relation between two algebraic expressions that are represented using inequality symbols. It is a concept similar to linear equations. However, in place of the equality sign, there are inequality symbols. For instance, consider:
(Linear equation)
(Linear inequality)
## Solving Linear Equations in One Variable
Solving linear inequalities is much similar to solving linear equations (i.e. step-by-step). While solving linear inequalities the following rules need to be considered:
Rule-1: Equal numbers may be added or subtracted from both sides of the considerable inequality without having any effect on the sign of inequality.
Rule-2: Both the sides of an inequality could be divided or multiplied to the same positive number. However, when this is done with a negative number, the sign of inequality would be reversed. For instance, – 8 < – 7, whereas (– 8) (– 2) > (– 7) (– 2) i.e. 16 > 14 and 3 > 2 while – 3 < – 2.
## Graphical representation of inequalities in one variable
The solutions of inequalities in one variable could be the graphed on the Euclidean plane. While dealing with inequalities in one variable, the graph would probably be a number line. Since there is only one variable, it would be represented on the number line. This process would be the same as plotting a number on the number line.
Steps for graphing a solution on the number line
• We would first need to solve the inequality, applying the rules of inequality to find the solution.
• Make a number line
• Plot the point on the number line
1. To represent strict inequality (which includes > (greater than) and < (less than) signs), use open circles. This would indicate that the points plotted on the number line are not to be included in the range of solutions for the considerable inequality.
2. For representing slack inequalities (which include ≥ (greater than or equal to) and ≤ (less than or equal to) signs) use closed or filled circles. This would be an indication that the points plotted on the number line are to be included in the solution range of the inequalities.
• Further, shade the number line according to the signs of inequality.
• For the signs > and ≥ shade the left of the point marked on the number line.
• For signs < and ≤, shade the right of the point marked on the number line.
Given below are a few examples for the same:
The shaded portion on the number line is known to represent the range of true values for the considerable inequality. | 584 | 2,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2020-05 | latest | en | 0.94897 |
https://oeis.org/A104340 | 1,718,307,979,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861488.82/warc/CC-MAIN-20240613190234-20240613220234-00534.warc.gz | 393,873,542 | 3,980 | The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A104340 Numbers n such that (digital reversal of n) - n = 9. 1
12, 23, 34, 45, 56, 67, 78, 89 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The sequence is complete: there is no other such number (in base 10). LINKS Table of n, a(n) for n=1..8. Tanya Khovanova, Non Recursions EXAMPLE 98-89=9 MATHEMATICA Select[Range[1000], FromDigits[Reverse[IntegerDigits[#]]]-#==9&] (* Harvey P. Dale, Aug 26 2011 *) Select[Range[1000], IntegerReverse[#]-#==9&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 28 2017 *) CROSSREFS Cf. A104341. Sequence in context: A088783 A029756 A346508 * A367362 A136414 A017401 Adjacent sequences: A104337 A104338 A104339 * A104341 A104342 A104343 KEYWORD base,nonn,fini,full AUTHOR Zak Seidov, Mar 02 2005 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified June 13 15:42 EDT 2024. Contains 373389 sequences. (Running on oeis4.) | 431 | 1,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-26 | latest | en | 0.725385 |
https://www.experts-exchange.com/questions/20526331/Use-of-Operator-in-C.html | 1,532,049,940,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591455.76/warc/CC-MAIN-20180720002543-20180720022543-00156.warc.gz | 900,417,066 | 16,943 | • C
# Use of Operator &= in C
Hi,
Can anyone please let me know the use/meaning of operator &=
eg from a book:
#define L_FIX
#define X_BIAS 0x0010
#define REG_2 2
#define l_tone_on 1
{
#ifdef L_FIX
if ((addr == REG_2) && (l_tone_on)) {
return;
}
}
#endif
}
Thanks
###### Who is Participating?
I wear a lot of hats...
"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.
Commented:
if u have any done electronics courses .. then u might have idea of what's the meaning of AND
like 1 AND 0 = 0
0 AND 1 = 0
0 AND 0 = 0
1 AND 1 = 0
this '&' operator in C also same thing .. the difference being it works on the binary representation of the number
2 in binary will be '10'
3 will be '11'
so 2 & 3 will be 11 & 10 = 10 = 2
and in above case when u do
a&=b; this means
a= a&b;
operate & operator on 'a' and 'b' and assign it to 'a'
& and | operators are used to set the particular bit ( specified in the MASK ) to 1 OR 0.. and also to know whats the bit value
i hope u can now understand how it is being used in above particular case..
0
Experts Exchange Solution brought to you by
Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.
Commented:
>>the difference being it works on the binary representation of the number
actually no difference :P
0
Author Commented:
Thanks for excellent explanation
0
###### It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
C
From novice to tech pro — start learning today.
Question has a verified solution.
Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.
Have a better answer? Share it in a comment. | 586 | 2,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-30 | latest | en | 0.906297 |
https://libraryofessays.com/cumulative-gpa-calculator | 1,600,467,622,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400189264.5/warc/CC-MAIN-20200918221856-20200919011856-00685.warc.gz | 505,562,198 | 21,061 | Free Cumulative GPA Calculator
Calculate Scores with Experts Tips
Learn How to Use Cumulative GPA Calculator Wit Pro Tips
Cumulative GPA is one of the major factors that determine the placement of college admission and award of scholarships. Due to that critical significance, many students are ever looking for ways to assess their score and if possible work toward boosting their overall GPA. As a student, you will most times receive a transcript for a particular semester with an indicated cumulative GPA. However, you might doubt its authenticity. In such a case, it would be advisable to use a GPA calculator to confirm if they are the correct values.
Generally, most calculations related to average grade point will be seeking to find either the unweighted or weighted GPA. Decades ago, it was practically effortless to calculate a cumulative GPA. Changes into the education sector introduced new methods of estimating for a student’s GPA. However, it would be factual to say that the new metrics proved confusing as each institution formulated their procedure of calculating the cumulative.
What Is a Cumulative GPA? Basics Factors Considered During Calculations
Want to know the GPA cumulative meaning? It is the overall grade point average generated from dividing the total quality points earned in all attempted courses by the aggregate credit hours earned in the entire program. Many schools deploy different cumulative GPA calculator online methods to calculate student's score. You should not worry though as you can convert any GPA values to the scale of your choice.
Many students confuse cumulative GPA with the overall GPA. Although the two share some attributes, they refer to different values. Overall GPA is defined as the average of all cumulative GPA earned in courses taken in all semesters. Another difference between the two metrics is that the overall GPA takes into consideration all the grades irrespective of the institution as opposed to cumulative GPA that takes into account only a particular learning facility.
The hours that a student dedicated to the studies each week translates to credits points. So, this means the more hours in class earns you more GPA points. The credit points are multiplied by the grade points earned for each course during the process of calculating the cumulative GPA.
No matter the grading scale used, a GPA calculator cumulative will help you to estimate your score. Consequently, you can conveniently track your performance. The grades indicate on the transcript is interpreted as a measure of performance. As such, you need to learn the factors that will affect the value of your GPA. If you score a cumulative GPA of 4, it means you have secured an A.
Unweighted cumulative GPA account for the average course grades awarded at the end of a particular semester. The calculations are based on the 4.0 grading scale against letter grades equivalents ranging from A-F. To find a weighted GPA score divide the sum of honor points by the number of semester classes accomplished.
On the other hand, the weighted cumulative GPA applies to students tested on honor, BI, or AP courses. 5.0 grading scale is used to establish the cumulative GPA of a student on the weighted approach. Calculations are done the same as that of cumulative GPA only that the points for the honor or BI are assigned a higher point value during calculations.
The number of credits points earned at the end of the term will drastically affect your final cumulative GPA. More credit hours per week will accord you a higher grade points average. Also, taking more classes in a specific semester means that the added points will be reflected in the results. You can calculate the cumulative GPA for all semesters using the provided calculator.
As indicated on the tool, you are provided with fields to enter your course credits, the number of semesters and grades. Once done filling the details in each row, authorize the calculator to initiate computation of your GPA. Furthermore, one can convert any grading scale values into a 4.0 scale as it is commonly regarded as the standard system in most high schools and states.
How to Calculate Cumulative GPA Using Different Grading Scales
Would you like to know how to calculate your cumulative GPA using online calculators? All that is required of you is the value of letter grades and credit hours for each course. Once you enter the figures in each provided field, you can get your result within seconds. You don’t have to spend hours performing manual arithmetic to know the academic score. Follow this step by step process to find a cumulative GPA using the calculator:
• Enter the credit and grade points earned for each course. You may click on the toggle button to add more rows to accommodate extra details.
• Select whether you want a ‘weighed’ or’ unweighted’ cumulative GPA.
• The calculator will compute the values automatically.
• Hover to the bottom and select, ‘calculate cumulative GPA’ button.
• Your current cumulative GPA will be displayed
To understand how the weighted and unweighted GPA is calculated, you must refer to the conversion table formulated for each of the two grading scales. You can skip the field, 'course number' as it is optional. Make sure you enter the correct values for your credit hours and grades points as they are the main elements factored in the calculation. Make use of the ‘+' and '-' buttons to add or remove extra fields or columns. You can add as many rows as possible.
We hope that our provided information has been beneficial in outlining all you need to know about cumulative GPA calculator. We expect that the detailed guidelines have effectively illustrated the step to follow to calculate your cumulative GPA using our free calculator. The in-built tool can assist you to find cumulative GPA without any effort on your side. Feel free to utilize its efficiency to approximate your academic scores.
New Essays | 1,125 | 5,965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-40 | latest | en | 0.950485 |
https://oeis.org/A014298 | 1,575,660,306,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540490743.16/warc/CC-MAIN-20191206173152-20191206201152-00166.warc.gz | 493,483,740 | 4,270 | This site is supported by donations to The OEIS Foundation.
Please make a donation to keep the OEIS running. We are now in our 55th year. In the past year we added 12000 new sequences and reached 8000 citations (which often say "discovered thanks to the OEIS"). We need to raise money to hire someone to manage submissions, which would reduce the load on our editors and speed up editing. Other ways to donate
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A014298 a(n) = 2^n * (3n)! / (2n+1)!. 1
1, 2, 24, 576, 21120, 1048320, 65802240, 5000970240, 446557224960, 45830873088000, 5316381278208000, 687893507997696000, 98231192942070988800, 15345895252950201139200, 2603510504983275503616000, 476694375041453927694336000, 93692112621783944697741312000 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS G. C. Greubel, Table of n, a(n) for n = 0..250 FORMULA From G. C. Greubel, Jun 12 2019: (Start) G.f.: Hypergeometric3F1(1/3, 2/3, 1; 3/2; 27*x/2). E.g.f.: sqrt(2/(3*x)) * sin( arcsin(sqrt(27*x/2))/3 ). E.g.f.: hypergeometric2F1(1/3, 2/3; 3/2; 27*x/2) (End) MATHEMATICA Table[2^n (3n)!/(2n+1)!, {n, 0, 20}] (* Harvey P. Dale, Mar 19 2016 *) PROG (PARI) a(n) = 2^n * (3*n)! / (2*n+1)! \\ Michel Marcus, Jun 24 2013 (MAGMA) [2^n*Factorial(3*n)/Factorial(2*n+1): n in [0..20]]; // G. C. Greubel, Jun 12 2019 (Sage) [2^n*factorial(3*n)/factorial(2*n+1) for n in (0..20)] # G. C. Greubel, Jun 12 2019 (GAP) List([0..20], n-> 2^n*Factorial(3*n)/Factorial(2*n+1) ) # G. C. Greubel, Jun 12 2019 CROSSREFS Sequence in context: A156525 A170904 A090732 * A280794 A090316 A128578 Adjacent sequences: A014295 A014296 A014297 * A014299 A014300 A014301 KEYWORD nonn,easy AUTHOR STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified December 6 14:15 EST 2019. Contains 329806 sequences. (Running on oeis4.) | 765 | 2,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-51 | latest | en | 0.652395 |
https://alex.state.al.us/lesson_view.php?&print=friendly&res_id=35634 | 1,656,108,269,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00587.warc.gz | 147,729,777 | 8,901 | # ALEX Lesson Plan
## Finding Suitable Solutions
You may save this lesson plan to your hard drive as an html file by selecting "File", then "Save As" from your browser's pull down menu. The file name extension must be .html.
This lesson provided by:
Author: Megan Nichols System: Chickasaw City School: Chickasaw City Elementary School
General Lesson Information
Lesson Plan ID: 35634 Title: Finding Suitable Solutions Overview/Annotation: Students will be exposed to three different scenarios. The scenarios will require that students hypothesize two solutions, test their hypotheses, document the results, and document the property that proved the effectiveness of the material chosen. An example of a scenario would be, “When provided toilet paper, tissue paper and paper towels, which material would be most effective in cleaning spilled water, and what property makes it so effective?” Students will then present the data collected in a Google Slides presentation. The lesson's total duration is about six days.This unit was created as part of the ALEX Interdisciplinary Resource Development Summit.
Associated Standards and Objectives
Content Standard(s):
Science SC2015 (2015) Grade: 2 2 ) Collect and evaluate data to determine appropriate uses of materials based on their properties (e.g., strength, flexibility, hardness, texture, absorbency).* Alabama Alternate Achievement Standards AAS Standard: SCI.AAS.2.2- Identify common materials and appropriate uses based on their physical properties (e.g., rubber bands stretch, sidewalks are hard, paper tears).
Local/National Standards:
Primary Learning Objective(s):
• Students will collaborate with their group to create two hypothesized solutions to three posed scenarios.
• Students will carry out experiments to test each scenario with the given materials provided by the teacher in the bags (For example, Scenario Bag 1 will be used to solve Scenario 1).
• Students will then work with their group to evaluate the effectiveness of the tested materials and document the results.
• Students will identify and analyze the properties that contributed to the effectiveness of the material.
• Students will work with their group members to create a Google Slide presentation consisting of at least three slides that describes the scenario, hypothesis, outcome of the test, and the property or properties that contributed to the effectiveness of the materials used. (Students may choose to put each scenario on an individual slide, or they may choose to place each task on an individual slide.) | 489 | 2,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-27 | latest | en | 0.894643 |
https://au.mathworks.com/matlabcentral/cody/problems/45201-check-if-integer-is-a-prime-number/solutions/3347738 | 1,606,338,952,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141184123.9/warc/CC-MAIN-20201125183823-20201125213823-00140.warc.gz | 208,154,066 | 17,112 | Cody
# Problem 45201. Check if integer is a prime number
Solution 3347738
Submitted on 23 Oct 2020 by Dyuman Joshi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
assert(isequal(isPrime(2),true))
2 Pass
assert(isequal(isPrime(1),false))
3 Pass
assert(isequal(isPrime(3),true))
4 Pass
assert(isequal(isPrime(11),true))
5 Pass
assert(isequal(isPrime(97),true))
6 Pass
assert(isequal(isPrime(199),true))
7 Pass
assert(isequal(isPrime(65),false))
8 Pass
assert(isequal(isPrime(93),false))
9 Pass
assert(isequal(isPrime(77),false))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 230 | 801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-50 | latest | en | 0.506359 |
https://www.bankersadda.com/reasoning-ability-quiz-for-sbi-ibps-prelims-2021-8th-april/ | 1,638,859,615,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363336.93/warc/CC-MAIN-20211207045002-20211207075002-00198.warc.gz | 731,150,521 | 32,597 | Latest Banking jobs » Reasoning Ability Quiz For SBI, IBPS...
# Reasoning Ability Quiz For SBI, IBPS Prelims 2021- 8th April
Directions (1-5): Study the following information and answer the questions given below:
There are seven friends are going for an exam in a week starting from Monday to Sunday (of the same week) but not necessarily in the same order. Only one person goes for exam on each day. All of them different heights.
A goes before Thursday. G is taller than only two persons. B is shorter than F. Three friends go for exam in between the days on which A and F goes. Two friends go for exam in between F and the one who is taller than G but shorter than F. Only one friend goes for exam in between B and G. Only one friend goes for exam in between the second tallest person and G. The third tallest person goes for exam on Saturday. Two friends go for exam in between D and C who is taller than E but shorter than B. D is taller than F but not the tallest person.
Q1. Who among the following person going to the exam on Thursday?
(a) The one who goes just before F
(b) The tallest person
(c)A
(d)C
(e)None of these
Q2. F goes on which day?
(a) Monday
(b) Friday
(c) Sunday
(d) Tuesday
(e) Saturday
Q3. How many persons give exam in between G and E?
(a) One
(b) Three
(c) None
(d) Two
(e) More than three
Q4. Who among the following is the tallest Person?
(a) The one who goes on Tuesday
(b) F
(c) G
(d) The one who goes on Friday
(e) C
Q5. How many Persons are taller than the one who goes for exam on Sunday?
(a) Four
(b) Three
(c) One
(d) Two
(e) None
Directions (6-10): Study the following statement carefully and answer the given questions. In a certain code
‘boost survey puzzle request’ is written as ‘pi li si xi’,
‘palace record cave request’ is written as ‘ti pi hi ch’,
‘survey cave record palace’ is written as ‘ti ch hi xi’
‘puzzle record survey dangerous’ is written as ‘xi ni ch li’.
Q6. What does ‘li’ stand for?
(a) boost
(b) palace
(c) request
(d) puzzle
(e) survey
Q7. What is the code for ‘cave’?
(a) hi
(b) li
(c) ti
(d) pi
(e) Either (a) or (c)
Q8. Which of the following represents ‘survey boost record puzzle’?
(a) si ch li pi
(b) xi si ch li
(c) si ch ni pi
(d) xi si ch hi
(e) pi li si ni
Q9. What does ‘ch’ stand for?
(a) Record
(b) cave
(c) palace
(d) dangerous
(e) survey
Q10. Which of the following may represent ‘puzzle record logical request’?
(a) li pi ni ch
(b) hi si xi li
(c) ti ni li mo
(d) li pi ch mo
(e) ni li pi ch
Directions (11-13): Study the information carefully and answer the questions given below.
Q is the wife of P. P is the grandfather of R. Q has only one child(son) who is married to T’s child. T has only two children one son and one daughter. X is grandson of T. S is brother in law of son of T. U and V are children of T. W is married to the son of T. X is nephew of U and he is W’s child. Only consider given persons.
Q11. If R is married to Y than how is Y related to S?
(a) Son
(b) Daughter
(c) Son in law
(d) Daughter in law
(e) Cannot be determined
Q12. How is S related to T?
(a) Son
(b) Daughter
(c) Son in law
(d) Daughter in law
(e) Husband
Q13. How is Q related to R?
(a) Grandfather
(b) Grandmother
(c) Uncle
(d) Aunt
(e) Either (c) or (d)
Q14. If all the letters in the word ‘COUNTRY’ are arranged in alphabetical order from left to right in such a way that vowels are arranged first followed by consonants, then how many letters are there in between U and T after the arrangement?
(a) Two
(b) One
(c) None
(d) Three
(e) Four
Q15. Among five friends A, B, C, D, and E, B is taller than C. E is taller than A. D is shorter than A but taller than B. Who among them is third tallest person?
(a) E
(b) C
(c) B
(d) A
(e) D
Practice More Questions of Reasoning for Competitive Exams:
###### Study Plan for IBPS and SBI Exams 2021
Solutions
Solutions (1-5) :
Sol.
S1. Ans.(d)
S2. Ans.(e)
S3. Ans.(b)
S4. Ans.(a)
S5. Ans.(c)
Solution (6-10):
Sol.
S6. Ans.(d)
S7. Ans.(e)
S8. Ans.(b)
S9. Ans.(a)
S10. Ans.(d)
Solutions (11-13):
Sol.
S11. Ans.(e)
S12. Ans.(c)
S13. Ans (b)
S14. Ans.(d)
Sol. Original Word- COUNTRY
After arrangement- OUCNRTY
S15. Ans.(e)
Sol. E > A > D > B > C
Practice with Online Test Series for SBI and IBPS Prelims 2021:
×
Thank You, Your details have been submitted we will get back to you.
Join India's largest learning destination
What You Will get ?
• Daily Quizzes
• Subject-Wise Quizzes
• Current Affairs
• Previous year question papers
• Doubt Solving session
OR
Join India's largest learning destination
What You Will get ?
• Daily Quizzes
• Subject-Wise Quizzes
• Current Affairs
• Previous year question papers
• Doubt Solving session
OR
Join India's largest learning destination
What You Will get ?
• Daily Quizzes
• Subject-Wise Quizzes
• Current Affairs
• Previous year question papers
• Doubt Solving session
Enter the email address associated with your account, and we'll email you an OTP to verify it's you.
Join India's largest learning destination
What You Will get ?
• Daily Quizzes
• Subject-Wise Quizzes
• Current Affairs
• Previous year question papers
• Doubt Solving session
Enter OTP
Please enter the OTP sent to
/6
Did not recive OTP?
Resend in 60s
Join India's largest learning destination
What You Will get ?
• Daily Quizzes
• Subject-Wise Quizzes
• Current Affairs
• Previous year question papers
• Doubt Solving session
Join India's largest learning destination
What You Will get ?
• Daily Quizzes
• Subject-Wise Quizzes
• Current Affairs
• Previous year question papers
• Doubt Solving session
Almost there
+91
Join India's largest learning destination
What You Will get ?
• Daily Quizzes
• Subject-Wise Quizzes
• Current Affairs
• Previous year question papers
• Doubt Solving session
Enter OTP
Please enter the OTP sent to Edit Number
Did not recive OTP?
Resend 60
By skipping this step you will not recieve any free content avalaible on adda247, also you will miss onto notification and job alerts
Are you sure you want to skip this step?
By skipping this step you will not recieve any free content avalaible on adda247, also you will miss onto notification and job alerts
Are you sure you want to skip this step? | 1,742 | 6,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-49 | latest | en | 0.935379 |
https://edurev.in/course/quiz/attempt/21426_Test-Single-Correct-MCQs-Moving-Charges-Magnetism-/d8cd2ef5-cb39-4a6a-bf7d-790ebe698f32 | 1,670,419,857,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00706.warc.gz | 241,580,169 | 69,542 | JEE > Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced
# Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced
Test Description
## 24 Questions MCQ Test Physics For JEE | Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced for JEE 2022 is part of Physics For JEE preparation. The Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced questions and answers have been prepared according to the JEE exam syllabus.The Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced below.
Solutions of Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced questions in English are available as part of our Physics For JEE for JEE & Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced solutions in Hindi for Physics For JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced | 24 questions in 50 minutes | Mock test for JEE preparation | Free important questions MCQ to study Physics For JEE for JEE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 1
### A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field such that is perpendicular to the plane of the loop. The magnetic force acting on the loop is
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 1
The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements. By Fleming's left hand rule, on element AB the direction of force will be leftwards and the magnitude will be
dF = I ( d l)Bsin 90°= I ( dl)B
On element CD, the direction of force will be towards right on the plane of the paper and the magnitude will be
dF = I ( d l)B.
These two forces will cancel out.
NOTE : Similarly, all forces acting on the diametrically opposite elements will cancel out in pair. The net force acting on the loop will be zero.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 2
### A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle q at the centre.The value of the magnetic induction at the centre due to the current in the ring is
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 2
Magnetic field at the centre due to current in arc ABC is
Magnetic field at the centre due to current in arc ADB is
Therefore net magnetic field at the centre
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 3
### A proton, a deuteron and an α - particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If rp, rd, and ra denote respectively the radii of the trajectories of these particles, then
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 3
KEY CONCEPT :
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 4
A circular loop of radius R, carrying current I, lies in x-y plane with its centre at origin. The total magnetic flux through x-y plane is
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 4
The magnetic lines of force created due to current will be in such a way that on x – y plane these lines will be perpendicular. Further, these lines will be in circular loops. The number of lines moving downwards in x – y plane will be same in number to that coming upwards of the x – y plane. Therefore, the net flux will be zero.
One such magnetic line is shown in the figure.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 5
A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 5
Force due to electric field will make the charged particle released from rest to move in the straight line (that of electric field). Since the force due to magnetic field is zero, therefore, the charged particle will move in a straight line.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 6
A particle of charge q and mass m moves in a circular orbit of radius r with angular speed ω . The ratio of the magnitude of its magnetic moment to that of its angular momentum depen ds on
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 6
The angular momentum L of the particle is given by
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 7
Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper, as shown. The variation of the magnetic field B along the line XX' is given by
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 7
The wires at A and B are perpendicular to the plane of paper and current is towards the reader. Let us consider certain points.
Point C (mid point between A and B) : The magnetic field at C due to A ( ) is in upward direction but magnetic field at C due to B is in downward direction.
Net field is zero.
Point E : Magnetic field due to A is upward and magnetic field due to B is downward but
∴Net magnetic field is in downward direction.
Net field upwards. Similarly,, other points can be considered.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 8
An infinitely long conductor PQR is bent to form a right angle as shown in Figure. A current I flows through PQR. The magnetic field due to this current at the point M is H1. Now, another infinitely long straight conductor QS is connected at Q so that current is I/2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now H2. The ratio H1/H2 is given by
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 8
Case 1 : Magnetic field at M due to PQ and QR is
Case 2 : When wire QS is joined.
H2 = (Magnetic field at M due to PQ) + (magnetic field at M due to QR) + (Magnetic field at M due to QS)
NOTE : The magnetic field due to an infinitely long wire carrying current at a distance R from the end point is half that at a distance R from the middle point.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 9
An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +x-direction and a magnetic field along the +z-direction, then
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 9
Case of positively charged particle : Two forces are acting on the positively charged particle (a) due to electric field in the positive x-direction. (b) Force due to magnetic field.
This forces will move the positively charged particle towards Y-axis.
Case of negatively charged particle.
Two forces are acting on the negatively charged particle (a) due to electric field in the negative X-direction. (b) due to magnetic field
Same direction as that of positive charge. (c) is the correct answer.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 10
A non-planar loop of conducting wire carrying a current I is placed as shown in the figure. Each of the straight sections of the loop is of length 2a. The magnetic field due to this loop at the point P(a, 0, a) points in the direction
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 10
NOTE : If we take individual length for the purpose of calculating the magnetic field in a 3-Dimensional figure then it will be difficult.
Here a smart choice is divide the loop into two loops.
One loop is ADEFA in y-z plane and the other loop will be ABCDA in the x – y plane.
We actually do not have any current in the segment AD. By choosing the loops we find that in one loop we have to take current from A to D and in the other one from D to A. Hence these two cancel out the effect of each other as far as creating magnetic field at the concerned point P is considered.
The point (a, 0, a) is in the X-Z plane.
The magnetic field due to current in ABCDA will be in + ve Z-direction.
NOTE : Due to symmetry the y-components and xcomponents will cancel out each other.
Similarly the magnetic field due to current in ADEFA will be in x-direction.
∴ The resultant magnetic field will be
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 11
Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB respectively and the trajectories are as shown in the figure. Then
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 11
KEY CONCEPT : When a charged particle is moving at right angles to the magnetic field then a force acts on it which behaves as a centripetal force and moves the particle in circular motion.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 12
A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at the center is
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 12
Let us consider a thickness dx of wire. Let it be at a distance x from the centre O.
Number of turns per unit length
∴ Number of turns in thickness dx =dx
Small amount of magnetic field is produced at O due to thickness dx of the wire.
On integrating, we get,
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 13
A particle of mass m and charge q moves with a constant velocity v along the positive x-direction. It enters a region containing a uniform magnetic field B directed along the negative z-direction, extending from x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b is
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 13
Width of the magnetic field region where 'R' is its radius of curvature inside magnetic field,
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 14
A long straight wire along the Z-axis carries a current I in the negative Z-direction. The magnetic vector field at a point having coordinates (x, y) in the Z = 0 plane is
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 14
The wire carries a current I in the negative z-direction.
We have to consider the magnetic vector field
Magnetic field is perpendicular to OP.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 15
The magnetic field lines due to a bar magnet are correctly shown in
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 15
NOTE : Magnetic lines of force form closed loops.
Inside a magnet, these are directed from south to north pole.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 16
For a positively charged particle moving in a x-y plane initially along the x-axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond P.
The curved path is shown in the x-y plane and is found to be non-circular. Which one of the following combinations is possible?
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 16
The velocity at P is in the X-direction (given).
After P, the positively charged particle gets deflected in the x – y plane toward – y direction and the path is non-circular.
Since in option (b), electric field is also present therefore it will also exert a force in the + X direction.
The net result of the two forces will be a non-circular path.
Only option (b) fits for the above logic. For other option, we get some other results.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 17
A conducting loop carrying a current I is placed in a uniform magnetic field pointing into the plane of the paper as shown. The loop will have a tendency to
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 17
KEY CONCEPT : Use Fleming's left hand rule. We find that a force is acting in the radially outward direction throughout the circumference of the conducting loop.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 18
A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III & IV arrange them in the decreasing order of Potential Energy
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 18
As we know that, potential energy of a magnet in a magnetic field
U=−m⋅B
=−mBcosθ
where, m= magnetic dipole moment of the magnet
B= magnetic field
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 19
An electron travelling with a speed u along th e positive xaxis enters into a region of magnetic field where B = –B0 kˆ (x > 0). It comes out of the region with speed v then
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 19
The force acting on electron will be perpendicular to the direction of velocity till the electron remains in the magnetic field. So the electron will follow the path as given.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 20
A magnetic field exists in the region a < x < 2a, and , in the region 2a < x < 3a, where B0 is a positive constant. A positive point charge moving with a velocity , where v0 is a positive constant, enters the magnetic field at x = a. The trajectory of the charge in this region can be like
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 20
Use the vector for m of B and v in the formulae to get the instantaneous direction of force at x = a and x = 2a.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 21
A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 21
Let us consider an elemental length dl subtending an angle dθ at the centre of the circle. Let FB be the magnetic force acting on this length. Then
FB = BI (dl) directed upwards as shown
= BI (Rdθ)
Let T be the tension in the wire acting along both ends of the elemental length as shown. On resolving T, we find that the components. cancel out and
the components. add up to balance FB.
At equilibrium
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 22
A long insulated copper wire is closely wound as a spiral of ‘N’ turns. The spiral has inner radius ‘a’ and outer radius ‘b’. The spiral lies in the XY plane and a steady current ‘I’ flows through the wire. The Z-component of the magnetic field at the centre of the spiral is
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 23
A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector kˆ is coming out of the plane of the paper. The magnetic moment of the current loop is
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 23
The magnetic moment of a current carrying loop is given by
the direction is towards positive z-axis.
Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 24
An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field, as a function of the radial distance r from the axis is best represented by
Detailed Solution for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced - Question 24
## Physics For JEE
257 videos|633 docs|256 tests
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code
Information about Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced Page
In this test you can find the Exam questions for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Single Correct MCQs: Moving Charges & Magnetism | JEE Advanced, EduRev gives you an ample number of Online tests for practice
## Physics For JEE
257 videos|633 docs|256 tests | 4,179 | 17,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-49 | latest | en | 0.796334 |
https://math.stackexchange.com/questions/1034039/basic-calculus-analysis-question-why-is-frac-dydx-dx-dy/1034081 | 1,701,641,418,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00073.warc.gz | 442,863,354 | 40,997 | basic calculus/analysis question. why is $\frac {dy}{dx} dx = dy$?
if $y$ is a function of $x$, why is $\frac {dy}{dx} dx = dy$? I have not learnt real analysis, but have done a bunch of math method courses at university and this has been bugging me. Why is it you can treat that as a fraction?
I would like the traditional calculus view first, then maybe differential forms view later. $dx$ to me is $\lim_{\Delta x \to 0} \Delta x$ and $\frac {dy}{dx} = \lim_{h\to 0} \frac {y(x+h)-y(x)}{h}$
I think $dy$ would be $\lim_{h\to 0} {y(x+h)-y(x)}$ but I'm not sure.
I don't see how you can just play around with these limits as if they were fractions. What proof is there that you can do so?
• Well, think about what $\int y'(x) \mathrm{d}x$ is equal to. It's really just formal a formal expression at the intro calc level though. Nov 22, 2014 at 19:24
• What does $dx$ even mean to you? Once you get to more formal definitions, it is basically the definition. Nov 22, 2014 at 19:25
• That equation can be thought of as defining dy as a function of x and dx (so dx is an independent variable). For example, if $y = x^2$, then $dy = 2x dx$.
– BigL
Nov 22, 2014 at 19:26
• Your view of $dx$ can't be useful; all you've said is that, to you, $dx$ is $0$. (actually, you probably have the wrong idea about limits, if you're imagining you said something different than $0$)
– user14972
Nov 22, 2014 at 19:39
• @user1564795 dx is not 0, but it goes to 0: $dx\to 0$ It represents a very, very small change of x. Nov 22, 2014 at 19:44
The way that differentials $dx, dy,$ etc are usually defined in elementary calculus is as follows:
Consider a $C_1$ function $f: \Bbb R \to \Bbb R$.
We define the tangent to the graph of $f(x)$ at the point $(c,f(c))$ to be $y=f'(c)(x-c) + f(c)$.
Here the value $x-c$ is called the increment of $x$, and is denoted $\Delta x$. So our tangent function can be written as $y=f'(c)\Delta x + f(c)$. Note that $\Delta x$ is actually evaluated at a specific value $c$ and thus should possibly be more correctly written as $\Delta x(c)$ or $\Delta x|_c$, but that'll get tedious so I'll leave off the explicit $c$ dependence.
Because the increment of $x$ can be just as easily defined on the graph of $f$ as on its tangent line $y$, we can also consider $\Delta x$ to be the differential of $x$, denoted $dx$. That is, we'll define $dx := \Delta x$. You can see that we could define all of the independent variables of a function in the same way for a function of multiple variables -- so for $g(x,y)$, we'd have $dx=\Delta x = x-c_1$ and $dy=\Delta y = y-c_2$.
The increment of the function $f$ is defined to be $\Delta f := f(c+\Delta x) - f(c)$. This is a measure of how much $f$ has changed given a change in $x$.
One of the fundamental ideas of calculus is that the increment $\Delta f$ should be approximately the same as the increment of its tangent $\Delta y$ for sufficiently small changes of $x$. From the above definition of the increment of a function, we can see that $\Delta y = y(c+\Delta x) - y(c) = [f'(c)((c+\Delta x) -c) + f(c)] - [f'(c)(c-c)+f(c)]$ $=f'(c)\Delta x + f(c) -0 -f(c) = f'(c)\Delta x$. Thus $\Delta y = f'(c)\Delta x$.
We define the differential of the function $f$ at $c$ to be equal to the increment of its tangent function. So $df|_c := \Delta y|_c$, or, dropping the $c$ dependence from our notation, $df:=\Delta y=f'(c)\Delta x$.
Notice that with these definitions, there is a fundamental difference between the differentials of independent variables -- like $dx$ -- and functions of those variables -- like $df$. Differentials of an independent variable is just the increment of that variable whereas the differential of a function is defined to be the increment of the tangent to that function, i.e. $df=\Delta y=f'(c)\Delta x=f'(c)dx$.
Also you should notice, that we define differentials (at least the differentials of functions) in terms of the derivative, not the other way around like early developers of calculus did. So we assume that a definition of the derivative has already been devised and agreed upon and we can then define our differentials in terms of that derivative.
NOTE: It is important that you realize that $\frac {df}{dx}$ is just a suggestive notation for $f'(x)$. It is NOT a fraction of differentials. In fact the only reason we use it is because it helps students learn things like the chain rule. But the derivative is actually defined by a limit, NOT by a ratio of infinitesimals (unless you're learning non-standard analysis).
So in response to your question: using the very first definitions given to students in elementary calculus. $df$ is literally defined by $df=f'(x)dx = \frac {df}{dx}dx$.
Now what are differentials used for? Differentials are an approximation of the change of a function for a very small change in the independent variable(/s). That is, $df \approx \Delta f$ when $\Delta x$ is sufficiently small.
In fact, the smaller that we make $\Delta x$, the smaller the difference between $df$ and $\Delta f$ becomes. The mathematical statement of this is that $\lim_{\Delta x \to 0} \frac {df}{\Delta f} = 1$. So $df$ and $\Delta f$ are "the same" for very, very small values of $\Delta x$.
Because OP seems to think that calculus is based on infinitesimals, I will try to elucidate this as fallacy. Notice that I did not use infinitesimals anywhere in this. $dx=x-c$ is finite. $df=f'(c)dx=f'(c)(x-c)$ is finite. There are no infinitesimals in this answer. And more generally in calculus, we don't use infinitesimals. Your first course in calculus you may have talked about them, but they are just meant to be a heuristic. No definitions actually require them.
One more thing to note: if you're familiar with Taylor's theorem, then let's look at the Taylor expansion of a $C_k$ function $f$ near the point $(c,f(c))$. We have $f(x) = f(c) + f'(c)(x-c) + \frac {f''(c)}{2!}(x-c)^2 + \cdots + \frac {f^{(k)}(c)}{k!}(x-c)^k + R$, where the remainder, $R$, is very small. Then you can see that the tangent function to the graph of $f$ is just the $1$st order Taylor expansion of $f$. And the differential $df$ is just the second term of Taylor expansion.
NOTE: IMO, a better, more useful definition of differentials is as differential forms. For a good reference, pick up the book Advanced Calculus: A Differential Forms Approach by Harold M. Edwards.
• You have shown that $\Delta f = \Delta y = f'(c)(x-c)=f'(c)\Delta x$ However $\Delta f$ could be large. For example $f'(x-c)$ was incredibly large, enough to to overpower $\Delta x$. So I think $df =/= \Delta f$ as $df$ should also be infinitesimal Nov 22, 2014 at 21:40
• Let me correct my above sentence.I meant: "For example if $f′(c)$ was incredibly large, enough to overpower Δx." Nov 22, 2014 at 21:49
• The condition that $\Delta x$ is very small isn't enough to ensure $df$ is very small. It's a hole in your argument that the definition of a derivative in my OP didn't have. You're basically saying $df=\Delta y = any size$. Nov 22, 2014 at 21:59
• Ok after some thought I agree with your proof. However one thing still bugs me a little. Doesn't this mean dx has to be infinitely small to get perfect accuracy? In which case the independent variables are always infinitesimals and the dependent is not. Right? Nov 22, 2014 at 22:18
If you pick two points on the function, let's say $$x_1$$ and $$x_2$$, the change in between is $$\Delta x = x_2-x_1$$. If you insert $$x_1$$ into the function $$f(x)$$, you will get $$f(x_1)=y_1$$. If you insert $$f(x_2)$$, you will get $$y_2$$. The change in between is $$y_2-y_1=\Delta y$$. The slope can be calculated with the formula: $$\frac{\mathrm{change\:in\:}y}{\mathrm{change\:in\:}x}$$, which is the same as $$\frac{\Delta y}{\Delta x}$$. Let's say that in the figure below $$\frac{\Delta y}{\Delta x}=2$$. If you go $$1$$ to the right, the $$\Delta x=1$$. Plugging that into the formula gives you $$\frac{\Delta y}{1}=2 \rightarrow \Delta y=2$$. So every step you go to the right, you will go two steps up.
To get the slope as accurate as possible, we need to make the change of $$x$$ as small as possible. Let's say that $$x_2=x_1+h$$. To make the change as small as possible, we need to make $$h$$ as small as possible.
The change of $$x$$ is $$\Delta x$$, which is $$h$$. So $$\Delta x = h$$. Now we need to get $$\Delta y$$, which is $$y_2-y_1$$. We know that $$y_2=f(x_2)$$ and $$x_2=x+h$$. So $$y_2=f(x+h)$$. We also know that $$y_1 = f(x)$$. $$\Delta y = y_2-y_1 = f(x+h)-f(x)$$. Plugging these in the formula $$\frac{\Delta y}{\Delta x}$$ makes:
$$\frac{\Delta y}{\Delta x}=\frac{f(x+h)-f(x)}{h}$$
Now the problem is that $$h\ne0$$, because division by $$0$$ is undefined. So we need to make $$h$$ as small as possible, approaching $$0$$. So:
$$\frac{\Delta y}{\Delta x}=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
• You mean dy/dx equals that limit, not change in y/ change in x. Also, this doesn't answer my question at all. Thanks for the effort. Nov 22, 2014 at 20:35 | 2,628 | 9,010 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-50 | latest | en | 0.950725 |
http://planetcalc.com/332/ | 1,521,768,909,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648113.87/warc/CC-MAIN-20180323004957-20180323024957-00027.warc.gz | 243,086,160 | 16,252 | homechevron_rightProfessionalchevron_rightComputers
# Conversion from decimal numeral system
This calculator converts given decimal value to other positional numeral system
This calculator was created for those who asked to convert decimal number to uncommon positional numeral systems, for example, with base of 3, 5, 7, 12.
To get result, you should enter decimal number and target numeral system base, i.e. 16.
For systems with base greater than 10 I traditionally use english alphabeth letters for denoting digits. Thus, 10 = A, 11 = B, 12 = C, and so on.
### Conversion from decimal numeral system
Result
Save the calculation to reuse next time or share with friends. | 150 | 679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-13 | latest | en | 0.810649 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/20%3A_Entropy_and_The_Second_Law_of_Thermodynamics/20.05%3A_The_Famous_Equation_of_Statistical_Thermodynamics | 1,713,629,879,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00231.warc.gz | 144,921,833 | 29,614 | # 20.5: The Famous Equation of Statistical Thermodynamics is S=k ln W
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
A common interpretation of entropy is that it is somehow a measure of chaos or randomness. There is some utility in that concept. Given that entropy is a measure of the dispersal of energy in a system, the more chaotic a system is, the greater the dispersal of energy will be, and thus the greater the entropy will be. Ludwig Boltzmann (1844 – 1906) (O'Connor & Robertson, 1998) understood this concept well, and used it to derive a statistical approach to calculating entropy. Boltzmann proposed a method for calculating the entropy of a system based on the number of energetically equivalent ways a system can be constructed.
Boltzmann proposed an expression, which in its modern form is:
$S = k_b \ln(W) \label{Boltz}$
$$W$$ is the number of available microstates in a macrostate (ensemble of systems) and can be taken as the quantitative measure of energy dispersal in a macrostate:
$W=\frac{A!}{\prod_j{a_j}!} \nonumber$
Where $$a_j$$ is the number of systems in the ensemble that are in state $$j$$ and $$A$$ represents the total number of systems in the ensemble:
$A=\sum_j{a_j} \nonumber$
Equation 20.5.1 is a rather famous equation etched on Boltzmann’s grave marker in commemoration of his profound contributions to the science of thermodynamics (Figure $$\PageIndex{1}$$).
##### Example $$\PageIndex{1}$$:
Calculate the entropy of a carbon monoxide crystal, containing 1.00 mol of $$\ce{CO}$$, and assuming that the molecules are randomly oriented in one of two equivalent orientations.
Solution:
Using the Boltzmann formula (Equation \ref{Boltz}):
$S = nK \ln (W) \nonumber$
And using $$W = 2$$, the calculation is straightforward.
\begin{align*} S &= \left(1.00 \, mol \cot \dfrac{6.022\times 10^{23}}{1\,mol} \right) (1.38 \times 10^{-23} J/K) \ln 2 \\ &= 5.76\, J/K \end{align*} \nonumber | 895 | 2,912 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-18 | latest | en | 0.596413 |
https://www.iqenergy.org.ua/answers/58199-15-less-than-half-a-number | 1,712,919,948,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296815919.75/warc/CC-MAIN-20240412101354-20240412131354-00205.warc.gz | 726,369,450 | 6,404 | # 15 less than half a number
Answer: 15 less than half a number
1/2x - 15
Answer: 100 ÷ 2 = 50
50 - 15 = 35
## Related Questions
Based on information from Harper’s Index, 37% of adult Americans believe in Extraterrestrials. Out of a random sample of 100 adults who attended college, 47 claim they believe in Extraterrestrials. Does this indicate that the proportion of individuals who attended college and believe in Extraterrestrials is higher than the proportion found by Harper’s Index? Use α = 0.05 for your Confidence Interval, show all steps, clearly explain your conclusion
z(s) is in the rejection zone , therefore we reject H₀
We have enough evidence to claim the proportion of individuals who attended college and believe in extraterrestrials is bigger than 37%
Step-by-step explanation:
We have a prortion test.
P₀ = 37 % P₀ = 0,37
sample size = n = 100
P sample proportion = P = 47 % P = 0,47
confidence interval 95 %
α = 0,05
One tail-test (right tail) our case is to show if sample give enough information to determine if proportion of individual who attended college is higher than the proportion found by Harper´s index.
1.-Hypothesis:
H₀ null hypothesis P₀ = 0,37
Hₐ alternative hypothesis P₀ > 0,37
2.-Confidence interval 95 %
α = 0,05 and z(c) = 1.64
3.-Compute of z(s)
z(s) = [ P - P₀ ] /√(P₀Q₀/n) ]
z(s) = [ ( 0,47 - 0,37 ) / √0.37*0,63/100
z(s) = 0,1 /√0,2331/100 ⇒ z(s) = 0,1 /0,048
z(s) = 2.08
4.-Compare z(s) and z(c)
z(s) > z(c) 2.08 > 1.64
5.-Decision:
z(s) is in the rejection zone , therefore we reject H₀
We have enough evidence to claim the proportion of individuals who attended college and believe in extraterrestrials is bigger than 37%
What is the ratio of 8 noisemakers to 6 party hats?
The ratio of noisemakers to party hats=8/6
=4/3
The ration will be 8:6 which can be reduced to 4:3
Trisha Oakley works at the telephone company. She earns an annual salary of \$26,000.00. If Trisha works. 50 hours a week and is paid biweekly, how much does she earn per hour? | 711 | 2,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-18 | latest | en | 0.758605 |
http://back2cloud.com/percent-error/percentage-error-in-calculation.php | 1,527,226,089,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867041.69/warc/CC-MAIN-20180525043910-20180525063910-00563.warc.gz | 24,006,856 | 5,822 | Home > Percent Error > Percentage Error In Calculation
# Percentage Error In Calculation
## Contents
Simply multiply the result, 0.1, by 100. You measure the dimensions of the block and its displacement in a container of a known volume of water. That's why the standard deviation can tell you how spread out the examples in a set are from the mean. The formula for calculating percentage error is simple:[1]'[(|Exact Value-Approximate Value|)/Exact Value] x 100 The approximate value is the estimated value, and the exact value is the real value. have a peek at these guys
Just add the percentage symbol to the answer and you're done. Email check failed, please try again Sorry, your blog cannot share posts by email. Determine, for each measurement, the error, percent error, deviation, and percent deviation. You look up the density of a block aluminum at room temperature and find it to be 2.70 g/cm3. other
## Percent Error Formula Chemistry
Get the best of About Education in your inbox. Tips Some teachers like the percent error to be rounded to a certain point; most people will be satisfied with the percent error rounded to three significant digits. Ways of Expressing Error in Measurement: 1. Examples: 1.
Answer this question Flag as... View all posts by Todd Helmenstine → Post navigation ← Direct Image Of Exoplanet Sets New Record Using Stem Cells and Herpes To Fight Brain Cancer → 3 thoughts on “Calculate Answer this question Flag as... Negative Percent Error It is used in chemistry and other sciences to report the difference between a measured or experimental value and a true or exact value.
These are the calculations that most chemistry professors use to determine your grade in lab experiments, specifically percent error. Please try again. Deviation -- subtract the mean from the experimental data point Percent deviation -- divide the deviation by the mean, then multiply by 100: Arithmetic mean = ∑ data pointsnumber of data The precision of a measuring instrument is determined by the smallest unit to which it can measure.
Live Traffic Stats Find jobs Our awards! Disclaimer Advertise on this site Facebook page Google+ page basic mathematics study group Illustrated fractions Educational math software Math Percent Error Worksheet However, after he carefully measured his height a second time, he found his real height to be 5 feet. When you calculate the density using your measurements, you get 8.78 grams/cm3. Skeeter, the dog, weighs exactly 36.5 pounds.
## Percent Error Calculator
About Todd HelmenstineTodd Helmenstine is the physicist/mathematician who creates most of the images and PDF files found on sciencenotes.org. http://www.wikihow.com/Calculate-Percentage-Error Did this article help you? Percent Error Formula Chemistry Create an account EXPLORE Community DashboardRandom ArticleAbout UsCategoriesRecent Changes HELP US Write an ArticleRequest a New ArticleAnswer a RequestMore Ideas... Percent Error Definition It is the difference between the result of the measurement and the true value of what you were measuring.
The theoreticalvalue (using physics formulas)is 0.64 seconds. http://back2cloud.com/percent-error/percentage-error-calculation-in-mass.php Step 2: Divide the error by the exact value (we get a decimal number) Step 3: Convert that to a percentage (by multiplying by 100 and adding a "%" sign) As Observed Value True Value RelatedPercentage Calculator | Scientific Calculator | Statistics Calculator In the real world, the data measured or used is normally different from the true value. Typically, you hope that your measurements are all pretty close together. Can Percent Error Be Negative
The smaller the unit, or fraction of a unit, on the measuring device, the more precisely the device can measure. If you are measuring a football field and the absolute error is 1 cm, the error is virtually irrelevant. Please enter a valid email address. check my blog The absolute value of the error is divided by an accepted value and given as a percent.|accepted value - experimental value| \ accepted value x 100%Note for chemistry and other sciences,
Ex: |-0.1| = 0.1 5 Multiply the result by 100. What Is A Good Percent Error Should the accepted or true measurement NOT be known, the relative error is found using the measured value, which is considered to be a measure of precision. The relative error expresses the "relative size of the error" of the measurement in relation to the measurement itself.
## About Today Living Healthy Chemistry You might also enjoy: Health Tip of the Day Recipe of the Day Sign up There was an error.
Flag as... Here is how to calculate percent error, with an example calculation.Percent Error FormulaFor many applications, percent error is expressed as a positive value. The absolute error of the measurement shows how large the error actually is, while the relative error of the measurement shows how large the error is in relation to the correct Percent Error Definition Chemistry For example, you measure a length to be 3.4 cm.
Flag as... What is the percent error? Chemistry Expert Share Pin Tweet Submit Stumble Post Share By Anne Marie Helmenstine, Ph.D. http://back2cloud.com/percent-error/percentage-error-calculation-formula.php In other words, amount of error = bigger − smaller Percent error word problem #1 A student made a mistake when measuring the volume of a big container.
The absolute value of a number is the value of the positive value of the number, whether it's positive or negative. A stopwatch has a circular dial divided into 120 divisions.time interval of 10 oscillation of a simple pendulum is measure as 25 sec.by using the watch Max. % error in the Also from About.com: Verywell & The Balance This site uses cookies. Copper's accepted density is 8.96 g/cm3.
If you measure the same object two different times, the two measurements may not be exactly the same. Then, convert the ratio to a percent. Calculate the percent error of your measurement.Subtract one value from the other:2.68 - 2.70 = -0.02 Depending on what you need, you may discard any negative sign (take the absolute value): 0.02This How to solve percentage error without the exact value given?
Kick Images, Getty Images By Anne Marie Helmenstine, Ph.D. While both situations show an absolute error of 1 cm., the relevance of the error is very different. Math CalculatorsScientificFractionPercentageTimeTriangleVolumeNumber SequenceMore Math CalculatorsFinancial | Weight Loss | Math | Pregnancy | Other about us | sitemap © 2008 - 2016 calculator.net Error in Measurement Topic Index | Algebra Index Write an Article 124 Homepage Math blog Homework helper!
Steps 1 Know the formula for calculating percentage error. Percent error or percentage error expresses as a percentage the difference between an approximate or measured value and an exact or known value. It is used in chemistry and other sciences to report the difference between a measured or experimental value and a true or exact value. Please try again.
If the object you are measuring could change size depending upon climatic conditions (swell or shrink), be sure to measure it under the same conditions each time. Tolerance intervals: Error in measurement may be represented by a tolerance interval (margin of error). We can do this multiplying both the numerator and the denominator by 20 We get (1 × 20)/(5 × 20) = 20/100 = 20% I hope what I explained above was What is the percent error the man made the first time he measured his height?
Since the measurement was made to the nearest tenth, the greatest possible error will be half of one tenth, or 0.05. 2. Todd also writes many of the example problems and general news articles found on the site. Two standard deviations, or two sigmas, away from the mean (the red and green areas) account for roughly 95 percent of the data points. | 1,670 | 7,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-22 | longest | en | 0.867304 |
http://icpc.njust.edu.cn/Problem/CF/764B/ | 1,591,180,794,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347432521.57/warc/CC-MAIN-20200603081823-20200603111823-00254.warc.gz | 59,786,561 | 10,235 | Timofey and cubes
Time Limit: 1 second
Memory Limit: 256 megabytes
Description
Young Timofey has a birthday today! He got kit of n cubes as a birthday present from his parents. Every cube has a number ai, which is written on it. Timofey put all the cubes in a row and went to unpack other presents.
In this time, Timofey's elder brother, Dima reordered the cubes using the following rule. Suppose the cubes are numbered from 1 to n in their order. Dima performs several steps, on step i he reverses the segment of cubes from i-th to (n - i + 1)-th. He does this while i ≤ n - i + 1.
After performing the operations Dima went away, being very proud of himself. When Timofey returned to his cubes, he understood that their order was changed. Help Timofey as fast as you can and save the holiday — restore the initial order of the cubes using information of their current location.
Input
The first line contains single integer n (1 ≤ n ≤ 2·105) — the number of cubes.
The second line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109), where ai is the number written on the i-th cube after Dima has changed their order.
Output
Print n integers, separated by spaces — the numbers written on the cubes in their initial order.
It can be shown that the answer is unique.
Sample Input
Input74 3 7 6 9 1 2Output2 3 9 6 7 1 4Input86 1 4 2 5 6 9 2Output2 1 6 2 5 4 9 6
Sample Output
None
Hint
Consider the first sample.
1. At the begining row was [2, 3, 9, 6, 7, 1, 4].
2. After first operation row was [4, 1, 7, 6, 9, 3, 2].
3. After second operation row was [4, 3, 9, 6, 7, 1, 2].
4. After third operation row was [4, 3, 7, 6, 9, 1, 2].
5. At fourth operation we reverse just middle element, so nothing has changed. The final row is [4, 3, 7, 6, 9, 1, 2]. So the answer for this case is row [2, 3, 9, 6, 7, 1, 4].
None | 595 | 1,834 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-24 | latest | en | 0.938318 |
https://mobilira.com/gambling/kelly-criterion/ | 1,726,550,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00311.warc.gz | 374,608,431 | 11,309 | # The Kelly Criterion
## Introduction
The Kelly Criterion is a bet-sizing technique which balances both risk and reward for the advantage gambler. The same principle would work for any investment with an expectation of being profitable. For the gambler/investor with average luck bankroll and a fixed bet size, the expected bankroll growth after one bet is:
For example, suppose a casino ran a promotion in craps where the 2 paid 3 to 1 and the 12 paid 4 to 1. A 3, 4, 9, 10, or 11 still pay 1 to 1 and every other total loses. The probability of a 2 or 12 is 1/36 each, of any even money win is 14/36 and of a loss is 20/36. Suppose also the player bets 1% of his bankroll every bet. Then the expected bankroll growth per bet would be:
(1 + (0.01*3))^(1/36) * (1 + (0.01*1)^(14/36) * (1 + (0.01*-1))^(20/36) * (1 + (0.01*4))^(1/36)) - 1 = 0.00019661.
This product is maximized by Kelly betting. Kelly betting also minimizes the expected number of bets required to double the bankroll, when bet sizing is always in proportion to the current bankroll.
The Kelly bet amount is the optimal amount for maximizing the expected bankroll growth, for the gambler with average luck. While betting more than Kelly will produce greater expected gains on a per-bet basis, the greater volatility causes long-term bankroll growth to decline compared to exact Kelly bet sizing. Betting double Kelly results in zero expected growth. Anything greater than double Kelly results in expected bankroll decline. What is more commonly seen is betting less than the full Kelly amount. While this does lower expected growth, it also reduces bankroll volatility. Betting half the Kelly amount, for example, reduces bankroll volatility by 50%, but growth by only 25%.
For simple bets that have only two outcomes, the optimal Kelly bet is the advantage divided by what the bet pays on a "to one" basis. For bets with more than one possible outcome, the optimal Kelly wager is that which maximizes the log of the bankroll after the wager. However, for bets with more than one outcome, that can be hard to determine. Most gamblers use advantage/variance as an approximation, which is a very good estimator. For example, if a bet had a 2% advantage, and a variance of 4, the gambler using "full Kelly" would bet 0.02/4 = 0.5% of his bankroll on that event. Remember that variance is the square of standard deviation, which is listed for many games in my Game Comparison Guide.
Let’s look at three examples.
Example 1: A card counter perceives a 1% advantage at the given count. From my Game Comparison Guide, we see the standard deviation of blackjack is 1.15 (which can vary according to the both the rules and the count). If the standard deviation is 1.15, then the variance is 1.152 = 1.3225. The portion of bankroll to bet is 0.01 / 1.3225 = 0.76%.
Example 2: A casino in town is offering a 5X points promotion in video poker. Normally the slot club pays 2/9 of 1% in free play. So at 5X, the slot club pays 1.11%. The best game is 9/6 Jacks or Better at a return of 99.54%. After the slot club points, the return is 99.54% + 1.11% = 100.65%, or a 0.65% advantage. The Game Comparison Guide shows the standard deviation of 9/6 Jacks or Better is 4.42, so the variance is 19.5364. The portion of bankroll to bet is 0.0065 / 19.5364 = 0.033%. By the way, this exact promotion is going on at the Wynn as I write this, for September 2 and 3, 2007.
Example 3: A sports wager has a 20% chance of winning, and pays 9 to 2. The advantage is 0.2×4.5 + 0.8×-1 = 0.1. The optimal Kelly wager is 0.1/4.5 = 2.22%.
Following is the exact math of example 3. Let x be optimal Kelly bet, with a bankroll of 1 before the bet. The expected log of the bankroll after the bet is...
f(x) = 0.2 × log(1+4.5x) + 0.8 × log(1-x)
To maximize f(x), take the derivative and set equal to zero.
f'(x) = 0.2 × 4.5 / (1+4.5x) - 0.8 / (1-x) = 0
0.9 / (1+4.5x) = 0.8/(1-x)
0.9 - 0.9x = 0.8 + 3.6 x
4.5x = 0.1
x = .1/4.5 = 1/45 = 2.22%
The math gets much messier when there is more than one possible outcome, such as in video poker. The method is still the same, but getting the solution for x is harder. The easiest way to solve for x in such cases, in my opinion, is experimenting with different values, using the higher and lower techniques (like the Clock Game on the "Price is Right"), until the f'(x) gets very close to zero.
I did this for two common video poker games with greater than 100% return. For "Full Pay Deuces Wild," with a return of 100.76%, the optimal bet size is 0.0345% of bankroll. For " 10/7 Double Bonus," with a return of 100.17%, the optimal bet size is 0.0062637% of bankroll. I have heard a rule of thumb that to make it in video poker you should have a bankroll of 3 to 5 times the royal amount you play for. If playing Full Pay Deuces wild, the exact amount is 3.66 royals. For 10/7 Double Bonus it is 19.96 royals.
## Simulations
To prove my statement that Kelly minimizes the number of bets to double the bankroll I assumed an even money bet with a 51% chance of winning, for a 2% advantage, and 2% Kelly bet size. Here is how many bets were required on average to double the bankroll at various bet sizes. If a winning wager would put the bettor over double the bankroll, he would only bet what was needed to exactly double the bankroll.
### Average Bets to Double Bankroll
Bet Size Average Bets
0.5% 7,901
1% 4,617
2% 3,496
3% 4,477
## Kelly Vs. Optimal Video Poker Strategy
In my Sep. 20, 2007 Ask the Wizard column I suggested the Kelly bettor should sometimes not play optimal video poker strategy. My reasons are explained there. | 1,598 | 5,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-38 | latest | en | 0.926473 |
https://www.rdocumentation.org/packages/MASS/versions/7.3-1/topics/loglm | 1,590,431,697,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347389309.17/warc/CC-MAIN-20200525161346-20200525191346-00010.warc.gz | 909,402,243 | 7,438 | # loglm
0th
Percentile
##### Fit Log-Linear Models by Iterative Proportional Scaling
This function provides a front-end to the standard function, loglin, to allow log-linear models to be specified and fitted in a manner similar to that of other fitting functions, such as glm.
Keywords
models, category
##### Usage
loglm(formula, data, subset, na.action, ...)
##### Arguments
formula
A linear model formula specifying the log-linear model.
If the left-hand side is empty, the data argument is required and must be a (complete) array of frequencies. In this case the variables on the right-hand side may be the na
data
Numeric array or data frame. In the first case it specifies the array of frequencies; in then second it provides the data frame from which the variables occurring in the formula are preferentially obtained in the usual way. This argu
subset
Specifies a subset of the rows in the data frame to be used. The default is to take all rows.
na.action
Specifies a method for handling missing observations. The default is to fail if missing values are present.
...
May supply other arguments to the function loglm1.
##### Details
If the left-hand side of the formula is empty the data argument supplies the frequency array and the right-hand side of the formula is used to construct the list of fixed faces as required by loglin. Structural zeros may be specified by giving a start argument with those entries set to zero, as described in the help information for loglin.
If the left-hand side is not empty, all variables on the right-hand side are regarded as classifying factors and an array of frequencies is constructed. If some cells in the complete array are not specified they are treated as structural zeros. The right-hand side of the formula is again used to construct the list of faces on which the observed and fitted totals must agree, as required by loglin. Hence terms such as a:b, a*b and a/b are all equivalent.
##### Value
• An object of class "loglm" conveying the results of the fitted log-linear model. Methods exist for the generic functions print, summary, deviance, fitted, coef, resid, anova and update, which perform the expected tasks. Only log-likelihood ratio tests are allowed using anova.
The deviance is simply an alternative name for the log-likelihood ratio statistic for testing the current model within a saturated model, in accordance with standard usage in generalized linear models.
##### Warning
If structural zeros are present, the calculation of degrees of freedom may not be correct. loglin itself takes no action to allow for structural zeros. loglm deducts one degree of freedom for each structural zero, but cannot make allowance for gains in error degrees of freedom due to loss of dimension in the model space. (This would require checking the rank of the model matrix, but since iterative proportional scaling methods are developed largely to avoid constructing the model matrix explicitly, the computation is at least difficult.) When structural zeros (or zero fitted values) are present the estimated coefficients will not be available due to infinite estimates. The deviances will normally continue to be correct, though.
##### References
Venables, W. N. and Ripley, B. D. (2002) Modern Applied Statistics with S. Fourth edition. Springer.
• loglm
##### Examples
# The data frames Cars93, minn38 and quine are available
# in the MASS package.
# Case 1: frequencies specified as an array.
sapply(minn38, function(x) length(levels(x)))
## hs phs fol sex f
## 3 4 7 2 0
minn38a <- array(0, c(3,4,7,2), lapply(minn38[, -5], levels))
minn38a[data.matrix(minn38[,-5])] <- minn38$fol fm <- loglm(~1 + 2 + 3 + 4, minn38a) # numerals as names. deviance(fm) ##[1] 3711.9 fm1 <- update(fm, .~.^2) fm2 <- update(fm, .~.^3, print = TRUE) ## 5 iterations: deviation 0.0750732 anova(fm, fm1, fm2) LR tests for hierarchical log-linear models Model 1: ~ 1 + 2 + 3 + 4 Model 2: . ~ 1 + 2 + 3 + 4 + 1:2 + 1:3 + 1:4 + 2:3 + 2:4 + 3:4 Model 3: . ~ 1 + 2 + 3 + 4 + 1:2 + 1:3 + 1:4 + 2:3 + 2:4 + 3:4 + 1:2:3 + 1:2:4 + 1:3:4 + 2:3:4 Deviance df Delta(Dev) Delta(df) P(> Delta(Dev) Model 1 3711.915 155 Model 2 220.043 108 3491.873 47 0.00000 Model 3 47.745 36 172.298 72 0.00000 Saturated 0.000 0 47.745 36 0.09114 # Case 1. An array generated with xtabs. loglm(~ Type + Origin, xtabs(~ Type + Origin, Cars93)) Call: loglm(formula = ~Type + Origin, data = xtabs(~Type + Origin, Cars93)) Statistics: X^2 df P(> X^2) Likelihood Ratio 18.362 5 0.0025255 Pearson 14.080 5 0.0151101 # Case 2. Frequencies given as a vector in a data frame names(quine) ## [1] "Eth" "Sex" "Age" "Lrn" "Days" fm <- loglm(Days ~ .^2, quine) gm <- glm(Days ~ .^2, poisson, quine) # check glm. c(deviance(fm), deviance(gm)) # deviances agree ## [1] 1368.7 1368.7 c(fm$df, gm$df) # resid df do not! c(fm$df, gm\$df.residual) # resid df do not!
## [1] 127 128
# The loglm residual degrees of freedom is wrong because of
# a non-detectable redundancy in the model matrix.
Documentation reproduced from package MASS, version 7.3-1, License: GPL-2 | GPL-3
### Community examples
Looks like there are no examples yet. | 1,425 | 5,164 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-24 | latest | en | 0.845939 |
http://spmath84109.blogspot.com/2009/11/jerics-rate-growing-post.html | 1,527,038,504,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865023.41/warc/CC-MAIN-20180523004548-20180523024548-00256.warc.gz | 283,496,233 | 20,007 | Sunday, November 1, 2009
%%%%% Jeric's Rate Growing Post
Definition:
Ratio: is where you compare two quantities measured in the same unit
Rate: is where you compare two quantities measured in different units
The difference between rate and ratio is that rate compares things from different units like comparing Canadian money to dirty American money and ratio compares things from the same unit like black bears to brown bears.
4.Determine the unit rate in each situation.
a) An orca swims 110 km in 2h.
First divide 110km by 2 to get 55 and you divide 2h by 2 to get 1. So the orca swam 55km/h
b)A Canada goose flies 800 km in 12.5h.
First you divide 800 by 12.5 to get 64 and you divide 12.5 by 12.5 to get 1. So the goose flew 64km/h The goose flew 64km/h
c) Cathy plants 45 daffodils in 30 min.
First you divide 45 by 30 to get 1.5 and you divide 30 by 30 to get 1. So 1 daffodils were planted in 1 min. 1daffodils/min
10.Trevor rode his mountain bike 84 km in 3h. Jillian rode 70 km in 2.5h. Who is the faster cyclist? How do you know ?
Trevor and Jillian are both equal speeds, because when you bring both their km and hours to the unit rate, they both equal 28km/h.
Coin Rate:
How many pennies would there be in 2 jars.
If there were 1000 pennies combined in 2 jars you would just need to find the unit rate ( divide each number by 6 ) and then you would get your answer. So there are 500pennies/jar
1) What would the recipe look like if it had to serve 10 people? Show your calculations.
The recipe would probably look like everything was double digits.
1/2 lb of ground beef 0.5 x 2.5 = 1.25
1 medium onion diced 1 x 2.5 = 2.5
1 celery finely diced 1 x 2.5=2.5
1 clove garlic 1 x 2.5 = 2.5
14oz can of tomatoes chopped 14 x 2.5= 35
1/2 of tomato paste 0.5 x 2.5 =1.25
1tsp parsley 1 x 2.5= 2.5
1 1/2 tsp basil 1.5 x 2.5= 3.75
1 tsp oregano 1 x 2.5 = 2.5
1 tsp sugar 1 x 2.5 = 2.5
1/2 worchestershire sauce 0.5 x 2.5= 1.25
1/2 tsp seasoning salt 0.5 x 2.5 = 1.25
1 bay leaf 1 x 2.5 = 2.5
2) What would the recipe look like if you had to only serve 1 person? Show your calculations.
The recipe would probably look like a recipe with lots of small numbers and decimal points.
1/2 lb of ground beef 0.5 / 4 = 0.125
1 medium onion diced 1 / 4 = 0.25
1 celery finely diced 1 / 4 = 0.25
1 clove garlic 1 / 4 = 0.25
14oz can of tomatoes chopped 14 / 4 = 3.5
1/2 of tomato paste 0.5 / 4 =0.125
1tsp parsley 1 / 4 = 0.25
1 1/2 tsp basil 1.5 / 4 = 0.375
1 tsp oregano 1 / 4 = 0.25
1 tsp sugar 1 / 4 = 0.25
1/2 worchestershire sauce 0.5 / 4 = 0.125
1/2 tsp seasoning salt 0.5 / 4 = 0.125
1 bay leaf 1 / 4 = 0.25
Technorati Tags
© Blogger templates Psi by Ourblogtemplates.com 2008 | 960 | 2,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-22 | latest | en | 0.943323 |
https://docs.astropy.org/en/latest/_sources/stats/ripley.rst.txt | 1,660,433,797,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-33/segments/1659882571989.67/warc/CC-MAIN-20220813232744-20220814022744-00574.warc.gz | 221,613,694 | 2,475 | .. _stats-ripley: ****************************** Ripley's K Function Estimators ****************************** Spatial correlation functions have been used in the astronomical context to estimate the probability of finding an object (e.g., a galaxy) within a given distance of another object [1]_. Ripley's K function is a type of estimator used to characterize the correlation of such spatial point processes [2]_, [3]_, [4]_, [5]_, [6]_. More precisely, it describes correlation among objects in a given field. The `~astropy.stats.RipleysKEstimator` class implements some estimators for this function which provides several methods for edge effects correction. Basic Usage =========== The actual implementation of Ripley's K function estimators lie in the method ``evaluate``, which take the following arguments: ``data``, ``radii``, and optionally, ``mode``. The ``data`` argument is a 2D array which represents the set of observed points (events) in the area of study. The ``radii`` argument corresponds to a set of distances for which the estimator will be evaluated. The ``mode`` argument takes a value on the following linguistic set ``{none, translation, ohser, var-width, ripley}``; each keyword represents a different method to perform correction due to edge effects. See the API documentation and references for details about these methods. Instances of `~astropy.stats.RipleysKEstimator` can also be used as callables (which is equivalent to calling the ``evaluate`` method). Example ------- .. EXAMPLE START Using Ripley's K Function Estimators To use Ripley's K Function Estimators from ``astropy``'s stats sub-package: .. plot:: :include-source: import numpy as np from matplotlib import pyplot as plt from astropy.stats import RipleysKEstimator z = np.random.uniform(low=5, high=10, size=(100, 2)) Kest = RipleysKEstimator(area=25, x_max=10, y_max=10, x_min=5, y_min=5) r = np.linspace(0, 2.5, 100) plt.plot(r, Kest.poisson(r), color='green', ls=':', label=r'\$K_{pois}\$') plt.plot(r, Kest(data=z, radii=r, mode='none'), color='red', ls='--', label=r'\$K_{un}\$') plt.plot(r, Kest(data=z, radii=r, mode='translation'), color='black', label=r'\$K_{trans}\$') plt.plot(r, Kest(data=z, radii=r, mode='ohser'), color='blue', ls='-.', label=r'\$K_{ohser}\$') plt.plot(r, Kest(data=z, radii=r, mode='var-width'), color='green', label=r'\$K_{var-width}\$') plt.plot(r, Kest(data=z, radii=r, mode='ripley'), color='yellow', label=r'\$K_{ripley}\$') plt.legend() .. EXAMPLE END References ========== .. [1] Peebles, P.J.E. *The large scale structure of the universe*. .. [2] Ripley, B.D. *The second-order analysis of stationary point processes*. Journal of Applied Probability. 13: 255–266, 1976. .. [3] *Spatial descriptive statistics*. .. [4] Cressie, N.A.C. *Statistics for Spatial Data*, Wiley, New York. .. [5] Stoyan, D., Stoyan, H. *Fractals, Random Shapes and Point Fields*, Akademie Verlag GmbH, Chichester, 1992. .. [6] *Correlation function*. | 793 | 2,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-33 | latest | en | 0.657511 |
https://lists.jboss.org/archives/list/rules-dev@lists.jboss.org/message/6X7MVU4ZZ4H4KDAJFJSXBSUGPYPP3J5R/attachment/2/attachment.html | 1,660,558,445,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572163.61/warc/CC-MAIN-20220815085006-20220815115006-00012.warc.gz | 347,018,430 | 2,635 | Otherwise has been dragging on since 2006. There are many skeletons in that cave.
I will believe it when I see it !
On Fri, Apr 1, 2011 at 7:25 AM, Michael Anstis wrote:
I bet Edson can't wait to refactor the parser for that ;)
On 31 March 2011 21:11, Mark Proctor wrote:
on a related note I do plan to add OTHERWISE support at a DRL level, just no time to do it right now. Once it's supported at a DRL level, you won't need to as much work on figuring out the inverse options etc.
Mark
On 31/03/2011 20:25, Michael Anstis wrote:
Hi,
I'm adding support for "otherwise" to (for the time being) the guided decision table in Guvnor.
The idea being if you set a cell to represent "otherwise" the generated rule is the opposite of the accumulation of the other cells; perhaps best explained with an example:-
Person( name == )
Mark
Kris
Geoffrey
<otherwise>
This would generate:-
Person(name not in ("Mark", "Kris", "Geoffrey")
Equals is the simple example, this is my thoughts for the other operators we might like to support:-
• != becomes "in (<list of the other cells' values)"
• < becomes ">= the maximum value of the other cells' values
For example:-
Person ( age < )
10
20
30
<otherwise>
Person ( age >= 30 )
• <= becomes "> the maximum value of the other cells' values
• > becomes "<= the minimum value of the other cells' values
• >= becomes "< the minimum value of the other cells' values
• "in" becomes "not in (<a list of all values contained in all the other cells' lists of values>)"
For example:-
Person ( name in )
Jim, Jack
Lisa, Jane, Paul
<otherwise>
Person ( name not in ("Jim", "Jack", "Lisa", "Jane", "Paul" ) )
• I'm not sure there is a simple solution for "matches" and "soundslike" but welcome advice, although a possibility might be to create a compound field constraint:-
Person ( name soundslike )
Fred
Phil
not Person ( name soundslike "Fred" || soundslike "Phil" )
Would this be considered the most suitable approach?
Inputs and thoughts welcome.
Thanks,
Mike
```
_______________________________________________
rules-dev mailing list
rules-dev@lists.jboss.org
https://lists.jboss.org/mailman/listinfo/rules-dev
```
_______________________________________________
rules-dev mailing list
rules-dev@lists.jboss.org
https://lists.jboss.org/mailman/listinfo/rules-dev
_______________________________________________
rules-dev mailing list
rules-dev@lists.jboss.org
https://lists.jboss.org/mailman/listinfo/rules-dev
--
Michael D Neale
home: www.michaelneale.net
blog: michaelneale.blogspot.com | 632 | 2,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-33 | latest | en | 0.890078 |
https://www.numbersaplenty.com/2645153 | 1,721,084,394,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00686.warc.gz | 813,289,648 | 3,120 | Search a number
2645153 = 7443853
BaseRepresentation
bin1010000101110010100001
311222101110122
422011302201
51134121103
6132410025
731324550
oct12056241
94871418
102645153
111547385
12a76915
13717ca4
144cbd97
15373b38
hex285ca1
2645153 has 8 divisors (see below), whose sum is σ = 3033408. Its totient is φ = 2259504.
The previous prime is 2645141. The next prime is 2645171. The reversal of 2645153 is 3515462.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 2645153 - 216 = 2579617 is a prime.
It is not an unprimeable number, because it can be changed into a prime (2645453) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 2675 + ... + 3527.
It is an arithmetic number, because the mean of its divisors is an integer number (379176).
22645153 is an apocalyptic number.
It is an amenable number.
2645153 is a deficient number, since it is larger than the sum of its proper divisors (388255).
2645153 is an equidigital number, since it uses as much as digits as its factorization.
2645153 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1303.
The product of its digits is 3600, while the sum is 26.
The square root of 2645153 is about 1626.3926340217. The cubic root of 2645153 is about 138.2983289118.
The spelling of 2645153 in words is "two million, six hundred forty-five thousand, one hundred fifty-three".
Divisors: 1 7 443 853 3101 5971 377879 2645153 | 488 | 1,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.864864 |
https://forum.bebac.at/forum_entry.php?id=16324&order=time | 1,718,275,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00049.warc.gz | 242,653,063 | 8,112 | ## Diagnostics: R and Phoenix [Study Assessment]
Hi ElMaestro et al.,
❝ Extract some model diagnostics: DF's and LogLikelihood, and compare to find out which result is the better candidate.
I can only provide the results of R and Phoenix:
R 3.2.5
library(nlme) Subj <- c(1, 2, 4, 5, 6, 4, 5, 6, 7, 8, 9, 7, 8, 9) Dose <- c(25, 25, 50, 50, 50, 250, 250, 250, 75, 75, 75, 250, 250, 250) AUC <- c(326.40, 437.82, 557.47, 764.85, 943.59, 2040.84, 2989.29, 4107.58, 1562.42, 982.02, 1359.68, 3848.86, 4333.10, 3685.55) Cmax <- c(64.82, 67.35, 104.15, 143.12, 243.63, 451.44, 393.45, 796.57, 145.13, 166.77, 296.90, 313.00, 387.00, 843.00) resp <- data.frame(Subj, Dose, Cmax, AUC) resp$Subj <- factor(resp$Subj) muddle <- lme(log(Cmax) ~ log(Dose), data=resp, random=~1|Subj) sum.muddle <- summary(muddle) CI.muddle <- intervals(muddle, level=0.9, which="fixed") print(sum.muddle); CI.muddle\$fixed[, ] Linear mixed-effects model fit by REML Data: resp AIC BIC logLik 14.24355 16.18317 -3.121774 Random effects: Formula: ~1 | Subj (Intercept) Residual StdDev: 0.3347319 0.1206792 Fixed effects: log(Cmax) ~ log(Dose) Value Std.Error DF t-value p-value (Intercept) 1.9413858 0.24314072 7 7.984618 1e-04 log(Dose) 0.7617406 0.04727976 5 16.111347 0e+00 Correlation: (Intr) log(Dose) -0.863 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -1.07547728 -0.35579449 -0.03301391 0.45088601 0.91853654 Number of Observations: 14 Number of Groups: 8 lower est. upper (Intercept) 1.4807366 1.9413858 2.4020350 log(Dose) 0.6664696 0.7617406 0.8570116
Phoenix 6.47.0.768
Model Specification and User Settings Dependent variable : logCmax Transform : None Fixed terms : int+logDose Random/repeated terms : Subject Denominator df option : satterthwaite Class variables and their levels Subject : 1 2 4 5 6 7 8 9 Final variance parameter estimates: Var(Subject) 0.112045 Var(Residual) 0.0145635 REML log(likelihood) -0.623363 -2* REML log(likelihood) 1.24673 Akaike Information Crit. 9.24673 Schwarz Bayesian Crit. 11.1864 Effect:Level Estimate StdError Denom_DF T_stat P_value Conf T_crit Lower_CI Upper_CI --------------------------------------------------------------------------------------------- int 1.9413858 0.2431407 9.2 7.98462 1.980E-5 90 1.829 1.4967592 2.3860125 logDose:logDose 0.7617406 0.0472798 5.9 16.11135 4.241E-6 90 1.949 0.6695783 0.8539029
Estimates and their SEs are exactly the same. CIs are not (due to different DFs?).
PS: An ideas how to weight by 1/log(Dose) in lme()? Suggested by Chow/Liu and gives me a better fit in Phoenix.
Dif-tor heh smusma 🖖🏼 Довге життя Україна!
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes | 1,161 | 3,069 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-26 | latest | en | 0.29481 |
https://the-avocado.org/2019/02/28/so-you-think-you-can-think-puzzle-19-letterboxes/ | 1,656,870,655,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104248623.69/warc/CC-MAIN-20220703164826-20220703194826-00015.warc.gz | 625,304,924 | 26,648 | # So You Think You Can Think Puzzle 19: Letterboxes
Hello folks, and welcome to another puzzly puzzle! This week, we’ve got something a little special, and a little copyright-infringey. Below, you can see ten little grids of letters, each 3-by-3. Well, you should see them as little mini Boggle boards, with just enough room to hold one person’s name in them. Your job is to discover the correct solution in each grid, moving from letter to letter to build your solution one letter at a time. Start on any letter you like in the grid, and move between letters vertically, horizontally, or diagonally, only one letter at a time. You can double back through letters you’ve already used, reusing them as often as you like. However, you cannot “stand” on a letter to add it twice in a row. Basic Boggle rules here apply.
But what is it you’re looking for? Well, each grid holds the name of a famous personAll ten names fall into a specific pop-culture category, and you’ll likely be familiar with all of them. Each person’s name is spelled out, first and last name, in one grid. And grid #10 has something a little special to it as well, still fitting into the overall category, just being a little more….complete.
What’s the category? What are the names you’re trying to find? Oh, you’ll have to discover that along the way yourself, braveocados!
```1. 2. 3.
G O I N K C R H S
S R K D A F O A P
W A U E P R N J E
4. 5. 6.
T I M C N Y E I T
J A R R V A D R S
S O R E D U N A B
7. 8. 9.
L F N A O H I B E
A O S N S W N R A
R U C D E R G M N
10.
J T H
O E A
L C N```
Last puzzle’s solutions:
Spoiler
1) Galaxy Queso
2) Yonder Woman
3) Single All the Way
4) Revenue of the Nerds
5) The Best Erotic Marigold Hotel
6) Gone with the Mind
7) The Last Pedi
8) The Taming of the Shrek
9) Octohussy
10) The Breakfast Clue
11) Super Rich Avians
12) That Touch of Oink
[collapse] | 568 | 2,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-27 | latest | en | 0.897447 |
http://www.algebra.com/algebra/homework/expressions/expressions.faq.question.545897.html | 1,369,281,770,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702777399/warc/CC-MAIN-20130516111257-00084-ip-10-60-113-184.ec2.internal.warc.gz | 318,858,438 | 4,586 | # SOLUTION: Evaluate: 32 + (8 - 2) · 4 - 6/3
Algebra -> Algebra -> Expressions -> SOLUTION: Evaluate: 32 + (8 - 2) · 4 - 6/3 Log On
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Algebra: Evaluation of expressions, parentheses Solvers Lessons Answers archive Quiz In Depth
Question 545897: Evaluate: 32 + (8 - 2) · 4 - 6/3Answer by Student A(4) (Show Source): You can put this solution on YOUR website!You start by solving what is in the parentheses : 8-2 = 6 Then you do multiplication or division depending on which comes first as you move from left to right. Here multiplication is first so : 6x4= 24. Then the question becomes 32 + 24 - 6/3 . 6/3 = 2 => 32+24-2=54. So your answer is 54. | 249 | 841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2013-20 | latest | en | 0.81621 |
https://brilliant.org/problems/luke-the-cat/ | 1,501,232,548,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549448146.46/warc/CC-MAIN-20170728083322-20170728103322-00582.warc.gz | 619,151,118 | 12,615 | # Luke the cat
Luke the cat likes jumping on steps. One day, Luke the cat finds a bowl of its favourite food, salmon, in a bowl on top of a flight of $$12$$ steps. However, Luke the cat is tired and hungry and can only jump up $$1$$ or $$2$$ steps at a time. How many ways are there for it to get to the bowl of salmon?
× | 86 | 323 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-30 | longest | en | 0.972723 |
https://www.gamedev.net/forums/topic/659814-convert-view-space-to-world-space-issues/ | 1,516,272,791,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887224.19/warc/CC-MAIN-20180118091548-20180118111548-00323.warc.gz | 917,202,651 | 37,750 | # Convert view space to world space issues
This topic is 1156 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hi all!
In a 3D game that I'm writing a (post process) pixel shader for I try to transform a view space to a world space coordinate. My HLSL code looks about like this:
float3 world_pos = mul(view_pos,(float3x3)m_WV) + camera_pos
This works, but only for certain view angles and camera positions. E.g. when I look "from south" to the position in question it looks as it should (I mark the position to be transformed on screen as a colored sphere), but when turning the camera more than about 20 degrees, or shifting the camera position so that I will look "from east" the transformation renders completely off.
I must be missing something here, but I don't know what. I've tried normalizing, transposing and some other basic mutations / additions to my code but didn't find any working solution.
Any hints?
##### Share on other sites
I think you need to transform it with inverse of view matrix to undo "view" part.
float3 world_pos = mul(view_pos, ViewInverse);
Maybe you don't even need to do this, why do you need world pos? maybe you can set all to work in world-view space?
This way you can save yourself a lot of inverting which is expensive.
Edited by belfegor
##### Share on other sites
Maybe you don't even need to do this, why do you need world pos? maybe you can set all to work in world-view space?
This way you can save yourself a lot of inverting which is expensive.
Thanks. I need to check whether a certain point in screen space / view space lies between two points from those I got the world space coordinates. It needs to a real "in between" in world space, so I probably can't simply do that all in screen space. Furthermore I need other geometric checks apart from "in between" such as distance-to, on the same line etc. Those checks and calculations can be done pretty straightforward in world-space but would become quite painfully and error-prone in view/screen space.
I searched that forum for inverting a matrix in HLSL but people didn't help much the ones asking for it and just kept saying "don't do that!" - so ignoring those advises against doing it, how WOULD I calculate the inverse of the view matrix (or whatever matrix I need here) in the HLSL shader? And no, transform doesn't do the trick here, I've already tried that.
Edited by Meltac
##### Share on other sites
Oh, you are the guy that makes STALKER mods?
Can i see the file where m_WV is defined, maybe the view-inverse is provided but under different name.
I have Clear Sky unpacked, and In hmodel.h file i see something that might be it:
uniform half3x4 m_v2w;
Since STALKER is using "raw" shaders, i think column major is default (unless it is overridden somewhere) so you need to swap matrix-vector for mul op:
//float3 V = mul( VECTOR, MATRIX );
float3 V = mul( MATRIX, VECTOR );
Edited by belfegor
##### Share on other sites
Oh, you are the guy that makes STALKER mods? smile.png
Yes I am
Can i see the file where m_WV is defined, maybe the view-inverse is provided but under different name.
Here are the file relevant contents / matrix definitions - I have literally tried them all:
uniform half3x4 m_W;
uniform half3x4 m_V;
uniform half4x4 m_P;
uniform half3x4 m_WV;
uniform half4x4 m_VP;
uniform half4x4 m_WVP;
uniform half4 timers;
uniform half4 fog_plane;
uniform float4 fog_params;
uniform half4 fog_color;
uniform float3 L_sun_color;
uniform half3 L_sun_dir_w;
uniform half3 L_sun_dir_e;
uniform half4 L_hemi_color;
uniform half4 L_ambient;
uniform float3 eye_position;
uniform half3 eye_direction;
uniform half3 eye_normal;
uniform float4 dt_params;
I have Clear Sky unpacked, and In hmodel.h file i see something that might be it:
uniform half3x4 m_v2w;
Yes that's in my hmodel.h as well but since I'm doing a post-process (i.e. working on a different shading stage) I doubt I could make use of that matrix (but I'll check).
Since STALKER is using "raw" shaders, i think column major is default (unless it is overridden somewhere) so you need to swap matrix-vector for mul op:
//float3 V = mul( VECTOR, MATRIX );
float3 V = mul( MATRIX, VECTOR );
Hmm, that's strange. I've seen lots of code parts in the default stalker shaders where it's done the way I have it - but I'll check as well. Haven't seen any pragma specifying row major so far, though.
Btw, what do you mean by "raw" shaders?
EDIT:
To come back to my initial post, would you say that I get the right results under *some* conditions using the code I've post is purely coincidence or why is that? I just need to make sure that it's me doing something wrong, and not the X-Ray engine providing wrong matrix data.
Edited by Meltac
##### Share on other sites
Here are the file relevant contents / matrix definitions - I have literally tried them all:
... //code snip
Nothing in there suggest what you need.
Yes that's in my hmodel.h as well but since I'm doing a post-process (i.e. working on a different shading stage) I doubt I could make use of that matrix (but I'll check).
m_v2w name suggest that it is "view to world" matrix, if you look below in same file in hmodel function:
...
half3 nw = mul( m_v2w, normal );
They take view-space normal (probably from g-buffer i guess) and calculate nw "normal in world-space".
The problem Is m_v2w available/passed in that "shading stage"?
Hmm, that's strange. I've seen lots of code parts in the default stalker shaders where it's done the way I have it - but I'll check as well. Haven't seen any pragma specifying row major so far, though.
Matrices might be passed transposed then this order would work:
float3 V = mul( VECTOR, MATRIX );
You need to check documentation (if any?) to see how they expose/pass their matrices.
Btw, what do you mean by "raw" shaders?
I meant they are not using "Effect framework" in which row-major is default.
##### Share on other sites
Ok, thanks.
I'll check the m_v2w matrix from hmodel, you might be right with that. And yes, I meant that the engine seems to pass that matrix in the geometry and/or lighting phase/stage of the graphics pipeline to the HLSL shaders, but not in the post-processing stage where I do my stuff.
You need to check documentation (if any?)
Hehe, that's one of the best STALKER jokes I've ever heard
(there is absolutely NO documentation about the shaders, neither officially nor in-officially - otherwise I wouldn't be here so often)
I'll post my progress when I got a chance to check these things.
Edited by Meltac
##### Share on other sites
Ok, I've tried this yesterday. *Something* worked - but I'm not yet sure about it.
I've replaced my previous view space to world space conversion with this:
float3 world_pos = mul( m_v2w, view_pos ).xyz + camera_pos
Indeed the engine seems to pass some value for m_v2w after adding it to my HLSL shader code:
uniform float3x4 m_v2w;
However the result is strange. I have this debug code to check whether the transformation is correct:
if (distance(world_pos, check_pos) < 1.0)
return float4(1,0,0,1);
This is supposed to render a red sphere 1 meter around the spot whose world space coordinate I want to check against (check_pos). That worked well before (when using the code in my initial post), but as mentioned only under specific conditions (within a limited camera position and direction range).
NOW the result is completely different. Instead of rendering a sphere around the check position, regardless of the camera position, the shader renders now a red circle around the player if and only if he is within 1 meter from the check position !?
The good news is that the circle stays there regardless of the camera direction - this wasn't the case before. But how to interpret the new result I don't know. My first thought was that adding the camera position to the transformation formula might not be necessary anymore and cause this output, as the new result is obviously depending on the players / camera position:
float3 world_pos = mul( m_v2w, view_pos ).xyz
But removing that camera_pos part from the code doesn't help either as there will be nothing rendered at all at the check position.
Any ideas???
Edited by Meltac
##### Share on other sites
float3 world_pos = mul( m_v2w, view_pos ).xyz + camera_pos
If the m_v2w is the view-to-world matrix, then you don't need to add the camera_pos since it is already in the matrix.
Cheers!
I didn't notice that you tried this already.
Edited by kauna
##### Share on other sites
If the m_v2w is the view-to-world matrix, then you don't need to add the camera_pos since it is already in the matrix.
Yes I have already tried that without success. We suppose that m_v2w is the view-to-world matrix (unfortunately there's no documentation for that engine):
m_v2w name suggest that it is "view to world" matrix, if you look below in same file in hmodel function:
...
half3 nw = mul( m_v2w, normal );
They take view-space normal (probably from g-buffer i guess) and calculate nw "normal in world-space".
So if this assumption should be correct, what could I still be doing wrong?
Edited by Meltac
##### Share on other sites
I don't know what might be wrong now, but i would suggest to try something but in view-space as it should give you same results as in world-space.
You can transform your check_pos in view-space and do the compare with view-space position:
float3 check_pos_vs = mul( m_V, check_pos ) .xyz;
if (distance(view_pos, check_pos_vs) < 1.0)
return float4(1,0,0,1);
and let me know if that works.
##### Share on other sites
Thanks for the suggestion. I just have tried that. Doesn't work with m_V, but using m_VP instead does!
At least approximately. Meanly, the check sphere is rendered correctly on the right spot, but it moves a bit with the players / camera movement. That's another reason why I wanted to do it all in world space. The engine seems to provide either the view space position sampler state (s_position), or the transformation matrix (m_VP) based on an approximate camera position with disregard to any applied camera physics effects such as head bobbing. That way I can't do my calculations precisely enough because the rendering generates graphical glitches. Having it all in world space shouldn't cause those side effects (I was hoping).
BTW, using m_V as intended instead of m_VP causes about the same effect as described before: The check sphere will be rendered when the player / camera is located within a range of < 1.0 meters from the check position. Does that ring a bell?
Edited by Meltac
##### Share on other sites
Head-bobbing does not matter at all, you need to debug and find actual problem elsewhere.
Maybe you should post whole shader so i can see whole picture. How do you obtain view_pos? From g-buffer?
##### Share on other sites
The whole shader contains way too much code irrelevant to the topic to post here, but I've extracted all parts of interest here:
uniform float3x4 m_W;
uniform float3x4 m_V;
uniform float4x4 m_P;
uniform float3x4 m_WV;
uniform float4x4 m_VP;
uniform float4x4 m_WVP;
uniform sampler2D s_position;
float3 pos_1_world = float3( 146.73, 0.70, -85.29);
float3 uv_posxy = tex2D(s_position, center).xyz;
float4 check_pos_vs = mul(m_VP, float4(pos_1_world, 1));
if (distance(uv_posxy, check_pos_vs) < 1.0)
final += float4(1,0,0,1);
That's all I got. All the uniform variables are just references to matrices or sampler states passed by the engine. How the engine calculates those and what buffers or registers might be involved I don't know. I basically just use what the engine gives me.
Whether or not camera (post-)effects such as head-bobbing matter or not I cannot say for sure, but when looking at the result I see that everything is fine and correct as long as the camera only turns around or moves slowly in one direction, but when doing bigger and/or abrupt movement changes such as sprinting, jumping, or leaning sidewards the result starts getting off.
Edited by Meltac
##### Share on other sites
float3 uv_posxy = tex2D(s_position, center).xyz;
The varibale name "center" used for uv coordinate suggest that it is constant for every pixel (center of screen?), tho it should not because you need to sample view-pos from g-buffer, how/is it passed from vertex shader?
##### Share on other sites
. The engine seems to provide either the view space position sampler state (s_position), or the transformation matrix (m_VP) based on an approximate camera position with disregard to any applied camera physics effects such as head bobbing.
One of the pros of view space is that camera's position is the origin of the space, (0,0,0), so you don't need any additional parameter. If you do your calculations in view space, head bobbing or not, camera's position will be (0,0,0).
##### Share on other sites
float3 uv_posxy = tex2D(s_position, center).xyz;
The varibale name "center" used for uv coordinate suggest that it is constant for every pixel (center of screen?), tho it should not because you need to sample view-pos from g-buffer, how/is it passed from vertex shader?
Sorry to have missed that when extracting the relevant parts from the shader code. "center" refers to the uv coordinate passed from the previous shader unit (which is in this case not a vertex but another pixel shader as I'm in post-processing stage), not the screen center. Misleading, I know.
Edited by Meltac
##### Share on other sites
You cannot have 2 pixel shaders run simultaneously, you probably meant "helper" function. If you don't want to post whole shader at least show me the whole "pipeline" for uv coordinates (how it is calculated/obtained). I suspect at this because the rest of code is correct.
It doesn't matter if you are in "post-processing stage", you must have vertex shader, even if you dont set it yourself it is possible that it is set for you behind scene/implicitly by their shader system.
##### Share on other sites
You are right, there must be some (implicit) vertex shader which has probably just not been made accessible / changeable by the devs.
I didn't mean running two pixel shaders simultaneously, nor did I mean "helper" function. I just wanted to say that I do not have (or have access) to any vertex shader on the post processing stage, so the buffer input I get in that post process shader is basically the output (i.e. color etc.) from the last pixel shader before the pipeline passed the data to the post processing - it doesn't matter in this case whether there's a "hidden" vertex shader inbetween because it would basically just handle over its own inputs without doing any vertex manipulations or the like.
So, here's the portion showing the input of the coordinates to the pixel shader:
struct v2p
{
float4 tc0: TEXCOORD0; // Center
float4 tc1: TEXCOORD1; // LT
float4 tc2: TEXCOORD2; // RB
float4 tc3: TEXCOORD3; // RT
float4 tc4: TEXCOORD4; // LB
float4 tc5: TEXCOORD5; // Left / Right
float4 tc6: TEXCOORD6; // Top / Bottom
};
float4 main(v2p I):COLOR {
float2 center=I.tc0.xy;
...
}
Normally quite all stuff (sampling color / position and the like) is done using the center coordinate (that's also the case in the original / unmodded ) shader. That's why I'm using that coordinate as input to all my stuff as well.
##### Share on other sites
That seems fine, there must be something else wrong then.
##### Share on other sites
That seems fine, there must be something else wrong then.
As I said, I suspect that the engine passes some strange / unusual / wrong data to the shader, maybe the transformation matrix is not intended to be used that way in the post-processing stage or something. Not sure, though. Unfortunately I've met nobody so far who could confirm or explain this.
Edited by Meltac
##### Share on other sites
Have you tried to invert view matrix at shader by hand?
##### Share on other sites
Have you tried to invert view matrix at shader by hand?
I have asked how to do that earlier in this thread but unfortunately nobody replied on it. And in other threads here people keep just replying " don't ! " when somebody asks for it. This said, I would have tried if I would know how...
##### Share on other sites
http://www.nigels.com/glt/doc/matrix4_8cpp-source.html There are some code how to do it. It's not veryt efficient to inverse matrix per pixel but if its only way then you maybe just need to do it. | 3,967 | 16,670 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-05 | latest | en | 0.949721 |
https://www.cfd-online.com/Forums/cfx/176580-vof-fluid-sloshing-tank-rolling-tank.html | 1,680,370,542,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950110.72/warc/CC-MAIN-20230401160259-20230401190259-00232.warc.gz | 773,296,138 | 18,834 | # vof fluid sloshing in tank with rolling tank
Register Blogs Members List Search Today's Posts Mark Forums Read
August 22, 2016, 02:44 vof fluid sloshing in tank with rolling tank #1 Member naveen kumar s Join Date: Aug 2016 Location: india, bengalore Posts: 51 Rep Power: 8 hello, I have simulated sloshing technique in fluid tank in cfx ansys 14, but my problem is i have used expression for angular velocity of using sinusoidal (expression name is "amp" and its image is in attachment), this expression i have put in domain motion , the rolling of tank is 6 degree but in expression is my equation (angle in radiens*omega in (rad/sec)*sin(omega*t)) is right , i have multiplied omega with roll angle in radians , just see my expression , can i multiply angle in radians to omega (which is frequency). and below domain motion there is alternate rotation model what it does alternate motion , please give is there any other expression can give for fluid motion. can i give motion in initiation, u,v,w is there please help me thank you Last edited by navi; August 22, 2016 at 03:24. Reason: i guess image dind uploaded so typed an equation
August 22, 2016, 06:49 #2 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,327 Rep Power: 138 I do not understand your question. Can you state your question again?
August 23, 2016, 00:35 #3 Member naveen kumar s Join Date: Aug 2016 Location: india, bengalore Posts: 51 Rep Power: 8 for tank motion i have to give expression and my tank motion is rolling in degrees , it is 6 degree , and i have given expression as (6*(pi/180))*omega*sin(omega*t) omega is in rad/sec in above equation i have multiplied omega * (rolling angle in radians) is my equation is right for tank simulation i am giving this expression in outline tree default domiain> basic setting > domain motion > rotating > angular velocity in same dialog box, there is alternate rotation model, what this alternate rotation model does exactly
August 23, 2016, 06:03 #4 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,327 Rep Power: 138 So you are using rotating frames of reference to model an oscillating rotation? Be aware that rotating frames of reference assumes constant domain velocity so the angular accelerations will not be accounted for. It is up to you to determine whether that is important or not. If you model this using moving mesh it will include the full motion equations but the simulation will be considerably slower.
August 23, 2016, 06:12 #5 Member naveen kumar s Join Date: Aug 2016 Location: india, bengalore Posts: 51 Rep Power: 8 how can i apply moving mesh for 3d tank, in tutorials there are only for 2d and when i try doing it by understanding 2d and applying same for 3d ,i need to give mass, moment of inertia , all this i how can i feed correctly
August 23, 2016, 06:48 #6 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,327 Rep Power: 138 If you are using mass and moments of inertia then you are using the rigid body solver. I think you need to go back to basics and make sure you are modelling the correct physics. Is the motion known in advance or is it coupled to the the fluid motion? Does the tank deform at all? Please show an image of the tank motion.
August 24, 2016, 00:54
#7
Member
naveen kumar s
Join Date: Aug 2016
Location: india, bengalore
Posts: 51
Rep Power: 8
i have attached image, for this type simulation, i have alone gone through basics but its lil bit difficult so getting lot of doubt
Attached Images
Capture.jpg (79.2 KB, 41 views) expresions.JPG (25.7 KB, 26 views)
June 2, 2018, 00:51 sloshing and ship motion #8 New Member Join Date: Nov 2016 Posts: 16 Rep Power: 8 how to get result of sloshing pressure on tank walls ?
June 2, 2018, 08:04 #9 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,327 Rep Power: 138 * Put a monitor point on the surface recording the pressure. Then you will get a time history of pressure at that point in Solver Manager. * Integrate the pressure over the surface to get a force using the force function. Then put that CEL function in as a monitor point and you will have a time history in solver manager. * You can also do this in post-processing for any time steps you saved transient results files for. kites likes this. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.
Tags expression, initiation, sinosoidal, sloshing, vof | 1,155 | 4,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-14 | longest | en | 0.903996 |
https://tw2823.com/qa/question-which-is-not-a-data-type.html | 1,620,533,855,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988955.89/warc/CC-MAIN-20210509032519-20210509062519-00587.warc.gz | 597,249,488 | 8,737 | # Question: Which Is Not A Data Type?
## What is a Boolean data type?
In computer science, the Boolean data type is a data type that has one of two possible values (usually denoted true and false) which is intended to represent the two truth values of logic and Boolean algebra.
It is named after George Boole, who first defined an algebraic system of logic in the mid 19th century..
## Is 0 True or false?
Zero is used to represent false, and One is used to represent true. For interpretation, Zero is interpreted as false and anything non-zero is interpreted as true. To make life easier, C Programmers typically define the terms “true” and “false” to have values 1 and 0 respectively.
## Who is Boolean?
George Boole, (born November 2, 1815, Lincoln, Lincolnshire, England—died December 8, 1864, Ballintemple, County Cork, Ireland), English mathematician who helped establish modern symbolic logic and whose algebra of logic, now called Boolean algebra, is basic to the design of digital computer circuits.
## What is data and its type?
Data is a set of values of subjects with respect to qualitative or quantitative variables. Data is raw, unorganized facts that need to be processed. When data is processed, organized, structured or presented in a given context so as to make it useful, it is called information. …
## What data type is real?
In some languages the REAL data type is known as SINGLE, DOUBLE or FLOAT instead….Data types.Data typeDescriptionSample dataREALStores numbers that contain decimal places/values and can also store integers17.65CHARACTERStores a single character which can be a letter, number or symbol\$3 more rows
## Why string is non primitive data type?
String is non-primitive because only class can have methods. Primitive can not. And String need many functions to be called upon while processing like substring, indexof, equals, touppercase. It would not have been possible without making it class.
## Which is not a data type in Java?
A String in Java is actually a non-primitive data type, because it refers to an object. The String object has methods that are used to perform certain operations on strings.
## What are 4 examples of non primitive data types?
Non-Primitive Data Types: These data types are not actually defined by the programming language but are created by the programmer….They are as follows:boolean data type.byte data type.char data type.short data type.int data type.long data type.float data type.double data type.
## What data type is string?
A string is generally considered a data type and is often implemented as an array data structure of bytes (or words) that stores a sequence of elements, typically characters, using some character encoding.
## What is user defined data type?
A user-defined data type (UDT) is a data type that derived from an existing data type. You can use UDTs to extend the built-in types already available and create your own customized data types. There are six user-defined types: … Structured type.
## What data type is zip code?
CHAR datatypeZip codes are always 5 characters, hence you would need a CHAR datatype, rather than VARCHAR.
## What is a double data type?
double: The double data type is a double-precision 64-bit IEEE 754 floating point. … This data type represents one bit of information, but its “size” isn’t something that’s precisely defined. char: The char data type is a single 16-bit Unicode character.
## What is data type example?
A data type is a type of data. For example, if the variable “var1” is created with the value “1.25,” the variable would be created as a floating point data type. … If the variable is set to “Hello world!,” the variable would be assigned a string data type.
## What types of data are there?
Understanding Qualitative, Quantitative, Attribute, Discrete, and Continuous Data TypesAt the highest level, two kinds of data exist: quantitative and qualitative.There are two types of quantitative data, which is also referred to as numeric data: continuous and discrete.More items…•
## Are strings immutable in Java?
In Java, String is a final and immutable class, which makes it the most special. It cannot be inherited, and once created, we can not alter the object.
## What are the 5 data types?
Common data types include:Integer.Floating-point number.Character.String.Boolean.
## What are the 4 data types?
Some are used to store numbers, some are used to store text and some are used for much more complicated types of data ….The data types to know are:String (or str or text). … Character (or char). … Integer (or int). … Float (or Real). … Boolean (or bool).
## Which is not primitive data type?
In Java, non-primitive or reference data types, unlike primitive data types, which include byte, int, long, short, float, double, and char, do not store values, but address or references to information.
## Which data type can hold any type of data?
A database data type refers to the format of data storage that can hold a distinct type or range of values. When computer programs store data in variables, each variable must be designated a distinct data type. Some common data types are as follows: integers, characters, strings, floating point numbers and arrays.
## What is Boolean example?
Boolean expressions use the operators AND, OR, XOR, and NOT to compare values and return a true or false result. These boolean operators are described in the following four examples: x AND y – returns True if both x and y are true; returns False if either x or y are false.
## What are the 8 primitive data types?
Primitive types are the most basic data types available within the Java language. There are 8: boolean , byte , char , short , int , long , float and double . These types serve as the building blocks of data manipulation in Java. Such types serve only one purpose — containing pure, simple values of a kind. | 1,265 | 5,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-21 | latest | en | 0.925904 |
https://www.physicsforums.com/threads/de-problem-can-it-be-that-easy.510780/ | 1,547,706,448,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658844.27/warc/CC-MAIN-20190117062012-20190117084012-00479.warc.gz | 879,243,985 | 12,774 | # DE problem, can it be that easy?
1. Jun 29, 2011
### somebodyelse5
1. The problem statement, all variables and given/known data
Here is the problem. Is it as easy as my attempted solution? I think its more difficult but I am completely lost.
3. The attempt at a solution
simply integrate 1/.8S dS = 1 dt
use 2009 as start time so t=0 and S=.6 to solve for C
2. Jun 29, 2011
### Dick
Sure, the price of sand has nothing to do with the price of glass so you are right. You can solve it that way. On the other hand it looks to me like the question isn't asking to you solve the equation. It's asking for an estimate of the solution using 'linearity principals'. That's different from solving. Are there more parts? They must have a question having something to do with glass later on, given the elaborate set up.
3. Jun 29, 2011
### somebodyelse5
nope, thats it, I scanned everything involved. this is the last problem.
I would guess linearity, but we covered it for like 20 minutes and moved on. If I were to use linearity, where would I start?
4. Jun 29, 2011
### Dick
You know dS/dt and S in 2009. The only linearity principal that comes to mind is to use that slope to extrapolate along a straight line to 2011. That gives you a pretty poor estimate of the actual solution you would get solving the ODE. And the reason why it's poor has nothing to do with the extraneous 'glass' stuff. It's just that the function isn't very closely approximated by a linear function. I really don't get what this question's point is. Maybe you should just solve the ODE and give that answer. | 411 | 1,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-04 | latest | en | 0.972208 |
http://mytopics.org/topic-Application-of-Mathematics-in-Robotics-full-report | 1,477,068,337,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718285.69/warc/CC-MAIN-20161020183838-00330-ip-10-171-6-4.ec2.internal.warc.gz | 171,909,310 | 10,757 | Application of Mathematics in Robotics full report
Thread Rating:
• 1 Vote(s) - 1 Average
• 1
• 2
• 3
• 4
• 5
project topics Active In SP Posts: 2,492 Joined: Mar 2010 13-04-2010, 11:27 AM Application of Mathematics in Robotics.ppt (Size: 147.5 KB / Downloads: 176) Robotics Geometric description of arm movement Prepared by: Kalina Mincheva Revi Panidha Arbnor Hasani Content Introduction Geometric Description Simple Problems Conclusion Main parts of Robot™s arm: Different Joints Segments Joints Planar revolute joints Prismatic joints Ball joints Screw joints Types of Joints Video Presentation General joints video Our framework 2 segments 2 joints Ball joints can rotate 360D Video of our framework The forward kinematics problem The forward kinematic problem for a given robot arm is a systematic description of the relative positions of the segments on either side of a joint, thus determining the position and orientation of the hand from the arm. We will consider a robot in R3, in particular the set of polynomial equations that constraint the motion of the robot arm. The forward kinematic problem We can consider that both segments (parts of the arm) can move in the hole space. The first segment has length 2 and the second has length 1 The forward kinematic problem We see that the robot™s arm can move in the hole space, which is actually a sphere The equation of a sphere is: X2 + y2 + z2= r2; The two equations that describe the previous spheres X2 + y2 = 22 (u “ x)2 + (v “ y)2 + (w “ z)2 = 1 Does this system of polynomial equations have real solutions If yes then: How to solve this system of polynomial equations Difficult! Groebner Basis is the tool we can use. The Groebner Basis can help us solve the problem whether the robot can reach a certain point with center at (a, b, c) Solution: In order to find the coordinates of the point we are interested in, we have to find all points the arm can reach and see if this is one of them Points that can be reached = points that satisfy the above equations Solve a system of polynomial equations “ find the real roots The forward kinematic problem Linear systems “ reduced row echelon form In our case “ polynomial equations Groebner basis “ the equivalent (used to present the solutions of the equations in a reduced way) The forward kinematic problem Problem: We have 3 dimensions: Looking for a point with coordinates (u, v, w) But we have 6 variables! (x, y, z, u, v, w) We need to get rid of (x, y, z) Solution: Elimination The forward kinematic problem The Groebner basis “ one of the polynomials looks like this: u2 + v2 + w2= 5 “ equation of a sphere Problem: Number of points presented by (x, y, z, u, v, w) is not equal to the number of points presented by the above equation Not all points that lie in that sphere can be actually reached The forward kinematic problem For example seen in a plane: The forward kinematic problem The forward kinematic problem Extension (another theorem J) Check for the points excluded If the point is not among these “ it is reachable. The problem is solved! Gröbner Bases Application areas of Gröbner Bases Introduced by Bruno Buchberger, in 1965 Named after Wolfgang Gröbner “ Buchberger™s PhD Thesis Advisor. The goal: Present algorithmic solution of some of the fundamental problems in commutative algebra (polynomial ideal theory, algebraic geometry ) The method (theory plus algorithms) of Gröbner Bases provides a uniform approach to solving a wide range of problems expressed in terms of sets of multivariate polynomials. algebraic geometry, commutative algebra , polynomial ideal theory invariant theory robotics coding theory integer programming partial differential equations symbolic summation statistics non-commutative algebra systems theory compiler theory (non-commutative algebra) Problems that can be solved by Groebner basis method solvability and solving of polynomial systems of equations ideal membership problem elimination theory implicitization effective computation in residue class rings modulo polynomial ideals linear diophantine equations with polynomial coefficients (syzygies) Hilbert functions algebraic relations among polynomials Why is Gröbner Bases Theory Attractive The main problem solved by the theory can be explained in 5 minutes (if one knows operations addition and multiplication of polynomials). The algorithm that solves the problem can be learned in 15 minutes The theorem on which the algorithm is based is nontrivial to invent and to prove. Many problems in seemingly quite different areas of mathematics can be reduced to the problem of computing Gröbner bases. How Can Gröbner Bases Theory is Applied Given a set F of polynomials in k[x1, ¦, xn] We transform F into another set G of polynomials with certain nice properties (called a Gröbner Basis) such that F and G are equivalent i.e. generate the same ideal How Can Gröbner Bases Theory is Applied Many problems that are difficult for general F are easy for Gröbner Bases G There is an algorithm transforming an arbitrary F into an equivalent Gröbner basis G The solution of the problem for G can often be easily translated back into a solution of the problem for F Groebner Basis Sources http://www.energid.com/site/actin_movies.htm mark.math.helsinki.fi/ Symbolinen%20laskenta/Notes/Groebner/Intro.ppt Ideal, Varieties and Algorithms by David Cox, John Little, Donal O'Shea Thank you for your attention Q & A Use Search at http://topicideas.net/search.php wisely To Get Information About Project Topic and Seminar ideas with report/source code along pdf and ppt presenaion
## Important Note..!
If you are not satisfied with above reply ,..Please
# ASK HERE
So that we will collect data for you and will made reply to the request....OR try below "QUICK REPLY" box to add a reply to this page
Tagged Pages: applications maths in robotics, applications of mathematics ppt, what is use of maths in robotics, appications of mathematics topics, application of engineering mathematics ppt, application of mathematics in roboitcs, math application in robotics,
Popular Searches: seminar of indian mathematics, differential geometry to financial mathematics, mathematics project report on mechanical, robotics its application ppt, challenges are what make life interesting, application of mathematics in robotics, smart camera application download,
Quick Reply
Message
Type your reply to this message here.
Image Verification
Please enter the text contained within the image into the text box below it. This process is used to prevent automated spam bots.
(case insensitive)
Possibly Related Threads... Thread Author Replies Views Last Post bomb detection robotics using embedded controller jaseelati 0 312 06-01-2015, 04:50 PM Last Post: jaseelati radar tracking system concept and application ppt jaseelati 0 235 01-01-2015, 02:50 PM Last Post: jaseelati software defined radio for rfid application ppt jaseelati 0 217 09-12-2014, 02:26 PM Last Post: jaseelati witricity full report project report tiger 28 38,075 30-08-2014, 02:26 AM Last Post: radiopodarok.ru ACCIDENT PREVENTION USING WIRELESS COMMUNICATION full report computer science topics 5 7,570 17-04-2014, 11:07 AM Last Post: seminar project topic silicon on plastic full report computer science technology 2 2,931 13-04-2014, 10:34 PM Last Post: 101101 Automatic Emergency Light full report seminar class 7 17,561 08-03-2014, 02:28 PM Last Post: seminar project topic ACD-Anti Collision Device full report seminar presentation 11 18,176 10-01-2014, 03:20 PM Last Post: seminar project topic wireless charger full report project topics 21 18,457 10-01-2014, 12:58 PM Last Post: seminar project topic EMBEDDED SYSTEMS full report computer science technology 34 29,964 10-11-2013, 07:51 PM Last Post: Guest | 1,870 | 7,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-44 | longest | en | 0.849074 |
https://www.geeksforgeeks.org/category/data-structures/linked-list/page/11/ | 1,653,307,641,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00542.warc.gz | 886,564,192 | 19,421 | Given a singly linked list and a position, delete a linked list node at the given position. Example: Input: position = 1, Linked List =… Read More
Given a singly linked list and a position, delete a linked list node at the given position. Example: Input: position = 1, Linked List =… Read More
Given a singly linked list and a position, delete a linked list node at the given position. Example: Input: position = 1, Linked List =… Read More
Given a singly linked list and a position, delete a linked list node at the given position. Example: Input: position = 1, Linked List =… Read More
Given a singly linked list and a position, delete a linked list node at the given position. Example: Input: position = 1, Linked List =… Read More
Given two sorted singly linked lists having n and m elements each, merge them using constant space. First, n smallest elements in both the lists… Read More
Given two sorted singly linked lists having n and m elements each, merge them using constant space. First, n smallest elements in both the lists… Read More
Write a removeDuplicates() function that takes a list and deletes any duplicate nodes from the list. The list is not sorted. For example if the linked… Read More
Write a removeDuplicates() function that takes a list and deletes any duplicate nodes from the list. The list is not sorted. For example if the linked… Read More
Write a removeDuplicates() function that takes a list and deletes any duplicate nodes from the list. The list is not sorted. For example if the linked… Read More
Write a removeDuplicates() function that takes a list and deletes any duplicate nodes from the list. The list is not sorted. For example if the linked… Read More
Write a removeDuplicates() function that takes a list and deletes any duplicate nodes from the list. The list is not sorted. For example if the linked… Read More
We have discussed Linked List Introduction and Linked List Insertion in previous posts on a singly linked list.Let us formulate the problem statement to understand… Read More
We have discussed Linked List Introduction and Linked List Insertion in previous posts on a singly linked list.Let us formulate the problem statement to understand… Read More
We have discussed Linked List Introduction and Linked List Insertion in previous posts on a singly linked list.Let us formulate the problem statement to understand… Read More | 515 | 2,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-21 | latest | en | 0.72693 |
http://www.jiskha.com/display.cgi?id=1394508836 | 1,455,438,252,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701171770.2/warc/CC-MAIN-20160205193931-00072-ip-10-236-182-209.ec2.internal.warc.gz | 489,777,762 | 3,691 | Sunday
February 14, 2016
# Homework Help: Math help, anyone?
Posted by Princess Anna on Monday, March 10, 2014 at 11:33pm.
calculate the slant height for the given square pyramid.
Round to the nearest tenth.
Pyramid base= 6cm
height=5 cm
6.2 cm
5.8 cm
7.8 cm
7.2 cm
Calculate the length of the diagonal for the given rectangular prism.
Round to the nearest tenth.
Length= 10 cm
widith= 4cm
height= 10cm
14.7 cm
10.8 cm
12.2 cm
15.6 cm
Calculate the length of the diagonal for the given rectangular prism.
Round to the nearest tenth.
Length= 14cm
widith= 3
height= 4 cm | 183 | 578 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2016-07 | longest | en | 0.838086 |
https://www.jagranjosh.com/articles/physics-waves-formula-sheet-1696050090-1?ref=sticky_footer | 1,701,238,713,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100056.38/warc/CC-MAIN-20231129041834-20231129071834-00790.warc.gz | 969,847,533 | 42,812 | # CBSE Class 11 Physics Waves: Formula List, Definitions, and More
Waves Formulas & Diagrams: Students can find a list of important diagrams and formulas for CBSE Class 11 Physics Chapter Waves. Use the PDF download link attached below to save the formula sheet. this formula sheet is quite handy and a physical print out of these can be easily carried by students to their schools and tuitions.
CBSE Class 11 Physics Waves Formulas: This article hands out the complete list of formulas for Class 11 Physics Waves. Here, students can also find a PDF download link at the bottom of the article to save the formula sheet for future reference. All the formulas presented here have been picked up from the Class 11 Physics NCERT Textbook, thus students can be absolutely carefree while referring to this formula sheet.
Along with the formulas, we have also brought to you important definitions from the chapter. These definitions will help you understand the topic, clear your concepts, and prepare you for your examinations. As we all know, Class 11 is the toughest grade to be in. Given the advancement of curriculum, and a huge jump in the complexities of numerical problems, it becomes really hard for students.
Generally, students with PCM combination spend a huge amount of their day practicing for examinations. With this formula sheet, you won’t have to keep referring back to the chapters for searching formulas or memorizing them. This handy formula page can be your partner in school, tuition, or revision time.
## Formula Sheet for Class 11 Physics Waves
Definitions:
Waves - patterns, that move without the actual physical transfer or flow of matter as a whole, are called waves.
Transverse Waves- If the constituents of the medium oscillate perpendicular to the direction of wave propagation, the wave is called a transverse wave.
Longitudinal Wave- If they oscillate along the direction of wave propagation, it is called a longitudinal wave.
Capillary Waves- They are ripples of fairly short wavelength not more than a few centimetres and the restoring force that produces them is the surface tension of water.
Gravity waves- Gravity waves have wavelengths typically ranging from several meters to several hundred meters. The restoring force that produces these waves is the pull of gravity, which tends to keep the water surface at its lowest level. The oscillations of the particles in these waves are not confined to the surface only but extend with diminishing amplitude to the very bottom.
Wavelength- The minimum distance between two points having the same phase is called the wavelength of the wave, usually denoted by λ.
Period- the period of oscillation of the wave is the time it takes for an element to complete one full oscillation.
Principle of Superposition of Waves- when two pulses of equal and opposite shapes move towards each other. When the pulses overlap, the resultant displacement is the algebraic sum of the displacement due to each pulse. This is known as the principle of superposition of waves.
Waveform- It is the sum of wave functions of individual waves
Refracted Wave- If a wave is incident obliquely on the boundary between two different media the transmitted wave is called the refracted wave.
Nodes- The points at which the amplitude is zero are nodes.
Antinodes- the points at which the amplitude is the largest are called antinodes.
Normal Modes- The system cannot oscillate with any arbitrary frequency (contrast this with a harmonic traveling wave), but is characterized by a set of natural frequencies or normal modes of oscillation.
First Harmonic- The lowest possible natural frequency of a system is called its fundamental mode or the first harmonic.
Formula:
Displacement-
Displacement of a wave travelling in negative direction-
Wavelength- where k is the angular wave number or propagation constant; its SI unit is radian per meter or rad/m.
Period- y= where ω represents the angle and T represents the time
Angular Frequency- ω is called the angular frequency of the wave. Its SI unit is rad s 1.
Frequency- ν is usually measured in hertz
Speed of a travelling wave-
Speed of a transverse wave on stretched string- where μ represents velocity of the string
Wavelength of a transverse wave on stretched string-
Speed of a longitudinal wave- - where B is Bulk Modulus
Speed of longitudinal waves in a solid bar- where Y is the Young Modulus
Speed of a longitudinal wave in an ideal gas (Newton’s Formula)-
Laplace Correction (Modification of Newton’s formula)-
Principle of Superposition of Waves-
Reflected Wave at a rigid boundary-
Reflected Wave at an open boundary-
Position of Nodes-
Position of Antinodes-
Wavelengths of Stationary Waves-
Wavelengths of Stationary Frequencies-
Length of an antinode-
Wavelengths of an antinode-
Frequencies of an antinode-
Beat Frequency-
This formula sheet is presented to you to ease your preparation for the examination. Given the toughness and criticality of Class 11, it becomes important for the student to save their time and utilize it for studying. The formula sheet will help you achieve the goal by saving a lot of your time. Students who refer to the formula sheet presented here will no longer have to keep searching for formulas in between the chapters while practicing as well as while solving the NCERT Solutions. Additionally, frequent visibility of the formulas might enhance your remembering power and might assist you in easily gulping these formulas down your throat, for memorizing.
Also Find:
Class 11 Physics Mechanical Properties of Fluids Formula Sheet
Class 11 Physics Mechanical Properties of Solids Formula Sheet
Class 11 Physics Thermal Properties of Matter Formula Sheet
Class 11 Physics Thermodynamics Formula Sheet
Class 11 Physics Kinetic Theory Formula Sheet
Class 11 Physics Oscillations Formula Sheet
## Related Categories
खेलें हर किस्म के रोमांच से भरपूर गेम्स सिर्फ़ जागरण प्ले पर | 1,282 | 5,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-50 | latest | en | 0.916375 |
https://www.jiskha.com/questions/561500/a-2-9-kg-pile-of-aluminum-2-70-g-cm3-cans-is-melted-and-then-cooled-into-a-solid | 1,597,274,093,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738950.31/warc/CC-MAIN-20200812225607-20200813015607-00038.warc.gz | 691,546,553 | 6,393 | # Physical Science
A 2.9-kg pile of aluminum ( = 2.70 g/cm3) cans is melted, and then cooled into a solid cube. What is the volume of the cube?
a. 931 cm3
b. 1074 cm3
c. 1365 cm3
d. 889 cm3
Thank you!
1. 👍 0
2. 👎 0
3. 👁 486
1. 2900 grams of Al
2900 grams (1 cm^3/2.7 grams) = 1074 cm^3
1. 👍 0
2. 👎 0
2. Thanks Damon!!
1. 👍 0
2. 👎 0
3. What is the volume of a piece of platinum (ñ = 22.6 g/cm3) that has a mass of 0.35 kg?
1. 👍 0
2. 👎 0
## Similar Questions
1. ### Chemistry
The density of aluminum is 2.70g/cm3. If a cube of aluminum weighs 13.5 grams whatis the length of the edge of the cube. I need help with the formula to figure this one out.
asked by Jessica on June 1, 2011
2. ### Related Rates
Sand is falling into a conical pile at the rate of 10 m3/sec such that the height of the pile is always half the diameter of the base of the pile. Find the rate at which the height of the pile is changing when the pile is 5 m.
asked by Dimple on January 23, 2016
3. ### calculus
sand is being dropped onto a conical pile in such a way that the height of the pile is always twice the base radius.what is the rate of change of the volume of the pile with respect to the radius when the radius of the pile is 12
asked by raye on March 2, 2019
4. ### Elementary Analysis
A company wants to manufacture cylindrical aluminum cans with a value of 1000cm^3 (1 Liter) What radius and height of the can be to minimize the amount of aluminum used? Please help! This is very hard. Thanks.
asked by JEZRELL on February 23, 2013
1. ### Chemistry Help!!
a kilogram of aluminum metal and a kilogram of water are each warmed to 75 degree Celsius and placed in two identical insulated containers. one hour later, the two containers are opened and the temperature of each substance is
asked by dani on January 19, 2011
A display of cans on a grocery shelf consists of 20 cans on the bottom, 18 cans in the next row, and so on in an arithmetic sequence, until the top row has 4 cans. How many cans, in total, are in the display?
asked by Jnr John on April 16, 2013
3. ### Chemisrty Lab
dimensions of a concrete pile are 14 inchesx14 inches x 15 feet. (1 meter=3.28 feet) the pile has 4 faces and 2 ends. 1>what is the surface area in square feet, of a single pile? exclude the ends of the pile. 2>What is the total
asked by NIcole on September 11, 2007
4. ### math
A recycling center pays 25 cents per aluminum can recycled at its facility. If Tracy recycles 5 aluminum cans there, how many dollars will she earn?
asked by B on April 29, 2018
1. ### physics
1500 cm3 of ideal gas at STP is cooled to -20°C and put into a 1000 cm3 container. What is the final gauge pressure? 11 kPa 40 kPa 113 kPa 141 kPa 240 kPa
asked by lila on May 23, 2014
2. ### chemistry
On average, earth's crust contains about 8.1% aluminum by mass. If a standard 12-ounce soft drink can contains approximately 15.0g of aluminum, how many cans could be made from one ton of the Earth's crust?
asked by Maria on January 17, 2011
3. ### Science Physics
A pile driver falls a distance of 1.75 m before hitting a pile. Find its velocity as it hits the pile. So its 1.75x9.80=17.15?
asked by Jay on October 27, 2014
4. ### Math
Tell whether the data collected is continuous or discrete. Rose recycles aluminum cans each week. She records the total number of cans she recycles each week. Jackson is at a crab race. He records the distance traveled by his pet
asked by Anonymous on November 13, 2012 | 1,040 | 3,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-34 | latest | en | 0.92574 |
https://www.topmarkessays.com/blog/foundations-of-business-computing/ | 1,721,514,939,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00681.warc.gz | 896,781,190 | 14,201 | Six Mile Creek Telecentre Background: Pretend you live at Six Mile Creek rural community in the outback and you are planning to start up a telecentre as a small business to serve the local community. The proposed telecentre will provide Internet access as well as a photocopy service. You have done some research and found that there is demand for these services in the community and that the business can be profitable. But in order to determine how profitable the business will be, you need to carry out a cost-benefit analysis. Your cost-benefit analysis will be about comparing anticipated costs with anticipated benefits of the telecentre’s operation over a period of 7 years. In this assignment you will:
• carry out two cost-benefit analysis exercises using Microsoft Excel and then
• write a brief report using Miscrosoft Word to explain the results of your analysis.
Thus this assignment requires you to produce 2 files: an Excel spreadsheet and a Word document. PART A: Cost-Benefit Analysis You will carry out 2 cost-benefit analysis exercises:
• A Payback analysis
• A Return on Investment (ROI) analysis
The Payback analysis will determine how long it will take for the telecentre to pay for itself – this period is referred to as the payback period. For this, you will need to estimate the initial development costs, the annual operation costs and the annual benefits. The initial development costs, monthly operation costs and monthly benefits have been supplied for you below. The Return on Investment (ROI) analysis calculates profit as a percentage of the total costs over the analysis period. The formula for calculating ROI is: ROI = ((total benefits – total costs) / total costs) * 100 COSTS: The anticipated costs for the telecentre are grouped into 2 categories: Development Costs and Operating costs. 1. Development costs 1.1 One-off costs (these costs are incurred only once) Office Renovation @ \$40,000 Furniture costs @ \$15,000 System Installation costs @ \$10,000 1.2 Equipment and software costs You have decided to purchase:
• 20 desktop computers @ \$1,500 each
• 2 Servers @ \$4,000 each
• 1 system backup set @ \$3,000
• 2 printers @ \$800 each
• 2 heavy duty photocopiers @ \$3,000 each
• Microsoft Open Licenses (for operating system and Office suite) @ \$2,000
It is anticipated that the physical hardware equipment will be replaced every 4 years. After 4 years the price of these equipment items is expected to decrease by 20%. Software license fees remain the same price over the analysis period but they have to be renewed every 2 years. 2. Operating costs 2.1 Fixed costs (costs that are relatively constant and not dependent on level of use) Office Space Rental @ \$500/month Internet Service Provider (ISP) charges @ \$100/month Systems Administrator Salary @500/month Manager Salary @ \$800/month Salaries are expected to increase by \$50 every year 2.2 Variable costs (these vary depending on level of use but assume the following average values) Electricity @ \$200/month Maintenance costs @ \$400/month
3
Telephone costs @ \$300/month Supplies (paper, toner, etc) @ 300/month Assume that the monthly rates will remain the same over the period of the analysis
BENEFITS
Benefits of the telecentre operations will come from providing Internet Access to customers as well as providing a photocopying service. 1. Internet Access service Each desktop computer can fetch \$350/month 2. Photocopying Service Photocopying service is expected to fetch \$500 a month Your tasks: The Spreadsheet Both Payback analysis and ROI will be done on the same spreadsheet as they will use the same data set. Your spreadsheet should contain 4 sheets appropriately named: E.g., Development Costs, Operating costs, Benefits and Cost Benefit Analysis. The idea is to store sets of related data in one place (sheet) so that they can be manipulated independently. As well, you need to store hardware and software details, monthly rates for operating costs and benefit details (the data supplied to you above) in the appropriate sheets. Storing these details will enable you to:
• Use formulae to generate other data needed in the sheets.
• Make changes to details only in one place whenever there is a change.
For instance by storing the monthly Internet Service charge in one cell, if there is an increase in the rate then you need only change the rate in that cell and the change will cascade to all other cells whose values depend on this rate (because all sheets are linked by formulae). In this spreadsheet exercise, you should make use of the following Excel features:
• Use a spreadsheet containing more than one sheet
• Use appropriate formulae to generate needed data
• Know how to reference and copy a value from a cell in one sheet to a cell in another sheet | 1,015 | 4,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-30 | latest | en | 0.917807 |
http://freetofindtruth.blogspot.com/2014/01/the-gematria-of-50-states.html | 1,586,108,580,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371606067.71/warc/CC-MAIN-20200405150416-20200405180916-00310.warc.gz | 71,015,893 | 25,883 | ## Friday, January 17, 2014
### The Numerology/Gematria of the 50 States
The following states have been decoded using the Pythagorean Method
The Original 13 Colonies
1. DELAWARE 4+5+3+1+5+1+9+5 = 33 = 6
2. PENNSYLVANIA = 7+5+5+5+1+7+3+4+1+5+9+1 = 53 = 8
3. NEW JERSEY = 5+5+5 1+5+9+1+5+7 = 15+28 = 43 = 7
4. GEORGIA = 7+5+6+9+7+9+1 = 44 = 8
5. CONNECTICUT = 3+6+5+5+5+3+2+9+3+3+2 = 46 = 10 = 1
6. MASSACHUSETTS = 4+1+1+1+1+3+8+3+1+5+2+2+1 = 33 = 6
7. MARYLAND = 4+1+9+7+3+1+5+4 = 34 = 7
8. SOUTH CAROLINA = 1+6+3+2+8 3+1+9+6+3+9+5+1 = 20+37 = 57 = 12 = 3
9. NEW HAMPSHIRE = 5+5+5 8+1+4+7+1+8+9+9+5 = 15+52 = 67 = 13 = 4
10. VIRGINA = 4+9+9+7+9+5+9+1 = 53 = 8
11. NEW YORK = 5+5+5 7+6+9+2 = 15+24 = 39 = 12 = 3
12. NORTH CAROLINA = 5+6+9+2+8 3+1+9+6+3+9+5+1 = 30+37 = 67 = 13 = 4
13. RHODE ISLAND = 9+8+6+4+5 9+1+3+1+5+4 = 32+23 = 55 = 10 = 1
• The sum of the thirteen colonies = 66
• 6+8+7+8+1+6+7+3+4+8+3+4+1 = 66
• First Highway across U.S. = Route 66
• NFL established '66
• Woman = 23+15+13+1+14 = 66
• 66 represents creation and completeness in numerology because...
• 1+2+3+4+5+6+7+8+9+10+11 = 66
• 6+6 = 12, the number that follows 11
• After this, the sequence is broken; thus 12 = completeness
The Following 37 States in Order of Statehood
14. VERMONT = 4+5+9+4+6+5+2 = 35 = 8
15. KENTUCKY = 2+5+5+2+3+3+2+7 = 29 = 11 = 2
16. TENNESSEE = 2+5+5+5+5+1+1+5+5 = 34 = 7
17. OHIO = 6+8+9+6 = 29 = 11 = 2
18. LOUISIANA = 3+6+3+9+1+9+1+5+1 = 38 = 11 = 2
19. INDIANA = 9+5+4+9+1+5+1 = 34 = 7
20. MISSISSIPPI = 4+9+1+1+9+1+1+9+7+7+9 = 58 = 13 = 4
21. ILLINOIS = 9+3+3+9+5+6+9+1 = 45 = 9
22. ALABAMA = 1+3+1+2+1+4+1 = 13 = 4
23. MAINE = 4+1+9+5+5 = 24 = 6
24. MISSOURI = 4+9+1+1+6+3+9+9 = 42 = 6
25. ARKANSAS = 1+9+2+1+5+1+1+1 = 21 = 3
26. MICHIGAN = 4+9+3+8+9+7+1+5 = 46 = 10 = 1
27. FLORIDA = 6+3+6+9+9+4+1 = 38 = 11 = 2
28. TEXAS = 2+5+6+1+1 = 15 = 6
29. IOWA = 9+6+5+1 = 21 = 3
30. WISCONSIN = 5+9+1+3+6+5+1+9+5 = 44 = 8
31. CALIFORNIA = 3+1+3+9+6+6+9+5+9+1 = 52 = 7
32. MINNESOTA = 4+9+5+5+5+1+6+2+1 = 38 = 11 = 2
33. OREGON = 6+9+5+7+6+5 = 38 = 11 = 2
34. KANSAS = 2+1+5+1+1+1 = 11 = 2
35. WEST VIRGINIA = 5+5+1+2 4+9+9+7+9+5+9+1 = 13+53 = 66 = 12 = 3
36. NEVADA = 5+5+4+1+4+1 = 20 = 2
37. NEBRASKA = 5+5+2+9+1+1+2+1 = 26 = 8
38. COLORADO = 3+6+3+6+9+1+4+6 = 38 = 11 = 2
39. NORTH DAKOTA = 5+6+9+2+8 4+1+2+6+2+1 = 30 + 16 = 46 = 10
40. SOUTH DAKOTA = 1+6+3+2+8 4+1+2+6+2+1 = 20+16 = 36 = 9
41. MONTANA = 5+1+1+8+1 = 16 = 7
42. WASHINGTON = 5+1+1+8+9+5+7+2+6+5 = 49 = 13 = 4
43. IDAHO = 9+4+1+8+6 = 28 = 10 = 1
44. WYOMING = 5+6+6+4+9+5+7 = 42 = 6
45. UTAH = 3+2+1+8 = 14 = 5
46. OKLAHOMA = 6+2+3+1+8+6+4+1 = 31 = 4
47. NEW MEXICO = 5+5+5 4+5+6+9+3+6 = 15+33= 48 = 12 = 3
48. ARIZONA = 1+9+9+8+6+5+1 = 39 = 12 = 3
49. ALASKA = 1+3+1+1+2+1 = 9
50. HAWAII = 8+1+5+1+9+9 = 33 = 6
• 33 is coded into six states
• 3+3 = 6, thus six states with the coding
• Delaware
• Connecticut
• Massachusetts
• Illinois
• New Mexico
• Hawaii
• Only three states sum to 33; Delaware, Massachusetts, Hawaii
• 44 is coded into two states
• Georgia, the 4th State
• Wisconsin
• 555 is coded into 8 states
• New Jersey
• Pennsylvania
• Connecticut
• New Hampshire
• New York
• Tennessee
• Minnesota
• New Mexico (555+33)
• 66 is coded into one state
• West Virginia
• One only state has a single digit value of 5 (even though 555 is clearly important)
• Utah
The United States of America was founded by several 33rd Degree Freemasons, including the first President, George Washington. In their careful creation of the country, they coded numbers within the names of the states. For example, the first state and fiftieth and final state, Delaware and Hawaii respectively, both equate to the number 33. Further, Delaware was founded on a date that sums to 33. That date was December 7, 1787.
• 12/7/1787 = 1+2+7+1+7+8+7 = 33
The Thirteen Colonies Better Explained
Further, the first 13 colonies, when broken down to a single digit, sum to 66. Route 66 was the first highway built across the nation as well. The number 66 signifies completeness and creation, for two special reasons. One is because the word woman, which also represents these qualities, sums to 66.
• WOMAN = 23+15+13+1+14 = 66
• 66 = 6+6 = 12
• Remember this information for the next point
If you sum every number one through eleven, elven being the number before 12, you get 66.
• 1+2+3+4+5+6+7+8+9+10+11 = 66
• 6+6 = 12
• 12 follows eleven
• Thus, the numerology of one through eleven summed, equals 66, equals 12, the next number in the sequence
• The sequence is broken, and does not work for 13, the original number of colonies
• Thus, 13 is the sequence breaker, the element of change
• Thus the 13 colonies
• Delaware, first colony established December 7, 1787
• Rhode Island, thirteenth colony, established May 29, 1790
• 12/7/1787 = 1+2+7+1+7+8+7 = 33
• 5/29/1790 = 5+2+9+1+7+9+0 = 33
1. Virat Sharrmaa is A Jaipur Based Astro – Numerologist who is born and braught up in a traditional brahmin family of jaipur rajasthan ." Numerologist Mumbai " .Visit Our Website : www.viratsharrmaa.com | celebrity numerologist Mumbai
2. WHY WE DO WHAT WE DO
Numerologist.com is your Number 1 online resource for Numerology Education. Our mission here is to help you make the most of your life and future by providing you with high-quality Numerology wisdom and insight.
By empowering you with the knowledge you need to live a life aligned with your desires and frequencies, represented by your unique number patterns, Numerologist.com serves to open doors that would otherwise remain closed, adding value to your life, the lives of those you touch, and to the world. | 2,629 | 5,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-16 | longest | en | 0.478417 |
https://www.physicsforums.com/threads/problem-understanding-operator-algebra.719090/ | 1,511,489,653,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807056.67/warc/CC-MAIN-20171124012912-20171124032912-00159.warc.gz | 840,008,266 | 13,686 | # Problem understanding operator algebra
1. Oct 27, 2013
### Luna=Luna
"It is left as a problem for the reader to show that if $[S,T]$ commutes with S and T, then $[e^{tT}, S] = -t[S,T]e^{tT}$
I'm not sure if i'm missing something here, but i dont even see how it is possible to arrive at this answer.
I get:
$$[e^{tT}, S] = e^{tT}S - Se^{tT}$$
Then using the fact that $[S,T]$ commutes with S and T this gives:
$SST-STS = STS-TSS$
and
$TST-TTS = STT-TST$
and see no way to go further.
One major thing is I don't even see how the factor of -t just appears in the identity?
$[e^{tT}, S] = -t[S,T]e^{tT}$
2. Oct 27, 2013
### MisterX
The typical way to do something like this would be to use a Taylor expansion of the function of the operator, in this case $e^{tT}$. What you were trying to show is an example of a more general result. | 276 | 843 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-47 | longest | en | 0.903257 |
https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_13&oldid=129322 | 1,669,755,402,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710711.7/warc/CC-MAIN-20221129200438-20221129230438-00198.warc.gz | 143,433,642 | 14,010 | # 2008 AIME II Problems/Problem 13
## Problem
A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$. Then the area of $S$ has the form $a\pi + \sqrt{b}$, where $a$ and $b$ are positive integers. Find $a + b$.
## Solution 1
If you're familiar with inversion, you'll see that the problem's basically asking you to invert the hexagon with respect to the unit circle in the Cartesian Plane using the Inversion Distance Formula. This works because the point in the Cartesian Plane's complex plane equivalent switches places with its conjugate but we can do that in the Cartesian plane too (just reflect a point in the Cartesian plane over the x-axis)! If you're familiar with inversion you can go plot the inverted figure's Cartesian Plane Equivalent. Then simply continue on with the figure shown in the below solution.
## Solution 2
If a point $z = r\text{cis}\,\theta$ is in $R$, then the point $\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)$ is in $S$ (where cis denotes $\text{cis}\, \theta = \cos \theta + i \sin \theta$). Since $R$ is symmetric every $60^{\circ}$ about the origin, it suffices to consider the area of the result of the transformation when $-30 \le \theta \le 30$, and then to multiply by $6$ to account for the entire area.
We note that if the region $S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace$, where $R_2$ is the region (in green below) outside the circle of radius $1/\sqrt{3}$ centered at the origin, then $S_2$ is simply the region inside a circle of radius $\sqrt{3}$ centered at the origin. It now suffices to find what happens to the mapping of the region $R-R_2$ (in blue below).
The equation of the hexagon side in that region is $x = r \cos \theta = \frac{1}{2}$, which is transformed to $\frac{1}{r} \cos (-\theta) = \frac{1}{r} \cos \theta =$2 . Let $r\text{cis}\,\theta = a+bi$ where $a,b \in \mathbb{R}$; then $r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}$, so the equation becomes $a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1$. Hence the side is sent to an arc of the unit circle centered at $(1,0)$, after considering the restriction that the side of the hexagon is a segment of length $1/\sqrt{3}$.
Including $S_2$, we find that $S$ is the union of six unit circles centered at $\text{cis}\, \frac{k\pi}{6}$, $k = 0,1,2,3,4,5$, as shown below.
$[asy] defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype("4 4"); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i < 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label("1/\sqrt{3}",(0,-0.5),W,fontsize(8)); [/asy]$ $\Longrightarrow$ $[asy] defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype("4 4")); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype("4 4")); draw(Circle((0,0),1),linetype("4 4")); label("\sqrt{3}",(0,-0.5),W,fontsize(8)); add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p); [/asy]$
The area of the regular hexagon is $6 \cdot \left( \frac{\left(\sqrt{3}\right)^2 \sqrt{3}}{4} \right) = \frac{9}{2}\sqrt{3}$. The total area of the six $120^{\circ}$ sectors is $6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}$. Their sum is $2\pi + \sqrt{27}$, and $a+b = \boxed{029}$. - Th3Numb3rThr33
## Solution 3 (Calculus)
One can describe the line parallel to the imaginary axis $x=\frac{1}{2}$ using polar coordinates as $r(\theta)=\dfrac{1}{2\cos{\theta}}$
so $z$ is equal to $z=(\dfrac{1}{2\cos{\theta}})(cis{\theta}) \rightarrow \frac{1}{z}=2\cos{\theta}cis(-\theta)$
Dividing the hexagon to 12 equal parts we get that
$Area = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}r^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}(2\cos\theta)^2 d\theta$
which is a routine computation:
$Area = 12\int_{0}^{\frac{\pi}{6}}2(\cos\theta)^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta$
$Area = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta = 12[\frac{1}{2}\sin{2\theta}+\theta]_0^{\frac{\pi}{6}}=12(\frac{\sqrt{3}}{4}+\frac{\pi}{6})=2\pi+3\sqrt{3}=2\pi + \sqrt{27}$
$a+b = \boxed{029}$.
2008 AIME II (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions | 1,921 | 5,077 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 50, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2022-49 | latest | en | 0.820073 |
http://cn.metamath.org/mpeuni/strfvd.html | 1,660,275,334,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571538.36/warc/CC-MAIN-20220812014923-20220812044923-00303.warc.gz | 11,183,140 | 3,948 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > strfvd Structured version Visualization version GIF version
Theorem strfvd 16111
Description: Deduction version of strfv 16114. (Contributed by Mario Carneiro, 15-Nov-2014.)
Hypotheses
Ref Expression
strfvd.e 𝐸 = Slot (𝐸‘ndx)
strfvd.s (𝜑𝑆𝑉)
strfvd.f (𝜑 → Fun 𝑆)
strfvd.n (𝜑 → ⟨(𝐸‘ndx), 𝐶⟩ ∈ 𝑆)
Assertion
Ref Expression
strfvd (𝜑𝐶 = (𝐸𝑆))
Proof of Theorem strfvd
StepHypRef Expression
1 strfvd.e . . 3 𝐸 = Slot (𝐸‘ndx)
2 strfvd.s . . 3 (𝜑𝑆𝑉)
31, 2strfvnd 16083 . 2 (𝜑 → (𝐸𝑆) = (𝑆‘(𝐸‘ndx)))
4 strfvd.f . . 3 (𝜑 → Fun 𝑆)
5 strfvd.n . . 3 (𝜑 → ⟨(𝐸‘ndx), 𝐶⟩ ∈ 𝑆)
6 funopfv 6376 . . 3 (Fun 𝑆 → (⟨(𝐸‘ndx), 𝐶⟩ ∈ 𝑆 → (𝑆‘(𝐸‘ndx)) = 𝐶))
74, 5, 6sylc 65 . 2 (𝜑 → (𝑆‘(𝐸‘ndx)) = 𝐶)
83, 7eqtr2d 2806 1 (𝜑𝐶 = (𝐸𝑆))
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1631 ∈ wcel 2145 ⟨cop 4322 Fun wfun 6025 ‘cfv 6031 ndxcnx 16061 Slot cslot 16063 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1870 ax-4 1885 ax-5 1991 ax-6 2057 ax-7 2093 ax-9 2154 ax-10 2174 ax-11 2190 ax-12 2203 ax-13 2408 ax-ext 2751 ax-sep 4915 ax-nul 4923 ax-pr 5034 This theorem depends on definitions: df-bi 197 df-an 383 df-or 835 df-3an 1073 df-tru 1634 df-ex 1853 df-nf 1858 df-sb 2050 df-eu 2622 df-mo 2623 df-clab 2758 df-cleq 2764 df-clel 2767 df-nfc 2902 df-ral 3066 df-rex 3067 df-rab 3070 df-v 3353 df-sbc 3588 df-dif 3726 df-un 3728 df-in 3730 df-ss 3737 df-nul 4064 df-if 4226 df-sn 4317 df-pr 4319 df-op 4323 df-uni 4575 df-br 4787 df-opab 4847 df-mpt 4864 df-id 5157 df-xp 5255 df-rel 5256 df-cnv 5257 df-co 5258 df-dm 5259 df-iota 5994 df-fun 6033 df-fv 6039 df-slot 16068 This theorem is referenced by: strssd 16116
Copyright terms: Public domain W3C validator | 1,028 | 1,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2022-33 | latest | en | 0.122066 |
https://gamedev.stackexchange.com/questions/80371/why-do-we-say-things-are-rotated-around-the-z-axis-in-a-space-graph | 1,627,507,263,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153791.41/warc/CC-MAIN-20210728185528-20210728215528-00409.warc.gz | 287,856,268 | 38,882 | # Why do we say things are rotated around the z-axis in a space graph?
I've embarked on a journey of game development. I need to learn lots of 3D terminology, since I want to make 3D games. I learned that X is to the right, Y is up, and Z is towards you. Then why when a spatial is rotated we say it is rotated around the z-axis? I couldn't find a specific answer online.
All 3D rotation must occur about some axis. That's what 3D rotation is. However, not all such rotation occurs about the Z axis. Rotation can be about any axis: X, Y, Z or an arbitrary axis not aligned with any of the principle axes of the coordinate system.
What it sounds like you're referring to is the convention of rotating 2D shapes in a 2D plane about "the Z axis." 2D rotation occurs about a point, not an axis. However, if you interpret the 2D plane in question as being embedded within a 3D space, the point about which you are rotating becomes a line, and line is perpendicular to the plane you are rotating in: the Z axis, since it is at right angles to the existing X and Y axes of the 2D plane.
That's just a convention, though. In fact, so is the idea that "+X is to the right." It's certainly possible to define a coordinate system where +X is to the left, or towards you. Indeed, even in the conventional system you described, +Z could be towards or away from you depending on the handedness of the system.
• Also, Blender is Z-up and Y-forward, instead of Z-forward and Y-up. – jzx Jul 17 '14 at 0:20
• If I could up vote this answer, then I would. Thank you very much! – Stync Jul 17 '14 at 0:30
They come from two different conventions for aligning the world axes. These conventions are known as Z-up and Y-up.
Z-up
In mathematics, engineering and most every field but computer graphics, the convention is for the X axis and Y axis to represent the ground plane and the Z axis to point up.
Y-up
When computer graphics was in its infancy, the X axis and Y axis were associated with the width and height of the screen. When the Z axis was added for 3D graphics, it was natural for graphics programmers to think of Z as being "in" and "out" of the screen.
This, in itself, is not incompatible with mathematics and engineering. It just means that the default direction that a camera looks is down towards the ground plane.
However, some of the early computer graphics programmers found it more intuitive to think of the default view of the camera to be parallel to the ground, similar to the way their monitor was arranged in their office. This led to them thinking of Z as being forward/back in the world and Y being up/down.
Modern games use both conventions and major content creation tools like Maya support both. | 637 | 2,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-31 | latest | en | 0.974836 |
https://studysoup.com/tsg/17303/university-physics-13-edition-chapter-4-problem-38p | 1,660,484,765,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572033.91/warc/CC-MAIN-20220814113403-20220814143403-00029.warc.gz | 502,923,539 | 14,903 | ×
Get Full Access to University Physics - 13 Edition - Chapter 4 - Problem 38p
Get Full Access to University Physics - 13 Edition - Chapter 4 - Problem 38p
×
# CP An oil tanker’s engines have broken down, and the wind
ISBN: 9780321675460 31
## Solution for problem 38P Chapter 4
University Physics | 13th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
University Physics | 13th Edition
4 5 1 416 Reviews
27
0
Problem 38P
CP? An oil tanker’s engines have broken down, and the wind is blowing the tanker straight toward a reef at a constant speed of 1.5 m/s (?Fig. P4.34?). When the tanker is 500 m from the reef, the wind dies down just as the engineer gets the engines going again. The rudder is stuck, so the only choice is to try to accelerate straight backward away from the reef. The mass of the tanker and cargo is 3.6 X 107 kg, and the engines produce a net horizontal force of 8.0 X 104 N on the tanker. Will the ship hit the reef? If it does, will the oil be safe? The hull can withstand an impact at a speed of 0.2 m/s or less. Ignore the retarding force of the water on the tanker’s hull.
Step-by-Step Solution:
Step 1 of 3
Solution 38P Horizontal force F = 8.0 × 10 N 4 Mass of the tanker m = 3.6 × 10 kg 7 Acceleration of the tanker a = F/m 4 7 2 a = 8.0 × 10 N/3.6 × 10 m/s 3 2 a = 2.22 × 10 m/s Initial speed of the tanker is u = 1.50 m/s If the tanker has to stop, its final speed (v) will be zero. Let us now calculate the distance travelled by the tanker before it stops. Let S be the distance moved by the tanker. 2 2 From the equation, v = u 2aS 0 = 1.50 2 × 2.22 × 10 3 × S S = 506 m Since 506 m > 500 m, the tanker will hit the reef. Given that, the hull can withstand an impact at a speed of 0.2 m/s or less. Let us now calculate the speed of the tanker (v ) whfn it travels 500 m. v f2 = (1.5) 2 × 2.22 × 10 3× 500 v 2 = 2.25 2.22 f v f 0.173 m/s Therefore, when the tank hits the reef 500 m away, its speed will be 0.173 m/s which is lower than the impact speed of 0.2 m/s which the same can withstand. Therefore, the oil will be safe.
Step 2 of 3
Step 3 of 3
## Discover and learn what students are asking
Calculus: Early Transcendental Functions : First-Order Linear Differential Equations
?In Exercises 5-14, solve the first-order linear differential equation. $$\frac{d y}{d x}+\left(\frac{1}{x}\right) y=6 x+2$$
Statistics: Informed Decisions Using Data : Estimating a Population Mean
?The procedure for constructing a t-interval is robust. Explain what this means.
#### Related chapters
Unlock Textbook Solution | 797 | 2,671 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-33 | latest | en | 0.835996 |
http://www.learnattic.com/arithematics/percentage-questions-and-answers-pg11/ | 1,606,731,315,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141213431.41/warc/CC-MAIN-20201130100208-20201130130208-00455.warc.gz | 135,327,576 | 20,852 | # Percentage Questions and Answers Pg11
0
294
1. A student scored 32% marks in science subjects out of 300. How much should he score in language papers out of 200 if he is to get overall 46% marks?
1. 67%
2. 60%
3. 72%
4. 66%
2. In a class 60% of the student pass in Hindi and 45% pass in Sanskrit. If 25% of them pass in at least one subject, what percentage of the students fail in both the subjects?
1. 75%
2. 25%
3. 80%
4. 20%
3. A candidate who scores 30 percent fails by 5 marks, while another candidate who scores 40 percent marks gets 10 more than minimum pass marks. The minimum marks required to pass are
1. 70
2. 150
3. 50
4. 100
4. 90% of the students in a school passed in English, 85% passed in mathematics and 150 students passed in both the subjects. If no students failed in both the subjects, find the total number of students.
1. 220
2. 300
3. 120
4. 200
5. In an examination, 19% students fail in mathematics and 10% students fail in English. If 7% of all students have failed in both subjects, then find the number of students passed in both subjects is
1. 64% of all students
2. 78% of all students
3. 36% of all students
4. 71% of all students | 350 | 1,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-50 | longest | en | 0.929183 |
https://toucharcade.com/community/threads/double-mind-for-iphone-a-new-twist-to-a-classic-logic-puzzle.66417/ | 1,548,200,706,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583875448.71/warc/CC-MAIN-20190122223011-20190123005011-00280.warc.gz | 640,907,920 | 17,992 | # iPad Double Mind for iPhone - A new twist to a classic logic puzzle
Discussion in 'iPhone and iPad Games' started by Maiko, Sep 7, 2010.
1. ### Maiko New Member
Sep 7, 2010
2
0
0
#1
Last edited: Sep 7, 2010
Video here!
Double Mind is a new concept of color code-breaking, which expands the classic Mastermind game to the second dimension.
Double Mind includes three different games:
Classic: given a number of tentatives depending on the difficulty level you chose, you place colored balls on an empty grid. At each tentative, the computer answers by counting the right and wrong positions of the colors, for each row and column of the grid. Your goal is to guess the solution according to this feedback.
Swap: similar to Classic, but you start with all the right colors randomly shuffled in your grid. You play in turns against the computer by swapping colors on the grid, and the computer will do the same with its grid. The first one who guesses wins.
Shift: introduces a Rubik-cube style of play. The solution is shown, and your goal is to shift the colors by rows and columns in your grid in order to match it. You play against the computer at the same time, and the first one who matches the colors wins.
Double Mind is currently available on App Store here:
http://itunes.apple.com/us/app/double-mind/id386485777?mt=8
Check Maiko Games website for more info:
http://maikogames.com
2. ### badmanj Well-Known Member
#2
Picked this up and have played a few games. Have to say, this is really good. Great puzzle game, challenging enough and really nicely implemented. An absolute steal at the price!
Jamie.
Jul 20, 2010
908
0
16
UK
#3
looks good.
4. ### thumbs07 Well-Known Member
Jul 20, 2010
908
0
16
UK
#4
Bought this-well worth money, good logic game and extension of mastermind. Only one drip- and that was a bug in the slide mode- kuyambi's screen bugged out and the solution was not attainable.
5. ### MisterDrgn Well-Known Member
Mar 14, 2009
333
0
0
#5
Any other impressions? This sounds like fun...
6. ### Maiko New Member
Sep 7, 2010
2
0
0
#6
Last edited: Sep 8, 2010
Thank you thumbs07, I will take a look. Do you have any more info about how this bug is reproducible>
Maiko | 583 | 2,214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-04 | latest | en | 0.924152 |
http://acs.ist.psu.edu/discrete-math/applets/factorizer/FactorizerApplet.html | 1,542,332,621,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742968.18/warc/CC-MAIN-20181116004432-20181116030432-00544.warc.gz | 10,034,231 | 1,801 | # FactorizerApplet
Enter a number in the text box, and hit Go! to get that number's prime factorization. Please keep your number under 2,147,483,647 (the limit for 32-bit integers) and refrain from using any arithmetic operators such as '(', ')', '+', '*', etc.
## Further Questions
1. Any factor returned that is less than 24,990,001 is guaranteed to be prime.
a) Why is this true? b) Furthermore, no guarantees can be made about results above this threshold. Why? Hint: Consider the factorization algorithm. What is its inherent limit? Be bold, examine the source code! c) Sketch how to fix this applet so that it can guarantee prime results up to 25,000,000. d) Implement the change.
2. Despite the above limitation, the implemented algorithm is very fast for small 'n'. However, it will not factor some larger numbers.
a) Suggest a new algorithm which is not limited by the size of 'n'. Implement it in pseudocode. b) Learn how the pros do it! Research "Elliptic Curve Method" and "Quadratic Sieve" on Google. These techniques easily factor numbers upwards of 20 and 30 digits.
Created by Andrew Freed, arf132@psu.edu on January 30, 2002.
Modified by Andrew Freed, arf132@psu.edu on April 15, 2002.
Development of this applet was sponsored by the Penn State Fund for Excellence in Learning and Teaching (FELT), project "Java-based Teaching of Mathematics in Information Sciences and Technology", supervised by Frank Ritter and David Mudgett. | 352 | 1,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-47 | latest | en | 0.908767 |
https://pwayblog.com/tag/buckling/ | 1,685,588,161,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647525.11/warc/CC-MAIN-20230601010402-20230601040402-00190.warc.gz | 532,606,882 | 23,243 | How do the rails buckle?
Disclaimer – this includes a back-of-the-envelope calculation. Take it with a pinch of salt. If a steel beam is exposed to an increased temperature, it will tend to expand. If there is nothing to oppose that expansion, then the beam increases in length by ΔL. If, however, the beam’s ends don’t allow this expansion then…
Rail thermal force calculations for jointed track
This is the link to the second article I wrote together with Levente Nogy – “Rail thermal force calculations for jointed track” – and published in the Permanent Way Institution Journal – Vol 135 Part 2 (April 2017). Article 2
Stress transition zones within CWR
The location of the stress transition zone is not only limited to the extremities of a continuous welded rail (CWR) track, the case presented in a previous article – CWR stress transition zone. A stress transition zone may also be present between two fixed zones, inside the CWR. These internal stress transition zones are shorter…
CWR stress transition zone
(prelude to a new PWI Journal article) A stress transition zone is any section of continuous welded rails (CWR) where the thermal force is variable, the longitudinal resistance (p) is active and rail movement occurs due to rail temperature variations. The most common (and well known) location of the stress transition zone is at the…
A day in the life of a jointed track
ΔG = αLΔT°. Free expansion For a free thermal expansion jointed track the rails expand and contract freely and the track components do not provide any resistance to oppose this rail length variation. The joint gap varies linearly relative to the rail temperature. The figure below presents the joint gap variation for a jointed track formed…
When a 20 m rail is 20 m long?
Perhaps I’m splitting hairs here, but it is a fair question to ask: When a 20 m rail is 20 m long? Please, have your say and feel free to comment below, after voting! And this is not a trap question like “Which weighs more: 1 kg of steel rail or 1 kg of feathers?”. Later edit: By…
Jointed track response to rail temperature variations
The thermal behaviour of the jointed track can be analysed throughout a full annual temperature variation and used to define the joint expansion gap variations. The joint gap varies between the maximum value and zero and this is related to the way the rails are allowed by the track resistance forces to contract and expand in…
Rail thermal forces for jointed and CWR track
This is the presentation I gave to the West of England Section of the Permanent Way Institution:
An introduction to rail thermal force calculations
I feel very honoured and proud to see the article “An introduction to rail thermal force calculations” I wrote with my friend, Levente Nogy, published and featured on the front cover of the Journal of the Permanent Way Institution.
Significance of jointed track parameter variation
Joint resistance The normal rail joints are designed to allow the rail length variation due to temperature. To do this the joints have a well-defined maximum gap and a set of installation parameters to provide an optimum behaviour at temperature variation and a good maintenance regime. Any modern rail joint has a standard bolt tightening torque… | 683 | 3,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-23 | latest | en | 0.906444 |
https://cs.stackexchange.com/revisions/65545/1 | 1,642,672,137,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301730.31/warc/CC-MAIN-20220120065949-20220120095949-00467.warc.gz | 255,054,734 | 16,110 | 1 of 4
# Two recurrance function complexity comparison
I have two function T(n), how do i compare which are asymptotically better
1 - T(n) = n^(1/2) T(n^(1/2)) + 3 n, T(1) = 1, T(2) = 1;
and
2 - T(n) = 3 T(n/3) + 2n log n, T(1) = 1, T(2) = 1.
For the first function i guess there is O(sqrt(n)*sqrt(n)) for loop, and O(n) for the c; which becomes O(n^2) totally For the second one, the Master's theorem is usable, but as I assume the complexity becomes O(n*nlogn) <==> O(n^2 * logn) => O(n) for loop, a O(nlogn) for c
So if comparing both O()'s we can totally see that
O(n^2) < O(n^2 * logn)
Was there some mistakes, or this is not how recurrance unrolling is being done? | 242 | 682 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-05 | latest | en | 0.911318 |
https://www.cmm.bristol.ac.uk/forum/viewtopic.php?f=1&t=2418&sid=79a0fe80a98db1790f401e4efd2805c3 | 1,621,002,566,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989526.42/warc/CC-MAIN-20210514121902-20210514151902-00469.warc.gz | 742,673,985 | 5,648 | ## variance in multilevel multinomial models
Welcome to the forum for MLwiN users. Feel free to post your question about MLwiN software here. The Centre for Multilevel Modelling take no responsibility for the accuracy of these posts, we are unable to monitor them closely. Do go ahead and post your question and thank you in advance if you find the time to post any answers!
Remember to check out our extensive software FAQs which may answer your question: http://www.bristol.ac.uk/cmm/software/s ... port-faqs/
Jillianeh
Posts: 4
Joined: Wed Mar 22, 2017 9:01 pm
### variance in multilevel multinomial models
Hello!
I am trying to calculate my Intraclass correlation coefficient from a three level multinomial model.
The variance of the constant has three parts (1[a],
2, 3[c])
I am assuming 1 is the 3rd level variance, 2 is the 2nd level variance, and 3 is the covariance? And the level 1 variance would be 3.29 as per the logistic model?
Then the 3rd level ICC would be= 1/(1+2+3.29)
2nd level ICC would be= 2/(1+2+3.29)
Thanks in advance for the clarification
Jillian | 282 | 1,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-21 | latest | en | 0.917496 |
https://gmatclub.com/forum/profile-evaluation-160173.html | 1,496,141,817,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463614620.98/warc/CC-MAIN-20170530085905-20170530105905-00341.warc.gz | 927,251,215 | 49,116 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 30 May 2017, 03:56
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Profile Evaluation
Author Message
Intern
Joined: 12 May 2013
Posts: 2
GMAT Date: 09-18-2013
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
20 Sep 2013, 04:18
Hi,
Can you pls tell me which bschools in USA can i target? Can i target any of top 10 Bschools?
Planning to apply for 2014 Fall season.
GMAT Score: 730 (Q:50, V:39)
My profile is as below:
Nationality :Indian
Sex: Female
Graduation: B.Tec in Industrial Engineering and Management
College: NIT
CGPA: 9.2/10.0
Department rank 2.
Work Experience till date: Total - 2 years
Manuacturing sector- Construction Equipment Industry
Marketing department
Extra cUrricular:
- Volunteer for MAD( Making a difference) -- Teach a class of 6 students every weekend.
- Executive member of MECHSOC (Society for mechanical engineers) during college.
- Executive committe member of PANGEA (Technical symposium) at college.
- Coordinated various events in different fests during college at national level.
- Active member of NSS (National Service Scheme) for 2 years.
- Presented technical paper on Pressure Sensitive Paints at college level.
- Choreographer of our dance group in college.
- Won prizes in skit and group dance in National cultural fest.
- Participated in many activities in fests.
- Got nominated and elected as Caption of One of the houses of school ( School as divided intp 4 houses) for 1 year.
- Got nominated and elected as Vice Caption of One of the houses of schoolVice Caption of One of the houses for 1 year.
- Won many prizes in sketching and arts and crafts.
- Won 1st prize in paper writing at state level during school.
- Was class representative during school time.
Also should i retake GMAT?
Thanks & Regards.
Joined: 20 Apr 2003
Posts: 5575
Location: Los Angeles CA
Followers: 78
Kudos [?]: 548 [0], given: 74
### Show Tags
30 Sep 2013, 18:29
anikagupta12 wrote:
Hi,
Can you pls tell me which bschools in USA can i target? Can i target any of top 10 Bschools?
Planning to apply for 2014 Fall season.
GMAT Score: 730 (Q:50, V:39)
My profile is as below:
Nationality :Indian
Sex: Female
Graduation: B.Tec in Industrial Engineering and Management
College: NIT
CGPA: 9.2/10.0
Department rank 2.
Work Experience till date: Total - 2 years
Manuacturing sector- Construction Equipment Industry
Marketing department
Extra cUrricular:
- Volunteer for MAD( Making a difference) -- Teach a class of 6 students every weekend.
- Executive member of MECHSOC (Society for mechanical engineers) during college.
- Executive committe member of PANGEA (Technical symposium) at college.
- Coordinated various events in different fests during college at national level.
- Active member of NSS (National Service Scheme) for 2 years.
- Presented technical paper on Pressure Sensitive Paints at college level.
- Choreographer of our dance group in college.
- Won prizes in skit and group dance in National cultural fest.
- Participated in many activities in fests.
- Got nominated and elected as Caption of One of the houses of school ( School as divided intp 4 houses) for 1 year.
- Got nominated and elected as Vice Caption of One of the houses of schoolVice Caption of One of the houses for 1 year.
- Won many prizes in sketching and arts and crafts.
- Won 1st prize in paper writing at state level during school.
- Was class representative during school time.
Also should i retake GMAT?
Thanks & Regards.
yes, you have a competitive profile for the top ten, especially for those programs friendlier to younger applicants. I [url=http://blog.accepted.com/2012/07/06/should-you-retake-the-gmat-2/]recommend against retaking the GMAT.
[/url]
Best,
Linda
_________________
Linda Abraham
Accepted ~ The Premier Admissions Consultancy
310-815-9553
Co-Author of: MBA Admission for Smarties: The No-Nonsense Guide to Acceptance at Top Business Schools
Subscribe to Accepted's Blog
Re: Profile Evaluation [#permalink] 30 Sep 2013, 18:29
Similar topics Replies Last post
Similar
Topics:
Profile evaluation 1 16 Aug 2007, 16:27
Profile Evaluation 3 21 Aug 2007, 16:20
profile evaluation 3 06 Jul 2007, 12:06
Profile evaluation 17 01 Apr 2008, 14:49
Profile evaluation 1 18 Jun 2007, 20:50
Display posts from previous: Sort by | 1,258 | 4,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-22 | longest | en | 0.92119 |
https://kerodon.net/tag/03TZ | 1,675,851,991,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500758.20/warc/CC-MAIN-20230208092053-20230208122053-00690.warc.gz | 344,777,405 | 4,767 | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
Variant 6.3.2.6. Let $\kappa$ be an uncountable cardinal, and let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a morphism of simplicial sets which exhibits $\operatorname{\mathcal{D}}$ as a localization of $\operatorname{\mathcal{C}}$ with respect to some collection of edges $W$ (Definition 6.3.1.9). If $\operatorname{\mathcal{C}}$ is essentially $\kappa$-small, then $\operatorname{\mathcal{D}}$ is essentially $\kappa$-small.
Proof. Without loss of generality, we may assume that $F$ is a monomorphism of simplicial sets. Choose a categorical equivalence of simplicial sets $u: \operatorname{\mathcal{C}}\rightarrow \overline{\operatorname{\mathcal{C}}}$, where $\operatorname{\mathcal{C}}'$ is $\kappa$-small, and form a pushout diagram of simplicial sets
6.11
$$\begin{gathered}\label{equation:localization-essentially-small} \xymatrix { \operatorname{\mathcal{C}}\ar [r]^{u} \ar [d]^{F} & \overline{\operatorname{\mathcal{C}}} \ar [d]^{\overline{F}} \\ \operatorname{\mathcal{D}}\ar [r]^{v} & \overline{\operatorname{\mathcal{D}}} } \end{gathered}$$
Then (6.11) is a categorical pushout square (Example 4.5.4.12), so $v$ is also a categorical equivalence (Proposition 4.5.4.10). Moreover, the morphism $\overline{F}$ exhibits $\overline{D}$ as a localization of $\overline{\operatorname{\mathcal{C}}}$ with respect to $u(W)$ (Corollary 6.3.4.3). We may therefore replace $F$ by $\overline{F}$, and thereby reduce to proving Variant 6.3.2.6 in the special case where $\operatorname{\mathcal{C}}$ is $\kappa$-small. In particular, set of edges $W$ is $\kappa$-small.
Let $Q$ be a contractible Kan complex which is equipped with a monomorphism $\Delta ^1 \hookrightarrow Q$ and has only countably many simplices. Form a pushout diagram of simplicial sets
$\xymatrix { \coprod _{w \in W} \Delta ^1 \ar [r] \ar [d] & \operatorname{\mathcal{C}}\ar [d]^{G} \\ \coprod _{w \in W} Q \ar [r] & \operatorname{\mathcal{C}}', }$
so that $\operatorname{\mathcal{C}}'$ is $\kappa$-small (Remark 5.4.4.6). It follows from Corollary 6.3.4.3 that the morphism $G$ exhibits $\operatorname{\mathcal{C}}'$ as a localization of $\operatorname{\mathcal{C}}$ with respect to $W$. Using Proposition 5.4.5.5, we can choose an inner anodyne morphism $\operatorname{\mathcal{C}}' \hookrightarrow \operatorname{\mathcal{C}}''$, where $\operatorname{\mathcal{C}}''$ is a $\kappa$-small $\infty$-category. Then $\operatorname{\mathcal{C}}''$ is also a localization of $\operatorname{\mathcal{C}}$ with respect to $W$, so Remark 6.3.2.2 supplies a categorical equivalence of simplicial sets $\operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}''$. It follows that $\operatorname{\mathcal{D}}$ is essentially $\kappa$-small, as desired. $\square$ | 931 | 2,843 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-06 | latest | en | 0.590385 |
https://www.physicsforums.com/threads/equation-to-find-the-initial-velocity-of-a-projectile-fired-from-above-ground-level.577317/ | 1,511,416,963,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806736.55/warc/CC-MAIN-20171123050243-20171123070243-00533.warc.gz | 765,677,691 | 15,560 | # Equation to find the Initial Velocity of a Projectile fired from above ground level.
1. Feb 13, 2012
### dotachin
I'm currently doing a Lab in my University Physics class to calculate the initial velocity of a ball right as it's fired from a cannon. I'm not particularly looking for the answers, just the equation needed (i think I copied the equation from the professor incorrectly).
Here's what I have:
vnought = v/cos(θ)(g/2(xtan(θ)+H)), x= range, H = height off the ground. v is what i believe i copied down wrong. the issue i'm running into is we didn't take velocity data, only the range, height off the ground, and angle of the cannon.
my question is a) is this right and i'm just blanking out in a way to find the v? or b) what is the correct equation
I have some sample data if it would help with calculations. we did three shots and averaged them together for a general range.
trial 1: θ=5°, xavg = 5.13m, H= 1.0m | 242 | 931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-47 | longest | en | 0.939819 |
https://dsp.stackexchange.com/questions/69546/iir-to-fir-is-a-best-fit-polynomial-usually-necessary | 1,726,529,215,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00524.warc.gz | 189,861,441 | 47,847 | # IIR to FIR, is a best fit polynomial usually necessary?
I'm messing around with IIR/FIR filters and want to convert the former to the latter.
I set up a classic impulse response calculation.
X[4] = 1.0
Y[0] = 0.0
Y[1] = 0.0
for n in range( 2, L ):
Y[n] = 0.5 * X[n] + 0.3 * X[n-1] + 0.2 * Y[n-1] + 0.1 * Y[n-2]
and (shout out to Dan B and Matt L) using the scipy "lfilter" and "dimpulse" functions. When using initial values of zeros, they match.
Y2 = sig.lfilter( [ 0.5 , 0.3 ], [ 1, -0.2, -0.1], X )
T3, Y3 = sig.dimpulse( ( [ 0.5 , 0.3 ], [ 1, -0.2, -0.1], 1 ) )
for n in range( 20 ):
print( "%4d %10.5f %10.5f %10.5f" % \
( n, Y3[0][n].real, Y2[n].real, Y[n].real ) )
Here are the values.
0 0.00000 0.00000 0.00000
1 0.50000 0.00000 0.00000
2 0.40000 0.00000 0.00000
3 0.13000 0.00000 0.00000
4 0.06600 0.50000 0.50000
5 0.02620 0.40000 0.40000
6 0.01184 0.13000 0.13000
7 0.00499 0.06600 0.06600
8 0.00218 0.02620 0.02620
9 0.00094 0.01184 0.01184
10 0.00041 0.00499 0.00499
11 0.00017 0.00218 0.00218
12 0.00008 0.00094 0.00094
13 0.00003 0.00041 0.00041
14 0.00001 0.00017 0.00017
15 0.00001 0.00008 0.00008
16 0.00000 0.00003 0.00003
17 0.00000 0.00001 0.00001
18 0.00000 0.00001 0.00001
19 0.00000 0.00000 0.00000
The obvious way to get the FIR coefficients directly is to do the polynomial division.
\begin{align} H(z) &= \frac{B(z)}{A(z)} \\ &= \frac{b_0 + b_1 z + b_2 z^2 ...}{ 1 + a_1 z + a_2 z^2 .... }\\ &= h[0] + h[1] z + h[2] z^2 .... \end{align}
So I did some searching and found numpy.polydiv( B, A ), but was disappointed it doesn't work the way I wanted. It stops at "whole values" instead of "calculating the fractional part".
I wrote a routine to do this (included here for anybody else's benefit).
import numpy as np
#=============================================================================
def main():
B = np.array( [ 0.5 , 0.3 ] )
A = np.array( [ 1, -0.2, -0.1] )
print( B )
print( A )
Q, R = DividePolynomials( B, A, 15 )
print( Q )
print( R )
#=============================================================================
def DividePolynomials( ArgNum, ArgDen, ArgLength ):
Q = np.zeros( ArgLength * 2, dtype=complex )
R = np.zeros( ArgLength * 2, dtype=complex )
S = np.zeros( ArgLength * 2, dtype=complex )
R[0:len(ArgNum)] = ArgNum
for d in range( ArgLength ):
rd = R[d] / ArgDen[0]
Q[d] = rd
S.fill( 0.0 )
S[d:d+len(ArgDen)] = rd * ArgDen
R -= S
return Q[0:ArgLength], R[ArgLength:]
#=============================================================================
main()
Here is the output:
[ 0.5 0.3]
[ 1. -0.2 -0.1]
[ 5.00000000e-01+0.j 4.00000000e-01+0.j 1.30000000e-01+0.j
6.60000000e-02+0.j 2.62000000e-02+0.j 1.18400000e-02+0.j
4.98800000e-03+0.j 2.18160000e-03+0.j 9.35120000e-04+0.j
4.05184000e-04+0.j 1.74548800e-04+0.j 7.54281600e-05+0.j
3.25405120e-05+0.j 1.40509184e-05+0.j 6.06423488e-06+0.j]
[ 2.61793882e-06+0.j 6.06423488e-07+0.j 0.00000000e+00+0.j
0.00000000e+00+0.j 0.00000000e+00+0.j 0.00000000e+00+0.j
0.00000000e+00+0.j 0.00000000e+00+0.j 0.00000000e+00+0.j
0.00000000e+00+0.j 0.00000000e+00+0.j 0.00000000e+00+0.j
0.00000000e+00+0.j 0.00000000e+00+0.j 0.00000000e+00+0.j]
The coefficients match nicely to the expected values that came from the impulse analysis and the remainder gives me an idea of how converged it is.
Of course, I did some searching and found this:
Is there a way to derive an FIR filter using an IIR filter?
In the linked question, the selected answer involved curve fitting, the other answers were consistent with what I was expecting. However, adding the criteria that you want to keep the filter order low, of course makes a better fit polynomial possible than a truncated $$H(z)$$. I didn't follow the paper references. IEEE papers are usually behind some paywall. But I see this as the identical math problem that we have had around here of "What is the best polynomial to fit $$\sin(x)$$ from $$a$$ to $$b$$" with the quotient of $$B(z)/A(z)$$ playing the role of the Taylor series.
• Question 1: Is there a polynomial division function I missed in numpy/scipy that does what I want. [Solved: See Olli's answer]
• Question 2: In "real life", what are typical FIR lengths for typical IIR to FIR conversions, and is this extra polynomial fitting step generally needed/beneficial?
I realize that I am dealing with a small rather well behaved IIR in my example.
• I think scipy.signal.dimpulse is what you're looking for. Commented Aug 1, 2020 at 8:01
• @MattL. Thanks Matt. I am actually working on the general X case, not just an impulse, so this is a special case, but good to know and I'll include it. I figured if I can find the exact IIR to any X and Y then I can find a corresponding FIR, so full solution, compact IIR or extended FIR. In the process, I wanted to confirm the principle illustrated in my question, and couldn't find a polynomial (we used to call it synthetic division) routine. I thought that odd. So now it turns out if I can find any IIR in general I can also solve the best fit IIR to any FIR. Correct? Commented Aug 1, 2020 at 10:57
• I'm a bit confused at your problem statement. In the above question you seem to be looking for a way to approximate a given IIR filter by an FIR filter. In your comment under Hilmar's answer, you seem to be looking for an IIR filter that produces a given output sequence from a given input sequence. Those two problems are different, aren't they? Commented Aug 1, 2020 at 16:36
• Yes, my question concerns a subset of my problem. I should be able to convert back and forth between a best fit IIR and a best fit FIR. My initial inclination was IIR->FIR easy, FIR->IIR hard, then I found the link and the IIR->FIR was made harder by the order constraint. I'm just trying to get an idea of the landscape of the arena I am in. I have heard of "thousands of FIR coefficients before", confirmed again by Hilmar. A truncated FIR->IIR is going to be harder than a longer one I would think. Numerically I can find a best fit IIR, exactly with the right rank set, but more freedom... Commented Aug 1, 2020 at 16:46
• ... and a's and b's "swap values" for a better local fit. I think I need to ensure wider bandwidth in by test signal. I can also find and exact FIR fit for a FIR filter, but I don't think that is anything special. Neither might my IIR technique be. Commented Aug 1, 2020 at 16:50
You can still use NumPy's polydiv, if you first zero-pad B. In Python, after your numpy import and A and B initialization:
print(np.polydiv(np.pad(B, (0, 10)), A)[0])
In older NumPy versions, numpy.pad needs an additional parameter mode='constant', which was made the default value since NumPy 1.17. Running the above prints a sequence of numbers that is identical to what you got by the other means:
[5.00000e-01 4.00000e-01 1.30000e-01 6.60000e-02 2.62000e-02
1.18400e-02 4.98800e-03 2.18160e-03 9.35120e-04 4.05184e-04]
• Okay, I put zeros in the "B =" line of my code and the resulting numbers do match. Thanks. Q1 solved. Commented Aug 1, 2020 at 14:31
• @CedronDawg, I think polydiv() is deconvolution like polymul(), if existed, is basically convolution. Am I missing something here?
– Royi
Commented Jul 16, 2023 at 9:01
Question 2: In "real life", what are typical FIR lengths for typical IIR to FIR conversions, and is this extra polynomial fitting step generally needed/beneficial?
That depends highly on what your IIR filter does. In my neck of the woods (audio) the answer is typically "a few thousand".
It really depends on the "frequency" resolution. At what frequency intervals does something "interesting" happen? Here is is a simple example: Let's say you have a 3rd order butter-worth highpass at 40 Hz sampled at 44.1kHz. Dividing the two gives 1000 which is in the ballpark. Turns out 1024 is pretty bad, 2048 is "ok" and 4096 is "good".
Speaking more formally: it really depends on the location of your poles. The lower the frequency and the higher the Q, the more FIR coefficients you need.
I don't think polynomial division helps much here. You either need to truncate the IIR impulse response (with potentially some windowing/tampering at the tail end) or do a straight FIR fit of the transfer function, where you can play around with the error criteria. Trying to match a specific IIR response may not be useful: cut out the "middle man" and design your FIR filter directly from the requirements
• Thanks, I am trying to "cut out the middle man" in the other direction, wondering if it had any practical value and the dimensions of real life systems. Suppose a system is governed by an unknown IIR. My shallow understanding (just starting to explore) is that the usual procedure is to find a FIR (H(z)) then use methods to derive the best fit IIR (A(z) and B(z)). I set about trying to find the best IIR directly from an arbitrary input and it's corresponding output. Some success so far. The polynomial division is part of a double check. Commented Aug 1, 2020 at 13:35
• In my experience you never use a middle man. If you want FIR you design this directly, if you want IIR, you design that directly. There is an interesting middle ground: "Warped FIR" it's an IIR but can be designed using FIR design methods in a "warped" frequency domain. Commented Aug 1, 2020 at 16:24
• I'm trying to solve the opposite black box problem. Given a stretch of the input and and corresponding output of a filter, figure out whether it is IIR or FIR and what the coefficients are. These considerations are part of that, real life parameters will give me targets to test against. So, first I have to generate some ideal noiseless inputs and outputs to work against. The above code is part of that. There can be noise in the signals, vs noise in my data collection for real data. Commented Aug 1, 2020 at 16:39
I am not sure why you are trying to approximate an IIR with an FIR, but an efficient way to do that is with Truncated IIR Filters.
Might be worth exploring.
• Trying to get a feel for ball park numbers. I am supposing some one is measuring something and modeling it with an IIR. Then they want to reproduce it over a fairly small range of frequencies, so a small FIR might be a better fit and safer bet for their operating region. What are typical real life, in your experience, solving real problems, sizes of filters and do you ever convert them back and forth like that? So, I appreciate the link, I'll remember where it is. Matt left a good one too. How many to model voice degradation in air over a distance of inches? Commented Aug 3, 2020 at 2:29 | 3,345 | 10,879 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-38 | latest | en | 0.61684 |
https://brainmass.com/business/finance/housing-loan-question-principle-interest-and-repayment-structure-20389 | 1,713,997,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819971.86/warc/CC-MAIN-20240424205851-20240424235851-00478.warc.gz | 124,653,849 | 7,423 | Purchase Solution
# Housing Loan Question : Principle, Interest and Repayment Structure
Not what you're looking for?
A housing loan is taken out for \$250,000. The rate of interest is 6% per annum and the loan is over a 25 year period. The repayments are made monthly.
The formula:
(See attached)
gives An the amount owing at the end of the nth time period. Here n is given in months and r, expressed as a decimal, is the monthly rate of interest on the loan. P is the monthly repayment and A is the amount borrowed.
1) Calculate P for the above data
2) Calculate amount owing on the loan after 10 years
3) At end of 10 years the house buyer inherits \$80,000 which he pays off the loan. Assuming that he continues to make the same monthly repayments, calculate how many months before the loan is fully paid off.
##### Solution Summary
Questions relating to a housing loan relating to principle, interest and repayment structure are answered in detail.
##### Solution Preview
From the condition, we know
A=\$250,000, r=0.06/12=0.005, 25 years is 25*12=300 months.
At the end of the 25 years or 300 months, A(n)=0.
Thus we can find P.
1) ...
##### Accounting: Statement of Cash flows
This quiz tests your knowledge of the components of the statements of cash flows and the methods used to determine cash flows.
##### Basic Social Media Concepts
The quiz will test your knowledge on basic social media concepts.
##### Understanding Management
This quiz will help you understand the dimensions of employee diversity as well as how to manage a culturally diverse workforce.
##### Motivation
This tests some key elements of major motivation theories.
##### Introduction to Finance
This quiz test introductory finance topics. | 384 | 1,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-18 | latest | en | 0.926243 |
https://www.r-bloggers.com/2011/11/maximizing-omega-ratio/ | 1,623,719,907,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487614006.8/warc/CC-MAIN-20210614232115-20210615022115-00605.warc.gz | 865,394,893 | 27,502 | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
The Omega Ratio was introduced by Keating and Shadwick in 2002. It measures the ratio of average portfolio wins over average portfolio losses for a given target return L.
Let x.i, i= 1,…,n be weights of instruments in the portfolio. We suppose that j= 1,…,T scenarios of returns with equal probabilities are available. I will use historical assets returns as scenarios. Let us denote by r.ij the return of i-th asset in the scenario j. The portfolio’s Omega Ratio can be written as
$\Omega(L) = \frac{E\left [ max(\sum_{i=1}^{n}r_{ij}x_{i} - L, 0) \right ]}{E\left [ max(L - \sum_{i=1}^{n}r_{ij}x_{i}, 0) \right ]}$
I will use methods presented in Optimizing Omega by H. Mausser, D. Saunders, L. Seco (2006) paper to construct optimal portfolios that maximize Omega Ratio.
The maximization problem (pages 5-6) can be written as
$\Omega^{*}(L) = max_{x,u,d}\frac{\frac{1}{T} \sum_{i=1}^{n}u_{i}}{\frac{1}{T} \sum_{i=1}^{n}d_{i}} \newline\newline \sum_{i=1}^{n}r_{ij}x_{i} - u_{j}+d_{j} = L, j=1,...,T \newline\newline u_{j},d_{j}\geq 0, j=1,...,T \newline\newline u_{j}*d_{j} = 0, j=1,...,T$
It can be formulated as a linear programming problem with following transformation
$t=\frac{1}{\frac{1}{T} \sum_{i=1}^{n}u_{i}} \newline\newline \Omega^{*}(L) = max_{\tilde{x},\tilde{u},\tilde{d},t}\frac{1}{T} \sum_{i=1}^{n}\tilde{u}_{i} \newline\newline \sum_{i=1}^{n}r_{ij}\tilde{x}_{i} - \tilde{u}_{j}+\tilde{d}_{j} = L, j=1,...,T \newline\newline \frac{1}{T}\sum_{i=1}^{n}\tilde{d}_{i} = 1 \newline\newline \tilde{u}_{j},\tilde{d}_{j}\geq 0, j=1,...,T$
This method will only work for , I will use a Nonlinear programming solver, Rdonlp2, which is based on donlp2 routine developed and copyright by Prof. Dr. Peter Spellucci. Following code might not properly execute on your computer because Rdonlp2 is only available for R version 2.9 or below.
max.omega.portfolio <- function
(
ia, # input assumptions
constraints # constraints
)
{
n = ia$n nt = nrow(ia$hist.returns)
constraints0 = constraints
omega = ia$parameters.omega #-------------------------------------------------------------------------- # Linear Programming, Omega > 1, Case #-------------------------------------------------------------------------- # objective : Omega # [ SUM 1/T * u.j ] f.obj = c(rep(0, n), (1/nt) * rep(1, nt), rep(0, nt), 0) # adjust constraints, add u.j, d.j, t constraints = add.variables(2*nt + 1, constraints, lb = c(rep(0,2*nt),-Inf)) # Transformation for inequalities # Aw < b => Aw1 - bt < 0 constraints$A[n + 2*nt + 1, ] = -constraints$b constraints$b[] = 0
# Transformation for Lower/Upper bounds, use same transformation
index = which( constraints$ub[1:n] < +Inf ) if( len(index) > 0 ) { a = rbind( diag(n), matrix(0, 2*nt, n), -constraints$ub[1:n])
constraints = add.constraints(a[, index], rep(0, len(index)), '<=', constraints)
}
index = which( constraints$lb[1:n] > -Inf ) if( len(index) > 0 ) { a = rbind( diag(n), matrix(0, 2*nt, n), -constraints$lb[1:n])
constraints = add.constraints(a[, index], rep(0, len(index)), '>=', constraints)
}
constraints$lb[1:n] = -Inf constraints$ub[1:n] = Inf
# [ SUM r.ij * x.i ] - u.j + d.j - L * t = 0, for each j = 1,...,T
a = rbind( matrix(0, n, nt), -diag(nt), diag(nt), -omega)
a[1 : n, ] = t(ia$hist.returns) constraints = add.constraints(a, rep(0, nt), '=', constraints) # [ SUM 1/T * d.j ] = 1 constraints = add.constraints(c( rep(0,n), rep(0,nt), (1/nt) * rep(1,nt), 0), 1, '=', constraints) # setup linear programming f.con = constraints$A
f.dir = c(rep('=', constraints$meq), rep('>=', len(constraints$b) - constraints$meq)) f.rhs = constraints$b
# find optimal solution
x = NA
sol = try(solve.LP.bounds('max', f.obj, t(f.con), f.dir, f.rhs,
lb = constraints$lb, ub = constraints$ub), TRUE)
if(!inherits(sol, 'try-error')) {
x0 = sol$solution[1:n] u = sol$solution[(1+n):(n+nt)]
d = sol$solution[(n+nt+1):(n+2*nt)] t = sol$solution[(n+2*nt+1):(n+2*nt+1)]
# Reverse Transformation
x = x0/t
}
#--------------------------------------------------------------------------
# NonLinear Programming, Omega > 1, Case
#--------------------------------------------------------------------------
# Check if any u.j * d.j != 0 or LP solver encounter an error
if( any( u*d != 0 ) || sol$status !=0 ) { require(Rdonlp2) constraints = constraints0 # compute omega ratio fn <- function(x){ portfolio.returns = x %*% t(ia$hist.returns)
mean(pmax(portfolio.returns - omega,0)) / mean(pmax(omega - portfolio.returns,0))
}
# control structure, fnscale - set -1 for maximization
cntl <- donlp2.control(silent = T, fnscale = -1, iterma =10000, nstep = 100, epsx = 1e-10)
# lower/upper bounds
par.l = constraints$lb par.u = constraints$ub
# intial guess
if(!is.null(constraints$x0)) p = constraints$x0
# linear constraints
A = t(constraints$A) lin.l = constraints$b
lin.u = constraints$b lin.u[ -c(1:constraints$meq) ] = +Inf
# find solution
sol = donlp2(p, fn, par.lower=par.l, par.upper=par.u,
A=A, lin.u=lin.u, lin.l=lin.l, control=cntl)
x = sol$par } return( x ) } First let’s examine how the traditional mean-variance efficient frontier looks like in the Omega Ratio framework. # load Systematic Investor Toolbox setInternet2(TRUE) source(gzcon(url('https://github.com/systematicinvestor/SIT/raw/master/sit.gz', 'rb'))) #-------------------------------------------------------------------------- # Create Efficient Frontier #-------------------------------------------------------------------------- ia = aa.test.create.ia() n = ia$n
# 0 <= x.i <= 0.8
constraints = new.constraints(n, lb = 0, ub = 0.8)
# SUM x.i = 1
constraints = add.constraints(rep(1, n), 1, type = '=', constraints)
# Omega - http://en.wikipedia.org/wiki/Omega_ratio
ia$parameters.omega = 13/100 ia$parameters.omega = 12/100
# convert annual to monthly
ia$parameters.omega = ia$parameters.omega / 12
# create efficient frontier(s)
ef.risk = portopt(ia, constraints, 50, 'Risk')
# Plot Omega Efficient Frontiers and Transition Maps
layout( matrix(1:4, nrow = 2, byrow=T) )
# weights
rownames(ef.risk$weight) = paste('Risk','weight',1:50,sep='_') plot.omega(ef.risk$weight[c(1,10,40,50), ], ia)
# assets
temp = diag(n)
rownames(temp) = ia$symbols plot.omega(temp, ia) # mean-variance efficient frontier in the Omega Ratio framework plot.ef(ia, list(ef.risk), portfolio.omega, T, T) Portfolio returns and Portfolio Omega Ratio are monotonically increasing as we move along the traditional mean-variance efficient frontier in the Omega Ratio framework. The least risky portfolios (Risk_weight_1, Risk_weight_10) have lower Omega Ratio for 13% threshold (target return) and the most risky portfolios (Risk_weight_40, Risk_weight_50) have higher Omega Ratio. To create efficient frontier in the Omega Ratio framework, I propose first to compute range of returns in the mean-variance framework. Next split this range into # Portfolios equally spaced points. For each point, I propose to find portfolio that has expected return less than given point’s expected return and maximum Omega Ratio. #-------------------------------------------------------------------------- # Create Efficient Frontier in Omega Ratio framework #-------------------------------------------------------------------------- # Create maximum Omega Efficient Frontier ef.omega = portopt.omega(ia, constraints, 50, 'Omega') # Plot Omega Efficient Frontiers and Transition Maps layout( matrix(1:4, nrow = 2, byrow=T) ) # weights plot.omega(ef.risk$weight[c(1,10,40,50), ], ia)
# weights
rownames(ef.omega$weight) = paste('Omega','weight',1:50,sep='_') plot.omega(ef.omega$weight[c(1,10,40,50), ], ia)
# portfolio
plot.ef(ia, list(ef.omega, ef.risk), portfolio.omega, T, T)
The Omega Ratio efficient frontier looks similar to the traditional mean-variance efficient frontier for expected returns greater than 13% threshold (target return). However, there is a big shift in allocation and increase in Omega Ratio for portfolios with expected returns less than 13% threshold.
The Omega Ratio efficient frontier looks very inefficient in the Risk framework for portfolios with expected returns less than 13% threshold. But remember that goal of this optimization was to find portfolios that maximize Omega Ratio for given user constraints. Overall I find results a bit radical for portfolios with expected returns less than 13% threshold, and this results defiantly call for more investigation.
To view the complete source code for this example, please have a look at the aa.omega.test() function in aa.test.r at github. | 2,477 | 8,606 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-25 | longest | en | 0.723962 |
http://mathhelpforum.com/pre-calculus/22529-exponential-growth-decay.html | 1,524,444,286,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945668.34/warc/CC-MAIN-20180422232447-20180423012447-00102.warc.gz | 208,565,429 | 11,125 | # Thread: Exponential Growth and Decay
1. ## Exponential Growth and Decay
I;m having trouble with some questions on exponential growth. All of them seem to end with the same type of equation, one that I cannot figure out.
They will; end something like this..
2=1.15^x (1.15 to the power of x)
On simpler questions like this, you had to find a common base.
ie. 27=3^x. 3^3=3^x therefore x=3
But I don't see how you can find a common base wth 2 and 1.15, or other such numbers (where one has many decimal places-for example, one question ended up as 9=1.076842^x).
The question that ended with 2=1.15^x was as follows: A bacteria grows at 15% per day. In how many days will it have doubled?
Anybody have any ideas?
2. Originally Posted by Piggins
I;m having trouble with some questions on exponential growth. All of them seem to end with the same type of equation, one that I cannot figure out.
They will; end something like this..
2=1.15^x (1.15 to the power of x)
On simpler questions like this, you had to find a common base.
ie. 27=3^x. 3^3=3^x therefore x=3
But I don't see how you can find a common base wth 2 and 1.15, or other such numbers (where one has many decimal places-for example, one question ended up as 9=1.076842^x).
The question that ended with 2=1.15^x was as follows: A bacteria grows at 15% per day. In how many days will it have doubled?
Anybody have any ideas?
$\displaystyle 2 = 1.15^x$
Take the log of both sides. It doesn't matter which base you choose. (Though I'd choose either base 10 or base "e", since you can do those directly on your calculator.) My preference is to use ln:
$\displaystyle ln(2) = ln \left ( 1.15^x \right )$
$\displaystyle ln(2) = x \cdot ln(1.15)$
$\displaystyle x = \frac{ln(2)}{ln(1.15)}$
which you can plug into your calculator.
-Dan
3. Originally Posted by topsquark
$\displaystyle 2 = 1.15^x$
Take the log of both sides. It doesn't matter which base you choose. (Though I'd choose either base 10 or base "e", since you can do those directly on your calculator.) My preference is to use ln:
$\displaystyle ln(2) = ln \left ( 1.15^x \right )$
$\displaystyle ln(2) = x \cdot ln(1.15)$
$\displaystyle x = \frac{ln(2)}{ln(1.15)}$
which you can plug into your calculator.
-Dan
Is there any other way to do it? We have not learned logs, I believe that is Grade 12 math.
4. Originally Posted by Piggins
Is there any other way to do it? We have not learned logs, I believe that is Grade 12 math.
No i can't think of another way. I could explain the basics of logs to you, if you want.
Say we have the following:
$\displaystyle 10^{x} = 100$
That can be expressed as:
$\displaystyle x = log_{10} 100$
And to solve it on your calculator, simply type $\displaystyle log 100$ divided by $\displaystyle log 10$
If the base of the log is not mentioned, it is assumed to be 10.
For example:
$\displaystyle log 2 = x$
Thus $\displaystyle log_{10} 2 = x$
Thus $\displaystyle \frac{ log 2 }{ log 10 } = x$
5. Originally Posted by Piggins
Is there any other way to do it? We have not learned logs, I believe that is Grade 12 math.
The only other way I can think of is to estimate it. For example:
$\displaystyle 1.15^4 \approx 1.74901$
$\displaystyle 1.15^5 \approx 2.01136$
So maybe try
$\displaystyle 1.14^{4.7} \approx 1.92877$
etc.
-Dan
6. Thanks for the help everyone.
1) Estimating was also the only method I could think of.
2) Like 10 people asked about it today, so the teacher explained the basics of logs to us, so now I would do it as explained by you guys. | 1,038 | 3,548 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-17 | latest | en | 0.949972 |
http://www.sqlservercentral.com/Forums/Topic1065933-392-1.aspx | 1,369,178,976,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700871976/warc/CC-MAIN-20130516104111-00032-ip-10-60-113-184.ec2.internal.warc.gz | 725,761,421 | 21,402 | Recent PostsRecent Posts Popular TopicsPopular Topics
Home Search Members Calendar Who's On
Every 3rd Friday of the Month Rate Topic Display Mode Topic Options
Author
Message
Posted Thursday, February 17, 2011 12:33 PM
Forum Newbie Group: General Forum Members Last Login: Thursday, February 17, 2011 12:31 PM Points: 4, Visits: 18
I'm trying to find every 3rd friday of the month The Following code retruns for the current monthSELECT CONVERT(CHAR(10),DATEADD(wk,2,(CURRENT_TIMESTAMP -Day(CURRENT_TIMESTAMP)+1)+(6-DATEPART(dw,(CURRENT_TIMESTAMP -Day(CURRENT_TIMESTAMP)+1)))),121) if the current day is passed the 3rd friday of the month , I need to show them the next month's 3rd Friday dateThanks
Post #1065933
Posted Thursday, February 17, 2011 1:13 PM
SSCertifiable Group: General Forum Members Last Login: Today @ 1:49 PM Points: 6,703, Visits: 11,733
What does your requirement say if the day passed is between the third Saturday of the month and the last day of the month?Ooops, misread your comment "current day passed is" vs "current day is passed". __________________________________________________________________________________________________There are no special teachers of virtue, because virtue is taught by the whole community. --PlatoBelieve you can and you're halfway there. --Theodore RooseveltEverything Should Be Made as Simple as Possible, But Not Simpler --Albert EinsteinThe significant problems we face cannot be solved at the same level of thinking we were at when we created them. --Albert Einstein1 apple is not exactly 1/8 of 8 apples. Because there are no absolutely identical apples. --Giordy
Post #1065959
Posted Thursday, February 17, 2011 1:17 PM
SSC Rookie Group: General Forum Members Last Login: Yesterday @ 11:51 AM Points: 30, Visits: 248
You should be able to accomplish what you want by using a CASE statement. Psuedocode as follows.SELECT CASE WHEN current_timestamp > (your 3rd FRIDAY of month calculation)THEN (altered 3rd FRIDAY of a month calculation based on DATEADD(MONTH, +1, currenttimestamp) instead of current_timestampELSE (your current 3rd FRIDAY of the month calculation)ENDBe careful when you use DW in a datepart function as the number that represents a particular day will change based on the value of @@DATEFIRST (Depending on your locale the ordinal position of a day in a week may be different than what you expect). Check books online for SET DATEFIRST and @@DATEFIRST for more info.-Mike
Post #1065963
Posted Thursday, February 17, 2011 2:01 PM
Grasshopper Group: General Forum Members Last Login: Friday, April 08, 2011 6:54 AM Points: 11, Visits: 60
dosskavi (2/17/2011)SELECT CONVERT(CHAR(10),DATEADD(wk,2,(CURRENT_TIMESTAMP -Day(CURRENT_TIMESTAMP)+1)+(6-DATEPART(dw,(CURRENT_TIMESTAMP -Day(CURRENT_TIMESTAMP)+1)))),121)For January 2011, that formula returns the 14th, when the third Friday is on the 21st.`DECLARE @LoopDate AS DATETIMESET @LoopDate = CAST('2011-01-01' AS DATETIME)SELECTCONVERT(CHAR(10),DATEADD(wk,2,(@LoopDate -Day(@LoopDate)+1)+(6-DATEPART(dw,(@LoopDate -Day(@LoopDate)+1)))),121)2011-01-14`
Post #1065993
Posted Thursday, February 17, 2011 4:42 PM
Grasshopper Group: General Forum Members Last Login: Friday, April 08, 2011 6:54 AM Points: 11, Visits: 60
abair34 (2/17/2011)SELECT CASE WHEN current_timestamp > (your 3rd week of month calculation)THEN (altered 3rd week of a month calculation based on DATEADD(MONTH, +1, currenttimestamp) instead of current_timestampELSE (your current 3rd week of the month calculation)ENDThe 3rd Friday of the month of January 2011 is in the 4th week.
Post #1066079
Posted Thursday, February 17, 2011 4:46 PM
SSC Rookie Group: General Forum Members Last Login: Yesterday @ 11:51 AM Points: 30, Visits: 248
Sorry by 3rd week of the month I meant 3rd friday of the month!! :) Guilty as charged!
Post #1066080
Posted Thursday, February 17, 2011 6:59 PM
SSC-Dedicated Group: General Forum Members Last Login: Today @ 1:51 PM Points: 32,906, Visits: 26,789
Someone like Peter Larsson will probably come up with something simpler, but this works...` SELECT CASE WHEN GETDATE() <= CAST(DATEDIFF(dd,-3,DATEADD(mm,DATEDIFF(mm,0,GETDATE()),0))/7*7+4+14 AS DATETIME) --3rd Friday Current Month THEN CAST(DATEDIFF(dd,-3,DATEADD(mm,DATEDIFF(mm,0,GETDATE()),0))/7*7+4+14 AS DATETIME) --3rd Friday Current Month ELSE CAST(DATEDIFF(dd,-3,DATEADD(mm,DATEDIFF(mm,0,GETDATE())+1,0))/7*7+4+14 AS DATETIME) --3rd Friday Next Month END` --Jeff Moden"RBAR is pronounced "ree-bar" and is a "Modenism" for "Row-By-Agonizing-Row".First step towards the paradigm shift of writing Set Based code: Stop thinking about what you want to do to a row... think, instead, of what you want to do to a column." For better, quicker answers on T-SQL questions, click on the following... http://www.sqlservercentral.com/articles/Best+Practices/61537/For better answers on performance questions, click on the following... http://www.sqlservercentral.com/articles/SQLServerCentral/66909/
Post #1066098
Posted Thursday, February 17, 2011 8:51 PM
Grasshopper Group: General Forum Members Last Login: Friday, April 08, 2011 6:54 AM Points: 11, Visits: 60
Jeff Moden (2/17/2011)` SELECT CASE WHEN GETDATE() <= CAST(DATEDIFF(dd,-3,DATEADD(mm,DATEDIFF(mm,0,GETDATE()),0))/7*7+4+14 AS DATETIME) --3rd Friday Current Month THEN CAST(DATEDIFF(dd,-3,DATEADD(mm,DATEDIFF(mm,0,GETDATE()),0))/7*7+4+14 AS DATETIME) --3rd Friday Current Month ELSE CAST(DATEDIFF(dd,-3,DATEADD(mm,DATEDIFF(mm,0,GETDATE())+1,0))/7*7+4+14 AS DATETIME) --3rd Friday Next Month END`This appears to report:1) The 3rd Friday in April 2011 as the 22nd, when it is the 15th.2) The 3rd Friday in July 2011 as the 22nd, when it is the 15th.
Post #1066133
Posted Thursday, February 17, 2011 8:59 PM
SSC-Dedicated Group: General Forum Members Last Login: Today @ 1:51 PM Points: 32,906, Visits: 26,789
chris_n_osborne (2/17/2011)Jeff Moden (2/17/2011)` SELECT CASE WHEN GETDATE() <= CAST(DATEDIFF(dd,-3,DATEADD(mm,DATEDIFF(mm,0,GETDATE()),0))/7*7+4+14 AS DATETIME) --3rd Friday Current Month THEN CAST(DATEDIFF(dd,-3,DATEADD(mm,DATEDIFF(mm,0,GETDATE()),0))/7*7+4+14 AS DATETIME) --3rd Friday Current Month ELSE CAST(DATEDIFF(dd,-3,DATEADD(mm,DATEDIFF(mm,0,GETDATE())+1,0))/7*7+4+14 AS DATETIME) --3rd Friday Next Month END`This appears to report:1) The 3rd Friday in April 2011 as the 22nd, when it is the 15th.2) The 3rd Friday in July 2011 as the 22nd, when it is the 15th.Dang it... you're absolutely correct, Chris. Back to the drawing board. --Jeff Moden"RBAR is pronounced "ree-bar" and is a "Modenism" for "Row-By-Agonizing-Row".First step towards the paradigm shift of writing Set Based code: Stop thinking about what you want to do to a row... think, instead, of what you want to do to a column." For better, quicker answers on T-SQL questions, click on the following... http://www.sqlservercentral.com/articles/Best+Practices/61537/For better answers on performance questions, click on the following... http://www.sqlservercentral.com/articles/SQLServerCentral/66909/
Post #1066134
Posted Thursday, February 17, 2011 9:18 PM
Grasshopper Group: General Forum Members Last Login: Friday, April 08, 2011 6:54 AM Points: 11, Visits: 60
This appears to work, at least for 2010 and 2011 (I verified the dates against the calendar). However, it uses a look-up rather than a calculation.`BEGIN -- Show each 3rd Friday for 2010 and 2011.DECLARE @LoopCounter AS INTEGERDECLARE @LoopDate AS DATETIMESET @LoopCounter = 1SET @LoopDate = CAST('2010-01-28' AS DATETIME) -- Whatever Date is Passed (month 1 for convenience)SET @LoopDate -- Set @LoopDate to the 1st of the month. = CAST(CAST(YEAR(@LoopDate) AS VARCHAR(4)) + '-' + CAST(MONTH(@LoopDate) AS VARCHAR(2)) + '-' + CAST('01' AS VARCHAR(2)) AS DATETIME)WHILE @LoopCounter <= 24 BEGIN SELECT @LoopDate ,CASE -- Determine the 3rd Friday based on the first day of the month. WHEN DATEPART(dw, @LoopDate) = 1 THEN 20 WHEN DATEPART(dw, @LoopDate) = 2 THEN 19 WHEN DATEPART(dw, @LoopDate) = 3 THEN 18 WHEN DATEPART(dw, @LoopDate) = 4 THEN 17 WHEN DATEPART(dw, @LoopDate) = 5 THEN 16 WHEN DATEPART(dw, @LoopDate) = 6 THEN 15 WHEN DATEPART(dw, @LoopDate) = 7 THEN 21 END AS ThirdFriday SET @LoopCounter = @LoopCounter + 1 SET @LoopDate = DATEADD(month, 1, @LoopDate) ENDENDgo`
Post #1066138
Permissions | 2,454 | 8,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2013-20 | latest | en | 0.85768 |
http://www.gurufocus.com/term/COGS/POOL/Cost%2Bof%2BGoods%2BSold/Pool%2BCorp | 1,490,880,453,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218194600.27/warc/CC-MAIN-20170322212954-00178-ip-10-233-31-227.ec2.internal.warc.gz | 518,255,358 | 28,088 | Switch to:
GuruFocus has detected 6 Warning Signs with Pool Corp \$POOL.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Pool Corp (NAS:POOL)
Cost of Goods Sold
\$1,830 Mil (TTM As of Dec. 2016)
Pool Corp's cost of goods sold for the three months ended in Dec. 2016 was \$317 Mil. Its cost of goods sold for the trailing twelve months (TTM) ended in Dec. 2016 was \$1,830 Mil.
Cost of Goods Sold is directly linked to profitability of the company through Gross Margin. Pool Corp's Gross Margin for the three months ended in Dec. 2016 was 28.7%.
Cost of Goods Sold is also directly linked to Inventory Turnover. Pool Corp's Inventory Turnover for the three months ended in Dec. 2016 was 0.67.
Definition
Cost of goods sold (COGS) refers to the Inventory costs of those goods a business has sold during a particular period.
Pool Corp Cost of Goods Sold for the trailing twelve months (TTM) ended in Dec. 2016 was 372.227 (Mar. 2016 ) + 648.153 (Jun. 2016 ) + 491.878 (Sep. 2016 ) + 317.458 (Dec. 2016 ) = \$1,830 Mil.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Cost of Goods Sold is directly linked to profitability of the company through Gross Margin.
Pool Corp's Gross Margin for the three months ended in Dec. 2016 is calculated as:
Gross Margin = (Revenue - Cost of Goods Sold) / Revenue = (445.235 - 317.458) / 445.235 = 28.7 %
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
A company that has a moat can usually maintain or even expand their Gross Margin. A company can increase its Gross Margin in two ways. It can increase the prices of the goods it sells and keeps its Cost of Goods Sold unchanged. Or it can keep the sales price unchanged and squeeze its suppliers to reduce the Cost of Goods Sold. Warren Buffett believes businesses with the power to raise prices have moats.
Cost of Goods Sold is also directly linked to another concept called Inventory Turnover:
Pool Corp's Inventory Turnover for the three months ended in Dec. 2016 is calculated as:
Inventory Turnover = Cost of Goods Sold / Average Inventory = 317.458 / 470.636 = 0.67
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Inventory Turnover measures how fast the company turns over its inventory within a year. A higher inventory turnover means the company has light inventory. Therefore the company spends less money on storage, write downs, and obsolete inventory. If the inventory is too light, it may affect sales because the company may not have enough to meet demand.
Usually retailers pile up their inventories at holiday seasons to meet the stronger demand. Therefore, the inventory of a particular quarter of a year should not be used to calculate inventory turnover. An average inventory is a better indication.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Pool Corp Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 COGS 1,398 1,268 1,090 1,142 1,262 1,387 1,488 1,603 1,687 1,830
Pool Corp Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 COGS 439 270 326 604 461 297 372 648 492 317
Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts
GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 893 | 3,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-13 | latest | en | 0.960339 |
http://www.varsitytutors.com/gre_math-help/how-to-find-the-solution-to-an-inequality-with-subtraction | 1,485,035,436,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281226.52/warc/CC-MAIN-20170116095121-00019-ip-10-171-10-70.ec2.internal.warc.gz | 760,657,998 | 34,345 | # GRE Math : How to find the solution to an inequality with subtraction
## Example Questions
### Example Question #1 : Inequalities
The cost, in cents, of manufacturing pencils is . The pencils sell for 50 cents each. What number of pencils would need to be sold so that the revenue received is at least equal to the manufacturing cost?
Explanation:
If each pencil sells at 50 cents, pencils will sell at . The smallest value of such that
### Example Question #1 : How To Find The Solution To An Inequality With Subtraction
Find the slope of the inequality equation
7
–1
1
–7
0
–1
Explanation:
From the equation we can see that the slope is –1.
### Example Question #2 : How To Find The Solution To An Inequality With Subtraction
and are both integers.
If , and , which of the following is a possible value of ?
Explanation:
Take the values of y that are possible, i.e. 2 and 3, and plug them into the first inequality. First, plug in 2. 2 – 3x > 21. Subtract 2 from both sides, and then divide by –3. Don't forget that when you divide or multiply by a negative number in an inequality you must flip the inequality sign. Thus, x < –19/3. Now plug in 3. We find, following the same steps, that when y=3, x < –6. Thus –7 is the correct answer.
### Example Question #3 : How To Find The Solution To An Inequality With Subtraction
Quantity A:
The value(s) for which the following function is undefined:
Quantity B:
Which of the following is true?
Quantity A is larger.
Quantity B is larger.
The two quantities are equal.
A comparison cannot be detemined from the given information.
Quantity B is larger.
Explanation:
This question is not as hard as it seems. Remember that for real numbers, square roots cannot be taken of negative numbers. Therefore, we know that this function is undefined for:
This is simple to solve. Merely add to both sides:
Then, divide by :
Therefore, quantity A is less than quantity B. This means that quantity B is greater than it.
### Example Question #4 : How To Find The Solution To An Inequality With Subtraction
Quantity A:
Quantity B:
Which of the following is true?
Quantity A is larger.
Quantity B is larger.
The two quantities are equal.
A comparison cannot be detemined from the given information.
A comparison cannot be detemined from the given information.
Explanation:
Recall that when you have an absolute value and an inequality like
,
this is the same as saying that must be between and . You can rewrite it:
To solve this, you merely need to subtract from all three values:
Since is between and , it could be both larger or smaller than . Therefore, you cannot determine the relationship based on the given information. | 658 | 2,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2017-04 | latest | en | 0.87998 |
https://bylosingallgenerality.wordpress.com/2010/09/ | 1,532,269,484,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593302.74/warc/CC-MAIN-20180722135607-20180722155607-00599.warc.gz | 625,503,952 | 9,495 | ## Archive for September, 2010
### Sweet problem
September 19, 2010
You are on an island far far away and you arrive at an intersection. So you have to choose 1 out of 3 roads to continue on. At the end of one of the roads is a new car, another one leads to a goat and the third road just leads to a balance (which can only be used thrice) and 12 golden coins (one of which is fake). The moment you arrive at the intersection, three Gods appear. You know that one God either lies or doesn’t, one has 2 enveloppes for you and the third God is willing to shave you, since you don’t shave yourself. All three God’s have blue eyes and have to kill themselves if they know that.
With two questions, can you make sure to give the shaving God only golden coins so that he tells you how to get to your new car to be able to get a wolf and a cabbage to get the goat safely across the intersection, where, by then, all Gods are dead?
Btw1. I will try to post more often in the future
Btw2. Problem 1 from last post was silly | 261 | 1,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-30 | latest | en | 0.943868 |
https://brilliant.org/problems/cause-you-burn-with-the-brightest-flame/ | 1,490,751,811,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190134.67/warc/CC-MAIN-20170322212950-00454-ip-10-233-31-227.ec2.internal.warc.gz | 780,953,644 | 12,413 | # Cause you burn with the brightest flame
Geometry Level 3
A circle passes through the vertex $$C$$ of a square $$ABCD$$ and touches its sides $$AB$$ and $$AD$$ at $$M$$ and $$N$$ respectively .If the distance from $$C$$ to the line segment MN is equal to 5 units.Then the area of the rectangle $$ABCD$$ is:
× | 83 | 312 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-13 | longest | en | 0.847134 |
https://www.traditionaloven.com/building/food-glass/convert-gram-g-of-food-glass-to-food-glass-sphere-meter-1-m-d.html | 1,679,761,623,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945368.6/warc/CC-MAIN-20230325161021-20230325191021-00046.warc.gz | 1,132,838,540 | 11,784 | Food Grade Glass 1 gram mass to one-meter diameter spheres converter
Category: main menufood grade glass menuGrams
# food grade glass conversion
## Amount: 1 gram (g) of mass Equals: 0.00000000076 one-meter diameter spheres (∅ 1 m) in volume
Converting gram to one-meter diameter spheres value in the food grade glass units scale.
TOGGLE : from one-meter diameter spheres into grams in the other way around.
## food grade glass from gram to one-meter diameter sphere Conversion Results:
### Enter a New gram Amount of food grade glass to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other food grade glass measuring units - complete list.
Conversion calculator for webmasters.
## Glass grade for food
There is a 0.01 gram difference in mass density sense between glass which is used for food jars and window glass. Density of food grade glass is 2.52g/cm3
This glass type is being generally used for making glass bottles, drinking glasses, glass jars and bowls or containers for storing solid foods and beverages.
Where can the glass units converter be applied?
A glass mass versus volume units calculator can be used in specific situations. For example such as: certain weight of bottles plus various food grade glass fragments, all in one bulk amount, can be converted into an exact solid glass volume (in real meaning on paper without melting it in a gas or electric glass kiln or furnace first) and vise verse. Although with some efforts, the glass volume could be measured also without knowing its actual mass; by dipping it into a liquid and then measuring the created liquid excess above the initial level.
Convert food grade glass measuring units between gram (g) and one-meter diameter spheres (∅ 1 m) but in the other reverse direction from one-meter diameter spheres into grams.
conversion result for food grade glass: From Symbol Equals Result To Symbol 1 gram g = 0.00000000076 one-meter diameter spheres ∅ 1 m
# Converter type: food grade glass measurements
This online food grade glass from g into ∅ 1 m converter is a handy tool not just for certified or experienced professionals.
First unit: gram (g) is used for measuring mass.
Second: one-meter diameter sphere (∅ 1 m) is unit of volume.
## food grade glass per 0.00000000076 ∅ 1 m is equivalent to 1 what?
The one-meter diameter spheres amount 0.00000000076 ∅ 1 m converts into 1 g, one gram. It is the EQUAL food grade glass mass value of 1 gram but in the one-meter diameter spheres volume unit alternative.
How to convert 2 grams (g) of food grade glass into one-meter diameter spheres (∅ 1 m)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
7.5788068138998E-10 * 2 (or divide it by / 0.5)
QUESTION:
1 g of food grade glass = ? ∅ 1 m
ANSWER:
1 g = 0.00000000076 ∅ 1 m of food grade glass
## Other applications for food grade glass units calculator ...
With the above mentioned two-units calculating service it provides, this food grade glass converter proved to be useful also as an online tool for:
1. practicing grams and one-meter diameter spheres of food grade glass ( g vs. ∅ 1 m ) measuring values exchange.
2. food grade glass amounts conversion factors - between numerous unit pairs variations.
3. working with mass density - how heavy is a volume of food grade glass - values and properties.
International unit symbols for these two food grade glass measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for gram is:
g
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for one-meter diameter sphere is:
∅ 1 m
### One gram of food grade glass converted to one-meter diameter sphere equals to 0.00000000076 ∅ 1 m
How many one-meter diameter spheres of food grade glass are in 1 gram? The answer is: The change of 1 g ( gram ) mass unit of food grade glass measure equals = to volume 0.00000000076 ∅ 1 m ( one-meter diameter sphere ) as the equivalent measure within the same food grade glass substance type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in g - grams for food grade glass amount, the rule is that the gram number gets converted into ∅ 1 m - one-meter diameter spheres or any other food grade glass unit absolutely exactly.
Conversion for how many one-meter diameter spheres ( ∅ 1 m ) of food grade glass are contained in a gram ( 1 g ). Or, how much in one-meter diameter spheres of food grade glass is in 1 gram? To link to this food grade glass gram to one-meter diameter spheres online converter simply cut and paste the following.
The link to this tool will appear as: food grade glass from gram (g) to one-meter diameter spheres (∅ 1 m) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,202 | 5,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-14 | latest | en | 0.861179 |
https://mathhelpboards.com/threads/help-solve-a-problem.6841/ | 1,606,466,756,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141191511.46/warc/CC-MAIN-20201127073750-20201127103750-00039.warc.gz | 405,863,262 | 14,537 | # Number TheoryHelp solve a problem...
##### New member
I'd love to solve this problem and stun a friend...
Prove if a/b and a/c then a/(b+C)
let a,b,c (element of) (integers.)
I'm sure there's a genius that can figure this out.
Thanks!!
#### Bacterius
##### Well-known member
MHB Math Helper
[JUSTIFY]I take it that "a/b" means "a divides b". Then, if $a$ divides $b$, then there exists an integer $k_1$ such that $b = ak_1$. Similarly, if $a$ divides $c$, then there exists an integer $k_2$ such that $c = ak_2$. Then $b + c = ak_1 + ak_2 = a(k_1 + k_2) = ak_3$ for $k_3 = k_1 + k_2$, and so $a$ divides $b + c$.
More intuitively, if $a$ divides both $b$ and $c$, that means that $b$ and $c$ are both multiples of $a$ (by definition). So their sum must also be a multiple of $a$, and the result follows.[/JUSTIFY] | 270 | 821 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-50 | longest | en | 0.829339 |
https://prepp.in/rrb-ntpc-exam/syllabus | 1,623,799,069,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621627.41/warc/CC-MAIN-20210615211046-20210616001046-00475.warc.gz | 431,696,099 | 59,867 | RRB NTPC Detailed Syllabus and Exam Analysis
Updated On -
Jun 10, 2021
Sudeshna Bhar
Exams Prepmaster
RRB NTPC exam will be conducted to around 35,208 posts of Non-Technical Popular Categories, Para Medical Staff, Material & Isolated Category, and Level-1. The prescribed syllabus is divided into Mathematics, General Intelligence and Reasoning, and General Awareness. Also, check RRB Upcoming Exam Dates 2021
The topics of the RRB NTPC syllabus maintains the difficulty level of Class 10, 12, and graduation levels. The candidates who will qualify for the CBT-I will be eligible for the CBT-II of the RRB NTPC Exam. Read the exam pattern before.
RRB NTPC Math Syllabus
RRB NTPC Syllabus of Mathematic
Here are the important topics of Mathematics from which the question of the mathematics section will be asked.
Topics Topics
Number stem Decimals
Fractions LCM, HCF
Ration and Proportions Percentage
Mensuration Time and Work
Time and Distance Simple and Compound Interest
Profit and Loss Elementary Algebra
Geometry and Trigonometry Elementary Statistics
RRB NTPC Mathematics: Topic-wise Difficulty Level
The below-table on the difficulty level and the expected number of questions is also tentative which has been extracted from the previous year’s question papers.
Topic No of Questions(Tentative) Difficulty Level
SI/ CI 2 Moderate
Mensuration 2 Easy
Ratio & Proportion 3 Easy
Height & Distance 2 Easy-Moderate
Profit/Loss 3 Easy
Geometry 2 Easy-Moderate
Number System 4 Easy-Moderate
Simplification 3 Easy
Time and Work 2 Easy-Moderate
Statistics 1 Easy
Time, Speed, and Distance 2 Easy
Average 1 Easy-Moderate
DI (Tabular) 3 Easy-Moderate
Total 30 Easy-Moderate
RRB NTPC GK Syllabus
RRB NTPC: Syllabus of General Awareness
General Awareness is a vast topic and really hard to cover before exams. In the following tabular, a list of topics has been mentioned for better preparation.
Topics
Art and Culture of India Games and Sports Current Events of National and International Importance
Indian Polity and Governance- Constitution and political stem General Science Monuments and Places of India
Indian Literature Physical, Social and Economic Geography of India and World UN and Other important World Organizations
History of India and Freedom Struggle General Scientific and Technological Developments including Space and Nuclear Program of India Basics of Computers and Computer Applications
Transport Systems in India Environmental Issues Concerning India and World at Large Famous Personalities of India and World
Flora and Fauna of India Common Abbreviations Flagship Government Programs
Indian Economy Important Government and Public Sector Organizations of India etc
RRB NTPC General Awareness: Topic-wise Difficulty Level
The below-table on the difficulty level and the expected number of questions is also tentative which has been extracted from the previous year’s question papers.
Topic No. of Questions(Tentative) Difficulty Level
History 7 Moderate
Geography 1 Easy
Polity 2 Easy
Static 2 Moderate
Biology 7 Easy-Moderate
Chemistry 2 Easy
Physics 4 Easy-Moderate
Computers 4 Easy
Current Affairs 11 Easy-Moderate
Total 40 Easy-Moderate
RRB NTPC Reasoning Syllabus
RRB NTPC: Syllabus of General Intelligence and Reasoning
The syllabus for General Intelligence and Reasoning can seem hard to cover. Here the books for RRB NTPC which will help to cover the syllabus.
Topics
Analogies Coding and Decoding Similarities and Differences
Analytical Reasoning Jumbling Puzzle
Statement-Conclusion Decision Making Alphanumeric Series
Completion of number and alphabetical series Mathematical Operations Relationships
Syllogism Venn Diagrams Data Sufficiency
Statement- Courses of Action Maps, Interpretation of Graphs
RRB NTPC General Intelligence and Reasoning: Topic-wise Difficulty Level
As per the analysis of previous year papers, Coding-Decoding, Sentence arrangement, Syllogism, etc are very important for the RRB NTPC exam. Here is a table for complete exam analysis.
Topic No of Questions Difficulty Level
Coding-Decoding 4 Easy
Sentence arrangement 1 Easy-Moderate
Venn Diagram 3 Moderate
Puzzle (Linear) 3 Easy
Blood Relation 4 Moderate
Statement & Assumptions 2 Moderate
Statement & Conclusion 2 Easy-Moderate
Syllogism 3 Easy-Moderate
Analogy 3 Easy
Mathematical Operations 4 Moderate
Odd one out 1 Easy-Moderate
Total 30 Easy-Moderate
RRB NTPC Syllabus FAQs
The candidates have to clear the stages for the RRB NTPC Exams such as First Stage-CBT, Second stage-CBT, CBAT, and Typing Skill Test. RRB NTPC applied candidates follow the above-mentioned syllabus and paper analysis to prepare for both CBT 1 and CBT 2 of the RRB NTPC exam.
RRB NTPC FAQs
RRB NTPC Syllabus and Exam Analysis FAQs
Ques. What is the best way to prepare as per the syllabus?
Ans. The most efficient way is by practicing and solving questions. For this candidates can go for previous year’s’ question papers or for mock tests as well.
Ques. What are the important topics for the General Intelligence and Reasoning section?
Ans. Candidates need to prepare for Analytical Reasoning, Coding, and Decoding, Data Sufficiency, Decision Making, Puzzle, etc.
Ques. What are the scoring areas of General Awareness?
Ans. Prepare Current Affairs, important dates, events, awards, geography, politics, etc.
Ques. What are the important topics for Mathematics?
Ans. The important topis for this section are Number system, Fractions, Percentage, Time and Work, Simple and Compound Interest, etc.
*The article might have information for the previous academic years, please refer the official website of the exam.
V
vipasha
I heard that Coding-Decoding is an important section in RRB NTPC General Intelligence and Reasoning, is it true?
J
jaya garg
Hi. Yes it is an important chapter. You can expect minimum 4 questions from Coding-Decoding section whose difficulty level is Easy. Some other important topics include Sentence arrangement, Venn Diagram, Puzzle (Linear), etc.
For more details on RRB NTPC General Intelligence and Reasoning, refer here.
T
tanishka shah
Is history section included in the RRB NTPC General Awareness section difficult?
J
jaya garg
Hi. A total of 7-8 questions are asked in the History chapter. There difficulty level is Moderate. Maximum questions are asked from Current Affairs, a total of 11-12 questions are asked whose difficulty level is Easy to Moderate.
S
shivkumar dama
Is RRB NTPC’ General Awareness section important, what all can I expect in this section?
J
jaya garg
Hi. General Awareness is a huge topic and really hard to cover before exams. In case you follow a detailed strategy, you can get a good, score. Some of the important chapters that you must study include Art and Culture of India, Games and Sports, Current Events of National and International Importance, etc.
T
tanishka shah
Can you tell some important RRB NTPC topics for Mathematic, so that I can prepare accordingly?
J
jaya garg
Hi. Some of the important topics of Mathematics that you must include in your preparation include Number stem, Decimals, Fractions, LCM, HCF, Ration and Proportions, Percentage, Mensuration, Time and Work, etc. Check latest updates on RRB NTPC Syllabus of Mathematic here.
R
riyaz
Can you tell me the topics asked in the RRB NTPC exam?
J
jaya garg
Hi. The topics of the RRB NTPC syllabus maintains the difficulty level of Class 10, 12, and graduation levels. The candidates who will qualify for the CBT-I will be eligible for the CBT-II of the RRB NTPC Exam. Read the RRB NTPC exam pattern here. | 1,799 | 7,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-25 | latest | en | 0.856045 |
https://www.geeksforgeeks.org/gate-gate-cs-2003-question-9/ | 1,597,045,276,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738653.47/warc/CC-MAIN-20200810072511-20200810102511-00411.warc.gz | 682,717,959 | 17,691 | # GATE | GATE-CS-2003 | Question 9
Assuming all numbers are in 2’s complement representation, which of the following numbers is divisible by 11111011?
(A) 11100111
(B) 11100100
(C) 11010111
(D) 11011011
Explanation: Since most significant bit is 1, all numbers are negative.
2’s complement of divisor (11111011) = 1’s complement + 1 = 00000100 + 1 = 00000101
So the given number is -5
The decimal value of option A is -25
Quiz of this Question
My Personal Notes arrow_drop_up
Article Tags :
1
Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 172 | 594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-34 | latest | en | 0.817981 |
http://famsoft.org/examples.aspx?cat=dtd | 1,591,361,513,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348500712.83/warc/CC-MAIN-20200605111910-20200605141910-00303.warc.gz | 38,894,649 | 105,798 | Personal C Sharp By famsoft.org Home How To Start Examples-Desktop Examples-Web PC# Methods Reference-Desktop Reference-Web
EXAMPLE 2: If you are wondering why we need the polar coordinate system, some applications can be significantly simplified by using this system. Here is a figure which could be hard to draw using Cartesian coordinates. ========================================================================================= public class a : pcs { public override void init() { base.init(); } public override void run() { lf=of=200;gm("ced"); // Draw a circle, diam=200 pixels for (float a=0;a<360;a+=60) { // make a=angle values between 0,300 kb=true;jf=100;kf=a;lf=of=200; // using polar coord's draw arches with center jd=(double)a+120;kd=120; // at radius=100, angle=a, start angle=a+120 gm("cad"); // Extent angle=120 } } } =========================================================================================
TUTORIAL: What are we drawing on? ----------------------- What we have is the form surface on which we can install controls. We can also paint its background with any color or tile it with a background image. We can also draw on it except that we don't like to do so for a reason which we'll explain shortly. On top of the form is a transparent bitmap image which we draw on. Since the bitmap image is transparent, everything the form contains shows through it and combines with the drawings. In order to maximize simplicity, avoid some errors and insure that your program runs once only, we include all program code into method run() Whenever the Form's size is minimized, then restored back to original size, the .NET software redraws for us all controls and background paint or image on the form, but it does not redraw anything else which we could have drawn directly on the form using method run(). PC# redraws the bitmap image with all the drawings on its surface, so we get everything to show up again. A second advantage of using the bitmap image is that it allows us to save our drawings into a file with ease and reliability. We'll see an example on that shortly. One more advantage, is the compatibility between drawing on the form and drawing on client's browser which we'll see when we get into web developing. Finally, this way has simplified miximg class (pcs)'s drawings with class (pcs3)'s We'll see that when we get into the chapter of "Using the Windows Presentation Foundation". ========================================================================================= EXAMPLE 3: Now, we'll see a demonstration of what you can do with "Shear" and "Rotation". This is the easiest way to draw a diamond shaped object or a parallelogram. ========================================================================================= public class a : pcs { public override void init() { base.init(); } public override void run() { lf=260;of=100;id=-1;od=0;gm("crd"); // Draw a rect with horiz shear factor=-1 cls="b05";gm("sps"); // Create solid blue pen lf=of=60;ad=45;gm("crf"); // Create, fill a square, rotated 45 degrs cls="r0";gm("sps"); // Change color to red lf=of=200;id=0;od=-1;gm("ced"); // Draw a circle with vert shear factor=-1 } } ========================================================================================= TUTORIAL: Shear Factors (id,od): ---------------------- To learn the effect of shearing, assume we have a Rectangle and like to convert it into a Parallelogram by rotating and stretching either its vertical sides or its horizontal sides by an angle (a) The table below shows the possible assignments of the shear factors and the resulting Shape in each case: id od Transformation result ------ ------ ------------------------------------------------ tan(a) 0 The vertical sides rotate anti-clockwise by (a) -tan(a) 0 The vertical sides rotate clockwise by (a) 0 tan(a) The horizontal sides rotate clockwise by (a) 0 -tan(a) The horizontal sides rotate anti-clockwise by (a) A combination of shearing, and rotation can make you able to get the necessary parallelogram for your application. =========================================================================================
EXAMPLE 5: Drawing the diamond shape has been easy. How about drawing an "ace of hearts"? ========================================================================================= public class a : pcs { public override void init() { base.init(); } public override void run() { cls="S9";gm("sps"); // Set color to black. lf=224;of=300;gm("crd"); // Draw card outline rectangle cls="r0";gm("sps"); // Set color to pure red. fns="trp24"; // Set font:TimesRoman,plain, size=24 os="A";jf=-92;kf=135;gm("ctf"); // Draw the text "A" at top left corner os="A";jf=92;kf=-135;ad=180;gm("ctf"); // Draw same rotated 180 deg at bottom right createHeart(); // Run local method to create the heart (gpp) // It comes large in size, centered into form GraphicsPath gpp1=(GraphicsPath)gpp.Clone(); // Make a copy of (gpp) GraphicsPath gpp2=(GraphicsPath)gpp.Clone(); // Make a second copy of (gpp) //--------------------------- Drawing the center heart ------------------------------ jd=kd=0.125;kf=9;gm("stu"); // Create a unit Affine Transform to scale // (gpp) to 1/8th without location change gm("gtf"); // Xfrm then render-fill (gpp) //----------------------------- Drawing Corner hearts ------------------------------- gpp=gpp1; // Make (gpp) refer to 1st copy of object jd=kd=0.05;jf=-97;kf=123;gm("stu"); // Create Affine Xfrms to scale to 5% and move gm("gtf"); // it to corner then render & fill (gpp) gpp=gpp2; // Make (gpp) refer to 1st copy of object jd=kd=0.05;jf=88;kf=-109;ad=180;gm("stu"); gm("gtf"); // Do the same at other corner except that the } // Xform also rotates it 180 deg's this time //--------------------------- Creating the Heart object ----------------------------- void createHeart(){ // This method creates a large size heart JF[0]=0;KF[0]=0;OF[0]=0.1f; // symbol located at Form's center, using JF[1]=10;KF[1]=30;OF[1]=0.1f; // the "GraphicsPath". JF[2]=30;KF[2]=60;OF[2]=0.1f; JF[3]=60;KF[3]=70;OF[3]=0.1f; // To get the data, you sketch the heart on JF[4]=100;KF[4]=60;OF[4]=0.1f; // paper first and obtain the x,y coordinates JF[5]=130;KF[5]=30;OF[5]=0.1f; // of points on it's outline. JF[6]=150;KF[6]=0;OF[6]=0.1f; JF[7]=150;KF[7]=-30;OF[7]=0.1f; // Get as many points as you can for the curved JF[8]=135;KF[8]=-60;OF[8]=0.01f; // sections and only the start & end points for JF[9]=100;KF[9]=-100;OF[9]=0.01f; // "straight line" sections. JF[10]=0;KF[10]=-230;OF[10]=0.01f; JF[11]=-100;KF[11]=-100;OF[11]=0.01f; // JF[], KF[] are the x,y coordinates for the JF[12]=-135;KF[12]=-60;OF[12]=0.01f; // points. OF[] = tension. For a straight lines JF[13]=-150;KF[13]=-30;OF[13]=0.1f; // OF[]=0 At sections where you like the curve JF[14]=-150;KF[14]=0;OF[14]=0.1f; // not to bend too much or in other words to JF[15]=-130;KF[15]=30;OF[15]=0.1f; // be close to a straight line, assign small JF[16]=-100;KF[16]=60;OF[16]=0.1f; // value to the tension. JF[17]=-60;KF[17]=70;OF[17]=0.1f; JF[18]=-30;KF[18]=60;OF[18]=0.1f; JF[19]=-10;KF[19]=30;OF[19]=0.1f; JF[20]=0;KF[20]=0;OF[20]=0; // Make sure to make your start point also your // end point since it's a closed figure. oi=21;gm("cp"); // total rows=21, create path } } ========================================================================================= TUTORIAL: The comments supplied with the code actually explained it all. However, we need to clear a point here. All shape objects we are using (including text objects) are GraphicsPath objects. So what was new in this example? We needed to unify all shape objects in order to simplify working with them. The GraphicsPath is an object which you can add any shape object to. So, we have used this feature to unify them all. Whenever you call method gm() to create any shape, the method creates a new GraphicsPath, adds the shape you wanted to it and makes (gpp) a reference to that object. The new object represents your shape since it is the only item it contains. In this example, we needed to add some lines and curves to make a closed figure. So, we used the same GraphicsPath feature to do the job. So all shape objects, simple or complex are represented by the same object type. REMARK: Starting at PC# version 4.40, you have another way to generate (gpp) for the Heart sketch. You can scan the sketch and save it into file then supply the name of the file to method gm("bg") to create (gpp) for you. Example 11 of the "Imaging" chapter shows you how to do this job. Affine transform will be discussed in details at a later example. =========================================================================================
EXAMPLE 7: Here is an example on Special Effects - Depth. It shows how to display text in large size letters with 3-D look. ========================================================================================= public class a : pcs { // Always remember, class name = file name public override void init() { base.init(); // Should be last statement in init() } public override void run() { cm("fwc");float wf=of; // Get Form's Client width and height, cm("fhc");float hf=of; // Store them for future use cls="b2";gm("sps"); // Prepare light blue paint lf=wf;of=hf;gm("crf"); // Fill form's background with it. fns="trp260"; // Set the font to TimesRoman, plain,260 xs="PC#"; // The string to be drawn with sp effects. float pos=-250; // Horiz position of each char to be drawn char[] C=xs.ToCharArray(); // Scan the string for (int i=0;i< xs.Length;i++) { // Read the string characters one by one os=""+C[i];gm("ct"); // Get the shape object for each char cls="s9s0";jf=pos;kf=0;id=20; // cls=brightest-darkest colors, location,depth ad=30;ks="d";gm("grs"); // ad=3D angle, ks="d" means select "depth" pos+=200; // Make each 2 chars 200 pixels apart } } } ========================================================================================= After method gm("grs") draws the 3D object, the present GraphicsPath object (gpp) which it ends with is for the front surface of the drawing. After you study the chapter of imaging, you'll know that this means that you can color the front surface of the drawing with any special color like "gold" or "chrome". Here is a new version of example 7 which draws a golden "PC#" string (It requires PC# version 4.40 or later): public class a : pcs { // Always remember, class name = file name public override void init() { j=825;k=425;dm("s"); // Resize Form base.init(); // Should be last statement in init() } public override void run() { cm("fwc");float wf=of; // Get Form's Client width and height, cm("fhc");float hf=of; // Store them for future use cls="b6";gm("sps"); // Prepare light blue paint lf=wf;of=hf;gm("crf"); // Fill form's background with it. fns="trp260"; // Set the font to TimesRoman, plain,260 xs="PC#"; // The string to be drawn with sp effects. float pos=-250; // Horiz position of each char to be drawn char[] C=xs.ToCharArray(); // Scan the string for (int n=0;n< xs.Length;n++) { // Read the string characters one by one os=""+C[n];gm("ct"); // Get the shape object for each char cls="o8Y6";jf=pos;kf=0;id=20; // cls=brightest-darkest colors, location,depth ad=30;ks="d";gm("grs"); // ad=3D angle, ks="d" means select "depth" gm("gc"); // Set the front surface of character as clip area lf=of=400;fls="images\\gold.bmp";gm("blf"); // Start a new bip using "gold.bmp" image scaled to be jf=pos;gm("br"); // slightly larger than (gpp) and draw it over (gpp) gm("gcn"); // Cancel previous setup of Clip area pos+=200; // Make each 2 chars 200 pixels apart } } } =========================================================================================
EXAMPLE 10: This example shows how to make the object you are drawing look elevated up or pushed down. ========================================================================================= public class a : pcs { // Always remember, class name = file name public override void init() { base.init(); // Should be last statement in init() } public override void run() { //-------------------- Painting background -------------------------- cm("fwc");float wf=of; // Get Form's Client width and height, cm("fhc");float hf=of; // Store them for future use cls="s4";gm("sps"); // Prepare light gray paint lf=wf;of=hf;gm("crf"); // Fill form's background with it xs="Personal C Sharp"; // Prepare string to be drawn fns="trb84"; // Set font to TimesRoman, bold, 84 //-------------------- "Elevated up" String -------------------------- y=40; // Vert pos where String is to be drawn cls="s9";gm("sps"); // Set color to white os=xs;kf=y+2;gm("ctf"); // Draw text slightly above (y) cls="S9";gm("sps"); // Set color to black os=xs;kf=y-2;gm("ctf"); // Draw text slightly under (y) cls="s4";gm("sps"); // Set color to wanted foreground color os=xs;kf=y;gm("ctf"); // Draw text exactly at (y) //-------------------- "Pushed down" String -------------------------- y=-40; // Vert pos where String is to be drawn cls="S9";gm("sps"); // Set color to black os=xs;kf=y+2;gm("ctf"); // Draw text slightly above (y) cls="s9";gm("sps"); // Set color to white os=xs;kf=y-2;gm("ctf"); // Draw text slightly under (y) cls="s4";gm("sps"); // Set color to wanted foreground color os=xs;kf=y;gm("ctf"); // Draw text exactly at (y) } } ========================================================================================= TUTORIAL: Easy example. All we needed to do, was to display the string 3 times one time in black color, one time in white color and one time in the wanted color between the first two. The white colored string looks like a reflection and the black colored string looks like a shadow. Since we normally expect the light source to be at the top, when we see a shadow at the bottom and a reflection at the top, we feel that the object is elevated up and when we see a shadow at the top and a reflection at the bottom, we feel that the object is pushed down. =========================================================================================
EXAMPLE 11: Now, let us see how we can save our drawing into file. We are going to save the piece of jewelry created in example 8 into a "jpeg" format file named "x.jpg". ========================================================================================= public class a : pcs { // Always remember, class name = file name public override void init() { base.init(); // Should be last statement in init() } public override void run() { j=k=200;cm("fs"); // Resize form to fit object cls="s7";gm("sps"); // Prepare light gray paint lf=of=200;gm("crf"); // Fill form's background with it. //------------------------- Drawing the object ------------------------- lf=of=170;gm("ce"); // Create the circular gold plate cls="o5y5";gm("spl"); // Prepare linear gradient brush for it gm("grf"); // then render-fill the gold plate. lf=6;of=50;gm("c="); // Create hexagon shape object at center. cls="r0";ks="r";gm("grs"); // Draw the object using sp effects-refl at jf=45;cls="b0";ks="r";gm("grs"); // center in red, then repeat 6 times using jf=-45;cls="b0";ks="r";gm("grs"); // different colors and different locations jf=22;kf=40;cls="g0";ks="r";gm("grs"); jf=-22;kf=40;cls="m0";ks="r";gm("grs"); jf=22;kf=-40;cls="m0";ks="r";gm("grs"); jf=-22;kf=-40;cls="g0";ks="r";gm("grs"); lf=of=25;gm("ce"); // Create a circle at center (pearl) for (int x=0;x<20;x++) { // Draw it 20 times using sp effects-reflection jf=80;kf=18*x;kb=true; // at locations around the plate. Polar coord's cls="p0";ks="r";gm("grs"); // are used for specifying locations } //---------------------------- Saving (bio) --------------------------- bip=bio;fls="x.jpg";gm("bsj"); // Make (bip) refer to same object (bio) refers // to, then save it in "jpeg" format } } ========================================================================================= TUTORIAL: The code remained the same, we have only added some code at the top to reduce form's size (which automatically reduces the default graphical device bitmap (bio) to the same new size. We have then painted the background of (bio) with light gray paint. After drawing the object we saved it into a jpg file. As you know (bio) is transparent, we depend on the Form underneath it in adding background to our drawings. This works fine as long as we are only interested in drawing them to the screen. If we want to save (bio), we must add the missing background. The jpg format is great in memory saving. The bmp format uses a lot more memory but looks better. Reducing form's size is essential in order to reduce the file size. =========================================================================================
EXAMPLE 12: Now, let us read the file back and draw the image. ========================================================================================= public class a : pcs { // Always remember, class name = file name public override void init() { base.init(); // Should be last statement in init() } public override void run() { fls="x.jpg";gm("blf"); // Create a new bitmap object and load the // file into it at full scale. gm("br"); // Render (bip) to default graph. output device } } =========================================================================================
EXAMPLE 13: Let us now write a program to show how to handle a mixture of Controls and drawings and how to make them control each others. We are going to create two combo boxes and a button in between on the form. On the default graphical output bitmap (bio) above, we are going to draw 4 different shapes. The user selects a shape and a color by the combo boxes then click the button to start the execution. When this happens, the selected shape will be filled with selected color. Additionally, the jewel object which we have saved in a file will be used as the background image for the button. The button itself will be transparent so the user will be clicking on the drawing. Make sure that the jewel file "x.jpg" is available. If not, run Example 11 to generate it. ========================================================================================= public class a : pcs { public override void init() { bli=1; // Initialize at block 1 base.init(); } public override void setup() { cs="bt0";i=100;o=100;cls="s79s70";ib=true;ims="x.jpg";cm("i"); // Create bt0 with background image cs="ch0";cus="Select a Shape";j=-250;i=190;o=40;cls="b0s3";fns="trb16"; CIS=new string[] {"Triangle","Square","Circle","Pentagon"}; cm("i"); // Combo box ch0 with choices of shapes cs="ch1";cus="Select a Color";j=250;i=190;o=40;cls="b0s3";fns="trb16"; CIS=new string[] {"Red","Green","Blue","Black"}; cm("i"); // Combo box ch1 with choices of colors } public override void update() { // Var's: s=Shape object number // c=Color order number if ("bt0".Equals(cs)) { // if bt0 clicked cs="ch0";cm("gu"); // Get ch0 update (selected index) s=cui; // Assign selected shape index to (s) cs="ch1";cm("gu"); // Get ch1 update c=cui; // Assign selected color index to (c) bli=2;um("b"); // Goto block 2 for execution } } public override void run() { if (blp==1) { // Initialization block jf=-250;kf=90;lf=3;of=150;gm("c=d"); // Draw the triangle jf=250;kf=100;lf=of=120;gm("crd"); // Draw the square jf=-250;kf=-100;lf=of=130;gm("ced"); // Draw the circle jf=250;kf=-100;lf=5;of=150;gm("c=d");// Draw the pentagon } if (blp==2) { // Execution block cls="r0 g0 b0 S9 ".Substring(3*c,2); // Get (cls) for selected color gm("sps"); // Create solid pen/brush for the color if (s==0) { // If item 0 was selected, draw-fill triangle jf=-250;kf=90;lf=3;of=150;gm("c=f"); } else if (s==1) { // If item 1 was selected, draw-fill square jf=250;kf=100;lf=of=120;gm("crf"); } else if (s==2) { // If item 2 was selected, draw-fill circle jf=-250;kf=-100;lf=of=130;gm("cef"); } else if (s==3) { // If item 3 was selected, draw-fill pentagon jf=250;kf=-100;lf=5;of=150;gm("c=f"); } } } } ========================================================================================= TUTORIAL: Drawing the button: ------------------- We wanted both the foreground and the background colors of the button to be transparent. This is not completely allowed for controls. Foreground colors are always fully opaque even if we set them to be transparent. Background colors for most controls cannot be transparent either. Fortunately the button is one of few controls which can be of transparent background. The Label and Panel controls share this ability with the button. In method setup(), we made the assignments (ib=true;cls="s79s70") for the button. This means that we have requested the button to be "Flat" and to use for both foreground and background the color "s7" which is used for form's background and also for image's background. We selected the foreground to be fully opaque and the background to be fully transparent. Why did we want the button to be flat? This is because the standard botton has a three dimensional border which will prevent it from being invisible. The flat button also has a border except that its border is a rectangle drawn around it using the button's foreground color. So, we can make a flat button invisible by making its foreground color invisible. We can't make the foreground color transparent as mentioned before, but we can make it match the form's background color which will do the same job and this is what we have done. The background color can be made transparent by making its opacity digit "0". It also can be made invisible by selecting the color which matches the image's background color. We have done the two together which has actually been more than necessary. Making a drawing clickable: --------------------------- Using the same technique you can draw any object and make it clickable. This time we used an object which was saved into a file, but this is not necessary. You can draw the object on the present bitmap object (bip) and make (bip) the background image of a transparent button of equal size to its bounding rectangle. To set (bip) as the background image of any control assign "b" to (ims) and call cm("sg"). You can also do the same during setup using method cm("i") What could happen if an object is drawn on the top of a control? ---------------------------------------------------------------- This example opens the road for a new discussion. The controls and the drawings in the example have been apart from each other, so there has been no visibility problem for either one. What could happen if an object was drawn on the top of a control? What we have is the form with all the controls installed into it, and a transparent bitmap object above it containing the drawings. So, theoritically, the object drawn on the bitmap should cover the control making it invisible. In reality, this is not allowed. Although the object can hide the form's body, it cannot hide the controls it contains. The only way you can prevent a control from showing through your object, is by making the control invisible. The "text screen" is made of controls which cover the entire form, so if you turn it on, you'll have no chance to see any graphics. However, method tm() can make the text screen temporarely invisible allowing graphics to appear. To allow graphics to be visible, call tm("vg") To allow text to return back to visiblity call tm("vt") =========================================================================================
=============================================================================================== Scaling the image to be printed: ================================ Mode "prb" prints the present Bitmap object (bip) It allows us to set the position of the bitmap center relative to the center of the printable area of the paper in 1/100th of an inch unit. We do that by assigning the (x,y) components of (bip)'s position to (jf,kf) If we make no assignments to (jf,kf) the image will be centered into the page. Mode "prb" also allows us to set the width and height of the printed image by assigning them to (lf,of) They are also in 1/100th of an inch unit. Now, how can we calculate the necessary values of (lf,of)? Suppose that we like to match the width of the displayed image in inches with the width of the printed image (See the next Remark), how can we do it? Method gm("bR") can find for us (bip)'s resolution in Pixels/Inch over both the horizontal and vertical directions. So we can easily calculate the width in inches as follows: Width in inches = (Width in Pixels) / X-Resolution. Then all we need to do is to multiply this number by 100 and assign it to (lf) Similarly, we can calculate (of) Except that all this is unnecessary since it's the default. All you need to do if you like to match inch for inch is to supply (lf=of=0;) and method gm("prb") will do the calculations for you. If you like to assign values to (lf,of), we recommend assigning value to one of them only and assigning zero to the other one. This is because the method will replace the zero in this case with the value which makes the width to height ratio matches the width to height ratio of (bip) This should save you time and guarantee that the printed image will look like the original. REMARK: ------- The actual width of the image on your monitor may not match the calculated width in inches. This is because the width you see depends also on the monitor itself. The resolution amounts in pixels per inch which we use to calculate the image dimensions in inches come from the .NET class System.Drawing.Image. Printing large size images: --------------------------- Computer screens and print papers do not match in size. Standard paper size is 8.5" X 11" which means that paper's height exceeds its width. For computer screens, the opposite is true. Fortunately, we don't need to be concerned of the screen height since we don't draw on the Form directly, we draw on (bio) which can be of any height we choose and we can use vertical scrollbars. On the horizontal direction, we normally don't like to exceed the screen's width or install horizontal scrollbars although we can. The reason is that whatever exceeds the screen in width will normally not fit into standard size paper when printed. Whenever we have a large image to draw, the best to do is to create (bio) with same width as the form's and a height which makes the height/width ratio of the paper matches the height/width ratio of (bio) The outside dimensions of the "letter" size paper is 8.5 X 11 If you subtract 1 inch from all sides as margins, the net size would be 6.5 X 9. These are the numbers which you may like to use. In the next example, we are going to be using the "Printable Area" size which can be as large as 8.25 X 10.75. =============================================================================================== EXAMPLE 15: Create (bio) with full screen width and make it match printer's paper in height/width ratio. Draw something which can occupy (bio) in full then print it. =============================================================================================== public class a : pcs { // Always remember, class name = file name // Var's used: w,h = Width and height of printable area of page in 1/100th of an inch unit. // x,y = Calculated width and height of screen in pixels. // c = Used in color selection. public override void run() { //----------------------------- (a) Printing Text Lines ----------------------------- gm("dn"); // Stop display temporarely. gm("po"); // Open a new printing operation gm("pgw");om("ti");w=o; // Get paper PrintableArea's width, convert to (int) gm("pgh");om("ti");h=o; // Get paper PrintableArea's height, convert to (int) cm("fwc");om("ff");om("ti");x=o; // Get screen width, convert to (int) of=x*h/w;om("ff");om("ti");y=o; // Calculate how much screen height would be // to be proportional to paper. of=y;cm("fbv"); // Request vert scrollbar & (bio) height=y cls="s9";gm("ec"); // Paint (bio)'s background (necessary for scrollbar for (int k=-y/2;k< y/2;k+=100) { // Operation) Scan (bio) vertically and for (int j=-x/2;j< x/2;j+=100) { // horizontally and draw a colored circle each 100 c=(j+x/2)/100;if (c >7) c=0; // pixels. (c) is used to select one color for each cls="r0b0y0g0p0c0m0S9".Substring (c*2,2);gm("sps"); jf=j+25;kf=k;lf=of=25;gm("cef"); // circle. } } gm("dn"); // Resume display. lf=w;of=h;bip=bio;gm("prb"); // Assign (bio) to (bip) and print it so that it // covers entire printable area of page. gm("pc"); // Close printing operation. } } ==============================================================================================
DRAWING ON THE TEXT SCREEN ========================== The Text Screen is a very versatile display media. In addition to allowing both your program and the user to display text and images, file them and print them, it also allows the two to edit its contents. It can send any of its contents to the Clipboard and can also receive the Clipboard's content. This feature can easily be used to send your drawings to the Text Screen and insert them wherever you like there. Let us have an example. ========================================================================================= Example 16: Let us draw the same piece of jewelry which we have used in other examples on the text screen. In order to show that the Text Screen can contain combinations of text, pictures and dynamically generated drawings, we are going to display two text lines, one above the drawing and one below it. ========================================================================================= public class a:pcs { public override void init() { tia=toa="t"; // Use text screen for text input/output base.init(); // Initialize pcs } public override void run() { cls="r0";fns="trb18";os="Drawing on the Text Screen";tm();os="";tm(); // Display title cls="b0";fns="trb12";os="This text line comes before the drawing.";tm(); // Display first line on Text Screen ib=true;cm("fv"); // Make form invisible temporarely j=k=200;ib=true;cm("fs"); // Resize form to fit object cls="s7";gm("ec"); // Color background to match Text Screen. //--------------------- Drawing the object (No change in code) -------------------- lf=of=170;gm("ce"); // Create the circular gold plate cls="o5y5";gm("spl"); // Prepare linear gradient brush for it gm("grf"); // then render-fill the gold plate. lf=6;of=50;gm("c="); // Create hexagon shape object at center. cls="r0";ks="r";gm("grs"); // Draw the object using sp effects-refl at jf=45;cls="b0";ks="r";gm("grs"); // center in red, then repeat 6 times using jf=-45;cls="b0";ks="r";gm("grs"); // different colors and different locations jf=22;kf=40;cls="g0";ks="r";gm("grs"); jf=-22;kf=40;cls="m0";ks="r";gm("grs"); jf=22;kf=-40;cls="m0";ks="r";gm("grs"); jf=-22;kf=-40;cls="g0";ks="r";gm("grs"); lf=of=25;gm("ce"); // Create a circle at center (pearl) for (int x=0;x<20;x++) { // Draw it 20 times using sp effects-reflection jf=80;kf=18*x;kb=true; // at locations around the plate. Polar coord's cls="p0";ks="r";gm("grs"); // are used for specifying locations } //------------------------ Displaying (bio) on Text Screen ----------------------- imp=bio;sm("csi"); // Set (bio) into Clipboard cm("fsd"); // Return form to regular size. cm("fv"); // Return visibility to Form. os=" ";tm("d"); // Move cursor to wanted display position tm("ep"); // Text Screen edit-paste os="";tm(); // Move to next line cls="b0";os="And this one comes after the drawing.";tm(); } // Display second line on Text Screen. } ========================================================================================= TUTORIAL: While preparing the drawing, We wanted to reduce the form size to the size of the drawn object in order to eliminate the space around it. But since the form size also sets the Text Screen size, we had to resize the form back to original size after the drawing was complete. Also, we kept the Text Screen invisible during drawing for this same reason. The object was copied to the Clipboard while the form was small and pasted on the Text Screen after it has been returned to the original size. As you know each time the form is resized using cm("fs"), the Bitmap object (bio) with all the drawings on its surface are also resized by the same amounts horizontally and vertically. This means that if we have not changed visibility to the Text Screen we could have seen a larger elliptical piece of jewelry which would still be of no concern to us since we made our copy to the clipboard before the resizing took place. Displaying image files on the Text Screen is much easier, you assign to (ims) the image file name and call tm("dg") to display the image at the cursor.
=========================================================================================== CLIP AREA ========= (This section requires version 1.55 or higher) The GraphicsPath object (gpp) can be set as a clip area. A clip area is an area on the graphical output device which disallows drawing anything outside its borders. A negative clip area does the opposite. Drawings can appear only outside a negative clip area. To set (gpp) as a clip area call gm("gc") To set (gpp) as a negative clip area call gm("gc-") To add another (gpp) object to current clip area call gm("gc+") To modify current clip area by intersecting (gpp) with it, call gm("gc&") To modify it by xoring (gpp) with it call gm("gc^") and to reset the clip area call gm("gcn") ========================================================================================= Example 17: Create a large elliptical shape and make it a negative clip area, Create the graphical shape object of the text "PC#" and xor it with the clip area then paint the entire form with an image. See what you'll get. ========================================================================================= public class a : pcs { public override void init() { base.init(); } public override void run() { lf=700;of=300;gm("ce"); // Create an ellipse. gm("gc-"); // Make it a negative clip area. fns="trb260";os="PC#";kf=-20;gm("ct"); // Create (gpp) for the "PC#" text gm("gc^"); // Xor (gpp) with the clip area lf=60;of=30;fls="images\\icon.bmp";gm("blf"); gm("spt"); // Create a texture brush using icon.bmp image file lf=760;of=360;gm("cr"); // Create a form size rectangle gm("grf"); // Draw-fill the rectangle using texture brush } } | 8,555 | 34,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-24 | longest | en | 0.818948 |
https://plainmath.net/10828/you-can-rearrange-the-letters-in-the-word-sanction-how-many-times | 1,632,679,457,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057913.34/warc/CC-MAIN-20210926175051-20210926205051-00403.warc.gz | 508,768,555 | 8,384 | Question
# You can rearrange the letters in the word "Sanction" how many times?
Data distributions
You can rearrange the letters in the word "Sanction" how many times?
$$\displaystyle\frac{{{n}!}}{{{\left({\underset{{{1}}}{{{r}}}}!\right)}{\left({\underset{{{2}}}{{{r}}}}!\right)}\ldots{\left({\underset{{{k}}}{{{r}}}}!\right)}}}$$
$$\displaystyle\frac{{{8}!}}{{{2}!}}=\frac{{40320}}{{2}}={20160}$$ | 135 | 401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-39 | latest | en | 0.500564 |
http://www.briddles.com/2011/01/ | 1,513,428,688,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588072.75/warc/CC-MAIN-20171216123525-20171216145525-00720.warc.gz | 323,651,806 | 7,532 | ### #1 - Glass Problem
Given three bowls: 7, 4 and 3 liters capacity. Only the 7-liter is full. Pouring the water the fewest number of times, make the quantities of 2, 2, and 3 liters.
Three numerals in each number stand for litres in each bowl:
700 - 340 - 313 - 610 - 601 - 241 - 223 (overspilling 6 times)
### #2 - Number Problem
I just found a number with an interesting property:
When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.
It's not a small number, but it's not really big, either. When I looked for a smaller number with this property I couldn't find one.
Can you find it?
2519
### #3 - 3 RapidFire Problem
1.what gets hotter as it gets colder?
2.what is white when it's dirty turns black as you clean it?
3.what gets wetter as it dries?
1.what gets hotter as it gets colder?
The fireplace as winter approaches
2.what is white when it's dirty turns black as you clean it?
Blackboard
3.what gets wetter as it dries?
Towel
### #4 - Honestants and Swindlecants Problem
On this mysterious island there are 2 kinds of people - Honestants who always make true statements and Swindlecants who always make false statements.
In the pub the gringo met a funny guy who said: 'If my wife is an Honestant, then I am Swindlecant.' Who is this couple?
In this logical conditional (if-then statement) p is a hypothesis (or antecedent) and q is a conclusion (or consequent).
It is obvious, that the husband is not a Swindlecant, because in that case one part of the statement (Q) ... then I am Swindlecant. would have to be a lie, which is a conflict. And since A is an Honestant, the whole statement is true.
If his wife was an Honestant too, then the second part of statement (Q) ... then I am Swindlecant. would have to be true, which is a conflict again. Therefore the man is an Honestant and his wife is a Swindlecant.
### #5 - Tennis Ball
Given 27 table tennis balls, one is heavier than the others.
What is the minimum number of weighings (using a two-pan balance scale) needed to guarantee identifying the heavy one? Always. Of course, the other 26 balls weigh the same.
It is enough to use a pair of scales 3 times.
Divide the 27 balls to 3 groups, 9 balls in each. Compare 2 groups the heavier one contains the ball. If there is equilibrium, then the ball is in the third group. Thus we know the 9 suspicious balls.
Divide the 9 balls to 3 groups of 3. Compare 2 groups, and as mentioned above, identify the group of 3 suspicious balls.
Compare 2 balls (of the 3 possibly heavier ones) and you know everything.
So we used a pair of scales 3 times to identify the heavier ball.
### #6 - Trip Problem
If I went halfway to a town 60 km away at the speed of 30 km/hour, how fast do I have to go the rest of the way to have an average speed of 60 km/hour over the entire trip?
average speed = total distance / total time
60 = 60 / (1 + 30/r)
60(1+ 30/r) = 60
1 + 30/r = 1
30/r = 0
Which is impossible. (The rate of the second half cannot be 0)
### #7 - 6 Cup 1 move Problem
Here are six cups. Three of them are empty, three have wine in them. Your goal is to make the line of cups alternate (wine-empty-wine-empty or empty-wine-empty-etc). You can only touch one cup and that is the cup you move. You may not move any other cups with or without your hand except if you are pushing the cup you chose to move between two other cups. How do you do it?
Pour cup 4 into empty cup 1!
### #8 - CAT EXAM ProbleM
5+3+2 = 151022
9+2+4 = 183652
8+6+3 = 482466
5+4+5 = 202541
THEN ;
7+2+5 =
7*2 , 7*5 , 7*2 + 7*5 -2
### #9 - Tanya Date
Tanya wants to go on a date and prefers her date to be tall, dark and handsome.
Of the preferred traits - tall, dark and handsome - no two of Adam, Bond, Cruz and Dumbo have the same number.
Only Adam or Dumbo is tall and fair.
Only Bond or Cruz is short and handsome.
Adam and Cruz are either both tall or both short.
Bond and Dumbo are either both dark or both fair.
Who is Tanya's date?
Cruz is Tanya's date.
As no two of them have the same number of preferred traits - from (1), exactly one of them has none of the preferred traits and exactly one of them has all the preferred traits.
From (4) and (5), there are only two possibilities:
* Adam & Cruz both are tall and Bond & Dumbo both are fair.
* Adam & Cruz both are short and Bond & Dumbo both are dark.
But from (2), second possibility is impossible. So the first one is the correct possibility i.e. Adam & Cruz both are tall and Bond & Dumbo both are fair.
Then from (3), Bond is short and handsome.
Also, from (1) and (2), Adam is tall and fair. Also, Dumbo is the person without any preferred traits. Cruz is Dark. Adam and Cruz are handsome. Thus, following are the individual preferred traits:
Cruz - Tall, Dark and Handsome
Adam - Tall and Handsome
Bond - Handsome
Dumbo - None :-(
Hence, Cruz is Tanya's date.
### #10 - wherz dad Problem
Right now mum is 21 years older then her child. in 6 years her child will be 5 times younger than she.
Q: where is daddy?
on top of mom | 1,483 | 5,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-51 | latest | en | 0.943721 |
http://sourceforge.net/p/maxima/mailman/message/27422159/ | 1,432,985,563,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207930995.36/warc/CC-MAIN-20150521113210-00304-ip-10-180-206-219.ec2.internal.warc.gz | 228,754,602 | 6,869 | Learn how easy it is to sync an existing GitHub or Google Code repo to a SourceForge project!
## [Maxima-bugs] [ maxima-Bugs-3110242 ] bad serie sum
[Maxima-bugs] [ maxima-Bugs-3110242 ] bad serie sum From: SourceForge.net - 2011-04-28 06:20:51 ```Bugs item #3110242, was opened at 2010-11-16 18:45 Message generated for change (Comment added) made by andrejv You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3110242&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: None Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Emilio Suarez (folok) Assigned to: Andrej Vodopivec (andrejv) Summary: bad serie sum Initial Comment: %i1 sum((2^n)/n, n, 1, inf),simpsum; %i2 simplify_sum(%); %o3 -log(-1) Maxima uses de power series or log(1-x) which is valid only if |x|<1 and value it for x=2. The result sould be 'divergent' ---------------------------------------------------------------------- >Comment By: Andrej Vodopivec (andrejv) Date: 2011-04-28 08:20 Message: Fixed in git: (%i5) sum((2^n)/n, n, 1, inf)\$ (%i6) simplify_sum(%); Sum is divergent! #0: to_hypergeometric1(expr=2^(n+1)/(n+1),var=n,lo=1,hi=inf)(simplify_sum.mac line 757) #1: to_hypergeometric(expr=2^n/n,var=n,lo=1,hi=inf)(simplify_sum.mac line 743) #2: simplify_sum(expr='sum(2^n/n,n,1,inf))(simplify_sum.mac line 292) -- an error. To debug this try: debugmode(true); ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3110242&group_id=4933 ```
[Maxima-bugs] [ maxima-Bugs-3110242 ] bad serie sum From: SourceForge.net - 2010-11-16 17:45:24 ```Bugs item #3110242, was opened at 2010-11-16 18:45 Message generated for change (Tracker Item Submitted) made by folok You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3110242&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: None Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Emilio Suarez (folok) Assigned to: Nobody/Anonymous (nobody) Summary: bad serie sum Initial Comment: %i1 sum((2^n)/n, n, 1, inf),simpsum; %i2 simplify_sum(%); %o3 -log(-1) Maxima uses de power series or log(1-x) which is valid only if |x|<1 and value it for x=2. The result sould be 'divergent' ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3110242&group_id=4933 ```
[Maxima-bugs] [ maxima-Bugs-3110242 ] bad serie sum From: SourceForge.net - 2011-04-24 09:05:13 ```Bugs item #3110242, was opened at 2010-11-16 18:45 Message generated for change (Settings changed) made by andrejv You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3110242&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: None Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Emilio Suarez (folok) >Assigned to: Andrej Vodopivec (andrejv) Summary: bad serie sum Initial Comment: %i1 sum((2^n)/n, n, 1, inf),simpsum; %i2 simplify_sum(%); %o3 -log(-1) Maxima uses de power series or log(1-x) which is valid only if |x|<1 and value it for x=2. The result sould be 'divergent' ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3110242&group_id=4933 ```
[Maxima-bugs] [ maxima-Bugs-3110242 ] bad serie sum From: SourceForge.net - 2011-04-28 06:20:51 ```Bugs item #3110242, was opened at 2010-11-16 18:45 Message generated for change (Comment added) made by andrejv You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3110242&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: None Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Emilio Suarez (folok) Assigned to: Andrej Vodopivec (andrejv) Summary: bad serie sum Initial Comment: %i1 sum((2^n)/n, n, 1, inf),simpsum; %i2 simplify_sum(%); %o3 -log(-1) Maxima uses de power series or log(1-x) which is valid only if |x|<1 and value it for x=2. The result sould be 'divergent' ---------------------------------------------------------------------- >Comment By: Andrej Vodopivec (andrejv) Date: 2011-04-28 08:20 Message: Fixed in git: (%i5) sum((2^n)/n, n, 1, inf)\$ (%i6) simplify_sum(%); Sum is divergent! #0: to_hypergeometric1(expr=2^(n+1)/(n+1),var=n,lo=1,hi=inf)(simplify_sum.mac line 757) #1: to_hypergeometric(expr=2^n/n,var=n,lo=1,hi=inf)(simplify_sum.mac line 743) #2: simplify_sum(expr='sum(2^n/n,n,1,inf))(simplify_sum.mac line 292) -- an error. To debug this try: debugmode(true); ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3110242&group_id=4933 ``` | 1,614 | 5,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2015-22 | latest | en | 0.600232 |
https://it.mathworks.com/matlabcentral/cody/problems?sort=&term=group%3A%22Indexing+III%22 | 1,601,311,085,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401601278.97/warc/CC-MAIN-20200928135709-20200928165709-00183.warc.gz | 409,534,710 | 16,945 | Cody
# Problems
1 – 23 of 23
Problem Title Likes Solvers
#### Problem 2235. Back to basics - mean of corner elements of a matrix
Created by: Swapnali Gujar
1
331
#### Problem 2129. Sum of odd numbers in a matrix
Created by: Koteswar Rao Jerripothula
0
325
#### Problem 574. Sort numbers by outside digits
Created by: Michael Katz
1
165
#### Problem 2145. Find the index of n in magic(n)
Created by: Koteswar Rao Jerripothula
Tags magic, matrix
4
161
#### Problem 2228. Time Expansion
Created by: Abdullah Caliskan
3
150
#### Problem 1923. Find the maximum two numbers of every column of a matrix
Created by: Binbin Qi
Tags max
1
144
#### Problem 1812. Tridiagonal
Created by: Disha
Tags matrices
1
116
#### Problem 1986. Another colon problem
Created by: Jan Orwat
2
113
#### Problem 2200. counting groups!
Created by: Chris E.
1
104
#### Problem 2476. Max Change in Consecutive Elements
Created by: Chandan singh
1
97
#### Problem 2153. Replace pattern 0 1 0 and 1 0 1
Created by: Anette
1
88
Created by: G K
1
86
#### Problem 2528. expand intervals
Created by: Guillaume
3
81
#### Problem 1985. How unique?
Created by: Jan Orwat
1
80
#### Problem 2091. return row and column indices given 2 values which define a range
Created by: Marcel
1
73
#### Problem 2225. Three...is a magic number.
Created by: James
3
72
#### Problem 1896. Index one element in each vector of an array along a given dimension
Created by: Oliver Woodford
Tags indexing
6
71
#### Problem 2560. expand intervals vol.2
Created by: Jan Orwat
0
64
#### Problem 1961. Finding neighbors of [-1:1] in a matrix....
Created by: Chris E.
2
63
#### Problem 2158. Largest territory
Created by: CHEONNO LEE
1
62
#### Problem 2456. remove single elements
Created by: rifat
1
61
#### Problem 2545. compress sequence into intervals
Created by: Guillaume
1
50
#### Problem 2561. expand intervals vol.3
Created by: Jan Orwat
1
48
1 – 23 of 23 | 589 | 1,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-40 | latest | en | 0.656104 |
https://www.7thbhambb.co.uk/forbes/marshall/diesel/fired/boiler/agent/efficiency/calculation/1485/ | 1,652,845,264,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00305.warc.gz | 718,147,770 | 5,455 | ## forbes marshall diesel fired boiler Agent efficiency calculation
### biomass boiler efficiency calculation – Horizontal Coal
Additional suggestions for Boiler software by our robot: Free only. Categories. System Tools (15) A software tool for professional chimney calculation according to EN 13384. fired boiler as fan efficiency, boiler efficiency, tool, humidification, chiller, boiler, electrical, and. File – boiler efficiency calculation software found marcusxr.
### Forbes Marshall | SteamPedia | Indirect boiler efficiency
Forbes Marshall | SteamPedia | Indirect boiler efficiency
### Forbes Marshall | SteamPedia | Variation in Boiler
Boiler Efficiency: Introduction and Methods of Calculation
### thermax boiler efficiency calculation
In the direct method, efficiency is calculated by dividing energy delivered by the boiler by energy input as fuel, using the equation: % Efficiency = F (hs - hw) / NCV X f where F = Steam generation in Kg/hr hs = enthalpy of steam at operatin pressure in Kcal/Kg
### indirect method to calculate boiler efficiency
Calculating Boiler Efficiency - Forbes Marshall. This method calculates boiler efficiency by using the basic efficiency formula- η= (Energy output)/ (Energy input) X 100 In order to calculate boiler efficiency by this method, we divide the total energy output of a boiler by total energy input given to the boiler, multiplied by hundred.
### Industrial Boiler Efficiency Calculation & Factors
Pressure range of the boiler is about 0.7 MPa to 2 MPa and efficiency is 65 to 70%. Fuel in these boilers is added into the grate which heats the gases. Hot gases enter the front section of the boiler and leave the boiler from back and then enter the bottom flue and start moving to front section of boiler.
### diesel boiler efficiency based on the thermal heat
Calculating Boiler Efficiency - Forbes Marshall The indirect efficiency of a boiler is calculated by finding out the individual losses taking place in a boiler and then subtracting the sum from . This method involves finding out the magnitudes of all the measurable losses taking place in a boiler …
### How to convert diesel based boilers to gas fired boilers
forbes marshall diesel fired boiler efficiency calculation. PROJECT DETAIL. Product:forbes marshall diesel fired boiler efficiency calculation Standard: ASME, ISO,IBR Packaging Detail: Regular packing or nude packing, or upon customers requirement. Transportation: by land or by sea, depended on the exporting area; Email: [email protected
### Boilers : Types and Classification - Forbes Marshall
forbes marshall diesel fired boiler efficiency calculation. Calculating Boiler Efficiency - Forbes Marshall. The indirect efficiency of a boiler is calculated by finding out the individual losses taking place in a boiler and then subtracting the sum from . This method involves finding out the magnitudes of all the measurable losses taking place
and diesel oil under the conditions of randomly changed flow rates of blast-furnace and diesel is considered. Key Words: Blast Furnace Gas, Boiler, efficiency, etc. 1. INTRODUCTION: Steam generation system or boiler is used for changing the state of water i.e. from water to stem using thermal energy produced from various methods.
### oiler efficiency calculations-Industrial boiler manufacturer
Calculating Boiler Efficiency - Forbes Marshall. Thermal efficiency; Apart from these efficiencies, there are some other losses which also play a role while deciding the boiler efficiency and hence need to be considered while calculating the boiler efficiency. Combustion Efficiency. The combustion efficiency of a boiler is … [PDF] 1.
### Boiler Efficiency and Combustion | Spirax Sarco
No. 2 Oil: \$2.50 per one gallon at 140,000 btu/unit. Propane: \$1.00 per one gallon at 91,600 btu/unit. All you need to know to be able to use this calculator is the boiler horsepower, fuel type, boiler efficiency and yearly operation to determine your annual fuel costs.
### Steam Systems - Energy Efficiency by Design
Heating Time 55 minutes 40 minutes FO Consumption 800 Litres / day 450 Litres/ day Production (For batch of 100 ml) Steam : Fuel Ratio 9.3 : 1 14.0 : 1 Boiler Efficiency 66 % 84 % Fuel Consumption 25,000 Litres / month 13,000 Litres / month Dryness Fraction 60 % 98 % Fuel Furnace Oil Furnace Oil Boiler Capacity 400 kg/hr x 2, 600 kg/hr x 1 1100 kg/hr x 1
### Improving Direct Efficiency of Boilers - Forbes Marshall
Improving Direct Efficiency of Boilers - Forbes Marshall
### forbes marshall boilers - nordseeferienhaus-becker.de
Calculating Boiler Efficiency - Forbes Marshall. In order to calculate boiler efficiency by this method, we divide the total energy output of a boiler by total energy input given to the boiler, multiplied by hundred. Calculation of direct efficiency- E= [Q (H-h)/q*GCV]*100. Efficiency Calculations of Bagasse Fired Boiler …
### Gas Boiler Efficiency Calculation
Forbes Marshall offers a wide range of oil and gas fired boilers, solid fuel fired boilers,biomass boilers and waste heat recovery boilers which are efficient and offer safe operation. This website uses cookies.
### Spirax Sarco Boiler Thermal Efficiency
forbes marshall diesel fired boiler efficiency calculation このページを. boiler efficiency increasing mechanical projects 15 Ways To Increase Boiler Efficiency | Rasmussen MechanicalAuthor: Tim SebergerBoiler Efficiency and Cooling Tower Efficiency MechanicalDec 07, 2011 · The Boiler Efficiency and Cooling… View More
### Boiler Efficiency & Boiler Safety Solutions | Forbes Marshall
Forbes Marshall's Boiler metering and efficiency monitoring solutions together with our control systems and heat recovery units help keep track of critical parameters and recover every bit of energy that might be lost in the boilerhouse itself. | 1,208 | 5,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-21 | latest | en | 0.882701 |
https://www.onlinemath4all.com/how-to-memorize-multiplication-tables-fast.html | 1,558,553,370,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256948.48/warc/CC-MAIN-20190522183240-20190522205240-00451.warc.gz | 885,392,069 | 7,652 | # HOW TO MEMORIZE MULTIPLICATION TABLES FAST
## About "How to memorize multiplication tables fast"
"How to memorize multiplication tables fast" is the stuff much needed to the children who study math in schools.
Students are able to remember answer for
"Single digit number x Single digit number"
Example:
6 x 7 = 42 8 x 9 = 72 4 x 9 = 36
Always it is difficult for them to remember answer for
"Two digit number x Single digit number"
and
"Two digit number x Two digit number"
Here we have given shortcuts to find answer for the above multiplications which come in times table.
## Two digit number x Single Digit Number
This method is explained for "1" at the ten's place of two digit number (Times Table)
Steps to be followed to understand the above shortcut :
Step 1 :
First take the single digit number and multiply it by 10.
(7x10 = 70 in the above example)
Step 2 :
Multiply the single digit and unit digit of the two digit number
(7 x 2 = 14 in the above example)
Step 3 :
Add step 1 value and step 2 value.
(70+14 = 84 in the above example)
That's it.
Try yourself :
14 x 8 = ? 19 x 8 = ? 7 x 18 = ?
In this way you can try a lot.
## Two digit number x Two digit number
This method is explained for "1" at the ten's place of both the two digit numbers (Times Table)
Method 1 :
Method 2 :
Steps to be followed to understand both the methods:
Step 1 :
First take one of the two digit numbers and multiply it by 10.
Step 2 :
Multiply the unit digit of other two digit number by 10.
Step 3 :
Multiply unit digits of both the numbers.
Step 4 :
That's it
Try yourself:
17x18 = ? 18x19 = ? 15x19 = ?
In this way, you can try a lot.
## Beware!
To understand the stuff "how to memorize multiplication tables fast", you must have much practice on this. Otherwise, the shortcuts explained above will not work properly and you may not be able to get correct answer. So, please do much practice on these shortcuts using different numbers.
One more caution on these shortcuts:
We can take only "1" at the ten's place of two digit numbers and not any number.
Apart from the stuff "how to memorize multiplication tables fast", to know more about other math shortcuts, please click on the links given below.
Multiplication Tips and Tricks | 580 | 2,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2019-22 | latest | en | 0.792691 |
https://slidetodoc.com/chapter-37-central-limit-theorem-normal-approximations-to/ | 1,632,845,202,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00709.warc.gz | 567,879,541 | 9,660 | Chapter 37 Central Limit Theorem Normal Approximations to
• Slides: 16
Chapter 37: Central Limit Theorem (Normal Approximations to Discrete Distributions – 36. 4, 36. 5) http: //nestor. coventry. ac. uk/~nhunt/binomia http: //nestor. coventry. ac. uk/~nhunt/poisson l/normal. html 2
Continuity Correction - 1 http: //www. marin. edu/~npsomas/Normal_Binomial. htm 3
Continuity Correction - 2 W~N(10, 5) X ~ Binomial(20, 0. 5) 4
Continuity Correction - 3 Discrete a < X a ≤ X X < b X ≤ b Continuous a + 0. 5 < X a – 0. 5 < X X < b – 0. 5 X < b + 0. 5 5
Normal Approximation to Binomial 6
Example: Normal Approximation to Binomial (Class) The ideal size of a first-year class at a particular college is 150 students. The college, knowing from past experience that on the average only 30 percent of these accepted for admission will actually attend, uses a policy of approving the applications of 450 students. a) Compute the probability that more than 150 students attend this college. b) Compute the probability that fewer than 130 students attend this college. 7
Chapter 33: Gamma R. V. http: //resources. esri. com/help/9. 3/arcgisdesktop/com/gp_toolref /process_simulations_sensitivity_analysis_and_error_analysis_modeling /distributions_for_assigning_random_values. htm 8
Gamma Distribution • Generalization of the exponential function • Uses – probability theory – theoretical statistics – actuarial science – operations research – engineering 9
Gamma Function (t + 1) = t (t), t > 0, t real (n + 1) = n!, n > 0, n integer 10
Gamma Distribution: Summary • 11
Gamma Random Variable http: //en. wikipedia. org/wiki/File: Gamma_distribution_pdf. svg 12
Chapter 34: Beta R. V. http: //mathworld. wolfram. com/Beta. Distribution. html 13
Beta Distribution • This distribution is only defined on an interval – standard beta is on the interval [0, 1] – The formula in the book is for the standard beta • uses – modeling proportions – percentages – probabilities 14
• Beta Distribution: Summary 15
Shapes of Beta Distribution http: //upload. wikimedia. org/wikipedia/commons/9/9 a/Beta_distribution_pdf. png 16 X
Other Continuous Random Variables • Weibull – exponential is a member of family – uses: lifetimes • lognormal – log of the normal distribution – uses: products of distributions • Cauchy – symmetrical, flatter than normal 17 | 643 | 2,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-39 | latest | en | 0.75348 |
http://perplexus.info/show.php?pid=924 | 1,586,255,584,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371700247.99/warc/CC-MAIN-20200407085717-20200407120217-00509.warc.gz | 148,496,356 | 4,758 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Tug-o-war (Posted on 2003-06-21)
There is a tug-o-war competition consisting of rounds upon round that all end in a draw meaning both sides end in equal strength. There are several people of exactly equal strength and are represented by the same symbol. (i.e. all * pull the same, but a * and a @ must be different) Each team is representeed by a series of symbols followed by a dashed line being the rope. On the other side of the rope is the team that was equally matched in strength. Here are some of those rounds:
*%-----\$
\$\$\$-------@%*%@
@-------!*
Again remembering that ll of the above are ties, and assume that position on the rope doesn't matter, who will win the following match?
!@!!@!-------%\$*\$
See The Solution Submitted by Jon Rating: 3.4545 (11 votes)
Subject Author Date This is my solution Eileen 2005-03-14 12:34:57 Cool!! ratsnstuff 2003-10-09 10:06:44 Solution Lawrence 2003-08-24 13:03:58 No Subject snapp 2003-08-20 13:52:53 Solution FatBoy 2003-08-14 08:43:26 GOT IT! PRIVATE 2003-07-17 01:14:42 No Subject helena 2003-07-11 09:50:22 re: Yaman's solution Jim C 2003-07-03 07:35:14 re: solution Jim C 2003-07-03 07:33:42 *!\$@# puzzle! Jim C 2003-07-03 07:05:50 Possible solution Daniel Ciguenza 2003-06-26 15:24:57 solution yaman 2003-06-26 13:09:31 solution Chii 2003-06-25 22:11:14 solution Rajeev 2003-06-23 21:39:11 simpler solution? Hank 2003-06-23 05:00:47 possible soloution Chris 2003-06-23 00:40:53 Different Solution DJ 2003-06-22 23:34:56 Solution Lewis 2003-06-21 08:17:12 A start DJ 2003-06-21 08:05:20
Search: Search body:
Forums (0) | 565 | 1,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-16 | latest | en | 0.835384 |
http://forum.allaboutcircuits.com/threads/will-i-deplete-my-battery-fast-if-i-make-a-magnet.103541/ | 1,484,638,322,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279489.14/warc/CC-MAIN-20170116095119-00563-ip-10-171-10-70.ec2.internal.warc.gz | 104,114,233 | 16,267 | # Will I deplete my battery fast if I make a magnet?
Discussion in 'General Electronics Chat' started by Yaşar Arabacı, Nov 14, 2014.
1. ### Yaşar Arabacı Thread Starter Member
Nov 11, 2014
32
0
Hi,
I want to try wrapping cables around a steel bar and connecting 9V battery to it in order to experiment with electro-magnets. Should I be connecting some resistance to cable or does making an electro-magnet act like a resistance in the circuit?
2. ### MrChips Moderator
Oct 2, 2009
12,623
3,451
A 9V battery cannot provide a lot of current.
3. ### Yaşar Arabacı Thread Starter Member
Nov 11, 2014
32
0
Do you mean I won't see the effects of magnetism even without using any resistor? How many amps do I need in order to observe the effects of electromagnetism? I have a feeling that 220V AC 50Hz would be too much of a current to play with.
4. ### ISB123 Well-Known Member
May 21, 2014
1,240
531
You are basically short circuiting the wire its going to get very hot.You can use 4x1.5V AA batteries to play with small electromagnets.
5. ### joeyd999 AAC Fanatic!
Jun 6, 2011
2,756
2,916
I think you can get many amps for a short time (seconds?) via a dead short across a new 9V Alkaline. It won't last long, though, and it will get hot.
6. ### joeyd999 AAC Fanatic!
Jun 6, 2011
2,756
2,916
Get yourself some fine magnet wire and run *lots* of turns. You can compute the DC current by measuring the DCR of the coil, and dividing that into the battery voltage. The higher the coil resistance, the longer your battery will last.
7. ### BobTPH Active Member
Jun 5, 2013
806
121
I have actually put a multimeter on the 10A scale across a new 9V and it only gave about 1.5A.
Edit:
Magnetic field = Current x turns.
To get a good magnetic field from a 9V, put many many turns. Enough to create a resistance of about 18 Ohms, and it will draw 500mA, and be dead in less than 1 hour.
Bob
8. ### joeyd999 AAC Fanatic!
Jun 6, 2011
2,756
2,916
I just got 2.3A off a new Duracell.
9. ### ISB123 Well-Known Member
May 21, 2014
1,240
531
If you want you could buy 12V/3A transformer since those are pretty cheap and then make a simple power supply using rectifier bridge and a capacitor.
You cant use voltage regulators since they have short circuit protection and will over heat and die.
10. ### Yaşar Arabacı Thread Starter Member
Nov 11, 2014
32
0
I made a coil that created a resistance of 1 ohm and it drew 750mA current, but I didn't see any magnetism. I will try later with better equipment.
I guess my battery is about to be depleted
11. ### ISB123 Well-Known Member
May 21, 2014
1,240
531
Wire needs to be isolated it cannot be bare.
12. ### ian field Distinguished Member
Oct 27, 2012
4,447
791
The only thing a resistor will do is waste some of the power as heat.
If you're using a PP3 type 9V battery - they're amongst the lowest energy density batteries you can get.
Do a search for; "ampere-turns" - every turn of winding you put on has its magnetic field around it and contributes to the whole.
13. ### BobTPH Active Member
Jun 5, 2013
806
121
Tell us about your coil, what kind of wire, how many turns, and what the core is made of.
Bob
Jul 18, 2013
10,827
2,495
Try using an automotive battery, more current available.
Max.
15. ### alfacliff Well-Known Member
Dec 13, 2013
2,449
428
soft iron works best for that. hardened steel has different characteristics. I agree, more turns of fine wire, like #26 or so. a heavy cable will have way too little resistance to limit the current, and if you use a resistor, the current will be too low for the ampere turns. | 1,026 | 3,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-04 | latest | en | 0.908669 |
https://www.techylib.com/el/view/strawberrycokeville/convergence_of_empirical_means_with_alpha-mixing_input_sequences | 1,524,649,433,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947759.42/warc/CC-MAIN-20180425080837-20180425100837-00511.warc.gz | 869,323,541 | 21,076 | # Convergence of Empirical Means with Alpha-Mixing Input Sequences, and an Application to PAC Learning
Τεχνίτη Νοημοσύνη και Ρομποτική
7 Νοε 2013 (πριν από 4 χρόνια και 6 μήνες)
74 εμφανίσεις
Convergence of Empirical Means with
Alpha-Mixing Input Sequences,and an
Application to PAC Learning
M.Vidyasagar
Abstract—Suppose {X
i
} is an alpha-mixing stochastic
process assuming values in a set X,and that f:X →
R is bounded and measurable.It is shown in this note
that the sequence of empirical means (1/m)
m
i=1
f(X
i
)
converges in probability to the true expected value of the
function f(·).Moreover,explicit estimates are constructed
of the rate at which the empirical mean converges to the
true expected value.These estimates generalize classical
inequalities of Hoeffding,Bennett and Bernstein to the case
of alpha-mixing inputs.In earlier work,similar results have
been established when the alpha-mixing coefficient of the
stochastic process converges to zero at a geometric rate.
No such assumption is made in the present note.This
result is then applied to the problem of PAC (probably
approximately correct) learning under a fixed distribution.
I.I
NTRODUCTION
Suppose (X,S) is a measurable space,and let
{X
i
}
i=−∞
be a stationary two-sided stochastic process
assuming values in X,with the canonical representation.
Let
˜
P
0
denote the one-dimensional marginal probability
of
˜
P.Suppose f:X →[−F,F] is measurable and has
zero mean with respect to the measure
˜
P
0
.
1
Let {x
i
}
be a realization of the stochastic process {X
i
},and let
x denote (x
i
,i = −∞,...,∞) ∈ X
.Let us examine
the sequence of “empirical” means
ˆ
E
m
(f;x):=
1
m
m
i=1
f(x
i
).
One of the classical questions in the theory of empirical
processes is:When does the sequence of empirical
means converge to the true mean value of zero,and if
so,at what rate?
This question arises in a couple of contexts.First,
many problems in PAC (probably approximately correct)
learning theory can also be viewed as questions on the
convergence of empirical means to their true values,
the so-called ”law of large numbers” question.See [14]
for a discussion.Second,under certain circumstances,
1
If f does not have zero mean,we can replace f by f −E(f) and
apply the various results in the note.
the problems of system identification and stochastic
PAC learning theory.See [15] for a discussion.
More specifically,suppose we define the quantities
q
u
(m,;
˜
P):=
˜
P{x ∈ X
:
ˆ
E
m
(f;x) > }.
q
l
(m,;
˜
P):=
˜
P{x ∈ X
:
ˆ
E
m
(f;x) < −}.
q(m,;
˜
P):=
˜
P{x ∈ X
:|
ˆ
E
m
(f;x)| > }.
When is it the case that q(m,) → 0 as m → ∞?If
q(m,;
˜
P) →0 as m→∞,then it can be said that the
empirical means of f converge in probability to the true
mean.
There is a vast literature on the convergence of
empirical means the stochastic process consists of i.i.d.
random variables,that is,when
˜
P = (
˜
P
0
)
.See [10]
for proofs of these results.Hoeffding’s inequality states
that,for all m,,we have
q
l
(m,;(
˜
P
0
)
),q
u
(m,;(
˜
P
0
)
) ≤ exp(−2m
2
),
q(m,;(
˜
P
0
)
) ≤ 2 exp(−2m
2
).
Let σ
2
denote the variance of the function f.Then
Bennett’s inequality states that
q
u
(m,;(
˜
P
0
)
) ≤ exp
m
2
2
B(F/σ
2
)
,
where the function B(·) is defined by
B(λ):= 2
(1 +λ) ln(1 +λ) −λ
λ
2
.(1)
In particular,if we observe that B(λ) ≥ (1 + λ/3)
−1
whenever λ < 1,we get the Bernstein inequality,which
states that
q
u
(m,;(
˜
P
0
)
) ≤ exp
−m
2
2(σ
2
+F/3)
.
Each of these inequalities holds when f is replaced by
−f.Thus the estimate for the quantity q(m,;(
˜
P
0
)
)
is just twice the right side of each of these estimates.
Over the years several papers have addressed the
extension of the above (and other related) inequalities
Proceedings of the
44th IEEE Conference on Decision and Control, and
the European Control Conference 2005
Seville, Spain, December 12-15, 2005
MoA16.6
560
to the case where the stochastic process {X
i
} is not
necessarily i.i.d.In the present paper,it is shown that the
empirical means converge to zero when the stochastic
process {X
i
} is α-mixing.Moreover,each of the pre-
vious inequalities (Hoeffding,Bennett and Bernstein) is
extended to the case of α-mixing input sequences.It is
not assumed that the α-mixing coefficient converges to
zero at a geometric rate,as in earlier papers,notably [6],
[7].The estimates presented here improve upon those
in [14],Section 3.4.2.Once these estimates are derived,
they are applied to the problem of PAC (probably ap-
proximately correct) learning under a fixed distribution.
II.A
LPHA
-M
IXING
S
TOCHASTIC
P
ROCESSES
In this section,a definition is given of the notion of α-
mixing,and a fundamental inequality due to Ibragimov
is stated without proof.
Given the stochastic process {X
i
},let Σ
0
−∞
denote
the σ-algebra generated by the random variables X
i
,i ≤
0;similarly let Σ
k
denote the σ-algebra generated by
the random variables X
i
,i ≥ k.Then the alpha-mixing
coefficient α(k) of the stochastic process is defined by
α(k):= sup
A∈Σ
0
−∞
,B∈Σ
k
|
˜
P(A∩B) −
˜
P(A)
˜
P(B)|.
Clearly α(k) ∈ [0,1] for all k.Moreover,since Σ
k+1
Σ
k
,it is obvious that α(k) ≥ α(k +1).Thus {α(k)}
is nonincreasing and bounded below.The stochastic
process is said to be α-mixing if α(k) →0 as k →∞.
One of the most useful inequalities for α-mixing
processes is the following,due to Ibragimov [5].
Theorem 1:Suppose {X
i
} is an α-mixing process
on a probability space (X
,S
,
˜
P).Suppose f,g:
X
→R are essentially bounded,that f is measurable
with respect to Σ
0
−∞
,and that g is measurable with
respect to Σ
0
.Then
|E(fg,
˜
P)−E(f,
˜
P) E(g,
˜
P)| ≤ 4α(k) f
· g
.
(2)
For a proof,see [5] or [3],Theorem A.5.The proof
is also reproduced in [14],Theorem 2.2.
Since in this note we shall be taking expectations
and measures of the same function or set with respect
to different probability measures,we use the notation
E(f,
˜
P) to denote the expectation of f with respect to
the measure
˜
P.
Upon applying an inductive argument to the above
Corollary 1:Suppose {X
i
} is an α-mixing stochas-
tic process.Suppose f
0
,...,f
l
are essentially bounded
functions,where f
i
depends only on X
ik
.Then
E
l
i=0
f
i
,
˜
P
l
i=0
E(f
i
,
˜
P)
≤ 4lα(k)
l
i=0
f
i
.
(3)
III.M
AIN
R
ESULTS
In this section we state and prove the main results.
In particular,it is shown that empirical means converge
to the true mean value of zero,and explicit quantitative
estimates are given for the rate of convergence.These
estimates generalize the classical inequalities of Ho-
effding,Bennett and Bernstein to the case of α-mixing
inputs.
Theorem 2:Suppose f:X → [−F,F] has zero
mean and variance no larger than σ
2
.Suppose {X
t
}
is a stationary stochastic process with the law
˜
P,and
define q(m,;
˜
P) as before.Given an integer m,choose
k ≤ m,and define l:= m/k
.Define
B
Hoeffding
:= exp[−
2
l/2F
2
] +4α(k)l exp[l/F],
B
Bennett
:= exp
l
2
2
B(F/σ
2
)
+4α(k)l
1 +F
σ
2
l
,
where the function B(·) is defined in (1).
B
Bernstein
:= exp
−l
2
2(σ
2
+F/3)
+4α(k)l
1 +F
σ
2
l
.
Then we have the following inequalities:Hoeffding-
type:
q
l
(m,;
˜
P),q
u
(m,;
˜
P) ≤ B
Hoeffding
,(4)
q(m,;
˜
P) ≤ 2B
Hoeffding
.(5)
Bennett-type:
q
l
(m,;
˜
P),q
u
(m,;
˜
P) ≤ B
Bennett
,(6)
q(m,;
˜
P) ≤ 2B
Bennett
.(7)
Bernstein-type:
q
l
(m,;
˜
P),q
u
(m,;
˜
P) ≤ B
Bernstein
,(8)
q(m,;
˜
P) ≤ 2B
Bernstein
.(9)
Finally,suppose α(k) → 0 as k → ∞.Then
q(m,;
˜
P) →0 as m→∞.
The proof of the theorem makes use of the following
technical lemma.
Lemma 1:Suppose β(k) ↓ 0 as k → ∞,and h:
Z
+
→ R is strictly increasing.Then it is possible to
choose a sequence {k
m
} such that k
m
≤ m,and with
l
m
= m/k
m
we have
l
m
→∞,β(k
m
)h(l
m
) →0 as m→∞.
Proof:Though the function β is defined only for
integer-valued arguments,it is convenient to replace it by
another function defined for all real-valued arguments.
Moreover,it can be assumed that β(·) is continuous and
monotonically decreasing,so that β
−1
is well-defined,
by replacing the given function by a larger function if
561
necessary.With this convention,choose any sequence
{a
i
} such that a
i
↓ 0 as i →∞.Define
m
i
:= i β
−1
(a
i
/h(i)) .
Clearly a
i
/h(i) ↓ 0,so β
−1
(a
i
/h(i)) ↑ ∞.Therefore
−1
(a
i
/h(i)) ↑ ∞.Thus {m
i
} is a monotonically in-
creasing sequence.Given an integer m,choose a unique
integer i = i(m) such that m
i
≤ m < m
i+1
.Define
l
m
= i(m),and choose k
m
as the largest integer such
that l
m
= m/k
m
.Note that i(m) →∞as m→∞,so
that l
m
→∞.Next,since i β
−1
(a
i
/h(i)) = m
i
≤ m,
it follows that
k
m
≥ β
−1
(a
i
/h(i)) .
So
β(k
m
) ≤ β( β
−1
(a
i
/h(i)) )
≤ β[β
−1
(a
i
/h(i))] = a
i
/h(i).
Since l
m
= i,we have h(l
m
) = h(i).Finally
β(k
m
)h(l
m
) ≤ a
i
.
Since a
i
→0 as i →∞,the result follows.
Proof of the theorem:Given integers m,k,l,let r:=
m−kl,and define the sets of integers
I
i
:= {i,i +k,...,i +lk},1 ≤ i ≤ r,
I
i
:= {i,i +k,...,i +(l −1)k},r +1 ≤ i ≤ k.
Define p
i
:= |I
i
|/m,and note that
|I
i
| = l +1 for 1 ≤ i ≤ r,|I
i
| = l for r +1 ≤ i ≤ k,
k
i=1
p
i
= 1.
Next,define the random variables
a
m
(x):=
1
m
m
i=1
f(x
i
),
b
i
(x):=
1
|I
i
|
j∈I
i
f(x
j
),i = 1,...,k.
Then
a
m
(x) =
n
k=1
p
i
b
i
(x).
Step 1:Suppose γ > is arbitrary.It is claimed that
E[exp(γa
m
),
˜
P] ≤
k
i=1
p
i
E[exp(γb
i
),
˜
P].(10)
Note that exp(γ·) is a convex function.Therefore,for
each x,we have
exp(γa
m
(x)) ≤
k
i=1
p
i
exp(γb
i
(x)).
Taking expectations of both sides with respect to
˜
P
establishes the claim.
Step 2:It is claimed that
E[exp(γb
i
),
˜
P] ≤ {E[exp(γf/|I
i
|),
˜
P
0
]}
|I
i
|
+ 4α(k)(|I
i
| −1)e
γF
.(11)
Note that b
i
(x) depends only on x
i+jk
for j ranging
from 0 through |I
1
| −1.Thus the indices of the various
x’s are separated by k.Now apply Theorem 1.
2
This
shows that
E[exp(γb
i
),
˜
P] ≤ E[exp(γb
i
),(
˜
P
0
)
]+4α(k)(|I
i
|−1)e
γF
.
Next,we have
exp(γb
i
) =
j∈I
i
exp[γf(x
j
)/|I
i
|],
and under the probability measure (
˜
P
0
)
the random
variables f(x
j
) are independent.Therefore
E[exp(γb
i
),(
˜
P
0
)
] =
j∈I
i
E[exp(γf/|I
i
|),
˜
P
0
]
= {E[exp(γf/|I
i
|),
˜
P
0
]}
|I
i
|
.
Combining these inequalities establishes the claim.
Step 3:In this step,the quantity E[exp(γa
m
),
˜
P]
is estimated in three different ways,which lead re-
spectively to the Hoeffding-type,Bennett-type and
Bernstein-type inequalities.As these estimates are used
in the proofs of the “classical” versions of these inequal-
ities (i.e.,in the case of i.i.d.stochastic processes),only
very sketchy proofs are given.
Hoeffding-type:Note that f has zero mean and
assumes values over an interval of width 2F.Therefore
(see for example [2],p.122)
E[exp(γf/|I
i
|),
˜
P
0
] ≤ exp(γ
2
F
2
/2|I
i
|
2
).
Substituting this bound into (11) leads to
E[exp(γb
i
),
˜
P] ≤ exp(γ
2
F
2
/2|I
i
|) +4α(k)(|I
i
| −1)e
γF
≤ exp(γ
2
F
2
/2l) +4α(k)le
γF
,
since l ≤ |I
i
| ≤ l +1.Substituting this bound into (11)
shows that
E[exp(γa
m
),
˜
P] ≤ exp(γ
2
F
2
/2l) +4α(k)le
γF
,(12)
since
p
i
= 1.
Next,by Markov’s inequality,for any > 0 we have
˜
P{a
m
> } =
˜
P{exp(γa
m
) > e
γ
}
≤ E[exp(γa
m
),
˜
P]e
−γ
≤ exp(−γ +γ
2
F
2
/2l) +4α(k)le
γF−γ
≤ exp(−γ +γ
2
F
2
/2l) +4α(k)le
γF
2
Since the stochastic process is stationary,the fact that the indices
do not begin with zero is of no consequence.
562
since exp(−γ) ≤ 1.
The above inequality is valid for every choice of γ >
0.Now let us choose γ so as to minimize the exponent
of the first term.This choice of γ is
γ =
l
F
2
,−γ +γ
2
F
2
/2l = −
l
2
2F
2
.
This finally leads to the desired inequality
˜
P{a
m
> } ≤ exp(−l
2
/2F
2
) +4α(k)le
l/F
.
Note that the right side is B
Hoeffding
as defined earlier.
This establishes the Hoeffding type inequalities.
Bennett-type:If Y is a zero-mean random variable
bounded above by M and with variance σ
2
,then (see
e.g.,[10])
E[e
tY
,
˜
P
0
] ≤ exp[σ
2
g(t,M)],
where
g(t,M):=
j=2
t
j
j!
M
j−2
=
e
tM
−1 −tM
M
2
.
Now apply this inequality with Y = f,M = F and
t = γ/|I
i
|.This shows that
E[exp(γf/|I
i
|),
˜
P
0
] ≤ exp[σ
2
g(γ/|I
i
|,F)],
E[exp(γb
i
),
˜
P] ≤ exp[|I
i
2
g(γ/|I
i
|,F)]
+ 4α(k)(|I
i
| −1)e
γF
.
Now let us examine the exponent in the first term.Since
l ≤ |I
i
| ≤ l +1,we have that
|I
i
2
g(γ/|I
i
|,F) = σ
2
j=2
γ
j
j!|I
i
|
j−1
F
j−2
≤ σ
2
j=2
γ
j
j!l
j−1
F
j−2
= lσ
2
g(γ/l,F).
Therefore
E[exp(γb
i
),
˜
P] ≤ exp[lσ
2
g(γ/l,F)].
So
˜
P{a
m
> } ≤ exp
2
g
γ
l
,F
−γ
+4α(k)le
γF−γ
.
(13)
The above inequality is valid for every value of γ.Now
let us choose γ so as to minimize the first exponent.Let
c(γ):= lσ
2
g
γ
l
,F
−γ.
Then a routine calculation shows that c(·) is minimized
when
exp[γF/l] −1 = F/σ
2
,or γ =
l
F
ln
1 +
F
σ
2
.
With this choice of γ,we have
c(γ) = −
l
2
2
·
2
F
2
σ
4
B(F/σ
2
),
where B(·) is defined in (1).
Next,to estimate
˜
P{a
m
> },it is permissible to
replace γF −γ by the larger number γF in (13).This
finally leads to the upper bound
˜
P{a
m
> } ≤ exp
l
2
2
·
2
F
2
σ
4
B(F/σ
2
)
+ 4α(k)l exp
l ln
1 +
F
σ
2
.
Note that the right side is B
Bennett
defined earlier.This
establishes the Bennett type inequalities.
Bernstein-Type:As in the classical proof we have
that
B(λ) ≥ (1 +λ/3)
−1
∀λ.
Substituting this bound in the Bennett estimates leads to
the Bernstein type estimates.
The above bounds hold for any stochastic process
generating the samples.To show that q(m,;
˜
P) →0 as
m → ∞ whenever the stochastic process is α-mixing,
apply Lemma 1 with
β(k):= α(k),h(l):= 4l exp[4/lF].
Then it is always possible to choose a sequence {k
m
}
such that,with l
m
:= m/k
m
,we have
l
m
→∞,4α(k
m
)l
m
exp[4/l
m
F] →0 as m→∞.
Applying this fact to any of the proven bounds leads to
the desired conclusion that q(m,) →0 as m→∞.
.
Remarks:In the case where the stochastic process is
i.i.d.,it is clear that α(k) = 0 for all k ≥ 1.Hence,
given m,we can choose k
m
= 1 and l
m
= m.With this
choice,each of the inequalities in the theorem reduces
to its well-known counterpart for i.i.d.processes.
IV.A
N
A
PPLICATION TO
PAC L
EARNING
In this section,the estimate derived in the preceding
section is applied to a problem in fixed-distribution PAC
(probably approximately correct) learning.In particular,
it is shown that if a concept class is learnable with i.i.d.
inputs,it remains learnable with α-mixing inputs.
The reader is referred to Chapter 3 of [13],[14] for
detailed definitions and discussions of PAC learning;
only very brief descriptions are given here.
563
A.The PAC Learning Problem Formulation
Suppose as before that (X,S) is a measurable space,
and let F ⊆ [0,1]
X
consist of functions that are
measurable with respect to S.Such a family F is said
to be a function family.In case F consists solely of
binary-valued functions,i.e.,in case F ⊆ {0,1}
X
,then
F is said to be a concept class.
In the so-called ‘fixed distribution’ PAC learning
problem,there is a fixed (and known) stationary prob-
ability measure
˜
P on (X
,S
),and a fixed but un-
known function f ∈ F,called the ‘target’ function.
Let
˜
P
0
denote the one-dimensional marginal probability
corresponding to
˜
P.Suppose {x
i
}
i=−∞
is a sample path
of a stationary stochastic process {X
i
}
i=−∞
with the
law
˜
P.For each sample x
i
∈ X,an ‘oracle’ returns the
value f(x
i
) of the unknown function f at the sample
x
i
.Based on these ‘labelled samples,’ an algorithm
returns the ‘hypothesis h
m
(f;x).The goodness of the
hypothesis is measured by the so-called ‘generalization
error,’ defined as
d
˜
P
0
(f,h
m
):=
X
|f(x) −h
m
(x)|
˜
P
0
(dx).
Given an ‘accuracy’ > 0,the quantity
r(m,;
˜
P):= sup
f∈F
˜
P{x ∈ X
:d
˜
P
0
[f,h
m
(f;x)] > }
is called the ‘learning rate’ function.The algorithm is
said to be PAC (probably approximately correct) to
accuracy if r(m,;
˜
P) → 0 as m → ∞,for a fixed
> 0.The algorithm is said to be PAC if it is PAC for
every fixed > 0,i.e.,if r(m,;
˜
P) →0 as m→∞for
all > 0.The pair (F,
˜
P) is said to be PAC learnable
if there exists a PAC algorithm.
B.Known Results for the Case of I.I.D.Samples
Next we introduce the notion of covering numbers
and the finite metric entropy condition.Given a number
> 0,the -covering number of F with respect to
the pseudometric d
˜
P
0
is defined as the smallest number
of balls of radius with centers in F that cover F,
where the radius is measured with respect to d
˜
P
0
.The
-covering number is denoted by N(,F,d
˜
P
0
).In case
the set F cannot be covered by a finite number of balls
of radius ,the covering number is taken as infinity.The
set F is said to satisfy the finite metric entropy condition
with respect to d
P
if
N(,F,d
˜
P
0
) < ∞∀ > 0.
For the fixed distribution learning problem with i.i.d.
inputs,the following results are known.
Theorem 3:Suppose the stochastic process {X
i
} is
i.i.d.,i.e.,that
˜
P = (
˜
P
0
)
.Suppose the function
family F satisfies the finite metric entropy condition
with respect to d
˜
P
0
.Then the pair (F,(
˜
P
0
)
) is PAC
learnable.In case F is a concept class,the finite metric
entropy condition is also necessary for PAC learnability.
The proof of the theorem can be found in [1],or [14],
Theorem 6.7,p.238.
In case the function family F has finite metric en-
tropy,the following ‘minimal empirical risk’ (MER) al-
gorithm can be shown to be PAC.Again,the details can
be found in the above two references.Given F and an
accuracy > 0,find a minimal /2-cover {g
1
,...,g
N
)
for F.Given the sample sequence x
1
,...,x
m
,define
the empirical error
ˆ
J
i
:=
1
m
m
j=1
|f(x
j
) −g
i
(x
j
)|.
Note that the above quantity is computable since the
values f(x
j
) are available from the oracle.Also,
ˆ
J
i
is just the empirical estimate for the generalization
error d
˜
P
0
(f,g
i
) based on the sample x.Choose as the
hypothesis h
m
one of the g
i
such that
ˆ
J
i
is as small as
possible.This algorithm is called the ‘minimal empirical
risk’ algorithmbecause it generates a hypothesis h
m
that
matches the data as closely as possible on the samples
x
1
,...,x
m
.The learning rate for the minimal empirical
risk algorithmis given by (see [1] for the case of concept
classes or [14],Theorems 6.2 and 6.3 for the general
case)
r(m,;(
˜
P
0
)
) ≤ N exp(−m
2
/8)
if F is a function class,or
r(m,;(
˜
P
0
)
) ≤ N exp(−m/32)
if F is a concept class.
C.Fixed Distribution Learning with Alpha-Mixing Input
Sequences
With this brief introduction,we are in a position
to study the same problem when the learning sample
sequence {x
i
} is not i.i.d.,but is α-mixing.
Theorem 4:Suppose the stochastic process {X
i
} is
α-mixing with the law
˜
P,and that the function family
F satisfies the finite metric entropy condition with
respect to
˜
P
0
.Then the pair (F,
˜
P) is PAC learnable.
Specifically,suppose > 0 is a given accuracy,and
let N equal the /2-covering number of F with respect
to d
˜
P
0
.Let {g
1
,...,g
N
} be a minimal /2-cover,and
apply the minimal empirical risk algorithm.For any
integer m,let k ≤ m and let l:= m/k
.Then
r(m,;
˜
P) ≤ N[exp(−2l
2
) +4α(k) exp(2l)].
Proof:The proof closely follows that in the case of
i.i.d.inputs.Let all symbols be as above,and suppose
564
f ∈ F be the unknown target function.Renumber the
/2-cover in such a way that
d
˜
P
0
(f,g
1
) ≤ /2,d
˜
P
0
(f,g
i
) ≤ ,i = 2,...,k,
d
˜
P
0
(f,g
i
) > ,i = k +1,...,N.
It is clear that k ≥ 2.
Recall that
ˆ
J
i
is just an empirical estimate of the
distance d
˜
P
0
(f,g
i
) based on the sample x.Hence
d
˜
P
0
(f,h
m
) > only if
ˆ
J
1
−d
˜
P
0
(f,g
1
) > /4,and
∃i ∈ {k +1,...,N} s.t.d
˜
P
0
(f,g
i
) −
ˆ
J
i
> /4.
If the conditions in the above equation fail to hold,then
on the MER algorithm g
1
outperforms all the functions
g
k+1
,...,g
N
.Hence the hypothesis h
m
will equal one
of g
1
,...,g
k
and as a result d
˜
P
0
(f,h
m
) ≤ .Now the
probability of each of the events above is bounded,from
(4) and (5),by e
−2l
2
+4α(k)e
2l
.
3
Therefore
r(m,;
˜
P) ≤ N(−k +1)e
−2l
2
+4α(k)e
−2l
≤ N[exp(−2l
2
) +4α(k) exp(2l)].
This proves the estimate.Moreover,by Lemma 1,it is
always possible to choose a sequence {k
m
} such that
r
α
(m,) →0 as m→∞.
It is clear that,if all functions in F have a known
bounded variance,then one can also derive bounds of
the Bennett or Bernstein-type,instead of the Hoeffding-
type bounds as above.
Observe that if F is a concept class,then the finite
metric entropy condition is also necessary for PAC
learnability with i.i.d.inputs.This leads to the following
observation.
Corollary 2:Suppose F is a concept class,and
˜
P
a stationary probability measure.If the pair (F,
˜
P)
is PAC learnable with i.i.d.inputs with the law
˜
P
0
,
then it remains PAC learnable with an α-mixing input
sequence.
V.D
ISCUSSION AND
C
ONCLUSIONS
In this paper,we have shown that empirical means of a
function converge in probability to the true mean,when
the underlying sample process is α-mixing.Compared
with the earlier results of [6],[7],the main improvement
in the present case is that the law of large numbers is
established without assuming that the α-mixing coeffi-
cient decays to zero at a geometric rate.We have also
applied this result to show that if a concept class is PAC
3
Note that,since all the functions in F assume values in the interval
[0,1] which has width one,we should put F = 0.5 in each of the
above equations.
learnable with i.i.d.inputs,then it remains PAC learnable
with α-mixing samples.
Note that the present results (as well as earlier results)
apply only to a single function.By contrast,if the sample
process is β-mixing,uniform laws of large numbers
can be proven even for infinitely many functions.See
[9] for the result and [4] for estimates of the rate
of convergence.In [16],the author states that in her
opinion,the corresponding statement is not true for α-
mixing processes.However,this question is still open as
of now.
R
EFERENCES
[1] G.M.Benedek and A.Itai,“Learnability by fixed distribu-
tions,” Proc.First Workshop on Computational Learning Theory,
Morgan-Kaufmann,San Mateo,CA,80-90,1988.
[2] L.Devroye,L.Gyorfi and G.Lugosi,A probabilistic theory of
pattern recognition,Springer,1996.
[3] P.Hall and C.C.Heyde,Martingale Limit Theory and Its
[4] R.L.Karandikar and M.Vidyasagar,“Rates of convergence
of empirical means under mixing processes,” Stat.and Probab.
Letters,2002.
[5] I.A.Ibragimov,“Some limit theorems for stationary processes,”
Thy.Prob.Appl.,7,349-382,1962.
[6] D.S.Modha and E.Masry,“Minimum complexity regression
estimation with weakly dependent observations,” IEEE Trans.
Info.Thy.,42(6),2133-2145,November 1996.
[7] D.S.Modha and E.Masry,“Memory-universal prediction of
stationary randomprocesses,” IEEE Trans.Info.Thy.,44(1),117-
133,Jan.1998.
[8] K.Najarian,G.A Dumont,M.S.Davies and N.E.Heckman,
“PAC learning in non-linear FIR models,” Int.J.Adaptive
Control and Signal Processing,15,37-52,2001.
[9] A.Nobel and A.Dembo,“A note on uniform laws of averages
for dependent processes,” Stat.& Probab.Letters,17,169-172,
1993.
[10] D.Pollard,Convergence of Stochastic Processes,Springer-
Verlag,1984.
[11] V.N.Vapnik and A.Ya.Chervonenkis,“On the uniform con-
vergence of relative frequencies to their probabilities,” Theory of
Probab.Appl.16(2),264-280,1971.
[12] V.N.Vapnik and A.Ya.Chervonenkis,“Necessary and and
sufficient conditions for the uniform convergence of means to
their expectations,” Theory of Probab.Appl.,26(3),532-553,
1981.
[13] M.Vidyasagar,A Theory of Learning and Generalization,
Springer-Verlag,London,1997.
[14] M.Vidyasagar,Learning and Generalization with Application
to Neural Networks,(Second Edition),Springer-Verlag,London,
2003.
[15] M.Vidyasagar and R.L.Karandikar,“A learning theory ap-
proach to system identification and stochastic adaptive control,”
in Probabilistic and Randomized Methods for Design Under
Uncertainty,G.Calafiore and F.Dabbene (Eds.),Springer-
Verlag,London,pp.265-302,2005.
[16] B.Yu,“Rates of convergence of empirical processes for mixing
sequences,” Annals of Probab.,22(1),94-116,1994.
565 | 9,159 | 23,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-17 | latest | en | 0.868225 |
https://socratic.org/questions/5a72fca07c01493648fc0970 | 1,590,456,744,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390437.8/warc/CC-MAIN-20200525223929-20200526013929-00014.warc.gz | 556,941,746 | 6,491 | # Question c0970
Feb 1, 2018
$1.63 \text{mol}$ of sodium chloride.
#### Explanation:
Remember that the formula for moles, $n$, is $n = \frac{m}{M}$, where $m$ is the mass of the substance, and $M$ its molar mass.
Therefore, we must first calculate the mass of $N a C l$ present.
To do this, we can use the formula for density, $\rho$, when $\rho = \frac{m}{V}$, where $V$ is the volume of liquid.
The density of sodium chloride is $2.16 \frac{g}{m L}$.
So input: $2.16 = \frac{m}{44}$
$2.16 \cdot 44 = \frac{m}{44} \cdot 44$
$m = 95.04 g$ of sodium chloride.
Now we may use $n = \frac{m}{M}$.
The molar mass of sodium chloride is $58.4 \frac{g}{\text{mol}}$.
So simply input: $n = \frac{95.04}{58.4}$
$n = 1.63 \text{mol}$ of sodium chloride.
Feb 1, 2018
$\approx 1.63 m o l$
#### Explanation:
Well, we'll need a few steps here...
First, know that $1 m L = 1 c {m}^{3}$, so
$44 m L = 44 c {m}^{3}$
Now, we can use the density equation to find the mass of the sodium chloride:
$\rho = \frac{m}{V}$ or $m = \rho \cdot V$
The density of sodium chloride is $2.16 g \text{/} c {m}^{3}$
$\therefore \text{mass"=2.16g"/} c {m}^{3} \cdot 44 c {m}^{3} = 95.04 g$
So, we have $95.04 g$ of sodium chloride.
Next, we have to use the molar mass formula to find the amount of moles.
$\text{moles"="mass"/"molar mass}$
The molar mass of sodium chloride is $58.4 g \text{/} m o l$.
:."moles"=(95.04cancelg)/(58.4cancelg"/"mol)~~1.63mol#
So, our final answer is $1.63$ moles of sodium chloride. | 550 | 1,507 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 29, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2020-24 | latest | en | 0.726987 |
https://www.jiskha.com/questions/83493/hey-im-a-little-confused-with-this-question-2-n-x-4-n-1-8-n-2-i-have-to-change-all | 1,611,838,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704843561.95/warc/CC-MAIN-20210128102756-20210128132756-00656.warc.gz | 806,683,240 | 5,251 | # agebra
Hey, im a little confused with this question:
2^n x 4^(n+1)
/
8^(n-2)
I have to change all bases to 2 then simplify fully.
Thanks everoyne.
1. 👍
2. 👎
3. 👁
1. 4^(n+1) = 2^[2*(n+1)]= 2^(2n+2)
8^(n-2)= (2^3)^(n-2) = 2^(3n-6)
If the 8^(n-2) is supposed to be in a denominator,
2^n * 4^(n+1)
____________
8^(n-2)
= 2^(n + 2n + 2 -3n +6)
= 2^8 = 256
no matter what n is. Pick a value of n and prove it for yourself.
1. 👍
2. 👎
## Similar Questions
Hey guys! I have a hard time on this question. I would really appreciate it if you could help me :) The question is: How did conflict and cooperation within early civilizations relate to the use of resources? The hard part is
2. ### Trig
Theta = x Question: Given angle x, where 0degrees
3. ### chemistry
Predict the precipitate that forms: HCl + AgNO3 --> ??? i don't know what the answer is but i think it is Cing04 AgCl. You need to learn the solubility rules. If you don't have a web site I can give you one. hey watz up can you
4. ### Math
if x is an even number, what is the quotient of the next greater even number divided by the next greater odd number? i have no idea where to even start with this problem. i would really appriciate some help. Thanks! What can make
1. ### English Language Arts
Has anyone done the English semester exam? I am confused with the essay question. I chose prompt a, can anyone simplify this for me, by giving me different questions that I need to answer for this question?
2. ### science
Chemistry Hey guys, i have a question about how to figure out how much mass of a compound i need to find 130ppm here is the question. What mass of Fe(NH4)2(SO4)2*6H2O is required to prepare 2.25 L of a 130 ppm iron solution? Now i
3. ### English
Hey, Yesterday I asked a question about "the" and Writeacher helped me a lot. But now I have a further question. I have a rule in my grammar book, which says, that Nationalities with the suffix -ese. -ch, -sh, -ss; goes with the.
The question is Is the point (-3, -2) a solution of the intersection of the following set of quadratic equations: Y < -X^2 X^2 + Y^2 < 16 I guess I am somewhat confused by the way it's written. Would I be graphing this to find
1. ### calculus
If d/dx(f(x))=g(x) and d/dx(g(x))=f(x^2) , then d^2/dx^2(f(x^3)) , then d2dx2(f(x3))= A f(x^6) B g(x^3) C 3x^2g(x^3) D 9x^4f(x^6) + 6xg(x^3) E f(x^6) + g(x^3) I understood what the first half of the question said, but then I was
2. ### Language arts 2 question!??
Hey... one question from the story "Thank You, Ma'am."???: 2. In "Thank You, Ma'am," repeating certain words and phrases helps the author (1 point) complete the evidence review unknown skills establish the story's tone become a
3. ### For DrBob
Hey DrBob, I posted an answer to you question on my lab report just to let you know what the y= and r^2 is
4. ### Health
Hey! Random question - is the study of personal fitness considered a physical science? | 906 | 2,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-04 | latest | en | 0.936182 |
https://math.stackexchange.com/questions/2111879/if-f-is-holomorphic-except-for-z-0-then-lim-n-to-infty-fraca-na-n | 1,716,752,365,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00785.warc.gz | 326,713,838 | 39,497 | # If $f$ is holomorphic except for $z_0$, then $\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0$
The question is from Stein & Shakarchi - Complex Analysis Chapter 2, Exercise 14.
Suppose that $$f$$ is holomorphic in an open set $$\Omega$$ containing the closed unit disc, except for a pole at $$z_0$$ on the unit circle. Show that if $$f$$ is given by a power series expansion $$f(z)=\sum^\infty_{n=0}a_n z^n$$ in the unit disc $$D_1(0)$$, then $$\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0.$$
I solved this problem by using the pole formula $$f(z)=(z-z_0)^{-m}g_0(z)\implies f^{(n)}(z)=(z-z_0)^{-m-n}g_n(z)$$ where $$g_0$$ and $$g_n$$ are holomorphic and not zero at $$z_0$$, and $$m$$ is a positive integer. These are defined on $$D_{1+\epsilon}(0)\subset\Omega$$, which contains $$z_0$$ and $$\epsilon$$ is sufficiently small. I've got below : $$\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}}=\lim_{n\to\infty} \dfrac{\frac{f^{(n+1)}(0)}{(n+1)!}}{\frac{f^{(n)}(0)}{n!}}=\lim_{n\to\infty}\dfrac{1}{n+1}\left(\dfrac{-m-n}{0-z_0}+H(0)\right)=\dfrac{1}{z_0}.$$ where $$H(z)$$ is holomorphic in the disc.
Actually, the pole formula I've used appears in Chapter 3 so I should not use this, but I think it is worth to try. Is there something wrong about this solution?
Very thanks.
• I don't understand the question. If $f$ is given by a power series valid in a nbhd of the closed unit disk then it has no poles on the unit circle. Jan 24, 2017 at 20:17
• @juanarroyo OH, I've made a horrible mistake. Thank you for pointing out that. I edited the question. The power series is vaild for $f$ only in the disc. Jan 24, 2017 at 20:29
• Shouldn't that be "$f(z)=(z-z_0)^{-m}g_1(z)\implies f^{(n)}(z)=(z-z_0)^{-m-n}g_{n+1}(z)?$"
– zhw.
Jan 24, 2017 at 20:40
• @zhw. I wrote that two function $g_1$ and $g_2$ for just a notation. Maybe $g_2$ consists of $(z-z_0)$ and $g_1$, but the form will be very complicated. But your notation seems pretty nice. Thanks. Jan 24, 2017 at 20:43
First note that the proof you gave works perfectly well for $f(z)=\frac{1}{z+1} + \frac{1}{z-1}$. Your functions $g_n$ will not be defined in $D_{1+\epsilon}(0)$, but they are defined on an open set containing $0$, which is enough for your proof to go through. This however would yield a contradiction, since then you would have shown that $-1=\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=1$.
So what is the flaw in your proof? I see two. First, you assumed that none of the $a_n$ are $0$. Indeed if you look at the Taylor series for the function $f$ I defined, you will see that every even term is $0$. Thus the limit doesn't even exist.
Second, I'm not sure how you got that function $H$. Your proof is missing a lot of details, and when I try to reproduce it instead of $H$ I get
$$\frac{g_n'(0)}{g_n(0)}\frac{1}{n+1}$$
If you're looking for solutions, this has been asked before.
• Thanks very much. I know other good sols and I'm trying a different approach. I think $f$ you defined isn't satisfy a condition of the question : it has two poles in $\Omega$ (on the unit circle), so it's not a good counterexample. I didn't consider of $a_n$ being zero, since there was no description about that in the original question. I just have predicted that none of them will be zero since the author didn't mention about that. That maybe a critical point. For $H(z)/(n+1)$, your formula is true and I found there is a logical gap since I assumed $g_n(0)\neq 0$. Thanks. Jan 25, 2017 at 6:47
• "it has two poles in $\omega$": but that's precisely the point. Your proof did not use the fact that $f$ has only one pole, so I can just as well apply it to my $f(z)$ and achieve a contradiction. Part of proving that $lim\frac{a_n}{a_{n+1}} = z_0$ is proving that the limit exists in the first place. Jan 25, 2017 at 14:01
• I totally misunderstood that. Thanks for kind explanation! Could I stick to this approach with some modification to prove the question? Jan 25, 2017 at 14:27
• I'm not sure, I thought about it for a bit but I couldn't figure out a way to proceed past $\frac{g_n'(0)}{g_n(0)}\frac{1}{n+1}$. Jan 25, 2017 at 14:39
• I'm stuck too.. OK, Thanks for helping me!! Jan 25, 2017 at 15:00
I see you are trying a different approach. It's good.
However your argument does not work since you do not use the condition that $f$ is holomorphic in an open set $\Omega$ containing the closed unit disc, except for a pole at $z_0$ on the unit circle.
Since your argument is a local one, if $g_0(z)$ is defined on $D_1(0)\cup \{|z-z_0|<\delta \}$, holomorphic and not zero at $z_0$, your argument goes and concludes $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\frac{1}{z_0}$.
If your argument is correct, you can prove:
Suppose that $f$ is holomorphic in an open set $\Omega$ containing the closed unit disc, except for two poles at $z_0$ and $z_1$ on the unit circle. Show that if $f$ is given by a power series expansion $$f(z)=\sum^\infty_{n=0}a_n z^n$$ in the unit disc $D_1(0)$, then $$\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0\quad \text{and}\quad \lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_1.$$
Proof. Consider $f(z)=(z-z_0)^{-m}g_0(z)$ on $D_1(0)\cup \{|z-z_0|<\delta \}$. Then $$\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0.$$ Consider $f(z)=(z-z_1)^{-m}g_1(z)$ on $D_1(0)\cup \{|z-z_1|<\delta \}$. Then $$\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_1.$$
Do you recognize your argument is wrong ?
• Yes, I got it. Thanks very much! Jan 25, 2017 at 14:29
Hint: Suppose the pole of order $d\geq 1$. Then $(z-z_0)^d f(z)$ is holomorphic in a disc of radius $1+\epsilon$. Show that: $$f(z) = \sum_{j=1}^d \frac{p_j}{(z-z_0)^j} + g(z)$$ with $p_d\neq 0$ and $g$ holomorphic in a disc of radius $1+\epsilon$. Now, expand each term in a powerseries at $0$ and show that the series for $p_d(z-z_0)^{-d}$ (a negative binomial expansion) will dominate all other terms and give the provided limit ratio. | 2,004 | 5,849 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-22 | latest | en | 0.857238 |
http://kenhung.me/2017-08-23-thoughts-on-the-signal-and-the-noise.html | 1,529,595,279,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864191.74/warc/CC-MAIN-20180621153153-20180621173153-00545.warc.gz | 170,597,329 | 3,427 | Posts
Kenneth Hung | Department of Mathematics | UC Berkeley
# Thoughts on "The signal and the noise"
I came across Nate Silver’s name in two past occasions: once when I was taking Stat 215a at Berkeley, another time during the 2016 elections. With some luxury of free time when I was interning in Chicago, I took the time to finally read this book.
First the good things. Primarily due to my line of research, I have always been more fond of the frequentist approach. The book gave a very convincing argument for the Bayesian approach: as long as we do not exclude the truth, then with more data, we will be “less and less and less wrong”. While it might be odd to assign probabilities to when Newton’s Second Law does not hold, it most certainly is a valid approach for the more contemporary research, where repeated experiments show very large deviations in effect sizes; Or using his example on climate change, while climate change is irrefutable, the scale and severity of its effect might not be so agreed upon if the estimates are showing rather large variances.
The book also gave a very intuitive explanation of overfitting with the earthquake example, especially when it comes to extrapolation (vs. intrapolation). For exceptionally rare events that are rarely or never seen, an estimate for the probability of such events might be unreliable (this can go both ways), but the payoff (or penalty) can be orders of magnitude, not unlike the main focus in Talen’s book Black Swan.
I have some confusion about some of the examples used, however. In arguing that more information available can make estimation worse, the book mentioned National Journal panelists’ predictions about the 2010 midterm elections. For all 435 races, the predictions from conservatives and liberals differed by 6 seats, while for the race in a few specific states (Nevada, Illinois, Pennsylvania, Florida and Iowa) they differed by 5 out of 11. Differing by 6 out of 435 is definitely a sign of consistency in estimation compared to 5 out of 11. However, giving accurate prediction out of 11 (idealized) coin flips is definitely a harder job that doing so for 435 flips. The small number of coin flips forbids us to make full use of law of large numbers, making the prediction innately difficult. The difference in the estimates by conservatives and liberals need to be put in perspective by proper scaling (such as $\sqrt{n}$). While I agree with the sentiment, I am not sure if this example is so appropriate.
In Chapter 10, the book analyzed a poker game against a mythical opponent “the Lawyer”, and walked us through a step-by-step usage on Bayesian analysis: a prior was setup (an initial read that my fellow interns in Chicago like to talk about in poker nights) followed by Bayesian updates to our prior as the game unfolds. Each of the updates are objective as poker is a fairly mathematical game. This is a benefit of poker game with clean-cut probabilities that does not translate so well to more complicated systems, say basketball games. With a subjective prior, and Bayesian updates being subjective by nature, I am worried that it is possible that the subjectivity can add up, disallowing us to converge to the truth. However since I mostly work on more theoretical problems, I believe that other statisticians may know the tradeoffs better in practice.
Nate Silver pointed out many pitfalls and their solutions in statistical applications, some of such solutions mutually opposing, leaving room for data analysts to strike a balance. The abundance of charts and examples also made it a great summer read for me to take a break — a break from statistics by reading about statistics. | 757 | 3,687 | {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-26 | latest | en | 0.957737 |
https://chem.libretexts.org/Courses/Grand_Rapids_Community_College/CHM_120_-_Survey_of_General_Chemistry/8%3A_Acids_and_Bases/8.05%3A_Autoionization_of_Water_and_pH | 1,563,525,932,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526153.35/warc/CC-MAIN-20190719074137-20190719100137-00501.warc.gz | 350,019,593 | 22,568 | # 8.5: Autoionization of Water and pH
Skills to Develop
• To understand the autoionization reaction of liquid water.
• To know the relationship among pH, pOH, and $$pK_w$$.
As you learned previously acids and bases can be defined in several different ways (Table $$\PageIndex{1}$$). Recall that the Arrhenius definition of an acid is a substance that dissociates in water to produce $$H^+$$ ions (protons), and an Arrhenius base is a substance that dissociates in water to produce $$OH^−$$ (hydroxide) ions. According to this view, an acid–base reaction involves the reaction of a proton with a hydroxide ion to form water. Although Brønsted and Lowry defined an acid similarly to Arrhenius by describing an acid as any substance that can donate a proton, the Brønsted–Lowry definition of a base is much more general than the Arrhenius definition. In Brønsted–Lowry terms, a base is any substance that can accept a proton, so a base is not limited to just a hydroxide ion. This means that for every Brønsted–Lowry acid, there exists a corresponding conjugate base with one fewer proton. Consequently, all Brønsted–Lowry acid–base reactions actually involve two conjugate acid–base pairs and the transfer of a proton from one substance (the acid) to another (the base). In contrast, the Lewis definition of acids and bases, focuses on accepting or donating pairs of electrons rather than protons. A Lewis base is an electron-pair donor, and a Lewis acid is an electron-pair acceptor.
Table $$\PageIndex{1}$$: Definitions of Acids and Bases
Definition Acids Bases
Arrhenius $$H^+$$ donor $$OH^−$$ donor
Brønsted–Lowry $$H^+$$ donor $$H^+$$ acceptor
Lewis electron-pair acceptor electron-pair donor
Because this chapter deals with acid–base equilibria in aqueous solution, our discussion will use primarily the Brønsted–Lowry definitions and nomenclature. Remember, however, that all three definitions are just different ways of looking at the same kind of reaction: a proton is an acid, and the hydroxide ion is a base—no matter which definition you use. In practice, chemists tend to use whichever definition is most helpful to make a particular point or understand a given system. If, for example, we refer to a base as having one or more lone pairs of electrons that can accept a proton, we are simply combining the Lewis and Brønsted–Lowry definitions to emphasize the characteristic properties of a base.
## Acid–Base Properties of Water
Recall that because of its highly polar structure, liquid water can act as either an acid (by donating a proton to a base) or a base (by using a lone pair of electrons to accept a proton). For example, when a strong acid such as HCl dissolves in water, it dissociates into chloride ions ($$Cl^−$$) and protons ($$H^+$$). The proton, in turn, reacts with a water molecule to form the hydronium ion ($$H_3O^+$$):
$\underset{acid}{HCl_{(aq)}} + \underset{base}{H_2O_{(l)}} \rightarrow \underset{acid}{H_3O^+_{(aq)}} + \underset{base}{Cl^-_{(aq)}} \label{16.3.1a}$
In this reaction, $$HCl$$ is the acid, and water acts as a base by accepting an $$H^+$$ ion. The reaction in Equation \ref{16.3.1a} is often written in a simpler form by removing $$H_2O$$ from each side:
$HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} \label{16.3.1b}$
In Equation \ref{16.3.1b}, the hydronium ion is represented by $$H^+$$, although free $$H^+$$ ions do not exist in liquid water as this reaction demonstrates:
$H^+_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)}$
Water can also act as an acid, as shown in Equation \ref{16.3.2}. In this equilibrium reaction, $$H_2O$$ donates a proton to $$NH_3$$, which acts as a base:
$\underset{acid}{H_2O_{(aq)}} + \underset{base}{NH_{3(aq)}} \rightleftharpoons \underset{acid}{NH^+_{4 (aq)}} + \underset{base}{OH^-_{(aq)}} \label{16.3.2}$
Water is thus termed amphiprotic, meaning that it can behave as either an acid or a base, depending on the nature of the other reactant. Notice that Equation $$\ref{16.3.2}$$ is an equilibrium reaction as indicated by the double arrow and hence has an equilibrium constant associated with it.
## The Relationship among pH, pOH, and $$pK_w$$
As mentioned on a previous page, water can undergo the self-ionization reaction called autoprotolysis:
$$H_2O_{(l)} \; + \; H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)}$$
The equilibrium constant for this reaction is
$K_w = [H_3O^+][OH^-] = 1.0 x 10^{-14} \; at \; 25^oC \label{16.3.8}$
Thus, at 25 oC, for pure water, the [H3O+] = [OH-] = 1.0 x 10-7M. If a substance is added to water so that the [H3O+] > [OH-], the solution is considered acidic. If a substance is added to water so that the [H3O+] < [OH-], the solution is considered basic. This is a perfectly reasonable set of relationships to use to decide if a solution is acidic or basic. However, scientists in the early 1900's decided to add a second set of relationships to describe the acidity of a solution, using logarithms to convert the small [H3O+] and [OH-] concentrations to a more 'manageable' number. This second set of relationships involves the calculation of pH and pOH values, where pH and the $$H^+$$ ($$H_3O^+$$) concentration are related as follows:
$pH=−\log_{10}[H^+] \label{16.3.9}$
$[H^+]=10^{−pH} \label{16.3.10}$
Because the scale is logarithmic, a pH difference of 1 between two solutions corresponds to a difference of a factor of 10 in their hydronium ion concentrations. Using the pH scale, the pH of a neutral solution is 7.00 ($$[H_3O^+] = 1.0 \times 10^{−7}\; M$$), whereas acidic solutions have pH < 7.00 (corresponding to $$[H_3O^+] > 1.0 \times 10^{−7}$$) and basic solutions have pH > 7.00 (corresponding to $$[H_3O^+] < 1.0 \times 10^{−7}$$).
Similar notation systems are used to describe many other chemical quantities that contain a large negative exponent. For example, chemists use an analogous pOH scale to describe the hydroxide ion concentration of a solution. The pOH and $$[OH^−]$$ are related as follows:
$pOH=−\log_{10}[OH^−] \label{16.3.11}$
$[OH^−]=10^{−pOH} \label{16.3.12}$
The constant $$K_w$$ can also be expressed using this notation, where $$pK_w = −\log\; K_w$$.
Because a neutral solution has $$[OH^−] = 1.0 \times 10^{−7}$$, the pOH of a neutral solution is 7.00. Consequently, the sum of the pH and the pOH for a neutral solution at 25 °C is 7.00 + 7.00 = 14.00. We can show that the sum of pH and pOH is equal to 14.00 for any aqueous solution at 25 °C by taking the negative logarithm of both sides of Equation \ref{16.3.8}:
\begin{align} −\log_{10} K_w &= pK_w \\[5pt] &=−\log([H_3O^{+}][OH^{−}]) \\[5pt] &= (−\log[H_3O^{+}])+(−\log[OH^{−}])\\[5pt] &= pH+pOH \label{16.3.13} \end{align}
Thus at any temperature, $$pH + pOH = pK_w$$, so at 25 °C, where $$K_w = 1.0 \times 10^{−14}$$, pH + pOH = 14.00. More generally, the pH of any neutral solution is half of the $$pK_w$$ at that temperature. The relationship among pH, pOH, and the acidity or basicity of a solution is summarized graphically in Figure $$\PageIndex{1}$$ over the common pH range of 0 to 14. Notice the inverse relationship between the pH and pOH scales.
Figure $$\PageIndex{1}$$: The Inverse Relationship between the pH and pOH Scales. As pH decreases, $$[H^+]$$ and the acidity increase. As pOH increases, $$[OH^−]$$ and the basicity decrease. Common substances have pH values that range from extremely acidic to extremely basic.
For any neutral solution, pH + pOH = 14.00 (at 25 °C) with pH=pOH=7.
Example $$\PageIndex{1}$$
The Kw for water at 100 °C is $$4.99 \times 10^{−13}$$. Calculate $$pK_w$$ for water at this temperature and the pH and the pOH for a neutral aqueous solution at 100 °C. Report pH and pOH values to two decimal places.
Given: $$K_w$$
Asked for: $$pK_w$$, $$pH$$, and $$pOH$$
Strategy:
1. Calculate $$pK_w$$ by taking the negative logarithm of $$K_w$$.
2. For a neutral aqueous solution, $$[H_3O^+] = [OH^−]$$. Use this relationship and Equation \ref{16.3.8} to calculate $$[H_3O^+]$$ and $$[OH^−]$$. Then determine the pH and the pOH for the solution.
Solution:
A
Because $$pK_w$$ is the negative logarithm of Kw, we can write
$pK_w = −\log K_w = −\log(4.99 \times 10^{−13}) = 12.302$
The answer is reasonable: $$K_w$$ is between $$10^{−13}$$ and $$10^{−12}$$, so $$pK_w$$ must be between 12 and 13.
B
Equation \ref{16.3.8} shows that $$K_w = [H_3O^+][OH^−]$$. Because $$[H_3O^+] = [OH^−]$$ in a neutral solution, we can let $$x = [H_3O^+] = [OH^−]$$:
\begin{align*} K_w &=[H_3O^+][OH^−] \\[5pt] &=(x)(x)=x^2 \\[5pt] x&=\sqrt{K_w} \\[5pt] &=\sqrt{4.99 \times 10^{−13}} \\[5pt] &=7.06 \times 10^{−7}\; M \end{align*}
Because $$x$$ is equal to both $$[H_3O^+]$$ and $$[OH^−]$$,
\begin{align*} pH &= pOH = −\log(7.06 \times 10^{−7}) \\[5pt] &= 6.15 \,\,\, \text{(to two decimal places)} \end{align*}
We could obtain the same answer more easily (without using logarithms) by using the $$pK_w$$. In this case, we know that $$pK_w = 12.302$$, and from Equation \ref{16.3.13}, we know that $$pK_w = pH + pOH$$. Because $$pH = pOH$$ in a neutral solution, we can set $$pH = pOH = y$$. Solving to two decimal places we obtain the following:
\begin{align*} pK_w &= pH + pOH \\[5pt] &= y + y \\[5pt] &= 2y \\[5pt] y &=\dfrac{pK_w}{2} \\[5pt] &=\dfrac{12.302}{2} \\[5pt] &=6.15 =pH=pOH \end{align*}
Exercise $$\PageIndex{1}$$
Humans maintain an internal temperature of about 37 °C. At this temperature, $$K_w = 3.55 \times 10^{−14}$$. Calculate $$pK_w$$ and the pH and the pOH of a neutral solution at 37 °C. Report pH and pOH values to two decimal places.
• $$pK_w = 13.45$$
• $$pH = pOH = 6.73$$
## Summary
Water is amphiprotic: it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion ($$H_3O^+$$). The autoionization of liquid water produces $$OH^−$$ and $$H_3O^+$$ ions. The equilibrium constant for this reaction is called the ion-product constant of liquid water (Kw) and is defined as $$K_w = [H_3O^+][OH^−]$$. At 25 °C, $$K_w$$ is $$1.0 \times 10^{−14}$$; hence $$pH + pOH = pK_w = 14.00$$.
• For any neutral solution, $$pH + pOH = 14.00$$ (at 25 °C) and $$pH = 1/2 pK_w$$.
• Ion-product constant of liquid water: $K_w = [H_3O^+][OH^−] \nonumber$
• Definition of $$pH$$: $pH = −\log10[H^+] \nonumber$ or $[H^+] = 10^{−pH} \nonumber$
• Definition of $$pOH$$: $pOH = −\log_{10}[OH^+] \nonumber$ or $[OH^−] = 10^{−pOH} \nonumber$
• Relationship among $$pH$$, $$pOH$$, and $$pK_w$$: $pK_w= pH + pOH \nonumber$
## Contributors
Modified by Tom Neils (Grand Rapids Community College) | 3,422 | 10,594 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-30 | latest | en | 0.890742 |
https://www.coursehero.com/file/6119929/sols-wk3/ | 1,529,578,319,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864139.22/warc/CC-MAIN-20180621094633-20180621114633-00149.warc.gz | 773,337,827 | 58,902 | sols_wk3
# sols_wk3 - Homework 3 due Thursday October 28 before 13:30...
This preview shows page 1. Sign up to view the full content.
Homework 3 due Thursday October 28 before 13:30 in corre- sponding group folder on door of office CM110 Consider the sequence { p n } generated by p 0 = 0 , p n +1 = - ( p 3 n + 2) / 6 , n = 0 , 1 , 2 , ... Show that for all n , - 1 3 p n 0 and that the sequence converges to a limit. Give a bound for its rate of convergence. Let p n +1 = g ( p n ) := - ( p 3 n + 2) / 6. We plan to apply the convergence theorem on the interval [ - 1 3 , 0]. Notice that g 0 ( x ) = - x 2 2 0 for x [ - 1 3 , 0] . Hence, we conclude that max x [ - 1 3 , 0] | g 0 ( x ) | ≤ 1 18 < 1 and that as g is monotone decreasing so that g : [ - 1 3 , 0] [ g (0) , g ( - 1 3 )] = [ - 1 3 , - 53 162 ] [ - 1 3 , 0] , i.e. g maps the interval [ - 1 3 , 0] into itself. Hence we can apply the convergence theorem which shows that - 1 3 p n 0 for all n and that the sequence converges to a unique limit, p . As for a bound on its rate of convergence, since the limit p 6 = 0, convergence is linear, since g 0 ( p ) 6 = 0, and so | p n +1 - p | ≤ 1 18 | p n - p | . For the iteration function g ( x ) = 1 + 1 /x + 1 /x 2 , 1. verify that g satisfies the conditions of the contraction mapping theorem on the interval [1 . 75 , 2 . 0]; 2. carry out a few iterations with the starting value p 0 = 1 . 825 and find an upper bound on the magnitude of the error in p
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern | 750 | 2,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-26 | latest | en | 0.889264 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.