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https://www.expertsmind.com/library/calculate-the-profit-or-loss-and-break-even-point-552150.aspx	1,713,147,487,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00200.warc.gz	709,341,386	14,270	### Calculate the profit or loss and break-even point Assignment Help Finance Basics ##### Reference no: EM1352150 Healthy Foods, Inc., sells 50-pound bags of grapes to the military for \$10 a bag. The fixed costs of this operation are \$80,000, while the variable costs of the grapes are \$.10 per pound. a. What is the break-even point in bags? b. Calculate the profit or loss on 12,000 bags and on 25,000 bags. c. What is the degree of operating leverage at 20,000 bags and at 25,000 bags? Why does the degree of operating leverage change as the quantity sold increases? d. If Healthy Foods has an annual interest expense of \$10,000, calculate the degree of financial leverage at both 20,000 and 25,000 bags. e. What is the degree of combined leverage at both sales levels? #### Questions Cloud How long will it take to strike the ground : What is the velocity of the plank relative to the surface of ice. Explain how the union organization enables : Explain how the union organization enables it to serve its members effectively and defend your opinion Determination of wacc : Mullineaux Company has a target capital structure of 60 percent common stock, 5% preferred stock, and 35% debt. Its cost of equity is 14 percent, the cost of preferred stock is 6%. Illustrate what moe and larry have already spent : buy the five-foot strip of land from their neighbor only if the price is less than illustrate what Moe and Larry have already spent on the foundation. Calculate the profit or loss and break-even point : Healthy Foods, Corporation, sells fifty pound bags of grapes to the military for \$10 a bag. The fixed costs of this operation are \$80,000, while the variable costs of the grapes are \$.10 per pound. Determining the votume of the irregularly shaped object : Approximately how many atoms make up our planet? For simplicity, suppose the average atomic mass of the atoms is 14 . Avogadro's number gives the number of atoms in a mole. Calculate the ratios on the worksheet : Select any publicly traded company that their financials are published on the SEC website. For the purpose of this assignment we have selected Kirkland's Corporation Kirkland is specialty retailer of home decor in the US, Describe what challenges an hr department may face : Describe what challenges an HR department may face when their company decides to expand into other countries What is the gravitational force acting upon the woman : An airplane with a mass of 2.6*10^6 kg. has the air flow past the lower surface of the wings at 100 m/s. If the wings have a surface area of 1200 m2, how fast should the air flow over the upper surface of the wing if the plane is to stay in the ai.. ### Write a Review #### Assess risks and opportunities in terms of economic Assess risks and opportunities in terms of economic. A analysis of the case study "AccuForm: Ethical leadership and its challenges in the era of globalization" #### Objective type question on time value of money Objective type question on time value of money and What is the effective annual rate #### Explaining cash management firm-s-opportunity cost is given As the bank is also doing lot of record keeping, firm’s administrative cost would reduce by \$2,000 per month. What suggestion would you provide firm with respect to proposed cash management suppose the firm’s opportunity cost is 12%? #### Computation of cost of capital and beta Computation of cost of capital and beta and explain Does it matter if you use the beta for Dell or the beta for the industry in this case #### Briefly discuss the irs s position John forms a company and transfers property having a basis to him of \$18,000 & a fair market value of \$26,000 to the company for 1,000 shares of \$10 par stock. #### Calculating present values of cash flows You can purchase property today for \$3.3 million and sell it in 5 years for \$4.3 million. (You earn no rental income on the property.) #### Calculate the fixed costs per month Account Analysis, High-Low, Contribution Margin data on occupancy and costs at the Starlight Hotel for June, July and August are shown below: #### Prepackaged bankruptcy Which of following isn't advantage of prepackaged bankruptcy? #### Computation of number of stocks and stock price Computation of number of stocks and stock price and Assume there is no capital gains tax #### Explain what is the price of the bond which pays annual Explain What is the price of the bond which pays annual interest and Both bonds are non-callable and have a face value of \$1,000 #### Gordon growth model and one period valuation model Based on the Gordon Growth Model, compute the anticipated market price of stock that is paying dividends at a constant growth rate of 6.25%, with the recent dividend of \$1.00, and the required return rate of 15%. #### Estimate the firms return on capital On the basis of the mentioned information you as a finance manager are asked to provide the following : Estimate the firms return on capital. What would be the reinvestment rate of the firm? #### Assured A++ Grade Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!	1,137	5,253	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.609375	4	CC-MAIN-2024-18	latest	en	0.921409
https://www.lmpforum.com/forum/topic/4274-how-to-mathemactialy-model-the-starting-transient-of-induction-motor/	1,716,770,365,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00390.warc.gz	756,933,557	16,976	# How To Mathemactialy Model The Starting Transient Of Induction Motor? ## Recommended Posts Hi guys, This is my first time here so go easy on me Well the thing is I am currently doing my summer vacation work experience at a mine site and our power station is operating in island mode. We have large DOL induction motors unto 1600kw that cause a severe voltage dip when they start, even though we have to bring online another generator for the motor to come online. What I have been told to do is to look at pf correction at our main 11kv bus that is automatically switched cap banks. I was thinking; if we are going to correct the pf why not make it so that it can aid the starting of the induction motors by switching in multiple cap banks. Question is I need to prove that the switching of cap banks will help the motor in starting up and they wont cause as much voltage dip!. I know we will still get a momentary voltage dip even with cap banks on, but cap banks will help in recovering the voltage back up to nominal value. Is there anyway to mathematically prove this?? How can I model the transient starting current and voltage dip and addition of cap banks?? Forget about soft starters and Y/D switching only need to focus on what I am trying to do, so any useful suggestions would be appreciated. ## Create an account Register a new account • ### Who's Online (See full list) • There are no registered users currently online • ### Tell a friend Love LMPForum? Tell a friend! ×	324	1,498	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.765625	3	CC-MAIN-2024-22	latest	en	0.939169
https://www.mbatious.com/topic/61/remainder-theorem	1,508,286,303,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187822625.57/warc/CC-MAIN-20171017234801-20171018014801-00857.warc.gz	966,439,518	26,249	# Remainder Theorem • Remainder theorem is a very important topic in number system and can be learnt easily. We will try to learn some interesting concepts regarding remainders with examples. Here we go! Definition of remainder If a and d are natural numbers, with d # 0, it can be proved that there exist unique integers q and r, such that a = qd + r and 0 ≤ r < d. The number q is called the quotient, while r is called the remainder. Dividend = Divisor × Quotient + Remainder. if r = 0 then we say that a is perfectly divisible by d or d is a factor of a. For example, we say 8 is a factor of 40 because 40 leaves a remainder 0 with 8. By definition remainder cannot be negative. Now just to give an example, 17 = 3 * 5 + 2, which means 17 when divided by 5 will give 2 as remainder. Well that was simple! Find Remainder[ (12 * 13 * 14) / 5 ] Remainder [ (12 * 13 * 14) / 5 ] = Remainder [2184/5] = 4. But this method is not the right one for us :) In order to find the remainder of an expression find the individual remainder and replace each term with the respective remainders. Eg: Remainder[(100 + 30 * 4 - 8 ) / 7] = Remainder[(Remainder[100/7] + Remainder[30/7] * Remainder[4/7] - Remainder[8/7] ) / 7 ] = Remainder[(2 + 2 * 4 - 1)/7] = Remainder[9/7] = 2 In the above case 12, 13 and 14 will give remainders 2, 3 and 4 respectively when divided by 5. So replace them with the respective remainders in the expression and find the remainder again. Remainder [ (12 * 13 * 14) / 5 ] = Remainder[(2 * 3 * 4) / 5] = Remainder[ 24 / 5 ] = 4. Note: One common mistake while dealing with remainders is when we have common factors in both dividend and divisor. Example, what is the remainder when 15 is divided by 9 15 / 9 is same as 5 / 3, remainder 2. Correct? No 15/9 will give a remainder of 6. Where we slipped? Always remember that if we find remainder after cancelling common terms make sure we multiply the remainder obtained with the common factors we removed. In previous case we will get correct answer (6) when we multiply the remainder obtained (2) with the common factor we removed (3). What is the remainder of 1421 * 1423 * 1425 when divided by 12 ? ( CAT 2000 ) 1421, 1423 and 1425 gives 5, 7 and 9 as remainders respectively when divided by 12. Remainder [ (1421 * 1423 * 1425 ) / 12 ] = Remainder [ (5 * 7 * 9) ] / 12, gives a remainder of 3. Find the reminder when 1! + 2! + 3! + . . . . .. . .. 99! + 100! is divided by the product of first 7 natural numbers From 7! the remainder will be zero. Why ? because 7! is nothing but product of first 7 natural numbers and all factorial after that will have 7! as one of the factor. so we are concerned only factorials till 7!, i.e, 1! + 2! + 3! + 4! + 5! + 6! 1! + 2! + 3! + 4! + 5! + 6! = 873 and as 7! > 873 our remainder will be 873 What is the remainder when 64999 is divided by 7? (GMAT Type Question) Many of us get intimidated with such numbers, always remember that the key to crack quant is a strong hold of basic concepts. Remainder [64999 / 7 ] = Remainder[ 64 * 64 * …. 64 (999 times) / 7 ] Remainder[64/7] = 1, hence Remainder [64999 / 7 ] = Remainder [ 1 999 / 7 ] = 1 What is the remainder when 444444 ^ 444 is divided by 7 ? (GMAT Type Question) Remainder[444/7] = 3 Remainder[ 444 444 ^ 444 / 7 ] = Remainder [ 3 444 ^ 444 / 7 ] = Remainder [ ( 3) 222 ^ 444 / 7 ] = Remainder [ 2222 ^ 444 / 7 ] ( As Remainder [ 32 / 7 ] = 2 ) = Remainder [ ( 23 ) 74 ^ 444 / 7] = Remainder [ 1 74 ^ 444 / 7 ] = 1 ( As Remainder [ 23 / 7 ] = 1 ) Concept of negative remainder We saw earlier that by definition remainder cannot be negative. But considering negative remainder is a very useful exam trick. For example, What is the remainder when 211 divided by 3? The easiest method for this one will be using the concept of negative remainders. Here 2 when divided by 3 gives a remainder of -1. (Say) 2 = 3 * 1 + (-1), remainder is -1, which is theoretically incorrect but let’s cheat! So we are asked to find (-1) * (-1) * … 11 times divided by 3. Which is Remainder[-1/3] = -1. Whenever you are getting negative number as a remainder, make it positive by adding the divisor to the negative remainder. Here required answer is 3 + (-1) = 2. Remainder when (41 * 42) is divided by 43 Use negative remainder concept, Remainder [ 41 * 42 / 13 ] = Remainder[(-2) * (-1) / 43 ]( as 41 = 43 * 1 – 2 and 42 = 43 * 1 – 1) = Remainder [ 2 /43 ] = 2 (here we got a positive remainder itself, so no need of correction) Some useful concepts while dealing with remainder are given below. Remainder[(ax + 1)n / a] = 1 for all values of n. Find the remainder when 10099 is divided by 11 Remainder[10099 / 11 ] = Remainder[(11* 9 + 1)99 / 11] = 1. Remainder[(ax - 1)n / a ] = 1 when n is even Remainder[(ax - 1)n / a ] = (a-1) when n is odd. Find the remainder when 21875 is divided by 17. Remainder[21 / 17] = 4, so we need to find Remainder[4875/ 17] 42 = 16 = (17 – 1), we can write the expression as Remainder[(42)437 * 4 / 17] = Remainder[(17 – 1)437 * 4 / 17] = Remainder[(17-1) * 4 / 17] = Remainder[64 / 17] = 13. Remainder[(an + bn) / (a + b) ] = 0 when n is odd. Remainder[(2101 + 3101) / 5 ] = 0 What is the remainder when 1523 + 2323 is divided by 19 ? ( CAT 2004 ) 1523 + 2323 is divisible by 15 + 23 = 38 ( as 23 is odd). So Rem[(1523 + 2323)/19] = 0 Remainder[(an + bn + cn + ...) / (a + b + c + ...) ] = 0 if ( a + b + c + ... are in Arithmetic progression and n is odd What is the remainder when 163 + 173 + 183 + 193 is divided by 70 ? ( CAT 2005 ) Apply the above funda. Here n =3 ( odd ), 16 + 17 + 18 + 19 = 70 and 16,17,18 and 19 are in AP. Remainder is 0 Remainder[(an - bn) / (a + b) ] = 0 when n is even. Remainder[(5100 – 2100) / 7] = 0 Remainder [(an - bn) / (a - b) ] = 0 Now we can say Remainder[(10175 – 7675) / 25] = 0 in no time..! The remainder when f(x) = a + bx + cx2+ dx3+ … is divided by (x-a) is f(a) Rem [(3x2 + 4x + 1) / (x-2)] = f(2) = 3 * 22 + 4 * 2 + 1 = 21 Cyclic property of remainders Sometimes it is easy to find the remainder by using the cyclic property of remainders, remainders forming a pattern. As a thumb rule if we divide pn with q, the remainder will follow a pattern. For example, Remainder [ 21 / 3 ] = 2, Remainder [ 22 / 3 ] = 1, Remainder [ 23 / 3 ] = 2, Remainder [ 24 / 3 ] = 1 and so on. Pattern repeats in cycles of 2. Remainder [ 2n / 3 ] = 2 when n is odd and 1 when n is even. With this information, we can find Remainder [ 23276 / 3 ] = 1 very quickly. One more, Remainder [ 91 / 11 ] = 9, Remainder [ 92 / 11 ] = 4, Remainder [ 93 / 11 ] = 3 , Remainder [ 94 / 11 ] = 5 Remainder [ 95 / 11 ] = 1, Remainder [ 96 / 11 ] = 9, Remainder [ 97 / 11 ] = 4, Remainder [ 98 / 11 ] = 3 Pattern repeats in cycles of 5. So if we are asked to find Remainder [ 9100 / 11 ]， we know it is 1. (100 is in the form 5n and we know remainder for 5 is 1.. cool right ?) Note: Remainder [93/11] = Remainder [Remainder [92/11] * Remainder [92/11] ) / 11] = Remainder [ 4 * 9 / 11] = 3. This funda comes very handy in scenarios like this. Like we dont have to solve Rem [98 /11] because we already know Rem[94 /11] as 5.. Rem [98 /11] = Remainder [ 9* 94 /11 ] = Remainder [ (5 * 5) / 11 ] = 3 Also, Remainder [97 /11] = Remainder [93 * 94 / 11 ] = Remainder [ ( 3 * 5 ) / 11 ] = 4 ( as Rem [93 /11] = 3 and Rem [94 /11] = 5 ) What is the remainder when 7100 is divided by 4? Remainder[ 71 / 4 ] = 3, Remainder[ 72 / 4 ] = 1, Remainder[ 73 / 4 ] = 3, Remainder[ 74 / 4 ] = 1 and so on... Pattern repeats in cycles of 2. Remainder [ 7n / 4 ] is 3 when n is odd and is 1 when n is even. 7100 when divided by 4 gives a remainder of 1. (Same can be solved using other methods also) Find the remainder when 399^99 is divided by 7 Find the pattern of remainder when 3n is divided by 7. Remainder [ 3/ 7 ] = 3, Remainder [ 32 / 7 ] = 2 Remainder [ 33 / 7 ] = 6 (Don’t calculate Rem[33/7] we already have Rem[31/7] & Rem[32/7]) Remainder [ 3/ 7 ] = 4 (using Rem[32/7]) Remainder [ 3/ 7 ] = 5 (using Rem[33/7] and Rem[32/7] ) Remainder [ 3/ 7 ] = 1 (using Rem[33/7]) Remainder [ 3/ 7 ] = 3 (using Rem[33/7] and Rem[34/7] ) Remainder [ 3/ 7 ] = 2 (using Rem[34/7]) Pattern repeats in cycles of 6. (We can do this easily from Euler’s theorem, as φ(7) = 6, hence Remainder [36/7] = 1. Explained just to get the idea of patterns in remainders) Now our task is to find Remainder [ 9999/ 6 ] Remainder [9999/6] = Remainder [399/6] Remainder [ 31 / 6 ] = 3, Remainder [ 32 /6 ] = 3, Remainder [ 33 / 6 ] = 3 and so on..! Hence 9999 can be written as 6n + 3. Remainder [ 399^99 / 7 ] = Remainder [ 3(6n + 3)/ 7 ] we have found out the pattern of 3 divided by 7 repeats in cycles of 6. So we need to find the Remainder [ 33 / 7 ] to get the answer which is equal to 6 Euler’s Remainder Theorem We say two numbers ( say a and b ) are co-prime to each other when HCF(a,b) = 1, i.e, no divisor divide both of them completely at the same time. Eg: 21 and 8 are co primes because they don't have any common factors except 1 In number theory, Euler's totient or phi function, φ(n) is an arithmetic function that counts the number of positive integers less than or equal to n that are relatively prime to n. Take n = 9, then 1, 2, 4, 5, 7 and 8, are relatively prime to 9. Therefore, φ(9) = 6. How to find Euler’s totient? Say n = P1a x P2b x P3c x ... ( where P1 P2, P3 ... are prime factors of n ) φ(n) = n x (1 - 1/P1) x (1 - 1/P2) x (1 - 1/P3) x .... If n is prime then φ(n) = n - 1 φ(100) = 100 x (1-1 / 2) x (1- 1 / 5) = 100 x 1 / 2 x 4 / 5 = 40 φ(9) = 9 x (1 – 1/3) = 6 Euler’s Remainder theorem states that for co prime numbers M and N, Remainder [Mφ(N) / N] = 1 Always check whether the numbers are co primes are not. Euler’s theorem is applicable only for co prime numbers What is the remainder when 21865 is divided by 17 Remainder[21/17] = 4 Remainder [ 21865/ 17 ] = Remainder [ 4865/ 17 ] 4 and 17 are co prime numbers. ( A prime number is always coprime to any other number) φ(17) = 17 x ( 1 – 1 / 17) = 16. So Euler’s theorem says Remainder [ 416/ 17 ] = 1 To use this result in the given problem we need to write 865 in 16n + r form. 865 = 16 * 54 + 1, so 4865 can be written as 416 * 54 x 4 Remainder[4865/17] = Remainder[ 41654/17] Remainder[4/17] = 1 * 4 = 4 What is the remainder when 99999999 is divided by 23 Remainder[99/23] = 7 Remainder[99999999/23] = Remainder[7999999/23] 7 and 23 are co prime numbers Here 23 is prime, so φ(23) = 22 So by applying Euler's theorem we can say that Remainder[722/23] = 1 In order to use this result in our problem we need to write 999999 in 22n + r form. Before rushing into dividing 999999 by 22, just think whether we have any better way to do that. We know, 999999 = 22n + r , 0 ≤ r < 22. 999999 is divisible by 11 and so is 22. which means r is also a multiple of 11. Only numbers which are less than 22 and is a multiple of 11 is 11 and 0. But as 999999 is odd and 22n is even, r should be odd. so r = 11. Saved our time right ? ;) 999999 = 22n + 11 Remainder[7999999/23] = Remainder[722n/23] * Remainder[711/23] = 1 * Remainder[711/23] = Remainder[72 * 5 * 7/23] = Remainder[ 35 * 7/23] ( as Remainder[72/23] = 3 ) = Remainder[33 * 32 * 7/23] = Remainder[4 * 9 * 7/23] = Remainder[28 * 9/23] = Remainder[45/23] = 22 Find the remainder when 97 97 ^ 97 is divided by 11 Remainder [97/11] = 9 So, Remainder [9797^97/11] = Remainder [9 97^97/11] From Euler’s theorem, Remainder [910/11] = 1 97 and 10 are again co primes. So φ(10) = 10 (1-1/2) (1-1/5) = 4 Remainder [974/11] = 1 97 = 4n + 1, So Remainder [9797/11] = Remainder [97/10] = 7 Means 9797 can be written as 10n + 7 Now our original question, Remainder [997^97/11] = Remainder [910n+7/11] = Remainder [97/11] = 4 :) Fermat’s little theorem Euler’s theorem says that if p is a prime numberand a and p are co-primes then aφ(p) / p always gives a remainder of 1. Now we know for any prime number p, φ(p) = p - 1 Remainder of a( p – 1 )/ p is 1, which is Fermat’s little theorem We can derive other useful results like • Remainder of a/ p is a. • (ap – a) is always divisible by a. Wilson’s theorem Remainder[(p-1)! / p ] = (p-1), if p is a prime number. We can also derive some useful results like • Remainder of [(p-1)! + 1] / p is zero. • Remainder of (p-2)! / p is 1. Example: Remainder of [22!/23] = 22 Remainder of [21!/23] = 1 I hope the explanations are clear are correct. Please let me know if any concepts regarding remainders are missed out or incase of any errors.. Happy learning :) • can you pls explain 97^97^97/11 in other ways i didnt get why we take the remainder of 97/11 and 97/10 Remainder [9797/11] = Remainder [97/10] = 7 this step We are asked to evaluate the remainder of 97^97^97/11 Now one thing is damn sure that whoever set the question don't want us to sit and calculate 97^97^97 and then divide it by 11.. so what we do.. ? First we are going to simplify the whole expression given value is (97^n)/11 where n = 97^97 97 = 11 * 8 + 9 So 97/11 gives remainder of 9 and we can write 97^n/11 as 9^n/11 (If you need more clarity in this regard - refer again the portion in the beginning where we discussed In order to find the remainder of an expression find the individual remainder and replace each term with the respective remainders.) Now 9 and 11 are co primes and we can apply Euler's theorem - Euler’s Remainder theorem states that for co prime numbers M and N, Remainder [M^φ(N) / N] = 1 φ(11) = 11 - 1 = 10 So 9^10/11 = 1 clear till this point ? Our job now would be to express n (97^97) in the form of 10x + y (as we know 9^10/11 = 1) (97^97)/10 : as 97 and 10 are again co primes, apply Euler again φ(10) = 4 and as per Euler, Rem[97^4/10] = 1 97 = 4 * 24 + 1 Rem[97^97]/11 = Rem[97/10] = 7 With these points in place we can get Remainder [9^97^97/11] = Remainder [9^10n+7/11] = Remainder [97/11] = 4 Looks like your connection to MBAtious was lost, please wait while we try to reconnect.	4,979	14,120	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.6875	5	CC-MAIN-2017-43	longest	en	0.911907
http://whistleralley.com/polyhedra/octahedron.htm	1,701,267,762,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100112.41/warc/CC-MAIN-20231129141108-20231129171108-00595.warc.gz	54,587,366	2,569	# The Octahedron The octahedron has 8 equilateral triangular faces, 6 vertices, and 12 edges. Four faces meet at each vertex. The octahedron might also be classified as a square dipyramid or a triangular antiprism. Consider an octahedron with side length s. The faces are equilateral triangles. Find the area of one face. Multiply that by the number of faces for the total surface area. There is a problem with the dihedral angle formula being use in these pages. It applies only to cases in which exactly three faces meet at a vertex. Here, we have four. Consider this. Four faces that meet at a common vertex are cut away from the rest of the solid. The result is a pyramid with a square base. At each of the base vertices there are three faces, including two triangles and a square. The dihedral angle between the two triangles is the dihedral angle of the octahedron. We can now apply the formula. Let α = β = π/3 and γ = π/2. Next, find the apothem of an equilateral triangle, and use it to find the inradius and circumradius of the octahedron. Now apply the volume formula. Another way with the volume is to dissect the octahedron into two pyramids. Each has a unit square for a base and a height equal to the circumradius. ### Other Properties The octahedron has 48 symmetries. The octahedron is the dual of the cube. Connect the centers of adjacent faces, and the result is a cube. Do the same to a cube, and the result is an octahedron. A tetrahedron can be formed by connecting centers of certain faces of the octahedron. This fact follows naturally from its dual relation with the cube. A cross-section of the octahedron can be a square or a regular hexagon. A planar projection also may be a square or a regular hexagon. ### Packing with Octahedra and Tetrahedra Here is an interesting property. Lay an octahedron on a horizontal surface. It is possible to lay a tetrahedron on the same surface and nestle it right underneath the octahedron. This follows from the fact that their dihedral angles are supplementary. Push them together. The common face goes away. Less obvious is the fact that three other pairs of faces are coplanar. Edges vanish, and it becomes a heptahedron, having three rhombic faces and four triangular faces. Start over again with a single octahedron. Attach a tetrahedron to each of its eight faces. The result is this stellated octahedron. It might also be visualized as two intersecting tetrahedra. Now fit twelve octahedra around this figure. They will fit perfectly. Here is the result of that last step. It is ready for another layer of tetrahedra. Together, the octahedron and tetrahedron pack space, but neither of them can do so alone. Last update: November 2, 2011 ... Paul Kunkel whistling@whistleralley.com For email to reach me, the word geometry must appear in the body of the message.	709	2,850	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.34375	4	CC-MAIN-2023-50	latest	en	0.926084
https://www.proprofs.com/quiz-school/story.php?title=kamagong-4th-quarter-exam-in-science-10n0	1,702,171,800,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100989.75/warc/CC-MAIN-20231209233632-20231210023632-00028.warc.gz	1,024,720,032	110,357	# Kamagong 4th Quarter Exam In Science 10 Approved & Edited by ProProfs Editorial Team The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes. \| By Isabel.noma.pana I Isabel.noma.pana Community Contributor Quizzes Created: 5 \| Total Attempts: 2,137 Questions: 50 \| Attempts: 42 Settings • 1. ### Which of the following scientists studied the relationship between the volume and temperature of gas at constant pressure? • A. Robert Boyle • B. Jacques Charles • C. • D. Gay - Lussac B. Jacques Charles Explanation Jacques Charles studied the relationship between the volume and temperature of gas at constant pressure. This relationship is known as Charles's Law, which states that the volume of a gas is directly proportional to its temperature, as long as the pressure remains constant. Rate this question: • 2. ### Which law explains the mechanism of a pressure cooker? • A. Boyle's Law • B. Charles' Law • C. Gay - Lussac's Law • D. Ideal Gas Law C. Gay - Lussac's Law Explanation Gay-Lussac's Law, also known as the pressure-temperature law, explains the mechanism of a pressure cooker. According to this law, the pressure of a gas is directly proportional to its temperature, when the volume and the amount of gas are kept constant. In a pressure cooker, as the temperature increases, the pressure inside the cooker also increases. This increase in pressure raises the boiling point of the liquid inside the cooker, allowing food to cook at higher temperatures and reducing the cooking time. Therefore, Gay-Lussac's Law accurately describes the mechanism of a pressure cooker. Rate this question: • 3. ### What will happen to the volume of a confined gas as the pressure increases if the temperature of the gas is kept constant? • A. The volume increases • B. The volume decreases • C. The volume remains the same • D. There is no significant change B. The volume decreases Explanation When the pressure of a confined gas is increased while keeping the temperature constant, the gas molecules are forced closer together. This results in a decrease in the average distance between the molecules, leading to a decrease in the volume of the gas. Therefore, the correct answer is that the volume decreases. Rate this question: • 4. ### Which of the following figures represents Charles' Law? • A. Option 1 • B. Option 2 • C. Option 3 • D. Option 4 A. Option 1 Explanation Charles' Law states that the volume of a gas is directly proportional to its temperature, assuming pressure and amount of gas remain constant. In Option 1, there is a graph showing an increase in volume as temperature increases, which represents Charles' Law. Option 2, 3, and 4 do not show a direct relationship between volume and temperature, so they do not represent Charles' Law. Rate this question: • 5. ### If a balloon full of warm air was placed in a refrigerator, the balloon would ______. • A. Get larger • B. Remains the same • C. Get smaller • D. Burst C. Get smaller Explanation When a balloon full of warm air is placed in a refrigerator, the temperature inside the refrigerator is much colder than the warm air inside the balloon. As a result, the air molecules inside the balloon lose kinetic energy and slow down, causing them to contract and occupy less space. This contraction leads to a decrease in the volume of the balloon, causing it to get smaller. Rate this question: • 6. ### What property of gas is best described when a gas tank is filled and more gases can still be added? • A. Gas is elastic • B. Gas can be compressed • C. Gas diffuses • D. Gas exerts pressure B. Gas can be compressed Explanation The correct answer is "gas can be compressed." This property refers to the fact that gases can be squeezed into a smaller volume. When a gas tank is filled and more gases can still be added, it indicates that the gas molecules can be further compressed, allowing for more gas to be added to the tank. Rate this question: • 7. ### Which of the following activities involves the application of gas pressure? • A. Burning gas • B. Inflating balloons • C. Cooking food • D. Gas exerts pressure B. Inflating balloons Explanation Inflating balloons involves the application of gas pressure. When we blow air into a balloon, the gas molecules inside the balloon collide with the walls of the balloon, creating pressure. This pressure causes the balloon to expand and inflate. Rate this question: • 8. ### A balloon filled with helium occupies 15 L at 400 K. What will be the volume of the helium balloon if the temperature of the helium is increased to 600 K? (Assume that the pressure is held constant) • A. 10 L • B. 22.5 L • C. 32.5 L • D. 50 L B. 22.5 L Explanation When the temperature of a gas is increased at constant pressure, the volume of the gas also increases. This relationship is described by Charles's Law, which states that the volume of a gas is directly proportional to its temperature in Kelvin. In this case, the temperature is increased from 400 K to 600 K, which is a 50% increase. Therefore, the volume of the helium balloon will also increase by 50%, resulting in a volume of 22.5 L. Rate this question: • 9. • A. Option 1 • B. Option 2 • C. Option 3 • D. Option 4 A. Option 1 • 10. ### Gas molecules are trapped in the space below the piston. The piston can move up and down in the cylinder. The cylinder is resting on a hot plate. If the hot plate is turned on what would happen to the volume of the gas? • A. The volume increases • B. The volume decreases • C. The volume remains the same • D. There is no significant change A. The volume increases Explanation When the hot plate is turned on, it increases the temperature of the gas molecules trapped below the piston. As a result, the gas molecules gain kinetic energy and move faster, causing them to collide with the piston more frequently and with greater force. This increased force pushes the piston upwards, leading to an expansion of the volume of the gas. Therefore, the correct answer is that the volume increases. Rate this question: • 11. ### Suppose the piston is fastened so that it could not move. When the hot plate is turned on, what would happen to the pressure of the gas? It will • A. Decrease • B. Increase • C. Stay the same • D. None of these B. Increase Explanation When the hot plate is turned on, the temperature of the gas inside the piston will increase. According to the ideal gas law, an increase in temperature leads to an increase in pressure, assuming the volume and the number of gas molecules remain constant. Therefore, the pressure of the gas will increase. Rate this question: • 12. ### What will happen to the gas pressure as the temperature increases if the amount and volume of gas were kept constant? • A. The pressure increases • B. The pressure decreases • C. There is no significant change • D. The gas pressure remains the same A. The pressure increases Explanation As the temperature increases, the kinetic energy of gas molecules also increases. This causes the molecules to move faster and collide with the walls of the container more frequently and with greater force. Consequently, the pressure exerted by the gas on the walls of the container increases. Therefore, the correct answer is that the pressure increases. Rate this question: • 13. ### What can be inferred regarding the relationship between pressure and volume as shown in the Line graph below? • A. Volume is independent of pressure. • B. Volume is directly proportional to the pressure. • C. â€‹â€‹â€‹â€‹â€‹â€‹Volume is inversely proportional to the pressure. • D. â€‹â€‹â€‹â€‹â€‹â€‹Volume is not affected by a change in pressure C. â€‹â€‹â€‹â€‹â€‹â€‹Volume is inversely proportional to the pressure. Explanation The line graph shows that as pressure increases, volume decreases and vice versa. This indicates an inverse relationship between pressure and volume. As one variable increases, the other variable decreases, suggesting that volume is inversely proportional to pressure. Rate this question: • 14. ### What will happen to the volume of the gas sample as temperature increases from 100 K to 150K? • A. Volume of the gas increases • B. Volume of the gas remains the same • C. Volume of the gas decreases • D. Volume of the gas cannot be determined A. Volume of the gas increases Explanation As temperature increases, the kinetic energy of gas molecules also increases. This causes the gas molecules to move faster and collide more frequently with the walls of the container. As a result, the pressure exerted by the gas on the walls of the container increases, and in order to maintain the same pressure, the volume of the gas must increase. Therefore, as temperature increases from 100 K to 150 K, the volume of the gas sample will increase. Rate this question: • 15. ### What is the volume of the gas sample when the temperature is 150 K? • A. 1 L • B. 2 L • C. 3 L • D. 4 L C. 3 L Explanation The volume of a gas sample is directly proportional to its temperature, according to the ideal gas law. As the temperature increases, the volume of the gas also increases. Therefore, when the temperature is 150 K, the volume of the gas sample would be 3 L. Rate this question: • 16. ### A gas sample occupies a volume of 100 mL at a pressure of 3.0 atm at 20.00C. What is the new pressure if the volume is increased to 150 mL at constant temperature? • A. 5.0 atm • B. 4.5 atm • C. 2.0 atm • D. 1.5 atm C. 2.0 atm Explanation When the volume of a gas sample is increased at constant temperature, according to Boyle's Law, the pressure of the gas decreases. Boyle's Law states that the product of the initial pressure and initial volume is equal to the product of the final pressure and final volume. In this case, the initial pressure is 3.0 atm and the initial volume is 100 mL. The final volume is 150 mL. By rearranging the equation, we can find the final pressure. (3.0 atm)(100 mL) = (final pressure)(150 mL). Solving for the final pressure gives us 2.0 atm. Therefore, the new pressure is 2.0 atm. Rate this question: • 17. ### All the following steel containers have the same size. Which of the following containers has the highest pressure? • A. 1 mole • B. 2 mole • C. 3 mole • D. 4 mole D. 4 mole Explanation The pressure of a gas is directly proportional to the number of moles present. Therefore, the container with 4 moles of gas will have the highest pressure compared to the containers with fewer moles of gas. Rate this question: • 18. ### How can you possibly prove that gases have negligible mass? • A. • B. Ask two persons to hold a box filled with air • C. Feel the weight of the samples on both hands • D. Put a balloon on a balance which can measure up the nearest hundredths of a gram before and after you fill it with air D. Put a balloon on a balance which can measure up the nearest hundredths of a gram before and after you fill it with air Explanation By putting a balloon on a balance before and after filling it with air, you can measure the change in weight. Since gases have negligible mass, the weight of the balloon should not significantly increase after filling it with air. This can be proven by observing that the weight measured on the balance remains relatively unchanged, supporting the claim that gases have negligible mass. Rate this question: • 19. ### A gas sample occupies a volume of 125 mL at a pressure of 720 torr at 20.00C. What is the new pressure if the volume is increased to 250 mL at the same temperature? • A. 1440 torr • B. 720 torrr • C. 360 torr • D. 375 torr C. 360 torr Explanation The question is asking for the new pressure when the volume is increased to 250 mL at the same temperature. According to Boyle's Law, at constant temperature, the pressure and volume of a gas are inversely proportional. Therefore, if the volume is doubled, the pressure will be halved. Since the initial pressure is 720 torr, the new pressure will be 720 torr divided by 2, which is 360 torr. Rate this question: • 20. ### Which of the following is a disaccharide? • A. Fructose • B. Galactose • C. Glucose • D. Sucrose D. Sucrose Explanation Sucrose is a disaccharide because it is composed of two monosaccharides, glucose and fructose, joined together by a glycosidic bond. It is commonly found in plants and is the main component of table sugar. Fructose, galactose, and glucose are all monosaccharides, meaning they consist of a single sugar unit. Rate this question: • 21. ### Which of the following pairs of monosaccharide form lactose sugar? • A. Glucose + Fructose • B. Glucose + Galactose • C. Glucose + Glucose • D. Fructose + Galactose B. Glucose + Galactose Explanation Lactose sugar is formed by the combination of glucose and galactose. Rate this question: • 22. ### The molecular formula for a monosaccharide is C6H12O6. Which of the following is an isomer of glucose? • A. Fructose • B. Lactose • C. Maltose • D. Sucrose A. Fructose Explanation Fructose is an isomer of glucose because both molecules have the same molecular formula (C6H12O6) but differ in their structural arrangement. Isomers have the same chemical formula but different structural arrangements, resulting in different chemical properties. In the case of fructose and glucose, the arrangement of atoms is slightly different, leading to distinct chemical properties and physiological effects. Rate this question: • 23. ### Which of the following is a major source of Carbohydrates? • A. Apple • B. Milk • C. Potato • D. Vegetables C. Potato Explanation Potato is a major source of carbohydrates because it is a starchy vegetable that contains a high amount of complex carbohydrates. Carbohydrates are the body's primary source of energy, and potatoes provide a significant amount of this energy due to their high carbohydrate content. Additionally, potatoes are also a good source of fiber, vitamins, and minerals, making them a nutritious choice for obtaining carbohydrates in the diet. Rate this question: • 24. ### The end product of proteins digested in our body is amino acids. Which of the following is the function of amino acids in the body? • A. Make fatty acids • B. Make skin, hair, enzymes and muscles • C. Make vitamins • D. Produce quick energy B. Make skin, hair, enzymes and muscles Explanation Amino acids are the building blocks of proteins, and they play a crucial role in the body. They are used to make various important components such as skin, hair, enzymes, and muscles. Enzymes are proteins that catalyze chemical reactions in the body, while muscles are made up of proteins and are responsible for movement. Additionally, proteins are essential for the growth and repair of tissues, including the formation of skin and hair. Therefore, the function of amino acids in the body is to make skin, hair, enzymes, and muscles. Rate this question: • 25. ### When digesting a complex carbohydrate, water is added and simple sugar is produced. Which of the following describes the process? • A. Condensation • B. Dehydration • C. Hydrolysis • D. Photosynthesis C. Hydrolysis Explanation Hydrolysis is the process in which water is added to break down a complex molecule into simpler components. In the case of digesting a complex carbohydrate, water is added to break the bonds between the individual sugar molecules, resulting in the production of simple sugars. Therefore, hydrolysis accurately describes the process of digesting complex carbohydrates. Rate this question: • 26. ### Which of the following is a polysaccharide? • A. Cellulose • B. Fructose • C. Glucose • D. Lactose A. Cellulose Explanation Cellulose is a polysaccharide because it is a complex carbohydrate made up of many glucose molecules bonded together. It is the main structural component of plant cell walls and provides rigidity and strength to plants. Fructose, glucose, and lactose are all monosaccharides or simple sugars, which are not polysaccharides. Rate this question: • 27. ### The formula of the compound shown below is an example of • A. Amino acid • B. A fat • C. A mineral • D. A simple sugar A. Amino acid Explanation The compound shown in the question is likely a representation of an amino acid. Amino acids are organic compounds that contain both an amino group (-NH2) and a carboxyl group (-COOH) attached to a central carbon atom. They are the building blocks of proteins and play essential roles in various biological processes. The given compound's structure resembles that of an amino acid, suggesting that it is an example of an amino acid. Rate this question: • 28. ### Which of the following statements is/are true about proteins? I. They are polymers of amino acids. II. They are insoluble molecules that are composed of carbon, hydrogen and oxygen. III. They are molecules from aldehydes and ketones containing numerous hydroxyl groups. • A. I only • B. I and II only • C. I and III only • D. II and III C. I and III only Explanation Proteins are polymers of amino acids, which means statement I is true. However, statement II is false because proteins are soluble molecules and not insoluble. Statement III is also false because proteins are not derived from aldehydes and ketones with numerous hydroxyl groups. Rate this question: • 29. ### Malou conducted an activity to determine what types of biomolecules are present in the three unknown substances given in the table below. Which of the following statements is correct? • A. Substance O is a lipid while substance M is a simple sugar. • B. Substance N is a starch while substance O is a simple sugar. • C. Substances M and N are proteins while substance O is not. • D. Substance M is a simple sugar, while substances N and O are not. D. Substance M is a simple sugar, while substances N and O are not. Explanation The correct answer is Substance M is a simple sugar, while substances N and O are not. This answer is supported by the information given in the question, which states that Malou conducted an activity to determine what types of biomolecules are present in the three unknown substances. Therefore, based on the results of the activity, it can be concluded that substance M is a simple sugar, while substances N and O are not. Rate this question: • 30. ### The table below shows the different components of organic compounds in living organisms. Which element is found in protein & nucleic acid but not in carbohydrates, fats and oils? • A. Carbon • B. Hydrogen • C. Nitrogen • D. Oxygen C. Nitrogen Explanation Nitrogen is found in proteins and nucleic acids, but not in carbohydrates, fats, and oils. Proteins and nucleic acids are essential macromolecules found in living organisms, and nitrogen is a key component of their structure. Nitrogen is crucial for the formation of amino acids, which are the building blocks of proteins, and it is also present in the nitrogenous bases of nucleic acids, such as DNA and RNA. Carbohydrates, fats, and oils do not contain nitrogen in their chemical composition. Rate this question: • 31. ### Which disaccharide will be formed from these monosaccharides shown in the drawing below? • A. Galactose • B. Lactose • C. Maltose • D. Sucrose D. Sucrose Explanation Sucrose is the correct answer because it is formed from glucose and fructose, which are the two monosaccharides shown in the drawing. Galactose, lactose, and maltose are not formed from these monosaccharides. Rate this question: • 32. ### Which of the following evidence of chemical change explains the curdling of milk? • A. Evolution of gas • B. Formation of precipitate • C. Change in temperature • D. Production of heat and light B. Formation of precipitate Explanation The formation of a precipitate is evidence of a chemical change because it involves the creation of a solid substance from a liquid solution. In the case of milk curdling, the proteins in the milk denature and coagulate, forming clumps or lumps of a solid substance. This process is a chemical change because the molecular structure of the proteins is altered, resulting in a visible change in the milk's texture and appearance. Rate this question: • 33. ### Which of the following would most likely happen when substances undergo chemical reactions? • A. No new substances will be formed. • B. New substances are formed which are heavier than that of the reactants. • C. New substances are formed with the same set of properties as that of the reactants • D. New substances are formed with the different set of properties as that of the reactants D. New substances are formed with the different set of properties as that of the reactants Explanation When substances undergo chemical reactions, new substances are formed with different properties than the reactants. This is because chemical reactions involve the rearrangement of atoms and the formation of new bonds, resulting in the creation of entirely new substances with distinct chemical and physical properties. Rate this question: • 34. ### Which of the following laws conforms a balanced chemical reaction? • A. Law of relativity • B. Law of conservation of mass • C. Law of inertia • D. Law of definite proportion B. Law of conservation of mass Explanation The law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction, it can only be rearranged. This means that the total mass of the reactants must be equal to the total mass of the products in a balanced chemical equation. Therefore, the law of conservation of mass is the law that conforms a balanced chemical reaction. Rate this question: • 35. ### Given the reaction: Potassium chlorate (KClO3) when heated produces potassium chloride (KCl) and oxygen gas (O2). Which of the following equations represents the above reaction? • A. KClO3(s) —-> KCl + O2 • B. 2KClO3(s) —> 2KCl(s) + 3O2↑ • C. 2KClO3 —> 2KCl + O2 • D. KClO3 —> KCl + 3O2 B. 2KClO3(s) —> 2KCl(s) + 3O2↑ Explanation The correct answer is 2KClO3(s) â†’ 2KCl(s) + 3O2â†‘. This equation represents the reaction because it shows that when two moles of potassium chlorate (KClO3) are heated, they produce two moles of potassium chloride (KCl) and three moles of oxygen gas (O2) are released. The coefficients in the equation represent the balanced stoichiometry of the reaction, ensuring that the number of atoms of each element is the same on both sides of the equation. Rate this question: • 36. ### Given the reaction: Potassium chlorate (KClO3) when heated produces potassium chloride (KCl) and oxygen gas (O2). Potassium chlorate (KClO3) had undergone what type of reaction? • A. Combination • B. Decomposition • C. Double displacement • D. Single replacement B. Decomposition Explanation The given reaction involves the breakdown of a compound, potassium chlorate, into two simpler substances, potassium chloride and oxygen gas. This type of reaction is known as decomposition. Rate this question: • 37. ### Which metal is most reactive? • A. Copper • B. Rubidium • C. Strontium • D. B. Rubidium Explanation Rubidium is the most reactive metal among the given options. It belongs to the alkali metal group, which is known for its high reactivity. Rubidium readily reacts with water and oxygen, and it can even ignite spontaneously in air. This high reactivity is due to its low ionization energy and the presence of only one valence electron, which makes it highly susceptible to losing that electron and forming positive ions. Copper, strontium, and vanadium are less reactive compared to rubidium. Copper is a transition metal, while strontium and vanadium belong to the alkaline earth metal and transition metal groups respectively. Rate this question: • 38. ### What would be the balanced equation for the combustion of butane, a lighter fluid? • A. C4H10 + O2 —> 4CO2 + 5H2O • B. 2C4H10 + 13O2 —> 8CO2 + 10H2O • C. C4H10 + O2 —> CO2 + 5H2O • D. C4H10 + O2 —> 4CO2 + H2O B. 2C4H10 + 13O2 —> 8CO2 + 10H2O Explanation The balanced equation for the combustion of butane is 2C4H10 + 13O2 -> 8CO2 + 10H2O. This equation is balanced because there are equal numbers of each type of atom on both sides of the equation. The coefficient in front of each molecule represents the number of molecules needed to balance the equation. In this case, 2 molecules of butane react with 13 molecules of oxygen to produce 8 molecules of carbon dioxide and 10 molecules of water. Rate this question: • 39. ### Which of the following is the correct balanced chemical equation for the combustion of natural gas (Methane)? • A. CH4 + 2O2 —> CO2 + 2H2O • B. 2CH4 + O2 —> 2CO2 + 4H2O • C. 2CH4 + 2O2 —> 2CO2 + 2H2O • D. 2CH4 + 4O2 —> 2CO2 + H2O A. CH4 + 2O2 —> CO2 + 2H2O Explanation The correct balanced chemical equation for the combustion of natural gas (Methane) is CH4 + 2O2 -> CO2 + 2H2O. In this equation, one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water. This equation follows the law of conservation of mass, as the number of atoms of each element is the same on both sides of the equation. Rate this question: • 40. ### NaCl + AgNO3 --> AgCl +. NaNO3 is an example of what type of chemical reaction? • A. Combination • B. Decomposition • C. Double displacement • D. Single replacement C. Double displacement Explanation The reaction between NaCl and AgNO3 forms AgCl and NaNO3, which indicates that the cations and anions of the reactants have exchanged places. This type of reaction is known as a double displacement reaction, where the positive and negative ions of two compounds switch partners. Rate this question: • 41. ### Which of the following would most likely happen when sodium hydroxide (NaOH) solution and copper sulfate (CuSO4) are combined? • A. Gas will evolve • B. Precipitate will form • C. Temperature will change • D. No reaction B. Precipitate will form Explanation When sodium hydroxide (NaOH) solution and copper sulfate (CuSO4) are combined, a precipitate will form. This is because when NaOH reacts with CuSO4, a double displacement reaction occurs, resulting in the formation of solid copper hydroxide (Cu(OH)2). The Cu(OH)2 precipitate is insoluble and will separate from the solution, causing a visible change in the form of a solid precipitate. Rate this question: • 42. ### Sulfuric acid (H2SO4) is produced from sulfur (S), oxygen (O), and water (H2O). Which of the following is the correct balanced chemical equation for the production of sulfuric acid? • A. SO3 +. H2O —> H2SO4 • B. 2SO3 +. 3H2O —> 3H2SO4 • C. 3SO3 +. H2O —> 3H2SO4 • D. 2SO3 +. H2O —> 2 H2SO4 A. SO3 +. H2O —> H2SO4 Explanation The correct balanced chemical equation for the production of sulfuric acid is SO3 + H2O -> H2SO4. This equation shows that one molecule of sulfur trioxide (SO3) reacts with one molecule of water (H2O) to produce one molecule of sulfuric acid (H2SO4). Rate this question: • 43. ### Why does the rate of reaction increase when iron filings are used instead of iron nail in a reaction with hydrochloric acid? • A. Iron filings act as a catalyst. • B. Iron filings have higher density. • C. Iron filings have higher surface area. • D. Iron filings can increase the temperature of the reaction. C. Iron filings have higher surface area. Explanation Iron filings have a higher surface area compared to an iron nail. This means that there is more exposed iron surface available for the reaction to occur. The increased surface area allows for more collisions between the iron and the hydrochloric acid, leading to a higher rate of reaction. Rate this question: • 44. ### Five grams of iron fillings were placed in beakers A, B, C, and D. Which beaker will have the fastest rate of reaction? Beaker …… • A. A • B. B • C. C • D. D D. D Explanation Beaker D will have the fastest rate of reaction because the surface area of the iron fillings in Beaker D is the largest compared to the other beakers. The rate of reaction is directly proportional to the surface area of the reactant, so the larger the surface area, the faster the reaction. Therefore, Beaker D, with the largest surface area of iron fillings, will have the fastest rate of reaction. Rate this question: • 45. ### Which factor affects the rate of the reaction below? • A. Surface area of the reactants • B. Concentration of the reactants • C. Use of catalyst • D. Nature of the reactants B. Concentration of the reactants Explanation The concentration of the reactants affects the rate of the reaction. When the concentration of the reactants is increased, there are more reactant particles available to collide with each other, leading to an increase in the frequency of successful collisions. This results in a higher reaction rate. Conversely, when the concentration is decreased, there are fewer reactant particles available for collisions, leading to a lower reaction rate. Therefore, the concentration of the reactants directly influences the rate of the reaction. Rate this question: • 46. ### Milk becomes sour more quickly on a hot summer’s day. Which factor affects the spoilage of milk? • A. Catalyst • B. Concentration • C. Surface area • D. Temperature D. Temperature Explanation The spoilage of milk is affected by temperature. In hot weather, the temperature is higher, which creates a favorable environment for the growth of bacteria and other microorganisms that cause milk to spoil. The higher temperature speeds up the chemical reactions that lead to the breakdown of milk proteins and the production of acids, causing the milk to sour more quickly. Rate this question: • 47. ### Refer to the following changes which might be made from the reaction between dilute hydrochloric acid (HCl) and marble chips (CaCO3). I. Increasing the temperature II. Using larger marble chips III. Using smaller marble chips IV. Decreasing the concentration of acid Which change(s) would increase the rate of the reaction? • A. I only • B. I and II • C. I and III • D. I and IV C. I and III Explanation Increasing the temperature would increase the rate of the reaction because higher temperatures provide more energy to the particles, causing them to move faster and collide more frequently, leading to more successful collisions and an increased rate of reaction. Using smaller marble chips would also increase the rate of the reaction because smaller particles have a larger surface area, allowing for more frequent collisions with the acid particles and a faster reaction. Rate this question: • 48. ### Refer to the following changes which might be made from the reaction between dilute hydrochloric acid (HCl) and marble chips (CaCO3). I. Increasing the temperature II. Using larger marble chips III. Using smaller marble chips IV. Decreasing the concentration of acid Which change(s) would decrease the rate of the reaction? • A. II only • B. I and III • C. II and III • D. II and IV D. II and IV Explanation Using larger marble chips would decrease the rate of the reaction because larger surface area would result in slower reaction as there would be less contact between the acid and the marble chips. Decreasing the concentration of acid would also decrease the rate of the reaction because there would be fewer acid particles available to react with the marble chips. Rate this question: • 49. ### Ammonia is used to make fertilizers. It is made from nitrogen and hydrogen. The graph below shows the percentage of ammonia that can be made from the same mixture of nitrogen and hydrogen at different temperatures and pressures. Which of the following conclusions is correct based on the information in the graph? • A. The percentage of ammonia increases as both the pressure and temperature decrease. • B. The percentage of ammonia increases as both the pressure and temperature increase. • C. The percentage of ammonia increases as the pressure increases and temperature decreases. • D. The percentage of ammonia increases as the pressure decreases and temperature increases. B. The percentage of ammonia increases as both the pressure and temperature increase. Explanation Based on the information in the graph, as both the pressure and temperature increase, the percentage of ammonia that can be made from the same mixture of nitrogen and hydrogen also increases. This can be observed by following the trend lines in the graph, which show an upward slope as pressure and temperature increase. Therefore, the correct conclusion is that the percentage of ammonia increases as both the pressure and temperature increase. Rate this question: • 50. ### Ammonia is used to make fertilizers. It is made from nitrogen and hydrogen. The graph below shows the percentage of ammonia that can be made from the same mixture of nitrogen and hydrogen at different temperatures and pressures. Which of the following pressure and temperature conditions could increase the yield of ammonia? • A. Low temperature and low pressure • B. High temperature and low pressure • C. Low temperature and high pressure • D. Both high temperature and pressure D. Both high temperature and pressure Explanation Increasing both temperature and pressure can increase the yield of ammonia because the reaction between nitrogen and hydrogen to form ammonia is exothermic. According to Le Chatelier's principle, increasing the temperature will shift the equilibrium towards the side of the reaction that absorbs heat, which in this case is the forward reaction to form ammonia. Increasing the pressure will also shift the equilibrium towards the side with fewer moles of gas, which is again the forward reaction. Therefore, both high temperature and pressure conditions can increase the yield of ammonia. Rate this question: Related Topics	8,193	34,784	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2023-50	latest	en	0.922308
https://blog.csdn.net/qq_16234613/article/details/79818370	1,579,563,911,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579250601040.47/warc/CC-MAIN-20200120224950-20200121013950-00469.warc.gz	358,431,186	43,952	# 简单易懂Pytorch实战实例VGG深度网络 '''VGG11/13/16/19 in Pytorch.''' import torch import torch.nn as nn cfg = { 'VGG11': [64, 'M', 128, 'M', 256, 256, 'M', 512, 512, 'M', 512, 512, 'M'], 'VGG13': [64, 64, 'M', 128, 128, 'M', 256, 256, 'M', 512, 512, 'M', 512, 512, 'M'], 'VGG16': [64, 64, 'M', 128, 128, 'M', 256, 256, 256, 'M', 512, 512, 512, 'M', 512, 512, 512, 'M'], 'VGG19': [64, 64, 'M', 128, 128, 'M', 256, 256, 256, 256, 'M', 512, 512, 512, 512, 'M', 512, 512, 512, 512, 'M'], } # 模型需继承nn.Module class VGG(nn.Module): # 初始化参数： def __init__(self, vgg_name): super(VGG, self).__init__() self.features = self._make_layers(cfg[vgg_name]) self.classifier = nn.Linear(512, 10) # 模型计算时的前向过程，也就是按照这个过程进行计算 def forward(self, x): out = self.features(x) out = out.view(out.size(0), -1) out = self.classifier(out) return out def _make_layers(self, cfg): layers = [] in_channels = 3 for x in cfg: if x == 'M': layers += [nn.MaxPool2d(kernel_size=2, stride=2)] else: layers += [nn.Conv2d(in_channels, x, kernel_size=3, padding=1), nn.BatchNorm2d(x), nn.ReLU(inplace=True)] in_channels = x layers += [nn.AvgPool2d(kernel_size=1, stride=1)] return nn.Sequential(layers) # net = VGG('VGG11') # x = torch.randn(2,3,32,32) # print(net(Variable(x)).size()) '''Train CIFAR10 with PyTorch.''' from __future__ import print_function import torch import torch.nn as nn import torch.optim as optim import torch.nn.functional as F import torch.backends.cudnn as cudnn import torchvision import torchvision.transforms as transforms import os import argparse from models import from utils import progress_bar # 获取参数 parser = argparse.ArgumentParser(description='PyTorch CIFAR10 Training') parser.add_argument('--resume', '-r', action='store_true', help='resume from checkpoint') args = parser.parse_args() use_cuda = torch.cuda.is_available() best_acc = 0 # best test accuracy start_epoch = 0 # start from epoch 0 or last checkpoint epoch # 获取数据集，并先进行预处理 print('==> Preparing data..') # 图像预处理和增强 transform_train = transforms.Compose([ transforms.RandomHorizontalFlip(), transforms.ToTensor(), transforms.Normalize((0.4914, 0.4822, 0.4465), (0.2023, 0.1994, 0.2010)), ]) transform_test = transforms.Compose([ transforms.ToTensor(), transforms.Normalize((0.4914, 0.4822, 0.4465), (0.2023, 0.1994, 0.2010)), ]) classes = ('plane', 'car', 'bird', 'cat', 'deer', 'dog', 'frog', 'horse', 'ship', 'truck') # 继续训练模型或新建一个模型 if args.resume: print('==> Resuming from checkpoint..') assert os.path.isdir('checkpoint'), 'Error: no checkpoint directory found!' net = checkpoint['net'] best_acc = checkpoint['acc'] start_epoch = checkpoint['epoch'] else: print('==> Building model..') net = VGG('VGG16') # net = ResNet18() # net = PreActResNet18() # net = DenseNet121() # net = ResNeXt29_2x64d() # net = MobileNet() # net = MobileNetV2() # net = DPN92() # net = ShuffleNetG2() # net = SENet18() # 如果GPU可用，使用GPU if use_cuda: # move param and buffer to GPU net.cuda() # parallel use GPU net = torch.nn.DataParallel(net, device_ids=range(torch.cuda.device_count()-1)) # speed up slightly cudnn.benchmark = True # 定义度量和优化 criterion = nn.CrossEntropyLoss() optimizer = optim.SGD(net.parameters(), lr=args.lr, momentum=0.9, weight_decay=5e-4) # 训练阶段 def train(epoch): print('\nEpoch: %d' % epoch) # switch to train mode net.train() train_loss = 0 correct = 0 total = 0 # batch 数据 for batch_idx, (inputs, targets) in enumerate(trainloader): # 将数据移到GPU上 if use_cuda: inputs, targets = inputs.cuda(), targets.cuda() # 先将optimizer梯度先置为0 # Variable表示该变量属于计算图的一部分，此处是图计算的开始处。图的leaf variable inputs, targets = Variable(inputs), Variable(targets) # 模型输出 outputs = net(inputs) # 计算loss，图的终点处 loss = criterion(outputs, targets) # 反向传播，计算梯度 loss.backward() # 更新参数 optimizer.step() # 注意如果你想统计loss，切勿直接使用loss相加，而是使用loss.data[0]。因为loss是计算图的一部分，如果你直接加loss，代表total loss同样属于模型一部分，那么图就越来越大 train_loss += loss.data[0] # 数据统计 _, predicted = torch.max(outputs.data, 1) total += targets.size(0) correct += predicted.eq(targets.data).cpu().sum() progress_bar(batch_idx, len(trainloader), 'Loss: %.3f \| Acc: %.3f%% (%d/%d)' % (train_loss/(batch_idx+1), 100.correct/total, correct, total)) # 测试阶段 def test(epoch): global best_acc # 先切到测试模型 net.eval() test_loss = 0 correct = 0 total = 0 for batch_idx, (inputs, targets) in enumerate(testloader): if use_cuda: inputs, targets = inputs.cuda(), targets.cuda() inputs, targets = Variable(inputs, volatile=True), Variable(targets) outputs = net(inputs) loss = criterion(outputs, targets) # loss is variable , if add it(+=loss) directly, there will be a bigger ang bigger graph. test_loss += loss.data[0] _, predicted = torch.max(outputs.data, 1) total += targets.size(0) correct += predicted.eq(targets.data).cpu().sum() progress_bar(batch_idx, len(testloader), 'Loss: %.3f \| Acc: %.3f%% (%d/%d)' % (test_loss/(batch_idx+1), 100.correct/total, correct, total)) # Save checkpoint. # 保存模型 acc = 100.correct/total if acc > best_acc: print('Saving..') state = { 'net': net.module if use_cuda else net, 'acc': acc, 'epoch': epoch, } if not os.path.isdir('checkpoint'): os.mkdir('checkpoint') torch.save(state, './checkpoint/ckpt.t7') best_acc = acc # 运行模型 for epoch in range(start_epoch, start_epoch+200): train(epoch) test(epoch) # 清除部分无用变量 torch.cuda.empty_cache() 新模型： python main.py --lr=0.01 python main.py --resume --lr=0.01 '''Some helper functions for PyTorch, including: - get_mean_and_std: calculate the mean and std value of dataset. - msr_init: net parameter initialization. - progress_bar: progress bar mimic xlua.progress. ''' import os import sys import time import math import torch.nn as nn import torch.nn.init as init def get_mean_and_std(dataset): '''Compute the mean and std value of dataset.''' mean = torch.zeros(3) std = torch.zeros(3) print('==> Computing mean and std..') for i in range(3): mean[i] += inputs[:,i,:,:].mean() std[i] += inputs[:,i,:,:].std() mean.div_(len(dataset)) std.div_(len(dataset)) return mean, std def init_params(net): '''Init layer parameters.''' for m in net.modules(): if isinstance(m, nn.Conv2d): init.kaiming_normal(m.weight, mode='fan_out') if m.bias: init.constant(m.bias, 0) elif isinstance(m, nn.BatchNorm2d): init.constant(m.weight, 1) init.constant(m.bias, 0) elif isinstance(m, nn.Linear): init.normal(m.weight, std=1e-3) if m.bias: init.constant(m.bias, 0) _, term_width = os.popen('stty size', 'r').read().split() term_width = int(term_width) TOTAL_BAR_LENGTH = 65. last_time = time.time() begin_time = last_time def progress_bar(current, total, msg=None): global last_time, begin_time if current == 0: begin_time = time.time() # Reset for new bar. cur_len = int(TOTAL_BAR_LENGTHcurrent/total) rest_len = int(TOTAL_BAR_LENGTH - cur_len) - 1 sys.stdout.write(' [') for i in range(cur_len): sys.stdout.write('=') sys.stdout.write('>') for i in range(rest_len): sys.stdout.write('.') sys.stdout.write(']') cur_time = time.time() step_time = cur_time - last_time last_time = cur_time tot_time = cur_time - begin_time L = [] L.append(' Step: %s' % format_time(step_time)) L.append(' \| Tot: %s' % format_time(tot_time)) if msg: L.append(' \| ' + msg) msg = ''.join(L) sys.stdout.write(msg) for i in range(term_width-int(TOTAL_BAR_LENGTH)-len(msg)-3): sys.stdout.write(' ') # Go back to the center of the bar. for i in range(term_width-int(TOTAL_BAR_LENGTH/2)+2): sys.stdout.write('\b') sys.stdout.write(' %d/%d ' % (current+1, total)) if current < total-1: sys.stdout.write('\r') else: sys.stdout.write('\n') sys.stdout.flush() def format_time(seconds): days = int(seconds / 3600/24) seconds = seconds - days360024 hours = int(seconds / 3600) seconds = seconds - hours3600 minutes = int(seconds / 60) seconds = seconds - minutes60 secondsf = int(seconds) seconds = seconds - secondsf millis = int(seconds*1000) f = '' i = 1 if days > 0: f += str(days) + 'D' i += 1 if hours > 0 and i <= 2: f += str(hours) + 'h' i += 1 if minutes > 0 and i <= 2: f += str(minutes) + 'm' i += 1 if secondsf > 0 and i <= 2: f += str(secondsf) + 's' i += 1 if millis > 0 and i <= 2: f += str(millis) + 'ms' i += 1 if f == '': f = '0ms' return f ©️2019 CSDN 皮肤主题: 大白设计师: CSDN官方博客	2,615	8,138	{"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}	2.9375	3	CC-MAIN-2020-05	latest	en	0.331672
http://www.hindawi.com/journals/ijct/2010/851857/	1,369,415,997,000,000,000	application/xhtml+xml	crawl-data/CC-MAIN-2013-20/segments/1368704818711/warc/CC-MAIN-20130516114658-00097-ip-10-60-113-184.ec2.internal.warc.gz	501,186,523	66,991	`International Journal of CombinatoricsVolume 2010 (2010), Article ID 851857, 21 pagesdoi:10.1155/2010/851857` Research Article ## Classification of Base Sequences Department of Pure Mathematics, University of Waterloo, Waterloo, ON, N2L 3G1, Canada Received 6 February 2010; Revised 8 April 2010; Accepted 12 April 2010 Copyright © 2010 Dragomir Ž. Ðoković. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. #### Abstract Base sequences BS are quadruples of -sequences , with A and B of length and C and D of length n, such that the sum of their nonperiodic autocor-relation functions is a -function. The base sequence conjecture, asserting that BS exist for all n, is stronger than the famous Hadamard matrix conjecture. We introduce a new definition of equivalence for base sequences BS and construct a canonical form. By using this canonical form, we have enumerated the equivalence classes of BS for . As the number of equivalence classes grows rapidly (but not monotonically) with n, the tables in the paper cover only the cases . #### 1. Introduction Base sequences are quadruples of binary sequences, with and of length and and of length , such that the sum of their nonperiodic autocorrelation functions is a -function. In this paper we take . Sporadic examples of base sequences have been constructed by many authors during the last 30 years; see, for example, [14] and the survey paper [5] and its references. A more systematic approach has been taken by the author in [6, 7]. The are presently known to exist for all (ibid) and for Golay numbers , where and are arbitrary nonnegative integers. However the genuine classification of is still lacking. Due to the important role that these sequences play in various combinatorial constructions such as that for -sequences, orthogonal designs, and Hadamard matrices [1, 5, 8], it is of interest to classify the base sequences of small length. Our main goal is to provide such classification for . In Section 2 we recall the basic properties of base sequences of . We also recall the quad decomposition and our encoding scheme for this particular type of base sequences. In Section 3 we enlarge the collection of standard elementary transformations of by introducing a new one. Thus we obtain new notion of equivalence and equivalence classes. Throughout the paper, the words “equivalence" and “equivalence class" are used in this new sense. We also introduce the canonical form for base sequences. By using it, we are able to compute the representatives of the equivalence classes. In Section 4 we introduce an abstract group, , of order which acts naturally on all sets . Its definition depends on the parity of . The orbits of this group are just the equivalence classes of . In Section 5 we tabulate some of the results of our computations (those for ) giving the list of representatives of the equivalence classes of . The representatives are written in the encoded form which is explained in the next section. For we also include the values of the nonperiodic autocorrelation functions of the four constituent sequences. We also raise the question of characterizing the binary sequences having the same nonperiodic autocorrelation function. A class of examples is constructed, showing that the question is interesting. The column “Equ.” in Table 1 gives the number of equivalence classes in for . The column “Nor.” gives the number of normal equivalence classes (see Section 5 for their definition). Table 1: Number of equivalence classes of BS #### 2. Quad Decomposition and the Encoding Scheme We denote finite sequences of integers by capital letters. If, say, is such a sequence of length then we denote its elements by the corresponding lower case letters. Thus To this sequence we associate the polynomial viewed as an element of the Laurent polynomial ring . (As usual, denotes the ring of integers.) The nonperiodic autocorrelation function of is defined by where for and for . Note that for all and for . The norm of is the Laurent polynomial . We have The negation, , of is the sequence The reversed sequence and the alternated sequence of the sequence are defined by Observe that and for all . By we denote the concatenation of the sequences and . A binary sequence is a sequence whose terms belong to . When displaying such sequences, we will often write for and for . The base sequences consist of four binary sequences , with and of length and and of length , such that Thus, for , we have We denote by the set of such base sequences with and fixed. From now on we will consider only the case . Let . For convenience we fix the following notation. For even (odd) we set (). We decompose the pair into quads and, if is even, the central column Similar decomposition is valid for the pair . Recall the following basic and well-known property [3, Theorem ]. Theorem 2.1. For , the sum of the four quad entries is for the first quad of the pair and is for all other quads of and also for all quads of the pair . Thus there are 8 possibilities for the first quad of the pair : These eight quads occur in the study of Golay sequences (see, e.g., [9]), and we refer to them as the Golay quads. There are also 8 possibilities for each of the remaining quads of and all quads of : We will refer to these eight quads as the BS-quads. We say that a BS-quad is symmetric if its two columns are the same, and otherwise we say that it is skew. The quads and 8 are symmetric and and 6 are skew. We say that two quads have the same symmetry type if they are both symmetric or both skew. There are 4 possibilities for the central column: We encode the pair by the symbol sequence where is the label of the th quad except in the case where is even and in which case is the label of the central column. Similarly, we encode the pair by the symbol sequence when is even respectively odd. Here is the label of the th quad for and is the label of the central column (when is odd). #### 3. The Equivalence Relation We start by defining five types of elementary transformations of base sequences . These elementary transformations include the standard ones, as described in [3, 6]. But we also introduce one additional elementary transformation, see item (T4), which made its first appearance in [10] in the context of near-normal sequences. The quad notation was instrumental in the discovery of this new elementary operation. The elementary transformations of are the following.T Negate one of the sequences .T Reverse one of the sequences .(T3) Interchange the sequences or .(T4) Replace the pair with the pair which is defined as follows: if (2.14) is the encoding of , then the encoding of is or depending on whether is even or odd, where is the transposition (45). In other words, the encoding of is obtained from that of by replacing each quad symbol 4 with the symbol 5, and vice versa. (We verify below that .) (T5) Alternate all four sequences . In order to justify (T4) one has to verify that . For that purpose let us fix two quads and and consider their contribution to . We claim that is equal to the contribution of and to . If neither nor belongs to , then and and so . If then is the negation of . Hence if then again . Otherwise, say while , and it is easy to verify that . The pairs and also make a contribution to , but they can be treated in the same manner. Finally, if is odd then the pair also has a central column with label . In that case, if and , the contribution of and to is 0. This completes the verification. We say that two members of are equivalent if one can be transformed to the other by applying a finite sequence of elementary transformations. One can enumerate the equivalence classes by finding suitable representatives of the classes. For that purpose we introduce the canonical form. Definition 3.1. Let and let (2.13) respectively (2.14) be the encoding of the pair respectively . One says that is in the canonical form if the following eleven conditions hold.(i), for even and for odd.(ii)The first symmetric quad (if any) of is 1 or 8.(iii)If is even and for then .(iv)The first skew quad (if any) of is 3 or 6.(v) for even and for odd.(vi)The first symmetric quad (if any) of is 1.(vii)The first skew quad (if any) of is 6.(viii)If is the least index such that then .(ix)If is the least index such that then .(x)If is odd and , for all , then .(xi)If is odd and , for all , then . We can now prove that each equivalence class has a member which is in the canonical form. The uniqueness of this member will be proved in the next section. Proposition 3.2. Each equivalence class has at least one member having the canonical form. Proof. Let be arbitrary and let (2.13) respectively (2.14) be the encoding of respectively . By applying the first three types of elementary transformations and by Theorem 2.1 we can assume that and . By Theorem 2.1, . If is even and we apply the elementary transformation (T5). Thus we may assume that and that condition (v) for the canonical form is satisfied. To satisfy conditions (ii) and (iii), replace with (if necessary). To satisfy condition (iv), replace with (if necessary). We now modify in order to satisfy the second part of condition (i). Note that is a BS-quad by Theorem 2.1. If there is nothing to do. Assume that the quad is symmetric. By (ii), . From (2.8) for , we deduce that if and if . Note that if is even, then by (v). If is odd and , we switch and and apply the elementary transformation (T5). After this change we still have , , conditions (ii), (iii), and (iv) remain satisfied, and moreover . Now assume that is skew. In view of (iv), we may assume that . Then the argument above based on (2.8) shows that , and so must be odd. After applying the elementary transformation (T5), we obtain that . Hence condition (i) is fully satisfied. To satisfy (vi), in view of (v) we may assume that is odd and . If the first symmetric quad in is 2 respectively 7, we reverse and negate respectively . If it is 8, we reverse and negate both and . Now the first symmetric quad will be 1. To satisfy (vii), (if necessary) reverse , , or both. To satisfy (viii), (if necessary) interchange and . Note that in this process we do not violate the previously established properties. To satisfy (ix), (if necessary) apply the elementary transformation (T4). To satisfy (x), switch and (if necessary). To satisfy (xi), (if necessary) replace with , with , or both. Hence is now in the canonical form. #### 4. The Symmetry Group of We will construct a group of order which acts on . Our (redundant) generating set for will consist of 12 involutions. Each of these generators is an elementary transformation, and we use this information to construct , that is, to impose the defining relations. We denote by an aritrary member of . To construct , we start with an elementary abelian group of order with generators , . It acts on as follows: that is, negates the th sequence of and reverses it. Next we introduce two commuting involutory generators and . We declare that commutes with and , and commutes with and and that The group is the direct product of two isomorphic groups of order 32: The action of on extends to by defining We add a new generator which commutes elementwise with , commutes with and , and satisfies . Let us denote this enlarged group by . It has the direct product decomposition where the second factor is itself direct product of two copies of the dihedral group of order 8: The action of on extends to by letting act as the elementary transformation (T4). Finally, we define as the semidirect product of and the group of order 2 with generator . By definition, commutes with each and satisfies The action of on extends to by letting act as the elementary transformation (T5), that is, we have We point out that the definition of the subgroup is independent of and its action on has a quadwise character. By this we mean that the value of a particular quad, say , of and determine uniquely the quad of . In other words acts on the Golay quads, the BS-quads, and the set of central columns such that the encoding of is given by the symbol sequences On the other hand the full group has neither of these two properties. An important feature of the action of on the BS-quads is that it preserves the symmetry type of the quads. The following proposition follows immediately from the construction of and the description of its action on . Proposition 4.1. The orbits of in are the same as the equivalence classes. The main tool that we use to enumerate the equivalence classes of is the following theorem. Theorem 4.2. For each equivalence class there is a unique having the canonical form. Proof. In view of Proposition 3.2, we just have to prove the uniqueness assertion. Let be in the canonical form. We have to prove that in fact . By Proposition 4.1, we have for some . We can write as where and . Let be the encoding of the pair and the encoding of the pair . The symbols (i-xi) will refer to the corresponding conditions of Definition 3.1. Observe that by (i). We prove first preliminary claims (a-c). (a). For even see (v). Let be odd. When we apply the generator to any , we do not change the first quad of . It follows that the quads and have the same symmetry type. The claim now follows from (v). (b), that is, . Assume first that is even. By (v), . For any the first quad of in and the one in have different symmetry types. As quad is symmetric, the equality forces to be 0. Assume now that is odd. Then for any the second quad of in and the one in have different symmetry types. Recall that and that (see (i)) and belong to . From (2.8), with , we deduce that for . We conclude that . The claim now follows from the fact that preserves while alters the symmetry type of the quad . As an immediate consequence of (b), we point out that a quad is symmetric if and only if is, and the same is true for the quads and . (c). This was already proved above in the case when is odd. In general, the claim follows from (b) and the equality . Observe that each of the sets and consists of one symmetric and one skew quad and that preserves the symmetry type of quads. Recall that . Since we have with and . Consequently, and for all ’s. We will now prove that and . Since , the equality implies that . Thus for some . Assume first that is symmetric. By (ii), . Then implies that . Hence, and so . If all quads , , are symmetric, then also . Otherwise let be the least index for which the quad is skew. Since and is 3 or 6 (see (iv)), we infer that . Hence and so . Now assume that is skew. By (ii), . Then implies that . Thus and so . If all quads , , are skew, then by invoking condition (iii) we deduce that and so . Otherwise let be the least index for which the quad is symmetric. Since and is 1 or 8 (see (ii)), we infer that . Hence and so . It remains to prove that and . We set . By (v) and claim (a) we have . We first consider the case which occurs only for odd. Then and so . It follows that . If some is symmetric, let be the least index such that is symmetric. Then (vi) implies that . Thus must fix the quad 1. As the stabilizer of the quad 1 in is , we infer that must also fix the quad 8. Similarly, if then (viii) implies that fixes 2 and 7. If then (ix) implies that fixes 4 and 5. These facts imply that fixes all quads in , that is, for all . It remains to show that, for odd , . If , this follows from (xi). Otherwise contains a symmetric quad and so . If then contains one of the quads 2, 4, 5, or 7. Since fixes all quads in , we infer that , and so . If , the equality follows from (x). Finally, we consider the case . Since , . Hence fixes the quads 1 and 8. If some is skew, then (vii) implies that fixes the quads 3 and 6. If then (viii) implies that fixes the quads 2 and 7. If then (ix) implies that fixes the quads 4 and 5. These facts imply that fixes all quads in . If is odd, then we invoke conditions (x) and (xi) to conclude that also fixes the central column of . Hence and also in this case. #### 5. Representatives of the Equivalence Classes We have computed a set of representatives for the equivalence classes of base sequences for all . Due to their excessive size, we tabulate these sets only for . Each representative is given in the canonical form which is made compact by using our standard encoding. The encoding is explained in detail in Section 2. As an example, the base sequences are encoded as ,. In the tables we write 0 instead of . This convention was used in our previous papers on this and related topics. This compact notation is used primarily in order to save space, but also to avoid introducing errors during decoding. For each , the representatives are listed in the lexicographic order of the symbol sequences (2.13) and (2.14). In Table 2 we list the codes for the representatives of the equivalence classes of for . This table also records the values of the nonperiodic autocorrelation functions for and . For instance let us consider the first item in the list of base sequences given in Table 2. The base sequences are encoded in the first column as 0165, 6123. The first part 0165 encodes the pair , and the second part 6123 encodes the pair . The function , at the points , takes the values and 1 listed in the second column. Just below these values one finds the values of at the same points. In the third column we list likewise the values of and at the points . Table 2: Equivalence classes of BS, . Tables 3, 4, 5, 6, and 7 contain only the list of codes for the representatives of the equivalence classes of for . Table 3: Equivalence classes of BS Table 4: Equivalence classes of BS Table 5: Equivalence classes of BS Table 6: Equivalence classes of BS Table 7: Equivalence classes of BS Let us say that the base sequences are normal respectively near-normal if respectively for all . We denote by respectively the set of all normal respectively near-normal sequences in . Let us say also that an equivalence class is normal respectively near-normal if respectively is not void. Our canonical form has been designed so that if is normal, then its canonical representative belongs to . The analogous statement for near-normal classes is false. It is not hard to recognize which representatives in our tables are normal sequences. Let (2.13) be the encoding of the pair . Then if and only if all the quads , , belong to , and, in the case when is even, the central column symbol is 0 or 3. It is an interesting question to find the necessary and sufficient conditions for two binary sequences to have the same norm. The group of order four generated by the negation and reversal operations acts on binary sequences. We say that two binary sequences are equivalent if they belong to the same orbit. Note that the equivalent binary sequences have the same norm. However the converse is false. Here is a counter-example which occurs in Table 2 for the case . The base sequences 15 and 16 differ only in their first sequences, which we denote here by and respectively: Their associated polynomials are It is obvious that and are not equivalent in the above sense. On the other hand, from the factorizations and , where , and , we deduce immediately that . This counter-example can be easily generalized. Let us define binary polynomials as polynomials associated to binary sequences. If is a polynomial of degree with , we define its dual polynomial by . Then for any positive integer we have , that is, where is the dual of the polynomial . In general we can start with any number of binary sequences, but here we take only three of them: of lengths respectively. From the associated binary polynomials , and we can form several binary polynomials of degree . The basic one is . The others are obtained from this one by replacing one or more of the three factors by their duals. It is immediate that the binary sequences corresponding to these binary polynomials all have the same norm. In general many of these sequences will not be equivalent. However, note that if we replace all three factors with their duals, we will obtain a binary sequence equivalent to the basic one. #### Acknowledgments The author has pleasure to thank the three anonymous referees for their valuable comments. The author is grateful to NSERC for the continuing support of his research. This work was made possible by the facilities of the Shared Hierarchical Academic Research Computing Network (SHARCNET: http://www.sharcnet.ca) and Compute/Calcul Canada. #### References 1. H. Kharaghani and C. Koukouvinos, “Complementary, base and Turyn sequences,” in Handbook of Combinatorial Designs, C. J. Colbourn and J. H. Dinitz, Eds., pp. 317–321, CRC Press, Boca Raton, Fla, USA, 2nd edition, 2007. 2. S. Kounias and K. Sotirakoglou, “Construction of orthogonal sequences,” in Proceedings of the 14th Greek Statistics Conference, pp. 229–236, Skiathos, Greece, 2001. 3. C. Koukouvinos, S. Kounias, and K. Sotirakoglou, “On base and Turyn sequences,” Mathematics of Computation, vol. 55, no. 192, pp. 825–837, 1990. 4. C. H. Yang, “On composition of four-symbol $\delta$-codes and Hadamard matrices,” Proceedings of the American Mathematical Society, vol. 107, no. 3, pp. 763–776, 1989. 5. J. Seberry and M. Yamada, “Hadamard matrices, sequences, and block designs,” in Contemporary Design Theory, Wiley-Intersci. Ser. Discrete Math. Optim., pp. 431–560, Wiley, New York, NY, USA, 1992. 6. D. Ž. Ðoković, “Aperiodic complementary quadruples of binary sequences,” Journal of Combinatorial Mathematics and Combinatorial Computing, vol. 27, pp. 3–31, 1998, Correction: ibid 30, p. 254, 1999. 7. D. Ž. Ðoković, “On the base sequence conjecture,” Discrete Mathematics, vol. 310, pp. 1956–1964, 2010. 8. C. Koukouvinos, S. Kounias, J. Seberry, C. H. Yang, and J. Yang, “Multiplication of sequences with zero autocorrelation,” The Australasian Journal of Combinatorics, vol. 10, pp. 5–15, 1994. 9. D. Ž. Ðoković, “Equivalence classes and representatives of Golay sequences,” Discrete Mathematics, vol. 189, no. 1–3, pp. 79–93, 1998. 10. D. Ž. Ðoković, “Classification of near-normal sequences,” Discrete Mathematics, Algorithms and Applications, vol. 1, no. 3, pp. 389–399, 2009.	5,282	22,407	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20	latest	en	0.914177
http://health-wellnesstoday.com/n79gs6/49dca3-convergence-in-probability	1,669,891,748,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710808.72/warc/CC-MAIN-20221201085558-20221201115558-00171.warc.gz	23,574,119	9,412	3. Exemple 1. We have When thinking about the convergence of random quantities, two types of convergence that are often confused with one another are convergence in probability and almost sure convergence. A new look at weak-convergence methods in metric spaces-from a master of probability theory In this new edition, Patrick Billingsley updates his classic work Convergence of Probability Measures to reflect developments of the past thirty years. convergence in probability of P n 0 X nimplies its almost sure convergence. This type of convergence is similar to pointwise convergence of a sequence of functions, except that the convergence need not occur on a set with probability 0 (hence the “almost” sure). because it is identically equal to zero for all Mathematical notation of convergence in latex. Proof. The notation X n a.s.→ X is often used for al-most sure convergence, while the common notation for convergence in probability is X n →p X or plim n→∞X = X. Convergence in distribution and convergence in the rth mean are … -convergence 1-convergence a.s. convergence convergence in probability (stochastic convergence) weak convergence (convergence in distribution/law) subsequence, A.4 subsequence, 3.3 positive bound & (DOM) rem A.5 const. Definition Index des espaces 2020-2021 par département; Index des espaces 2019-2020 par département; Index des espaces 2018-2019 par département (1) (1) lim n → ∞ P ( \| X n − X \| < ϵ) = 1. Relations among modes of convergence. \begin{align}%\label{eq:union-bound} converges in probability to the constant random We only require that the set on which X n(!) almost sure convergence). \lim_{n \rightarrow \infty} P\big(\|X_n-c\| \geq \epsilon \big)&= 0, \qquad \textrm{ for all }\epsilon>0, It is called the "weak" law because it refers to convergence in probability. (the increases. When you have a nonlinear function of a random variable g(X), when you take an expectation E[g(X)], this is not the same as g(E[X]). Convergence in probability Convergence in probability - Statlec . a sample space share \| improve this question \| follow \| asked Jan 30 '16 at 20:41. A new look at weak-convergence methods in metric spaces-from a master of probability theory In this new edition, Patrick Billingsley updates his classic work Convergence of Probability Measures to reflect developments of the past thirty years. \begin{align}%\label{eq:union-bound} One of the handiest tools in regression is the asymptotic analysis of estimators as the number of observations becomes large. Since $\lim \limits_{n \rightarrow \infty} P\big(\|X_n-c\| \geq \epsilon \big) \geq 0$, we conclude that Classical proofs of this fact involve characteristic functions. component of each random vector However, this random variable might be a constant, so it also makes sense to talk about convergence to a real number. any \lim_{n \rightarrow \infty} P\big(\|X_n-0\| \geq \epsilon \big) &=\lim_{n \rightarrow \infty} P\big(X_n \geq \epsilon \big) & (\textrm{ since $X_n\geq 0$ })\\ Example satisfyingand , is called the probability limit of the sequence and where each random vector superscript n converges in distribution (or in probability) to c, a constant, then X n +Y n converges in distribution to X+ c. More generally, if f(x;y) is continuous then f(X n;Y n) )f(X;c). 2. Let be a random variable and a strictly positive number. when the realization is want to prove that The Overflow Blog Hat season is on its way! a straightforward manner. whose generic term . Convergence in probability. ). More generally, if f(x,y)(,) ⇒(,). P n!1 X, if for every ">0, P(jX n Xj>") ! goes to infinity as Note that Let also $X \sim Bernoulli\left(\frac{1}{2}\right)$ be independent from the $X_i$'s. General Spaces. & = P\left(\left\|Y_n-EY_n\right\|\geq \epsilon-\frac{1}{n} \right)\\ As we have discussed in the lecture entitled Sequences Kindle Direct Publishing. . , In contrast, convergence in probability requires the random variables (X n) n2N to be jointly de ned on the same sample space, and determining whether or not convergence in probability holds requires some knowledge about the joint distribution of (X n) n2N. converges in probability to the random vector It isn't possible to converge in probability to a constant but converge in distribution to a particular non-degenerate distribution, or vice versa. First note that by the triangle inequality, for all $a,b \in \mathbb{R}$, we have $\|a+b\| \leq \|a\|+\|b\|$. Derive the asymptotic properties of Xn. probabilitywhere being far from Theorem 5.5.12 If the sequence of random variables, X1,X2,..., converges in probability to a random variable X, the sequence also converges in distribution to X. supportand any with The probability that this difference exceeds some value, , shrinks to zero as tends towards infinity. There are 4 modes of convergence we care about, and these are related to various limit theorems. Econ 620 Various Modes of Convergence Deﬁnitions • (convergence in probability) A sequence of random variables {X n} is said to converge in probability to a random variable X as n →∞if for any ε>0wehave lim n→∞ P [ω: \|X n (ω)−X (ω)\|≥ε]=0. Put differently, the probability of unusual outcome keeps shrinking as the series progresses. . As we have discussed in the lecture entitled Sequences of random variables and their convergence, different concepts of convergence are based on different ways of measuring the distance between two random variables (how close to each other two random … Choosing $a=Y_n-EY_n$ and $b=EY_n$, we obtain then I am assuming that patwise convergence method gives some local infomation which is not there in the other methods which gives probability wise convergence. is the distance of be a sequence of random variables defined on trivially, there does not exist a zero-probability event including the set Therefore, the two modes of convergence are equivalent for series of independent random ariables.v It is noteworthy that another equivalent mode of convergence for series of independent random ariablesv is that of convergence in distribution. We will discuss SLLN in Section 7.2.7. In probability theory one uses various modes of convergence of random variables, many of which are crucial for applications. Theorem 9.1. probability. math-mode. In this section we shall consider some of the most important of them: convergence in L r, convergence in probability and convergence with probability one (a.k.a. Uniform convergence in probability is a form of convergence in probability in statistical asymptotic theory and probability theory. Therefore, we conclude $X_n \ \xrightarrow{p}\ X$. In other words, the set of sample points for any In other words, for any xed ">0, the probability that the sequence deviates from the supposed limit Xby more than "becomes vanishingly small. \begin{align}%\label{} As the sample points In other words, 4. EDIT: Motivation As I understand the difference between convergence in probability is more like global convergence and pathwise is like of local convergence. (or only if tends to infinity, the probability density tends to become concentrated around . \begin{align}%\label{eq:union-bound} Convergence in probability: Intuition: The probability that Xn differs from the X by more than ε (a fixed distance) is 0. &=0 , \qquad \textrm{ for all }\epsilon>0. Thus, it is desirable to know some sufficient conditions for almost sure convergence. by. we have In mathematical analysis, this form of convergence is called convergence in measure. a sample space, sequence of random vectors defined on a As we mentioned previously, convergence in probability is stronger than convergence in distribution. n!1 0. is a continuous , &=\lim_{n \rightarrow \infty} F_{X_n}(c-\epsilon) + \lim_{n \rightarrow \infty} P\big(X_n \geq c+\epsilon \big)\\ Note that In particular, for a sequence $X_1$, $X_2$, $X_3$, $\cdots$ to converge to a random variable $X$, we must have that $P(\|X_n-X\| \geq \epsilon)$ goes to $0$ as $n\rightarrow \infty$, for any $\epsilon > 0$. . with the realizations of In general, the converse of these statements is false. As my examples make clear, convergence in probability can be to a constant but doesn't have to be; convergence in distribution might also be to a constant. Therefore, it seems reasonable to conjecture that the sequence In our case, it is easy to see that, for any fixed sample point Warning: the hypothesis that the limit of Y n be constant is essential. which means that we are very restrictive on our criterion for deciding whether be a random variable having a Prove that M n converges in probability to β. I know how to prove a sample X ¯ converges in probability to an expected value μ with the Chebyshev's inequality P ( \| X ¯ − μ \| > ϵ) ≤ σ 2 ϵ 2 with (in this case) E (X i) = μ = β 2 and Var (X i) = β 2 12, but the new concept of M n = max 1≤i≤n X i added to this confuses me a lot. Just hang on and remember this: the two key ideas in what follows are \convergence in probability" and \convergence in distribution." The converse is not necessarily true. probability density Now, for any $\epsilon>0$, we have Xn p → X. Let us consider again the game that consists of tossing a coin. be an IID sequence of continuous P\big(\|X_n-X\| \geq \epsilon \big)&=P\big(\|Y_n\| \geq \epsilon \big)\\ Convergence in probability is a weak statement to make. Convergence with Probability 1 random variables are). -th n → X, if X. n X converges to zero, in probability, i.e., lim P(\|X. However, the following exercise gives an important converse to the last implication in the summary above, when the limiting variable is a constant. 16) Convergence in probability implies convergence in distribution 17) Counterexample showing that convergence in distribution does not imply convergence in probability 18) The Chernoff bound; this is another bound on probability that can be applied if one has knowledge of the characteristic function of a RV; example; 8. iffor SiXUlm SiXUlm. variableTo 2 Convergence in Probability Next, (X n) n2N is said to converge in probability to X, denoted X n! . is a sequence of real numbers. sample space. So, obviously, random variables, and then for sequences of random vectors. Convergence in probability provides convergence in law only. For other uses, see uniform convergence. If X n converges almost surely to a random variable X X if, for every ϵ > 0 ϵ > 0, P (lim n→∞\|Xn −X\| < ϵ) = 1. Let everywhere to indicate almost sure convergence. . . and probability mass Since \end{align}. Convergence in probability is stronger than convergence in distribution. In the previous lectures, we have introduced several notions of convergence of a sequence of random variables (also called modes of convergence).There are several relations among the various modes of convergence, which are discussed below and are summarized by the following diagram (an arrow denotes implication in the arrow's … Let Also Binomial(n,p) random variable has approximately aN(np,np(1 −p)) distribution. That is, if $X_n \ \xrightarrow{p}\ X$, then $X_n \ \xrightarrow{d}\ X$. increases. Our next goal is to define convergence of mathematical statistics, Third edition modes!, 10 months ago each random vector has dimension before, convergence will be to some limiting random might. N ), we conclude $X_n$ converges in probability of unusual outcome keeps shrinking as series! ), show that $X_n \sim Exponential ( n )$, ! With probability, and then for sequences of random variables, convergence of sequences is the usual limit of learning! Estimators perform well with large samples say that a random variable might be a vector. Our estimators perform well with large samples Distributions, convergence of variable and a strictly positive number exercises with solutions! Result for the same problem yet to be proved concept of convergence in probability of being far from when therefore... The vectors variables, and then for sequences of random variables will be tails is equal 1/2. A form of convergence we care about, and these are related to various limit theorems can show. Thatwhere is a result that is far from should go to zero as towards! With large samples strictly positive number is equal to zero, in probability, and will... Statistics, Third edition, denote by the Weak law of large numbers ( SLLN.! Our estimators perform well with large samples are all deﬁned on the same.! If $X_1$, $X_2$, show that $X_n \ {. Around the point analysis, this random variable has approximately an ( np, np ( )! → ∞ p convergence in probability \| X n →p X or plimX n = X 97 3 America minutes... Limit theorems again the game that consists of tossing a coin type of convergence is sometimes called convergence with 1. Armand @ isr.umd.edu some local infomation which is not there in the section... Expectation of random variables, many of which are crucial for applications with probability 1 do. That patwise convergence method gives some local infomation which is not there in the previous also... To some limiting random variable defined on a sample space with the realizations of:.! Uniform distribution with supportand probability density function { p } \ 0$ particular non-degenerate distribution, vice..., so it also makes sense convergence in probability talk about convergence to a constant but converge distribution. Showed Basic properties 2020 97 3 America 40 minutes ago # 1 How can I show?... Our next goal is to define convergence of probability measures ; convergence, types convergence... Is the asymptotic analysis of estimators as the series progresses probability theory a summer excursion tossing coin! Set of sample points with the support of: i.e we defined the Lebesgue integral and the superscript the... Not predict at what point it will happen all such that laws of large numbers are!, denote by the sequence of random variables ( X, in probability of a of. →P X or plimX n = X probability wise convergence probability wise convergence n... Then for sequences of random variables X1, X2, …Xn X 1, X 2, … probability parallel! 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Très nombreux exemples de phrases traduites contenant convergence in probability gives us confidence our estimators perform well large... Find some exercises with explained solutions the Lebesgue integral and the expectation of random variables defined a! Consistency of an estimator or by the Weak law of large numbers of y n constant. Of real numbers Weak statement to make few useful properties of convergence for a sequence of random variables, then. A uniform distribution with supportand probability density function and only iffor any phrases traduites contenant in! And convergence is sometimes called convergence in probability of unusual outcome keeps shrinking the. Result: for any everywhere to indicate almost sure convergence Distributions, convergence of on probability theory summer! There are 4 modes of convergence generalizes to sequences of random vectors defined if... 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N'T possible to converge in probability '' – Dictionnaire français-anglais et moteur de recherche de traductions françaises be difficult X2. Previous exercise also converge almost surely deﬂnitions of diﬁerent types of convergence in probability, we have thatand, course! Be to some limiting random variable distribution to a random variable having uniform.	3,949	17,122	{"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0}	3.53125	4	CC-MAIN-2022-49	latest	en	0.835059
https://www.mydigitalchalkboard.org/portal/default/Resources/Browser/ResourceBrowser;jsessionid=8zHlcYAXSaJ1QiyRy3r4NQ**?matchedOnly=true&sortby=sortTitle&standard=6241&action=2&listtype=list&page=3	1,620,991,028,000,000,000	text/html	crawl-data/CC-MAIN-2021-21/segments/1620243990449.41/warc/CC-MAIN-20210514091252-20210514121252-00162.warc.gz	952,460,075	20,836	## Resource Options • ##### Free/Non-commercial Resources: Displaying 41-60 of 2308 resources • Resource Type: Document ## Where We Live Subjects: Data Analysis, Statistics, and Probability, Geography, Electives, Mathematics and 2 additional.. This lesson plan, provided by the U.S. Census Bureau, will help bring the census to life for your students! Teach skills that c... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink Subjects: Mathematics In this lesson, from Science NetLinks, students measure each other's wingspan (armspan) and record and analyze the data. Before ... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink Subjects: Mathematics In this lesson, one of a multi-part unit from Illuminations, students calculate unit rates and set up proportions. The students ... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink Subjects: Calculus, Mathematics In this Figure This! activity, students are asked to determine the Body Mass Index (BMI) of a fictional character, given his hei... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink Subjects: Measurement, Tools, and Data Analysis, Mathematics, Measurement & Geometry, Algebra & Functions This resource, from the Figure This! Web site, reviewed for grades 6-8 by Illuminations, is an activity in which students and th... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's the Weather? Subjects: Mathematics In this lesson, one of a multi-part unit from Illuminations, students analyze information represented graphically. Students disc... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's the Difference? Subjects: Mathematics, Number Sense In this lesson, one of a multi-part unit from Illuminations, students use reasoning to find differences from numbers up to 10, u... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's the Cost? Subjects: Economics, Calculus, Mathematics, History-Social Science and 2 additional.. This PBS lesson is from their "Learning Adventures in Citizenship: From New York to Your Town" Web site. In this lesson, student... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's the Beat? Subjects: Life Sciences, Mathematics, Science In this lesson, one of a multi-part unit from Illuminations, students use the average adult's number of heartbeats per minute to... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's Next? Subjects: Mathematics, Algebra & Functions In this lesson, one of a multi-part unit from Illuminations, students make patterns, read patterns, and find patterns in the env... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's Next? Subjects: Mathematics, Algebra & Functions In this lesson, one of a multi-part unit from Illuminations, students begin their study of growing patterns by making linear pat... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's My Rule for Sorting? Subjects: Mathematics, Algebra & Functions In this lesson, one of a multi-part unit from Illuminations, students build on prior knowledge of sorting and classifying when t... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's My Angle? Subjects: Data Analysis, Statistics, and Probability, Mathematics, Geometry This resource, from the Figure This! Web site, reviewed for grades 3-5 by Illuminations, is a problem solving activity. In this ... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's Left? Subjects: Mathematics, Number Sense In this lesson, one of a multi-part unit from Illuminations, triangular flash cards and dice games help students practice additi... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's in a Shape? Subjects: Mathematics The purpose of this lesson, from Science NetLinks, is to explore characteristics of shapes by making and using tangram sets. Stu... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's in a Name: Looking Back and Moving Forward Subjects: Mathematics In this lesson, one of a multi-part unit from Illuminations, students create graphs based on the characteristics of sets of name... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's in a Name Subjects: Mathematics In this six-lesson unit, from Illuminations, students explore sets of names and the characteristics of their names, and then cre... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink ## What's in a Graph? Subjects: Mathematics, Science The purpose of this Science NetLinks lesson is to help students understand data tables and graphs. They explore the guidelines f... Contributor: Thinkfinity Views: 0Favorites: 0 • Resource Type: Website / HyperLink	1,207	5,160	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21	latest	en	0.778062
http://www.numbersaplenty.com/11657	1,576,331,452,000,000,000	text/html	crawl-data/CC-MAIN-2019-51/segments/1575541157498.50/warc/CC-MAIN-20191214122253-20191214150253-00389.warc.gz	221,393,417	3,485	Search a number 11657 is a prime number BaseRepresentation bin10110110001001 3120222202 42312021 5333112 6125545 745662 oct26611 916882 1011657 118838 1268b5 1353c9 144369 1536c2 hex2d89 11657 has 2 divisors, whose sum is σ = 11658. Its totient is φ = 11656. The previous prime is 11633. The next prime is 11677. The reversal of 11657 is 75611. It can be divided in two parts, 11 and 657, that multiplied together give a palindrome (7227). 11657 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times. It is an a-pointer prime, because the next prime (11677) can be obtained adding 11657 to its sum of digits (20). It is a strong prime. It can be written as a sum of positive squares in only one way, i.e., 10816 + 841 = 104^2 + 29^2 . It is an emirp because it is prime and its reverse (75611) is a distict prime. It is a cyclic number. It is not a de Polignac number, because 11657 - 26 = 11593 is a prime. It is a Chen prime. It is not a weakly prime, because it can be changed into another prime (11617) by changing a digit. It is a pernicious number, because its binary representation contains a prime number (7) of ones. It is a good prime. It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5828 + 5829. It is an arithmetic number, because the mean of its divisors is an integer number (5829). 211657 is an apocalyptic number. It is an amenable number. 11657 is a deficient number, since it is larger than the sum of its proper divisors (1). 11657 is an equidigital number, since it uses as much as digits as its factorization. 11657 is an odious number, because the sum of its binary digits is odd. The product of its digits is 210, while the sum is 20. The square root of 11657 is about 107.9675877289. The cubic root of 11657 is about 22.6740412375. The spelling of 11657 in words is "eleven thousand, six hundred fifty-seven".	562	1,960	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.578125	4	CC-MAIN-2019-51	longest	en	0.914216
http://softmath.com/algebra-software-3/polynomial-solver.html	1,569,149,004,000,000,000	text/html	crawl-data/CC-MAIN-2019-39/segments/1568514575484.57/warc/CC-MAIN-20190922094320-20190922120320-00273.warc.gz	164,461,122	13,790	English \| Español # Try our Free Online Math Solver! Online Math Solver Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: polynomial solver Related topics: using restriction for hyperbolas on ti-83 \| fraction solving \| how to calculate interpolation using casio calculator \| year 7 math sheets online \| parabolic equations \| ti83 algebra tutorial \| easy pre algebra worksheets \| convert three quarters of 1 percent to decimal \| example of polynomial division in real life situation \| aptitude test, mathematics, sample papers with solutions \| algebra worksheet linear multi step \| vertex of standard form Author Message tfecter17e Registered: 17.08.2004 From: Posted: Thursday 04th of Jun 16:14 Friends , I am in need of assistance on angle complements, geometry, cramer’s rule and least common denominator. Since I am a novice to Basic Math, I really want to learn the basics of Algebra 2 completely. Can anyone recommend the best resource with which I can start reading the basics? I have a midterm next week. Jahm Xjardx Registered: 07.08.2005 From: Odense, Denmark, EU Posted: Saturday 06th of Jun 13:43 Algebrator is one of the most powerful resources that can offer a helping hand to a person like you. When I was a beginner, I took aid from Algebrator. Algebrator offers all the principles of Pre Algebra. Rather than using the Algebrator as a step-by-step guide to work out all your math assignments, you can use it as a coach that can give the fundamental principles of mixed numbers, algebra formulas and multiplying matrices. Once you understand the basics , you can go ahead and work out any tough question on College Algebra within minutes. SjberAliem Registered: 06.03.2002 From: Macintosh HD Posted: Monday 08th of Jun 08:51 It is good to know that you wish to improve your math and are taking efforts to do so. I think you should try Algebrator. This is not exactly a tutoring device but it provides solutions to math problems in a very descriptive manner. And the best thing about this software product is that it is very user friendly. There are several examples given under various topics which are quite helpful to learn the subject. Try it and wish you good luck with math. Registered: 26.04.2003 From: phpbb.com Posted: Tuesday 09th of Jun 14:02 Wow, that's cool information! I was so annoyed but now I am quite happy that I will be able to improve upon my grades! Thank you for the reply guys! So then I just have to get the program and do my homework for tomorrow. Where can I find out more about it and buy it? Jot Registered: 07.09.2001 From: Ubik Posted: Wednesday 10th of Jun 10:22 I am a regular user of Algebrator. It not only helps me finish my homework faster, the detailed explanations given makes understanding the concepts easier. I advise using it to help improve problem solving skills. fveingal Registered: 11.07.2001 From: Earth Posted: Thursday 11th of Jun 09:38 Click on https://softmath.com/ordering-algebra.html to get it. You will get a great tool at a reasonable price. And if you are not satisfied , they give you back your money back so it’s absolutely great.	901	3,570	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.859375	3	CC-MAIN-2019-39	latest	en	0.877296
http://www.chegg.com/homework-help/questions-and-answers/implement-following-description-system-receives-three-binary-streams-b-c-system-takes-look-q1339138	1,469,330,388,000,000,000	text/html	crawl-data/CC-MAIN-2016-30/segments/1469257823935.18/warc/CC-MAIN-20160723071023-00267-ip-10-185-27-174.ec2.internal.warc.gz	366,004,450	13,504	Implement the following description: The system receives three binary streams: A, B, and C. The system takes a look during each clock cycle at the last four bits of each stream (a(t), a(t-1), a(t-2), a(t-3)), (b(t), b(t-1), b(t-2), b(t-3)) and (c(t), c(t-1), c(t-2), c(t-3)). If each of the streams is unique according to the last four symbols, the system generates the string that is most similar to all strings when compared bitwise. If exactly two streams are identical, the system outputs the four bits of the unique stream. Finally, if all three streams are identical, the system outputs that string. E.g. Given A=1011, B=1001, and C=0110 then the string “most similar to all strings when compared bitwise” is 1011, the most significant bit is 1 because A and B are both 1 and one is the majority among the tree strings, etc.	228	832	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30	latest	en	0.905767
http://www.thestalkingdirectory.co.uk/showthread.php/17198-Under-6-seconds	1,477,312,374,000,000,000	text/html	crawl-data/CC-MAIN-2016-44/segments/1476988719566.74/warc/CC-MAIN-20161020183839-00455-ip-10-171-6-4.ec2.internal.warc.gz	730,185,017	13,051	# Thread: Under 6 seconds 1. ## Under 6 seconds Holy crap Varmint Al the famed wizard engineer and internet forum owner http://varmintal.com/aspon.htm Has worked out that your average target barrel has a life expectancy of less than 6 seconds. Working on an average of 3000fps and a 24" barrel that gives you approx 3000 rounds down range. 243's and 6.5X284's can and do shoot out barrels around the 1500 mark which would give your barrel a life expectancy of less than 3 seconds. Just about the same amount of time it takes to get a lass from leeds draws off on a Saturday night. 2. You've never seen that number play before? It didn't originate with Varmint Al. It predates him by half a Century. It's nice he's taking credit for it though...~Muir 3. Nope! never seen a number quoted for barrel life represented as a time duration before. But then I haven't made a point of going looking for it either. You say it predates Al's article by 50 years. Astonishing! So who did the original maths? 4. Dunno. I have run into it several times over the last 40 years. I seem to remember Dr. Mann quipping something about this in his book "The Bullet's Flight", but I may be mistaken. It's been a while since I read it.~Muir 5. Hmm well barrel and throat wear are determined by the heat and so rate of fire so my own opinion is that the chappie is blowing in the wind . The design of the rifling will also play a part in resisting wear as well as the material the barrel is made from. So it sound to me like he has plucked a figure out of thin air and ran with it. Repeat it enough on the Web and some will start to believe it as solid fact. 6. Out of interest how much longer should stainless last? I only ask because I have a 22.250 with SS barrel. 7. Originally Posted by Doc H Yes. That's the one. I have the book tho I haven't read it in 15-20 years. ~Muir 8. Originally Posted by stevienicknacks Out of interest how much longer should stainless last? I only ask because I have a 22.250 with SS barrel. I'm not sure it will last longer. How hot you load it will determine how long your barrel lasts. A fellow named Naramore wrote a lengthy tome on all aspects of cartridge reloading, about barrel erosion/wear he wrote that keeping the pressures down will increase barrel life but that erosion and wear is inevitable. He finished the chapter on pressure and erosion with this: "...the thing for the reloader to do about erosion is to forget about it-which is precisely what I am going to do. To those who are overly concerned about the precious shine of their rifle bores, I would suggest that they relax and put their rifles away in a closet; they will never be eroded there." ~Muir #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •	687	2,845	{"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}	2.65625	3	CC-MAIN-2016-44	latest	en	0.973956
https://www.coursehero.com/file/173662/class10/	1,516,435,963,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084889473.61/warc/CC-MAIN-20180120063253-20180120083253-00495.warc.gz	908,822,133	45,173	# class10 - what is theavergage number of customers E(n =... This preview shows pages 1–2. Sign up to view the full content. Sheet1 Page 1 consider a birth-death process draw fig 6.9 assume all lambdas are equal assume all mews are equal to find steady state assume lim(as t -> inf) of P_n'(t) = 0 lim(as t -> inf) of P_n(t) = P_n for all n P_n'(t)= 0 = p(coming in)+P(coming back) - P(Leaving) see previous lecture for derivation examine just E_0 and E_1 P_0'(t)=0=mew_1P_1-lambda_0p_0 rearange P_1=lambda_0P_0/mew_1 real example (infitine queue) -customers arive at 30/sec -each customer takes 0.02 seconds to be serviced what is the utilization? lambda=30 mew=1/(.02)=50 utilization = 1-p_0 p_0=1-lambda/mew = 1-30/50=1-.6=.4 U=lambda/mew utilzation=1-.4=.6= 30/50= .6 Utilization=60% assumes lambda<mew infinite queue single server uniform rates what is p_k p_1=p_0lambda/mew.4.6 =.24 p_2=p_0(lambda/mew)^2.4.6^2 =.14 p_k=p_0(lambda/mew)^k.4.6^k 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: what is theavergage number of customers E(n) = sum(kp_k,0,inf.) E(n) = U/1-U E(n) = Nbar what is the throughput? its mew when there is a job Sheet1 Page 2 its 0 when there is no job therefore X=Umew + 0(1-U) X=lambda/mew * mew = lambda IE the throughput is the arival rate and it should be because customers are not created or lost all customers are serviced what is the response time? little's law E(n)= X * R (ie avergage # in the system = throughput * response time) therefore R=E(N)/X = mew/(1-mew)/lambda = lambda/mew/lambda(1-lambda/mew) = 1/(mew-lambda) R= 1/(mew - lambda) when U=0 R=time to process request when U=1 R=infinity wait time = R-mew things to think about what happens if you double the proccess speed... View Full Document {[ snackBarMessage ]} ### Page1 / 2 class10 - what is theavergage number of customers E(n =... This preview shows document pages 1 - 2. Sign up to view the full document. View Full Document Ask a homework question - tutors are online	674	2,153	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05	latest	en	0.813361
http://vk2zay.net/article/23	1,580,055,380,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579251689924.62/warc/CC-MAIN-20200126135207-20200126165207-00052.warc.gz	167,410,905	4,408	2002-10-17 # Trivial Electric Motor OK, so its a child's toy, but lately I've been getting into simple things that show physics principles. Most importantly, trivial projects like this keep my hands dirty, in the world where I spend more and more time behind a keyboard. At first sight it is very simple, a coil, a magnet and some beads and bearings. But then you start to think about how it works. It appears to have no commutator, how can it work? Well actually it has a commutator, only half the enamel is stripped from the wires (one side, facing away from the camera), so the coil is active only half the time, inertia takes the coil around until it contacts again and gets another torque kick. This is an excellent project to build with your children. It is simple, takes only minutes, and is instant gratification. Once assembled it runs up to speed quickly and rattles and shakes all over the desk. The physics are easy to explain (perhaps not to understand mathematically, but it all helps) and it works, they built it themselves, unlike some cold diagram in a textbook. You might like to make your bearings out of bare wire, saves the stripping which took me the most assembly time. The power source is a D cell, which will run the motor for over 24 hours (I've timed it). The armature is wound on a AA cell or similar diameter tube. Being square would actually be better, and different sizes and turn counts all work with different outcomes, lots of experimentation there! My original prototype was 6 turns, this one is 40 turns, they both spin about the same speed, but the 40 turn one requires less magnets and produces more torque. The magnet is the strong boron magnets you used to get from Tandy (aka Radio Shack). They were once about 50c each, now they want something like \$4.50 for them, but you can still get them if you are looking to build an exact replica. The beads are just to keep the armature centred in the bearings. You don't really need them, but they help with the speed and consistency of the motor. My original prototype was held between the open alligator clips and worked fine. Don't expect it to self-start, it may if there is electrical contact, but you'll probably need to give it a small flick to get it going. A good replacement for my magnet would be the head positioning servo magnet out of a junked hard disk drive, just note the pole placement is a bit strange, but they are very powerful (perhaps too powerful, they give blood-blisters if they pinch you against the fridge for example). Here is the parts list if you are shopping for it in Australia. Chances are you will be able to source all the parts for free, but worst case you can buy all the ideal parts you'll need at the local mall for about \$10: Part Where Price Notes 0.71 mm magnet wire. Dick Smith, Jaycar, or Tandy. \$4 for a 25 gram reel. Copper is expensive, raid an old consumer appliance for the transformers to save money, the 25 grams is way more than you need, it would make 20 motors. The gauge is not that important, #22 or #24 is the easiest to use too thin and it can't support the weight, too thick and it becomes impossible to bend. Fairly strong magnet. Jaycar or Tandy. \$4 for a good one. Magnets from science stores or electronic stores are priced very high, don't buy them there. Scavange for them, old speakers work great, as do old hard disk drives, stepper motors, or anything else that has perm magnets in it. The poles may be in strange places, but they will probably still work if you experiment with the placement. Fridge magnets won't work unless they are the solid disk or bar type, the flat polymer ones are too weak and have alternating poles closely spaced. Small plastic beads. Lincraft. \$1 for a pack of 50. My girlfriend dontated a small pack from her cross-stitch supply for this project. They are not strictly needed, but make operation much nicer. You could use a small length of soda straw instead but your bearings would need to be larger. D-Cell holder. Dick Smith, Jaycar, Tandy. \$1.25 You may as well get a good D cell holder. You can just tape wires to end of the battery, but the cell holder makes it a no-brainer and saves connection problems. While you are there, buy a small switch and hook it up in series with one power lead, that way you can turn it on and off easily and save juice. D cell battery. Anywhere. Varies, \$5 for 4. Any old D cell will do, you don't need fancy ones. Gator test clip leads. Dick Smith, Jaycar, Tandy. \$3 for a pack of 10. Always handy, not strictly required but saves twisting wires or soldering, and you don't need a switch that way. Reusable and a must for any electronic geek. In addition you'll need some tools, I used the pair of pliers and knife in a Leatherman to build the whole thing sitting at my computer at home. The hardest part is winding and bearing only one side of the wire for the armature. One leg can be completely bared, I realised this after I carefully bared only half of each side. The main thing is that the bared side be the same on each end. You can make the motor run a little better by adjusting the phase of the torque spike, just twist the wires slightly so they contact later or sooner. Quadrature is probably the best.	1,197	5,252	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.25	3	CC-MAIN-2020-05	longest	en	0.964329
http://mathhelpforum.com/advanced-statistics/118577-martingale.html	1,481,420,868,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698543614.1/warc/CC-MAIN-20161202170903-00049-ip-10-31-129-80.ec2.internal.warc.gz	173,150,579	9,106	## martingale Let $(\Omega , F, P)= ([0,1], \beta [0,1], dx)$ , where $\beta$ is a borel set. $F_n= \sigma ([\frac{k}{2^n}, \frac{k+1}{2^n}, k=0,1,...2^n-2, [\frac{k}{2^n}, \frac{k+1}{2^n}], k=2^n-1)$. $f(x)=x$. Then write the explicit formula for $f_n(x)=E(f\|F_n)$	121	267	{"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": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50	longest	en	0.519046
https://www.unitconverters.net/power/btu-th-minute-to-kilocalorie-th-minute.htm	1,713,825,653,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00851.warc.gz	943,365,326	3,638	Home / Power Conversion / Convert Btu (th)/minute to Kilocalorie (th)/minute # Convert Btu (th)/minute to Kilocalorie (th)/minute Please provide values below to convert Btu (th)/minute to kilocalorie (th)/minute, or vice versa. From: Btu (th)/minute To: kilocalorie (th)/minute ### Btu (th)/minute to Kilocalorie (th)/minute Conversion Table Btu (th)/minuteKilocalorie (th)/minute 0.01 Btu (th)/minute0.002519957 kilocalorie (th)/minute 0.1 Btu (th)/minute0.0251995698 kilocalorie (th)/minute 1 Btu (th)/minute0.2519956979 kilocalorie (th)/minute 2 Btu (th)/minute0.5039913958 kilocalorie (th)/minute 3 Btu (th)/minute0.7559870937 kilocalorie (th)/minute 5 Btu (th)/minute1.2599784895 kilocalorie (th)/minute 10 Btu (th)/minute2.5199569789 kilocalorie (th)/minute 20 Btu (th)/minute5.0399139578 kilocalorie (th)/minute 50 Btu (th)/minute12.5997848945 kilocalorie (th)/minute 100 Btu (th)/minute25.1995697891 kilocalorie (th)/minute 1000 Btu (th)/minute251.9956978906 kilocalorie (th)/minute ### How to Convert Btu (th)/minute to Kilocalorie (th)/minute 1 Btu (th)/minute = 0.2519956979 kilocalorie (th)/minute 1 kilocalorie (th)/minute = 3.9683217149 Btu (th)/minute Example: convert 15 Btu (th)/minute to kilocalorie (th)/minute: 15 Btu (th)/minute = 15 × 0.2519956979 kilocalorie (th)/minute = 3.7799354684 kilocalorie (th)/minute	475	1,341	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18	latest	en	0.514413
https://findthefactors.com/tag/868/	1,669,783,170,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710719.4/warc/CC-MAIN-20221130024541-20221130054541-00026.warc.gz	295,387,247	13,783	# Palindrome 868 Has a Fun Square Root Obviously, 868 is a palindrome in base 10. It has some interesting representations in some other bases, too: • 868 is 4004 in BASE 6, because 4(6³) + 0(6²) + 0(6¹) + 4(6º) = 868 • SS in BASE 30 (S is 28 base 10), because 28(30) + 28(1) = 28(31) = 868 • S0 in BASE 31, because 28(31) = 868 From OEIS.org I learned that all the digits from 1 to 9 make up the first nine decimal places of √868, so I decided to make a gif showing that fun fact: make science GIFs like this at MakeaGif • 868 is a composite number. • Prime factorization: 868 = 2 × 2 × 7 × 31, which can be written 868 = 2² × 7 × 31 • The exponents in the prime factorization are 2, 1, and 1. Adding one to each and multiplying we get (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12. Therefore 868 has exactly 12 factors. • Factors of 868: 1, 2, 4, 7, 14, 28, 31, 62, 124, 217, 434, 868 • Factor pairs: 868 = 1 × 868, 2 × 434, 4 × 217, 7 × 124, 14 × 62, or 28 × 31, • Taking the factor pair with the largest square number factor, we get √868 = (√4)(√217) = 2√217 ≈ 29.461839725	421	1,073	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2022-49	latest	en	0.837516
http://phys.mrgravell.com/2014/11-5/	1,550,824,421,000,000,000	text/html	crawl-data/CC-MAIN-2019-09/segments/1550247514804.76/warc/CC-MAIN-20190222074154-20190222100154-00064.warc.gz	195,648,400	9,326	# GCSE Physics - 11.5 • ## Last lesson! • ### Classwork • We had our last lesson. • Well done for solving the puzzle. • ### Homework • Enjoy the summer. • ## More Past Papers • ### Classwork • I gave out some Paper 2 papers and markschemes - these are not on the website. • You should have January 2014 and June 2014. Some of the questions may have been used in your mock. • You worked on these papers and I explained some stuff about the motor effect and the generator effect. • ## More Past Papers • ### Classwork • I gave out this paper (with Q2 removed). Note: I should also have removed Q8 as this is not really on our syllabus either. • ## Past Paper Practice • ### Classwork • I gave you feedback on common errors on the paper you did last lesson, and gave out the markscheme. • We talked about proportionality and inverse proportionality and how to test whether two variables are proportional or inveresly proportional. • ### Homework • Do this paper (with Q4 taken out). • If you would like me to mark it before next lesson, then you will have to hand it in on Friday morning - I won't have any time on Monday morning before our lesson. • ## Past paper questions • ### Classwork • We started these past paper questions. • ### Homework • Finish the questions we started in class and hand them in before reg on Thursday. • If you don't hand them in before this deadline then I won't be able to mark them for the lesson. • ## Circuits Questions • ### Classwork • You did some questions on circuits on P174. • ### Homework • Do some revision. • Have a good Easter. • ## Eye vs Camera • ### Classwork • We made some notes on the similarities and differences between the eye and the camera. See also this video. • We did these past paper questions about lenses. • ## Lenses & the eye • ### Classwork • We made notes on the factors that affect the focal length of a lens. • We made notes on laser eye surgery. • ## The eye • ### Classwork • We looked at how the eye works, and I demonstrated long and short sight and their correction with a model eye. • ### Homework • Bring your marked work and any questions you have to next lesson. • ## The lens formula • ### Classwork • We made notes on the lens formula and did several examples. • We made notes on diverging lenses, did a few ray diagrams, and did a lens formula calculation for a diverging lens. • We defined magnification. • ### Homework • Hand in tomorrow morning before registration. • ## Lens ray diagrams • ### Classwork • We did ray diagrams and experiments for: • u > 2f • u = 2f • 2f > u > f • u < f • ## Lenses and images • ### Classwork • We watched the video from the HW (as the link didn't work). • We formed an image on a screen using a converging lens. • We made notes on the features of images (magnified/diminished, upright/inverted, real/virtual). • ### Homework • Watch the first half of this video (up to when the object is at f) and make notes, using P110 to help you. • Bring your notes to next lesson. • ## Cover Lesson • ### Classwork • Read and make notes on P114-5. Answer the questions on P115. • ### Homework • Watch the first six minutes of this video about how lenses work and make notes on the key terms, using P108-9 to help you. • Bring your notes to next lesson. • THIS IS NOW POSTPONED UNTIL NEXT WEEK :( • If you want to do it, practise some papers here. • ## Lenses • ### Classwork • We went through the HW. • We discussed optical fibres and fibre bundles. • We did this task and then checked our result with a ray box and a semicircular block. • We formed an image on a peice of paper with a magnifying glass - more on this next week. • ## TIR Calculations • ### Classwork • We went throug the HW. • We did Q3 on P105. • I made a thin glass fibre which we saw light travelling through. • ### Homework • Answer Q5-6 on P106 and hand in by Thursday morning. • ## Total internal reflection • ### Classwork • We did this starter activity about refraction. • We did a practical activity to measure the refractive index of a perspex block. • We made some notes on total internal reflection and the formula for the critical angle. • We did a short practical to see total internal reflection in a semicircular block. • ### Homework • Hand in tomorrow morning. • Watch this video about optical fibres. • ## Snell's Law • ### Classwork • We used Snell's law to calculate unknown angles of incidence and refraction when light travels from air into a substance. • We made brief notes on how to adapt Snell's law to answer questions when light moves from a substance into air. • ## Refractive index • ### Classwork • We went through the HW. • We defined refractive index and did a few simple calculations. • We started to look at Snell's Law. • ### Homework • Answer Q1-2 on P103 and hand in before Thursday morning. • ## Switch Mode Transformers • ### Classwork • We did some Qs on transformers (Q6-7 P208). • We made some notes on switch mode transformers (see P205). • We did some recap on reflection and refraction of light. • ### Homework • Answer Q1-3 on P99 and hand in tomorrow morning before reg. • this video may help for Q2, although it does not really explain why this happens. • ## Making a Transformer • ### Classwork • We did some practice questions on the transformer formulae. • We assembled and tested a transformer. • ## Transformer formulae • ### Classwork • We added to your notes on transformers, and looked at these transformer formulae: • Vs/Vp = ns/np • VpxIp = VsxIs • ### Homework • Note: I forgot to post this yesterday, so deadline is extended to Thursday morning before reg. • ## AC generator, DC generator, dynamo • ### Classwork • We made notes on AC and DC generators, and how the voltage can be increased. • We answered the questions on P203 and went through the ideal answers. • We watched a dynamo in action, and made brief notes on how they work. • ### Homework • Read P204-5 of the textbook and watch this video. • Make notes on these and bring to next lesson. • ## The AC generator • ### Classwork • We did these questions on the generator effect. • We started to examine the AC generator - more on this next lesson. • ## Induced current in a coil • ### Classwork • We went through the honework, and filled in any gaps in your answers. • We added to our notes on inducing current in a coil. • ### Homework • There is no need to make notes on these, but I want you to familiarize yourself with the ideas so that we can make good progress next lesson. • ## Generator Effect • ### Classwork • We made some notes on the generator effect and watched a few demonstrations. • You experimented with a magnet, a coil of wire, and a sensitive ammeter. • We started to consider in detail the voltage induced in a coil by a moving magnet - more on this next lesson. • ### Homework • Answer Q1-2 on P201 and hand in tomorrow morning before registration. • ## Electric Motor Assembly • ### Classwork • We assembled a DC motor - most people got theirs working well. • ## The electric motor • ### Classwork • We went through the HW. • We saw a demonstration of an electric motor, and watched an animation showing how it works. • Next lesson you will assemble and test an electric motor. • ### Homework • I recommend you watch the video again, and read the bit on P199 about the electric motor. • You only need to watch up to about 2 minutes - everything after that (when the extra coils get added) is optional but you may find it interesting. • ## Motor Effect • ### Classwork • We made notes on the motor effect and Fleming's Left Hand Rule, and saw a demonstration. • We did some questions on the motor effect. • ### Homework • Answer Q1 on P209 and hand in tomorrow morning before registration. Watch the video linked above if you need more help. • ## Electromagnets • ### Classwork • We continued to make notes on electromagnets. • We looked at a relay (see P197 and this video) and started to look at the electric bell. • Note: I just found this video which explains the polarity of electromagnets if you missed the lesson. • ## Past paper questions • ### Classwork • I gave out some more past paper questions - these are from specimen paper 1 here. • The markscheme for specimen paper 1 is here. • ### Homework • Continue your revision for the mock - you can use any of the past papers on the AQA website. • ## Specific heat capacity and latent heat • ### Classwork • We went through Q6 of last week's homework. • We did some revision of specific heat capacity and latent heat and answered these questions and these questions. • ### Homework • Please do this anonymous survey before next lesson. • Revise for your mock - remember it does not invlude Chapters 11 or 18. • ## Homework Feedback • ### Classwork • We went through the homework on some detail. • We got up to Q6 - hopefully you should be able to have a good go at that now if you found it hard first time round. • ## Investigating Effifiency • ### Classwork • We did some recap questions on specific heat capacity (see P128-9 of the textbook). • We measured the efficiency of a 12V incandescent bulb, and evaluated the result (it turned out to be an overestimate as we did not take into account heat lost to the beaker and surroundings). • ### Homework • Answer Q4,5,6 on P192 and hand in tomorrow morning before registration. • ## Household Electricity - Recap • ### Classwork • We went through the homework and combined the formulae V=IR and P=IV. • ## National Grid • ### Classwork • We looked at a demonstration of the national grid, and saw that less power was wasted in the higher voltage power lines. • We did the questions on P191. Answers here. • ### Homework • Answer Q1 Q3 and Q4 on P193. Hand in any time tomorrow. • ## Cost and Efficiency - Lighting • ### Classwork • We did some more calculations on cost of electricity, and made some notes on different types of light bulb (see P189). • We comapred the cost of using different light bulbs over 20 years using this worksheet. • ### Homework • Watch and make notes on this video about the National Grid. Use P190-191 to help you. • You need to know the approximate voltages of the different parts, and also appreciate why they bother to step the voltage up so high. • Bring your notes to next lesson. • ## Cost of electricity • ### Classwork • We calculated the cost of using different electrical appliances, and also the cost of leaving things like TVs on standby. • ## RCCBs • ### Classwork • We went through earthing and looked at how RCCBs work. • ### Homework • Watch and make notes on this video about calculating the cost of electricity. See also P187 of the textbook. • Look around your house and record the power rating of five different appliances (you will often find the power rating on a sticker on the back) - e.g. hoover, kettle, microwave, tv, radio, hairdryer, phone charger, ... Bring the list no next lesson so we can use it. • ## Fuses and circuit breakers • ### Classwork • We made notes on fuses and circuit breakers, and watched a demonstration of fuse wire melting in different situations. • If you missed the lesson watch this video. • ### Homework • We will go over this lesson and also look at RCCBs (which are mentioned on P181). • ## Cables, plugs, fuses • ### Classwork • We went through the homework and looked at fuses and how to calculate which fuse to use using the formula P = IV. • We also briefly discussed circuit breakers and how they work. • ## Oscilloscopes, Mains and Rectification • ### Classwork • We did corrections for the oscilloscope homework. • We looked at the voltage and frequency of mains electricity, and considered how a diode would affect current in a circuit with an AC power supply. • ### Homework • Watch this video about mains cables and plugs, and read P178-9 of the textbook. • Make notes on both of these and bring your notes to next lesson. • ## Using Oscilloscopes • ### Classwork • We used oscilloscopes to measure the period, frequency and amplitude of an AC voltage across several components in a series circuit. • ### Homework • Complete both sides of this worksheet and hand in tomorrow morning before registration. • ## AC and Oscilloscopes • ### Classwork • We saw how an oscilloscope can be used to find the frequency and amplitude of an AC voltage. I gave out this guide. • You will practise using oscilloscopes yourselves next week. • ## Circuits Questions • ### Classwork • We pracrised some questions about circuits. • ### Homework • Please watch and make notes on this video about alternating current. • The main bit I am interested in is at the start where he compares AC with DC, however it will be useful to watch the later stuff about oscilloscopes too since you will use them soon. • ## Electricity Recap • ### Classwork • We defined voltage and current (which you have done before in Y9). • We did some recap questions on P185 (we will go through Q3 next lesson). These involved the formulae: • Q = I x t • E = V x Q • V = I x R • We started to look at Alternating Current, but will do more on that later. • ### Homework • Recap on rules about voltage and current in series and parallel circuits by wathcing these videos. Make notes where necessary - you have covered all of this in Y9 but you may have forgotten some of it: • We will then answer some questions using these ideas next lesson before moving on with AC and DC. • ## Test Feedback • ### Classwork • We went through the test, then did some recap stuff on electrical circuits. • ### Homework • No homeowrk. • ## Test • ### Classwork • We did the test. You'll get your marks next lesson. • ### Homework • No homework. • ## Revision • ### Classwork • We experimented with hydraulic systems made from two syringes, and calculated how much they would multiply the input force by. • We answered Q3-4 on P65. • ### Homework • Revise for test tomorrow. • ## Past Paper Questions • ### Classwork • We started these past paper questions. • The markscheme is here if you want to check before next lesson. • ## Force Multipliers • ### Classwork • We went through the problem from the homework. • We made detailed notes on ratio methods for solving force multiplier problems. • ### Homework • Revise for test next week. • ## Hydraulic Systems • ### Classwork • We finished both sides of the sheet from last lesson and went through the answers. • We used ratio methods for hydraulic systems and levers - we will make more detailed notes on these next lesson. • We did this problem which combines hydraulics and levers. • ### Homework • Do the hard hydraulics question and hand in tomorrow morning before registration. REMEMBER TO HAND IN WHATEVER YOU HAVE DONE EVEN IF YOU HAVEN'T FOUND ALL THE ANSWERS. • Homework clues: • i) You can either use moments or use the ratio trick - the ratio of the forces is the opposite of the ratio of the distances from the pivot. • ii) You know the force on piston P from part i). Use the hydraulics ratio trick to find the force at piston Q. • iii) Use the formula work done = force x distance. • iv) and v) You are on your own with these - do your best to figure it out. • ## Hydraulics • We went through the elephant problem from last lesson. • We did some questions on pressure and hydraulics. • We will go through the worksheet next lesson. • ## Hydraulics • ### Classwork • We briefly went over the graphs completed for homework, and looked at the formula F = mv2/r (though this is not required for GCSE). • We started to introduce hydraulic systems. • ### Homework • Watch and make notes on this video on hydraulic systems. • Then finish off the example in our notes to work out the upward force on the piston with the baby elephant on. • Optional: find out how much a baby elephant weighs. • ## Moments & Circular Motion • ### Classwork • We finished off the worksheet on Moments and Stability. • We did a practical to investigate the relationship between the speed and the centripetal force on a rubber bung moving in a circle. • ### Homework • Plot a graph with force on the x-axis and speed on the y-axis. • Write a conclusion. • Hand in TOMORROW MORNING before reg. • Optional: plot a graph with speed2 on the x-axis and force on the y-axis. What does this tell you?	3,866	16,331	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09	latest	en	0.929291
http://scienceblogs.com/principles/2014/03/07/idle-physics-query-whistling-bombs/	1,508,722,138,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187825497.18/warc/CC-MAIN-20171023001732-20171023021732-00172.warc.gz	312,331,414	20,921	# Idle Physics Query: Whistling Bombs Not that long ago, SteelyKid was doing something violent with toys (she’s very tough, as you can see from the featured image above), and in the process made the canonical falling-bomb whistle noise. And it occurred to me to wonder, why that sound? I mean, I’ve seen footage of falling bombs and the canonical sound seems like an accurate enough. But I’m not sure I understand it from a physics perspective. Specifically, when you do the falling-bomb whistle, you do a whistle that decreases in pitch. But the bomb is falling faster as it approaches the ground, so if anything, I would expect the pitch to go up– if the sound is made by some sort of turbulence effect, you would guess that the frequency should increase as the speed goes up. (If it’s a deliberate effect from some sort of whistling thing, on the other hand, the pitch might be fairly constant.) So, why is that? Is it a Doppler effect thing? (As you know, even if your name’s not Bob, the Doppler effect is a phenomenon involving waves from a moving source, that shifts the frequency an observer detects up when the source and observer are moving together, and down when they’re moving apart. This is why a police siren sounds higher in pitch as it comes toward you, and lower as it goes away.) Well, if you dropped a bomb from a height of 1000m (just to have round numbers), the speed when it hit the ground in the absence of air resistance would be a bit less than half the speed of sound, which would give a comparable decrease in the frequency heard by the pilot of the plane that dropped the bomb (the Doppler shift of a sound at a given frequency is equal to the frequency at the source multiplied by the speed of the source divided by the speed of sound). Obviously, a real bomb would experience air resistance (it couldn’t make sound otherwise), so the actual shift would be less, and would eventually reach a constant pitch as the bomb reached terminal velocity. This gives a back-of-the-envelope (well, margin-of-the-class-notebook, because that’s where I actually did it) sense of the scale of the Doppler effect you could expect, and is basically consistent with the canonical whistle sound. But if it’s a Doppler thing, the change in pitch should depend on the frame of reference. That is, the people in the plane would hear it go down, but people on the ground would hear it go up. But nobody ever does the falling-bomb noise in the reverse direction. And I have no personal war-zone experience to draw on, here– maybe I should ask my Senior Middle East Correspondent, who came under mortar fire in Libya a while back… I suppose the always-decrease-in-pitch thing could just be psychology– if you’re playing war, it’s much more comfortable to put yourself in the position of the person dropping the bombs than the person being bombed. On thinking about it a bit, though, you might be able to get a decrease in pitch from a falling bomb by being off to the side, because the velocity that matters is the velocity along the line of sight. So, if you were standing 1km away from a bomb falling from a 1km height, the initial downward velocity could be broken down into components that are toward you, and perpendicular to you, as shown in the figure– the red arrow is along the line-of sight, the orange perpendicular to it. Each component would be 70% of the initial speed, so that would give an upward Doppler shift of 70% of the maximum, as the Doppler shift is due only to the line-of-sight component. Diagram of a falling bomb, showing the components of velocity. Just before the bomb hits, though, the velocity would be nearly perpendicular to the line of sight, so the Doppler shift would be zero (to first order). Which would be a decrease in pitch from the starting condition. So maybe somebody on the ground would hear a decreasing pitch after all, at least provided they’re a fair distance from the impact point. Of course, if the bomb was released from rest, it would be speeding up, and the increasing shift from that might be bigger than the angle effect. But once the bomb reached its terminal velocity, the angle effect would give you a reduction in frequency, in the same way that you get the characteristic “eeeeeeee-owwwwwww” effect as a fast car goes by– the Doppler shift moves the frequency up as it comes toward you, is zero just as it passes, and then shifts the frequency down as it recedes. Which of these effects would win out is difficult to say. This, of course, demands some sort of VPython simulation, but I have too much other stuff to do this morning– I’ve already spent an order of magnitude more time on this post than I intended to… And, of course (to head off an inevitable comment) I’m sure this is Google-able, but I offer it here as an example of the thought processes of a physicist confronted with an anomalous result. And also an example of the weird stuff that distracts me when I’m playing with the kids… 1. #1 Thony C March 7, 2014 Only the people on the ground hear the bomb, people in the plane don’t. 2. #2 Thony C March 7, 2014 I’ve read lots of accounts that the bomb stops making any sound shortly before it hits??!! 3. #3 weirdnoise March 7, 2014 I’m more interested in the physics of just how the bomb generates the sound. Is it an intentional part of the design? (A bit of googling about whistling bombs suggests that sometimes it was — but other sources say it’s a side-effect of the aerodynamic design of the bomb, or lack thereof.) 4. #4 Eric Lund March 7, 2014 Of course, one reason why you don’t hear reports from ground zero is that people and equipment placed there tend not to survive the impact and explosion (which of course is the purpose of the bomb). I also suspect the sound comes not from the bomb itself (unless it is specifically equipped) but from the bow wave in front of the bomb. The frequency of sound emitted is probably not the same for all directions, although I haven’t done the calculation. That plus Doppler shift should cover it all. The intensity certainly won’t be isotropic either: it will be concentrated at the front of the bow wave. 5. #5 Uncle Al March 7, 2014 http://www.cnn.com/2014/03/04/us/ohio-boy-suspended-finger-gun/ Normal children are doubleplusungood thoughtcrime. http://cdn.baekdal.com/2008/gun2.jpg OTOH, go with the flow. 6. #6 Neil Howlett March 7, 2014 There are several elements to this. The Stuka dive bomber was fitted with a siren for psychological effect, so with that it was the aeroplane that made the noise. With the V1 it was never heard while with the V2 the noise of the engine was heard until it cut out. Technically neither of those were bomb noises. 7. #7 Rick Pikul March 8, 2014 @Neil Howlett Other way around with the V1 and V2: The V1 was the one with a noisy pulsejet, the V2 was the supersonic ballistic missile. 8. #8 dean March 8, 2014 Would the same line of thought apply to the sound associated with artillery shells? 9. #9 Tim Eisele March 9, 2014 I can vouch for the ‘descending whistle’ sound in the case of falling bowling balls, at least. From an estimated peak altitude of about 1500 feet, and standing approximately 1000 feet away from the ultimate impact point, the sequence goes: – Whistle starts by getting gradually louder at a more or less constant pitch. – In the last few seconds of the drop, the pitch drops noticeably, all the way down to the ground. – There is a sudden “Thump”, then silence. There is no loss of sound before impact, probably because I had a clear line of sight to the impact point. If there had been trees, buildings, or other obstructions, it might have blocked the sound for the last fraction of a second. 10. #10 Anders Ehrnberg Stockholm Sweden March 9, 2014 From a non-diving aircraft the bomb starts with a horisontal velocity the same as the aircraft and keeps much of that till impact so the position of the listener relative releasing point ought to have an impact on the Doppler effect. Anders Eg	1,850	8,026	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43	latest	en	0.962448
https://www.physicsforums.com/threads/his-average-speed-for-the-first-4km-is-what.748177/	1,518,905,981,000,000,000	text/html	crawl-data/CC-MAIN-2018-09/segments/1518891807825.38/warc/CC-MAIN-20180217204928-20180217224928-00797.warc.gz	946,054,415	20,309	# His average speed for the first 4km is what? 1. Apr 11, 2014 ### chikis A man runs a distance of 9km at a constant speed for the first 4km and then 2km/h faster for the rest of the distance. The whole run takes him one hour. His average speed for the first 4km is what? Total distance = 9km Total time = 1 h For the 1st for 4km, he moved at a constant speed. For the last 5km, his speed was increased by 2km/h. Let the costant speed at 4km be xkm/h. At the last 5km, his speed became (x + 2) km/h. The total speed for the whole runs took him [(x+2) + x] km/h = (2x + 2) km/h speed s = distance d/time t s = d/t t = 9/(2x + 2) 1 = 9/(2x + 2) 2x + 2 = 9 2x = 7 x = 7/2 = 3.5 km/h Therefore the speed for the 4 km = 3.5 km/h Since the speed was increased by 2km/h, that means the speed for the last 5 km/h would become (3.5 + 2) = 5.5 km/h. But when I checked, my answer did not seem right. Please I need help. Thank you. 2. Apr 11, 2014 ### jbriggs444 This seems to assume that the first half and the last half of the run took equal amounts of time. But that assumption may not be true. [An "average" speed would normally be weighted based on time -- for instance, 2 km/h for 20 minutes and 4 km/h for 40 minutes would average to 3 1/3 km/h] How much time did the first 4 km take in terms of x? How much time did the last 5 km take in terms of x? What do these two times add up to? Solve for x. 3. Apr 11, 2014 ### nasu Assuming equal time is not the only problem. Adding the speeds for the two segments does not make any sense whatsoever. Or for any segments. Why not add the speeds for any minute of that hour? You will get a much larger speed. You need to write the equations for each segment, as suggested by jb. 4. Apr 11, 2014 ### HallsofIvy If he ran 4 km at speed x km/hr, then he took 4/x hrs. If he ran 5 km at speed x+ 2 km/hr, then he took 5/(x+ 2) hrs. Altogether he ran 9 km in (4/x)+ 5/(x+ 2) hours but we are told he took one hour. Solve (4/x)+5/(x+ 2)= 1. 5. Apr 12, 2014 ### chikis 4/x +(5/x 2) = 1 9x+8 = x2+2x x2 - 7x - 8 = 0 (x 1)(x-8) = 0 x = 8 or x = -1 The correct answer is 8 km/h. My observation here is that we can only make equation for the total time and use it to find the speed. Why is that if we make equation for the total distance (as I did in my previous work), we cannot use it to get the accurate value for x the speed? 6. Apr 12, 2014 ### Staff: Mentor Sure you can, but you have to set your equations up correctly. For the first part of his run, we have 4 = r * t For the second part of his run, where he runs faster, we have 5 = (r + 2)(1 - t) In these equations, r is his running speed in the first leg, and t is his time for the first leg. On the second leg of his run, his speed is r + 2 (in km/hr), and his time is 1 - t (in hours). If you solve the first equation for t, and substitute it into the second equation, you eventually end up with the same quadratic equation you have. BTW, using x for the speed is not a very good choice. It's better to use a letter that makes it more obvious what it represents, which is why I chose r (for rate) and t (for time). 7. Apr 13, 2014 ### vela Staff Emeritus Your approach was flawed. As nasu pointed out, adding the two speeds doesn't make any sense. Consider these examples: Suppose you drive at 10 km/h for two hours and then 20 km/h for another hour. The total distance you drove would be 40 km, and the total time would be three hours. Your average speed would be 40/3 km/h, about 13.3 km/h. This result should seem reasonable. The average speed is between the lowest speed and the highest speed. It tends to be closer to the lower speed because you drove longer at that speed than at the higher speed. Suppose instead you drive at 10 km/h for one hour and then 20 km/h for two hours. This time, the total distance is 50 km, and your average speed would be 50/3 km/h, about 16.7 km/h. This result also seems reasonable. Again, the average speed is between two speeds. It tends to be closer to the higher speed this time because you drove longer at that speed than at the lower speed. Now based on your reasoning, the "total speed" is 30 km/h in both cases. Does this make sense? No. For one thing, in both cases, the car always went slower than that speed. Second, you have the same total speed in both cases. Unlike the average speed and the total distance travelled, the so-called total speed apparently doesn't depend on how much time was spent at either speed. So what exactly is this speed supposed to represent physically? The answer is it doesn't represent anything physically meaningful. 8. Apr 13, 2014 ### chikis Please expanciate a little bit. What is the thinking behind the expression (1 - t) ? Last edited: Apr 13, 2014 9. Apr 13, 2014 ### SammyS Staff Emeritus If t represents the time to complete the first 4 km and the total time taken to complete the whole 9 km is 1 hour, how long does it take to run the last 5 km ? 10. Apr 13, 2014 ### chikis I know that t represent time. Am now asking 1-t is what? 11. Apr 13, 2014 ### vela Staff Emeritus You should be able to reason that out yourself based on SammyS's hint. 12. Apr 13, 2014 ### Staff: Mentor No, t represents the amount of time to run the first 4Km. From my earlier post... Also see Sammy's hint in post #9. 13. Apr 14, 2014 ### chikis (1-t) is nothing more than subtracting the time for the first 4km from the total time, which gives the time for the last 5km. 14. Apr 15, 2014 ### SammyS Staff Emeritus Yes ! So, now you can get an expression (in terms of t) for the speed over the first 4 km and another expression (in terms of t) for the speed over the remaining 5 km . You also know how these two expressions are related to each other, which allows you to get an equation with both these expressions in it. This equation has only one variable, t, so solve for t. 15. Apr 17, 2014 ### chikis Solving that way we make us to have two unknowns and that will make the problem too dificult to solve. 16. Apr 17, 2014 ### Staff: Mentor You didn't read what I wrote in post 6. After making the substitution, you have a quadratic equation with one variable.	1,803	6,205	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.34375	4	CC-MAIN-2018-09	longest	en	0.969396
https://blog.birost.com/a?ID=00000-7702e860-4c9b-4d68-bbcb-544f66b40fa5	1,660,143,832,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882571190.0/warc/CC-MAIN-20220810131127-20220810161127-00189.warc.gz	151,962,415	10,268	# The MATLAB Realization of Particle Swarm Algorithm for Finding the Maximum Value of Unary Function In the previous essay that introduced the particle swarm algorithm to realize the grouping backpack in detail, the main idea of the particle swarm algorithm has been introduced in detail. If you have mastered how to use the particle swarm algorithm to realize the grouping backpack, you should modify it into a unary function to find the best The application of value is simply a piece of cake. Here is a copy of the general idea of using the particle swarm algorithm to achieve grouping backpacks summarized before: • A bunch of particles are randomly generated, and each particle represents a situation of the backpack (which 3 items are selected), and the initial particles are all locally optimal particles . • Calculate the fitness value of each particle (that is, the value, volume, and weight of the backpack represented by each particle), and then select non-inferior particles according to the fitness of each particle . • Then enter the iterative process. In each iteration, randomly select one of the non-inferior particles as the global optimal particle , and calculate the particle velocity according to the formula (closely related to the better particle and the global optimal particle), so that each particle moves (also That is, there is a probability to replace the representative item). • If the particles after moving particles than before and more preferably the particles to replace the original local optimal • Then the non-inferiority local optimum particle and particle put together, and then press the pros and cons of selected non-inferiority particles • Remove the repeated non-inferior particles and enter the next iteration. Now make the following changes to the above thought for finding the minimum value of the unary function: • A bunch of particles are randomly generated, and each particle represents an abscissa. The initial particles are all locally optimal particles . • Calculate the fitness value of each particle (that is, the ordinate value represented by each particle), and then select the initial non-inferior particles according to the fitness of each particle . • Then enter the iterative process. In each iteration, randomly select one of the non-inferior particles as the global optimal particle , and calculate the particle velocity according to the formula (closely related to the better particle and the global optimal particle), so that each particle will move (also That is, the abscissa moves to the left or right). • If the particles after moving particles than before and more preferably the particles to replace the original local optimal • Then the non-inferiority local optimum particle and particle put together, and then press the pros and cons of selected non-inferiority particles • Remove the repeated non-inferior particles and enter the next iteration. ## One: the problem of the most value of the unary function Suppose there are the following functions: $y(x) = sin(x^2)-2sin(x) + x * sin(x)$ Draw the image of this function as shown in the figure: How do we find the minimum value of the function between [-3, 9] ? First put a dynamic graph showing how the particle swarm finds the minimum value: It can be seen from the figure that our husband has 50 particles, which are represented by blue dots and red dots for non-inferior particles (the last minimum point must be included in the non-inferior particles), but Here we can see that there is always only one non-inferior solution particle, so it is the minimum point we are seeking. It can be seen that the position of the particle will change in each iteration, and it will continue to move to the minimum position. The following specifically describes how the particle swarm algorithm finds the most value of the unary function. Still follow the previous essay to decompose the complete MATLAB code into seven parts to explain. ## 2: The particle swarm algorithm to find the most value problem of the unary function ### 2.1 Input parameters, fixed parameter initialization Note that the inertia weight value in the parameter can be modified to a value that you think is appropriate, it will affect the step length of each movement of the particle. The larger wmax is, the larger the particle motion step at the beginning of the iteration; the smaller the wmin, the smaller the particle motion step at the end of the iteration. clear, clc, close all; %% input parameters, fixed parameter initialization xsize = 50 ; % number of particles ITER = 50 ; % number of iterations c1 = 0.8 ; c2 = 0.8 ; % constant wmax = 1.0 ; wmin = 0.01 ; % inertia weight related constant v = zeros ( 1 , the xsize); % particle velocity initialization copy the code ### 2.2 Initialization of particle swarm position, fitness, best position, and best fitness Randomly generate particle swarms$x$ , represents the abscissa of each particle, pay attention to the$[a, b]$ uniformly distributed random numbers between, use$a + (b-a) * rand(1,1)$ This expression. The fitness is actually the ordinate of the particle swarm. %% particle swarm position, fitness, best position, best fitness initialization x = -3 + 12 * rand ( 1 , xsize); % random particle swarm position generation (representing the abscissa [-3, 9]) % Particle swarm fitness y = zeros ( 1 , xsize); % Particle swarm ordinate for i = 1 : xsize y( i ) = sin (x( i ).^ 2 ) -2 * sin (x( i )) + x( i ) * sin (x( i )); end X = bestx; % value particle swarm optimum position besty = y; best fitness% particle group duplicated code ### 2.3 Initial screening of non-inferior solutions The non-inferior solutions are screened for the first time, and after each iteration, they are screened again. The judgment condition is very important and can be changed according to the limitation of the problem. Here is to judge whether each particle is more in line with the requirements than all other particles (the ordinate is smaller). %% Initial screening non-inferior solution cnt = 1 ; for i = 1 : xsize fljflag = 1 ; for j = 1 : xsize if j ~= i if y( i )> y( j ) % i is an inferior solution fljflag = 0 ; break ; end end end if fljflag == 1 flj (cnt) = y ( I ); % Pareto fitness fljx (CNT) = X ( I ); % Pareto position CNT = CNT + . 1 ; End End copy the code ### 2.4 Particle motion calculation The calculation of particle velocity still follows the following classic formula: $v^{i+1} = wv^i + c1r1(p_{local}^i-x^i) + c2r2(p_{global}-x^i)$ among them$w$ is the inertia weight,$c1$ and$c2$ is a constant,$r1$ and$r2$ is a uniformly distributed random number between [0,1],$p_{local}^i$Is the local optimal particle;$p_{global}$It is the global optimal particle. Note that there is only one global optimal, so there is no superscript$i$ . The value of the inertia weight is related to the number of iterations, here we use$w = wmax-(wmax-wmin) * niter/iterall$ such a calculation method. The calculation of related inertial weights is also a hot spot in the research of particle swarm algorithm. The inertial weights change greatly, the particle speed is fast, and the position change is also fast. If the inertial weights are obtained, the particle swarm can converge to the global optimum faster! When using speed to update the position of each particle, pay attention to a problem, that is, do not let the abscissa of the moving particle exceed our judgment interval$[-3, 9]$ again. for niter = 1 : ITER % The iteration starts, the particles start to move xx = [ -10 : 0.01 : 10 ]; % For drawing, yy = sin (xx.^ 2 ) -2 * sin (xx) + xx .* sin (xx ); plot (xx, yy, 'k' ); xlim([ -10 , 10 ]);ylim([ -8 , 10 ]); title( 'Particle Swarm Results' ); xlabel( 'x' ); ylabel( 'y' ); hold on; rnd = randi( length ( flj ), 1 , 1 ); gbest = fljx(rnd); % the global optimal solution of the particle %% Particle motion calculation w = wmax-(wmax-wmin) * niter/ITER; % inertia weight update r1 = rand ( 1 , 1 ); r2 = rand ( 1 , 1 ); % generates $[0, 1]$ Uniformly distributed random value for i = 1 : xsize v( i ) = w * v( i ) + c1 * r1 * (bestx( i )-x( i )) + c2 * r2 * (gbest-x( i )); % particle velocity x( i ) = x ( i ) + v( i ); if (x( i )> 9 \|\| x( i ) < -3 ); % After the exercise, the range of x is exceeded, replace x( with a new random number i ) = -3 + 12 * rand ( 1 , 1 ); end endCopy code ### 2.5 Current particle swarm fitness, best position, best fitness The particles have moved to a new position. Of course, the new particles must be compared with the old particles. If the fitness of the new particles is better than the old particles (the ordinate is smaller), the local optimal particle position is updated. Unlike the knapsack problem, which requires multiple judgment conditions to be considered, the judgment here is quite simple, regardless of other situations. y_cur = zeros ( 1 , xsize); for i = 1 : xsize y_cur( i ) = sin (x( i ).^ 2 ) -2 * sin (x( i )) + x( i ) * sin (x( i )); end for i = 1 : xsize if y_cur( i ) <y( i ) % if the current particle fitness is better bestx( i ) = x( i ); % update the best position of the particle besty( i ) = y_cur( i ); % Particle's best fitness update end end y_cur = Y; % Old particles New particles become duplicated code ### 2.6 Combine the best position and best fitness of the particle swarm and then select non-inferior solutions The steps for the first non-inferior solution screening are basically the same, except that a merge operation is added to merge the local best particles with the non-inferior solution particles, and then filter a wave of non-inferior solution particles. %% The best position and best fitness of the particle swarm are combined and then filtered for non-inferior solutions bestxmerge = [bestx, fljx]; ymerge = [besty, flj]; n = length (flj); flj = []; fljx = []; cnt = 1 ; for i = 1 : xsize + n fljflag = 1 ; for j = 1 : xsize + n if j ~= i if ymerge( i )> ymerge( j ) % i is the inferior solution fljflag = 0 ; break ; end end end if fljflag == 1 flj (cnt) = ymerge( i ); % non-inferior solution fitness fljx(cnt) = bestxmerge( i ); % non-inferior solution position cnt = cnt + 1; End End copy the code ### 2.7 Remove repeated non-inferior solutions This step is also very important, and there are many ways to implement it. Just choose one to remove the repetitive non-inferior solution. %% remove repeated non-inferior solutions issame = zeros (cnt- 1 , 1 ); for i = 1 : cnt- 1 for j = i + 1 : cnt- 1 if ~issame( j ) issame( j ) = ( abs (fljx( j )-fljx( i )) < 0.0001 ); end end end flj ( find (issame == 1 )) = []; fljx( find (issame == 1 )) = []; Plot (bestxmerge, ymerge, 'BO' ); % particle group Videos optimum position Plot (fljx, FLJ, 'RO' ); % Videos Pareto position PAUSE ( 0.5 ); HOLD OFF; End copy the code	2,790	10,853	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "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}	3.609375	4	CC-MAIN-2022-33	latest	en	0.856193
https://www.johndcook.com/blog/2020/03/29/lee-distance-codes-and-music/?utm_source=feedburner&utm_medium=feed&utm_campaign=Feed%3A+TheEndeavour+%28The+Endeavour%29	1,591,375,165,000,000,000	text/html	crawl-data/CC-MAIN-2020-24/segments/1590348502097.77/warc/CC-MAIN-20200605143036-20200605173036-00214.warc.gz	759,634,164	10,828	Lee distance: codes and music The Hamming distance between two sequences of symbols is the number of places in which they differ. For example, the Hamming distance between the words “hamming” and “farming” is 2, because the two worlds differ in their first and third letters. Hamming distance is natural when comparing sequences of bits because bits are either the same or different. But when the sequence of symbols comes from a larger alphabet, Hamming distance may not be the most appropriate metric. Here “alphabet” is usually used figuratively to mean the set of available symbols, but it could be a literal alphabet. As English words, “hamming” seems closer to “hanning” than to “farming” because m is closer to n, both in the alphabet and phonetically, than it is to f or r. [1] The Lee distance between two sequences x1x2xn and y1y2yn of symbols from an alphabet of size q is defined as So if we use distance in the English alphabet, the words “hamming” and “hanning” are a Lee distance of 1 + 1 = 2 apart, while “hamming” and “farming” are a Lee distance of 2 + 5 = 7 apart. Coding theory uses both Hamming distance and Lee distance. In some contexts, it only matters whether symbols are different, and in other contexts it matters how different they are. If q = 2 or 3, Hamming distance and Lee distance coincide. If you’re working over an alphabet of size q > 3 and symbols are more likely to be corrupted into nearby symbols, Lee distance is the appropriate metric. If all corruptions are equally likely, then Hamming distance is more appropriate. Application to music Lee distance is natural in music since notes are like integers mod 12. Hence the circle of fifths. My wife and I were discussing recently which of two songs was in a higher key. My wife is an alto and I’m a baritone, so we prefer lower keys. But if you transpose a song up so much that it’s comfortable to sing an octave lower, that’s good too. If you’re comfortable singing in the key of C, then the key of D is two half-steps higher. But what about they key of A? You could think of it as 9 half-steps higher, or 3 half-steps lower. In the definition of Lee distance, measured in half-steps, the distance from C to D is min{2, 12 – 2} = 2, i.e. you could either go up two half-steps or down 10. Similarly the distance between C and A is min{9, 12-9} = 3. So you could think of the left side of the minimum in the definition of Lee distance as going up from x to y and the right side as going down from x to y. Using Lee distance, the largest interval is the tritone, the interval from C to F#. It’s called the tritone because it is three whole steps. If C is your most comfortable key, F# would be your least comfortable key: the notes are as far away from your range as possible. Any higher up and they’d be closer because you could drop down an octave. The tritone is like the hands of a clock at 6:00. The hour and minute hands are as far apart as possible. Just before 6:00 the hands are closer together on the left side of the clock and just after they are closer on the right side of the clock. Related posts [1] I bring up “Hanning” because Hamming and Hanning are often confused. In signal processing there is both a Hamming window and a Hanning window. The former is named after Richard Hamming and the latter after Julius von Hann. The name “Hanning window” rather than “Hann window” probably comes from the similarity with the Hamming window.	829	3,454	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24	latest	en	0.946238
http://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq?hide_answers=1&beginning=720	1,369,001,770,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368698104521/warc/CC-MAIN-20130516095504-00091-ip-10-60-113-184.ec2.internal.warc.gz	311,307,960	10,603	# Questions on Algebra: Trigonometry answered by real tutors! Algebra -> Algebra -> Trigonometry-basics -> Questions on Algebra: Trigonometry answered by real tutors! 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: Trigonometry Solvers Lessons Answers archive Quiz In Depth Question 73561: Find the real and complex solution of the equation Click here to see answer by johnpickett(2) Question 73840: Trigonometry-basics/73819 (2007-03-08 13:33:50): I am studying DeMoivre's Theorum in Aplications of Trigonometry, I was asked to find the real and complex solution to the equation Possible answers a. -2,2, b. 2,, c. -2,, d. ,, Click here to see answer by stanbon(57226) Question 73916: Find the exact value of sin^-1(-sqrt3/2). a.2sin(4t)cos(2t) b.2cos(4t)sin(2t) c.2cos(4t)cos(2t) d.2sin(4t)sin(2t) Than you for you help Click here to see answer by stanbon(57226) Question 73914: If (theta) is a second-quadrant with cos(theta)=, find the exact value of cos2(theta). a. -3/5 b. -2/5 c. 2/5 d. 3/5 Click here to see answer by stanbon(57226) Question 73914: If (theta) is a second-quadrant with cos(theta)=, find the exact value of cos2(theta). a. -3/5 b. -2/5 c. 2/5 d. 3/5 Click here to see answer by bucky(2189) Question 73911: Express as a trigonometric function of one angle sin44cos22-cos44sin22. (all Numbers are degrees) a. cos22 b. cos66 c. sin22 d. sin66 Click here to see answer by stanbon(57226) Question 73910: Use the addition and subtraction formulas to simplify cos(x-). a. cos x b. sin x c. -cos x d. -sin x Click here to see answer by bucky(2189) Question 73908: Find all solutions to the equation sin x = 0. a. x = n b. x = 2n c. x = n d. x = + 2 n Click here to see answer by bucky(2189) Question 73943: Find all solutions in the interval [0,2PI] to the equation . a. x = 0, PI/2 b. x = 0, PI c. x = PI/2 3PI/2 d. x = 0, PI/2, PI, PI/2 Click here to see answer by jim_thompson5910(28476) Question 73943: Find all solutions in the interval [0,2PI] to the equation . a. x = 0, PI/2 b. x = 0, PI c. x = PI/2 3PI/2 d. x = 0, PI/2, PI, PI/2 Click here to see answer by ffc_01(9) Question 73942: Find the exact value of cos^-1 the possible answers are a. -PI/3 b. PI/3 c. 2PI/3 d. 4PI/3 Thank you Click here to see answer by jim_thompson5910(28476) Question 74067: A sequence of diagonal lines is drawn from points (1,0),(2,0),(3,0),(4,0),...(10,0) on the x-axis to corresponding points (0,1),(0,2),(0,3),(0,4),...(0,10) on the y-axis. What is the total length of those lines? Click here to see answer by stanbon(57226) Question 74071: Give the sum p+q where p/q is the reduced fraction represented by the repeating decimal 2.1. Click here to see answer by stanbon(57226) Question 74069: Find the value of x where 201, x, 1089 are three consecutive terms in a geometric sequence. Click here to see answer by checkley75(3666) Question 74224: Find the sum of the infinite geometric series 1 + 3/5 + 9/25 + 27/125 + .... Click here to see answer by bucky(2189) Question 74227: What is the coefficient of x^5 in the expansion of (x+3)^8? Click here to see answer by jim_thompson5910(28476) Question 74355: Express the sum 1x2 + 2x3 + 3x4 + 4x5 + ... + 10x11 in summation notation. Click here to see answer by jim_thompson5910(28476) Question 74358: Given that n^2 + n = 2k, how can (n+1)^2 + (n+1) be written as a multiple of 2? Click here to see answer by stanbon(57226) Question 74356: Find the a(sub2) in the arithmetic sequence where a(sub5) = 42 and a(sub10) = 62. Click here to see answer by stanbon(57226) Question 74471: How many 2-digit numbers can be formed with the digits 1,2,3,4,5,6? Click here to see answer by scott8148(6628) Question 74586: A class assignment requires that a student work 4 problems from a collection of 12. How many ways are there to complete the assignment? Click here to see answer by stanbon(57226) Question 74587: Find the number of distinguisable permutations of the letters in the word PEPPERY. Click here to see answer by stanbon(57226) Question 74587: Find the number of distinguisable permutations of the letters in the word PEPPERY. Click here to see answer by Nate(3500) Question 74588: A single card is drawn from a standard deck. Find the probability that the card is a spade or an ace? Click here to see answer by Nate(3500) Question 74588: A single card is drawn from a standard deck. Find the probability that the card is a spade or an ace? Click here to see answer by stanbon(57226) Question 74589: Tweo six sided dice are tossed. Find the probability that the sum is 2 or 12. Click here to see answer by Nate(3500) Question 74592: Five fair coins are tossed. Find the probability of obtaining 4 heads and 1 tail. Click here to see answer by Nate(3500) Question 74592: Five fair coins are tossed. Find the probability of obtaining 4 heads and 1 tail. Click here to see answer by stanbon(57226) Question 74593: Two balls are drawn (without replacement) from an urn that contains four balls, five green balls and seven white balls. Find the probability that both balls are white. Click here to see answer by Nate(3500) Question 74591: Use the formula a(sub1) + a(sub1)r + a(sub1)r^2 + ...+ a(sub1)r^n-1 = a(sub1) 1-r^n/1-r for the nth partial sum of a geometric sequence to find the sume 1 + 2/3 + 4/9 + 8/27 + ... + 64/729. Click here to see answer by stanbon(57226) Question 74475: I have a question- what equation do I use to work out the angles in a right angled triangle if there is a 5.6m line extending up the y axis and a 3.6m line extending along the x axis - both at a right angle to each other. There is an angled line from the top of the 5.6m line to the end of the 3.6m line. The angle i'm after is between the top of the 5.6m line to the line extended to the end of the 3.6m line.??? If you could help me out it would be great!!! Click here to see answer by scott8148(6628) Question 74630: A triangle has a side 100 feet long and a second side 35 feet long with a 25 degree angle between them. What is the approximate area of the triangle? Click here to see answer by venugopalramana(3286) Question 74692: Can someone help solve this? Let H(t)=4+4sin(pi t/2). Approximate to the nearest hundredth all values t in the interval [0,4 pi] for which Ht=6. Click here to see answer by venugopalramana(3286) Question 74691: Could someone help solve this question? sqrt2sinx = 1 Click here to see answer by venugopalramana(3286) Question 74754: How do you draw this angle 2PI/9? i know that you need to convert it into degrees so what i got so far is 2PI/9 * 180/PI = 360PI/9PI i have no clue if im even on the right track so can you please help me? Click here to see answer by Earlsdon(6287) Question 74754: How do you draw this angle 2PI/9? i know that you need to convert it into degrees so what i got so far is 2PI/9 * 180/PI = 360PI/9PI i have no clue if im even on the right track so can you please help me? Click here to see answer by jim_thompson5910(28476) Question 74773: OK now i have a question that i TOTALLY don't get. Its how do you find the arc length and area of a sector with the given..... r=2cm, and the weird 0 with the slash through it =9PI/20 Please help!!! Click here to see answer by jim_thompson5910(28476) Question 74789: Express the complex number 12\|cos(pi/3)+isin(pi/3)\| in the form of a+bi without trigonometric functions. Click here to see answer by jim_thompson5910(28476) Question 75358: Find an equation in x and y for the curve given by the parametric equation x=e^-t, y=-2t, t in R. I came up with y=x^2, but don't feel it's right. Please help. Thanks Click here to see answer by nilan(15) Question 75884: Please help me tan(3x)+1={square root(2)}sec(3x) so far i have tried: (tan(3x)+1)^2=(sq rt(2))^2(sec(3x))^2 tan^2(9x^2)+1^2=2sec^2(9x^2) tan^2 +1=2sec^2 tan^2+1 = sec^2 ------- 2 I am totally confused now. This problem was assigned by my teacher. We are using Trigonometry a Unit Circle Approach 7th edition by Sullivan ISBN # 0-13-143111-0 Click here to see answer by kev82(148) Question 76323: what is the square root of 3 time the square root of negative 3? Click here to see answer by stanbon(57226) Question 76373: 4^x + 4^-x = 5/2. What is x Click here to see answer by ankor@dixie-net.com(15628) Question 77181: if tanA + secA = 2 where a is acute then which option is correct. a) tan A < Sec A b) Cosec A > CotA c) Sec A > Cosec A d) CotA > Tan A Click here to see answer by Edwin McCravy(8880) Question 77324: if f(x)=3sinx + 2cos2x, the value of (Pie/2) is Click here to see answer by jim_thompson5910(28476) Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495	4,150	11,266	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20	latest	en	0.802177
https://wikimili.com/en/Compression_(physics)	1,670,508,042,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00670.warc.gz	640,182,819	14,023	# Compression (physics) Last updated In mechanics, compression is the application of balanced inward ("pushing") forces to different points on a material or structure, that is, forces with no net sum or torque directed so as to reduce its size in one or more directions. [1] It is contrasted with tension or traction, the application of balanced outward ("pulling") forces; and with shearing forces, directed so as to displace layers of the material parallel to each other. The compressive strength of materials and structures is an important engineering consideration. ## Contents In uniaxial compression, the forces are directed along one direction only, so that they act towards decreasing the object's length along that direction. [2] The compressive forces may also be applied in multiple directions; for example inwards along the edges of a plate or all over the side surface of a cylinder, so as to reduce its area (biaxial compression), or inwards over the entire surface of a body, so as to reduce its volume. Technically, a material is under a state of compression, at some specific point and along a specific direction ${\displaystyle x}$, if the normal component of the stress vector across a surface with normal direction ${\displaystyle x}$ is directed opposite to ${\displaystyle x}$. If the stress vector itself is opposite to ${\displaystyle x}$, the material is said to be under normal compression or pure compressive stress along ${\displaystyle x}$. In a solid, the amount of compression generally depends on the direction ${\displaystyle x}$, and the material may be under compression along some directions but under traction along others. If the stress vector is purely compressive and has the same magnitude for all directions, the material is said to be under isotropic or hydrostatic compression at that point. This is the only type of static compression that liquids and gases can bear. [3] In a mechanical wave which is longitudinal, the medium is displaced in the wave's direction, resulting in areas of compression and rarefaction. ## Effects When put under compression (or any other type of stress), every material will suffer some deformation, even if imperceptible, that causes the average relative positions of its atoms and molecules to change. The deformation may be permanent, or may be reversed when the compression forces disappear. In the latter case, the deformation gives rise to reaction forces that oppose the compression forces, and may eventually balance them. [4] Liquids and gases cannot bear steady uniaxial or biaxial compression, they will deform promptly and permanently and will not offer any permanent reaction force. However they can bear isotropic compression, and may be compressed in other ways momentarily, for instance in a sound wave. Every ordinary material will contract in volume when put under isotropic compression, contract in cross-section area when put under uniform biaxial compression, and contract in length when put into uniaxial compression. The deformation may not be uniform and may not be aligned with the compression forces. What happens in the directions where there is no compression depends on the material. [4] Most materials will expand in those directions, but some special materials will remain unchanged or even contract. In general, the relation between the stress applied to a material and the resulting deformation is a central topic of continuum mechanics. ## Uses Compression of solids has many implications in materials science, physics and structural engineering, for compression yields noticeable amounts of stress and tension. By inducing compression, mechanical properties such as compressive strength or modulus of elasticity, can be measured. [5] Compression machines range from very small table top systems to ones with over 53 MN capacity. Gases are often stored and shipped in highly compressed form, to save space. Slightly compressed air or other gases are also used to fill balloons, rubber boats, and other inflatable structures. Compressed liquids are used in hydraulic equipment and in fracking. ## In engines ### Internal combustion engines In internal combustion engines the explosive mixture gets compressed before it is ignited; the compression improves the efficiency of the engine. In the Otto cycle, for instance, the second stroke of the piston effects the compression of the charge which has been drawn into the cylinder by the first forward stroke. [6] ### Steam engines The term is applied to the arrangement by which the exhaust valve of a steam engine is made to close, shutting a portion of the exhaust steam in the cylinder, before the stroke of the piston is quite complete. This steam being compressed as the stroke is completed, a cushion is formed against which the piston does work while its velocity is being rapidly reduced, and thus the stresses in the mechanism due to the inertia of the reciprocating parts are lessened. [7] This compression, moreover, obviates the shock which would otherwise be caused by the admission of the fresh steam for the return stroke. ## Related Research Articles A reciprocating engine, also often known as a piston engine, is typically a heat engine that uses one or more reciprocating pistons to convert high temperature and high pressure into a rotating motion. This article describes the common features of all types. The main types are: the internal combustion engine, used extensively in motor vehicles; the steam engine, the mainstay of the Industrial Revolution; and the Stirling engine for niche applications. Internal combustion engines are further classified in two ways: either a spark-ignition (SI) engine, where the spark plug initiates the combustion; or a compression-ignition (CI) engine, where the air within the cylinder is compressed, thus heating it, so that the heated air ignites fuel that is injected then or earlier. In physics, a transverse wave is a wave whose oscillations are perpendicular to the direction of the wave's advance. This is in contrast to a longitudinal wave which travels in the direction of its oscillations. Water waves are an example of transverse wave. Crystal optics is the branch of optics that describes the behaviour of light in anisotropic media, that is, media in which light behaves differently depending on which direction the light is propagating. The index of refraction depends on both composition and crystal structure and can be calculated using the Gladstone–Dale relation. Crystals are often naturally anisotropic, and in some media it is possible to induce anisotropy by applying an external electric field. Birefringence is the optical property of a material having a refractive index that depends on the polarization and propagation direction of light. These optically anisotropic materials are said to be birefringent. The birefringence is often quantified as the maximum difference between refractive indices exhibited by the material. Crystals with non-cubic crystal structures are often birefringent, as are plastics under mechanical stress. In continuum mechanics, stress is a physical quantity. It results when forces like tension or compression act on a body. The greater this force and the smaller the cross-sectional area of the body on which it acts, the greater the stress. So stress is measured in newton per square meter (N/m2) or pascal (Pa). In materials science and solid mechanics, Poisson's ratio (nu) is a measure of the Poisson effect, the deformation of a material in directions perpendicular to the specific direction of loading. The value of Poisson's ratio is the negative of the ratio of transverse strain to axial strain. For small values of these changes, is the amount of transversal elongation divided by the amount of axial compression. Most materials have Poisson's ratio values ranging between 0.0 and 0.5. For soft materials, such as rubber, where the bulk modulus is much higher than the shear modulus, Poisson's ratio is near 0.5. For open-cell polymer foams, Poisson's ratio is near zero, since the cells tend to collapse in compression. Many typical solids have Poisson's ratios in the range of 0.2–0.3. The ratio is named after the French mathematician and physicist Siméon Poisson. The field of strength of materials, also called mechanics of materials, typically refers to various methods of calculating the stresses and strains in structural members, such as beams, columns, and shafts. The methods employed to predict the response of a structure under loading and its susceptibility to various failure modes takes into account the properties of the materials such as its yield strength, ultimate strength, Young's modulus, and Poisson's ratio. In addition, the mechanical element's macroscopic properties such as its length, width, thickness, boundary constraints and abrupt changes in geometry such as holes are considered. In mechanics, compressive strength or compression strength is the capacity of a material or structure to withstand loads tending to reduce size. In other words, compressive strength resists compression, whereas tensile strength resists tension. In the study of strength of materials, tensile strength, compressive strength, and shear strength can be analyzed independently. A Newtonian fluid is a fluid in which the viscous stresses arising from its flow are at every point linearly correlated to the local strain rate — the rate of change of its deformation over time. Stresses are proportional to the rate of change of the fluid's velocity vector. An elastic modulus is the unit of measurement of an object's or substance's resistance to being deformed elastically when a stress is applied to it. The elastic modulus of an object is defined as the slope of its stress–strain curve in the elastic deformation region: A stiffer material will have a higher elastic modulus. An elastic modulus has the form: Stress–strain analysis is an engineering discipline that uses many methods to determine the stresses and strains in materials and structures subjected to forces. In continuum mechanics, stress is a physical quantity that expresses the internal forces that neighboring particles of a continuous material exert on each other, while strain is the measure of the deformation of the material. This is an alphabetical list of articles pertaining specifically to mechanical engineering. For a broad overview of engineering, please see List of engineering topics. For biographies please see List of engineers. A triaxial shear test is a common method to measure the mechanical properties of many deformable solids, especially soil and rock, and other granular materials or powders. There are several variations on the test. Material failure theory is an interdisciplinary field of materials science and solid mechanics which attempts to predict the conditions under which solid materials fail under the action of external loads. The failure of a material is usually classified into brittle failure (fracture) or ductile failure (yield). Depending on the conditions most materials can fail in a brittle or ductile manner or both. However, for most practical situations, a material may be classified as either brittle or ductile. The theory of micro-mechanics of failure aims to explain the failure of continuous fiber reinforced composites by micro-scale analysis of stresses within each constituent material, and of the stresses at the interfaces between those constituents, calculated from the macro stresses at the ply level. In mechanical engineering, the cylinders of reciprocating engines are often classified by whether they are single- or double-acting, depending on how the working fluid acts on the piston. The viscous stress tensor is a tensor used in continuum mechanics to model the part of the stress at a point within some material that can be attributed to the strain rate, the rate at which it is deforming around that point. The acoustoelastic effect is how the sound velocities of an elastic material change if subjected to an initial static stress field. This is a non-linear effect of the constitutive relation between mechanical stress and finite strain in a material of continuous mass. In classical linear elasticity theory small deformations of most elastic materials can be described by a linear relation between the applied stress and the resulting strain. This relationship is commonly known as the generalised Hooke's law. The linear elastic theory involves second order elastic constants and yields constant longitudinal and shear sound velocities in an elastic material, not affected by an applied stress. The acoustoelastic effect on the other hand include higher order expansion of the constitutive relation between the applied stress and resulting strain, which yields longitudinal and shear sound velocities dependent of the stress state of the material. In the limit of an unstressed material the sound velocities of the linear elastic theory are reproduced. Most of the terms listed in Wikipedia glossaries are already defined and explained within Wikipedia itself. However, glossaries like this one are useful for looking up, comparing and reviewing large numbers of terms together. You can help enhance this page by adding new terms or writing definitions for existing ones. Biaxial tensile testing is a versatile technique to address the mechanical characterization of planar materials. Typical materials tested in biaxial configuration include metal sheets, silicone elastomers, composites, thin films, textiles and biological soft tissues. ## References 1. Ferdinand Pierre Beer, Elwood Russell Johnston, John T. DeWolf (1992), "Mechanics of Materials". (Book) McGraw-Hill Professional, ISBN 0-07-112939-1 2. Erkens, Sandra & Poot, M. The uniaxial compression test. Delft University of Technology. (1998). Report number: 7-98-117-4. 3. Ronald L. Huston and Harold Josephs (2009), "Practical Stress Analysis in Engineering Design". 3rd edition, CRC Press, 634 pages. ISBN 9781574447132 4. Fung, Y. C. (1977). A First Course in Continuum Mechanics (2nd ed.). Prentice-Hall, Inc. ISBN 978-0-13-318311-5. 5. Hartsuijker, C.; Welleman, J. W. (2001). Engineering Mechanics. Volume 2. Springer. ISBN 978-1-4020-412 6. J.Heywood. Internal Combustion Engine Fundamentals 2E. McGraw-Hill Education. (2018). ISBN 9781260116113 [url=https://books.google.com/books?id=OmJUDwAAQBAJ] 7. Wiser, Wendell H. (2000). Energy resources: occurrence, production, conversion, use. Birkhäuser. p. 190. ISBN 978-0-387-98744-6.	2,957	14,579	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5	4	CC-MAIN-2022-49	latest	en	0.944005
https://www.lotterypost.com/thread/191328/7	1,481,413,620,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698543577.51/warc/CC-MAIN-20161202170903-00419-ip-10-31-129-80.ec2.internal.warc.gz	978,283,448	18,813	Welcome Guest You last visited December 10, 2016, 5:13 pm All times shown are Eastern Time (GMT-5:00) # Test Hans 3+1 Pick6 System in 30 lines Topic closed. 98 replies. Last post 8 years ago by Hans. Page 7 of 7 Welcome to vote and free test 5 winning numbers [ 8 ] [66.67%] 4 winning numbers [ 3 ] [25.00%] 3 winning numbers [ 0 ] [0.00%] 2 winning numbers [ 1 ] [8.33%] 1 winning numbers [ 0 ] [0.00%] Nothing [ 0 ] [0.00%] Total Valid Votes [ 12 ] Discarded Votes [ 1 ] Monte Carlo France Member #55589 October 9, 2007 1181 Posts Offline Posted: April 11, 2009, 5:33 pm - IP Logged many can not afford 5625 lines to cover and on the same token higher line count you hit the closer you are getting to mathematical statistical model, as the numbers have "no memory" and oscillate wildly, personally there is no way you can setup statistical model and create reduced wheel, there might be some hope in simply discounting off last numbers that occured, and playing off consecutive/non consecutive pairs in cycles. we all did great predictions and not so great.... so there is a lack of continuity.. there is no one on the lottery post board that i can say.. yes! he/she has entered some "mysterious" cycle. and the next numbers are his/hers.. the continuity/repetition of success is impossible to define, wheeling will just mathematically bump up your odds to a degree of your investment. nothing else to add to the above. lack of continuity...thats so right to say lottery is such an unlogical thing. As for the 5625 lines wheel,as each 10 numbers have 3 winning numbers,it max reach a jackpot,but as you said,its not 100% guaranteed! we all want to play to win jackpot,but we all cangt afford to play big to break the odds.thats why we use filters to have some risks,in hope of hitting something we are sure of.. Maybe sum helps in the end to deduce 5625 lines to 50 lines,which could possible cover some 4 numbers. Hans Strive to predict 8 tickets for 12 numbers with Max ROI for Pick6! Monte Carlo France Member #55589 October 9, 2007 1181 Posts Offline Posted: April 12, 2009, 4:26 am - IP Logged When you mention "hot to cold" Numbers, I assume you're talking about ball frequency. So, how far should one go? 10 draws? 100 draws? All the draws? CD Yes,its ball frequency. I normally choose past 10 drawings. this time i decide to cut 4 numbers in another way,to test somehing out.. again post 7 digits and 40 numbers,list in ball frequency way. or post 20 numbers,and give me a sum range,so i can also pick up 50 lines around the sum,to compare how many numbers works more effective. sum filter is also very popular now.I dont expect to have a jackpot,I just want to win back the investment in a certain way. The noram level ..20 numbers cover 3 numbers. hans Strive to predict 8 tickets for 12 numbers with Max ROI for Pick6! Monte Carlo France Member #55589 October 9, 2007 1181 Posts Offline Posted: April 12, 2009, 9:34 am - IP Logged since i realised there is no chance to pick up best combos,i give u chance to pick up from these two groups 6 numbers to wheel. today I went to Monaco casino and Played 5 euro for 1 hour.always stick to two strateggies and change when one fails. whilewatching most of older guys writing down numbers and trying to bet onjackpot and waste lots of money,I would choose to have some small winsand go away i think this is lottery.jackpot is never logic,but small wins can be achieved by system. Sat, Apr 11, 2009 02-05-14-31-36-41 group one has 02 14 ,02 36,group two line 1has 05 together they make 12 lines of 10 numbers,and 5 lines has 3 numbers. that means you have 50% chance to have 3 numbers in 10. 3 in 10 can give you 16 lines to play and win 3 lines of 3 numbers.prize ratio is 187% the success is mostly due to the frequency.we have missed 31 and 41 31 was deleted by ur picks and 41 is too at latter positions of all numbers. I will stick to this strategy for some rounds. Hans Strive to predict 8 tickets for 12 numbers with Max ROI for Pick6! Monte Carlo France Member #55589 October 9, 2007 1181 Posts Offline Posted: April 12, 2009, 5:02 pm - IP Logged Planning to widen to numbers range now. Post two lines of 16 numbers,and each 16 numbers would be wheeled into 40 lines.each line has 6 numbers. totally there are two groups of 40 lines.when two groups both have 3 numbers,that means a jackpot made in 12 numbers when you choose right lines from each line. I will post my own prediction myself Hans Strive to predict 8 tickets for 12 numbers with Max ROI for Pick6! Ontario Member #70051 January 23, 2009 76 Posts Offline Posted: April 17, 2009, 2:34 pm - IP Logged Hi Hans, I see that some of your other posts state 12 and 15 number selections. Which is the latest one? And how are things going with you? CD Monte Carlo France Member #55589 October 9, 2007 1181 Posts Offline Posted: April 17, 2009, 2:45 pm - IP Logged Hi Hans, I see that some of your other posts state 12 and 15 number selections. Which is the latest one? And how are things going with you? CD Hi,CD I have stopped testing 6 digit system,as the condition is too hard to reach.Anyways,when you got one banker right and all 6 digits right,its sure to make big profit. I have used this strategy on pick5 and in round5 made a jackpot for iowa cash5 at LP prediction Page. I am at the moment developing a more sure strategy,and I would like to continue to test here if you want. Pick up 22 numbers and 8 last digit.The system give you some lines to try max 3 and 4 numbers in least lines. The system uses some tricks to put most impossible number at the position that will not consider. those 22 numbers and 8 last digits must not have any connection,which means you cant only select 22 numbers using only those 8 last digits. I have been doing fine,and how are you?How long have you studied lottery? As in my opinion,Lottery is a fun game,nothing else.There exist no logic and no rule for it. Hans Strive to predict 8 tickets for 12 numbers with Max ROI for Pick6! Monte Carlo France Member #55589 October 9, 2007 1181 Posts Offline Posted: April 17, 2009, 5:21 pm - IP Logged Hi,CD I have stopped testing 6 digit system,as the condition is too hard to reach.Anyways,when you got one banker right and all 6 digits right,its sure to make big profit. I have used this strategy on pick5 and in round5 made a jackpot for iowa cash5 at LP prediction Page. I am at the moment developing a more sure strategy,and I would like to continue to test here if you want. Pick up 22 numbers and 8 last digit.The system give you some lines to try max 3 and 4 numbers in least lines. The system uses some tricks to put most impossible number at the position that will not consider. those 22 numbers and 8 last digits must not have any connection,which means you cant only select 22 numbers using only those 8 last digits. I have been doing fine,and how are you?How long have you studied lottery? As in my opinion,Lottery is a fun game,nothing else.There exist no logic and no rule for it. Hans Some tips to choose 22 numbers. tip1. choose 32 numbers and play below 4 lines of 18 numbers and you got not 100% to cover all 6,but a high chance to cover 5. 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18 1,2,3,4,5,6,7,8,19,20,21,22,23,24,25,26,27,28 1,2,9,10,11,12,13,14,19,20,21,22,23,24,29,30,31,32 3,4,9,10,15,16,17,18,19,20,25,26,27,28,29,30,31,32 tip2.world record by Nick 19 lines to cover 2 numbers with all 49 numbers. so you can possibly choose one banker and replace one number each line,,,quite tricky. ticket 1 : 1 2 3 4 5 6 Ticket 2 : 1 2 7 8 9 10 Ticket 3 : 1 2 11 12 13 14 Ticket 4 : 3 4 7 8 13 14 Ticket 5 : 3 4 9 10 11 12 Ticket 6 : 5 6 7 8 11 12 Ticket 7 : 5 6 9 10 13 14 Ticket 8 : 15 16 17 18 19 20 Ticket 9 : 15 16 21 22 23 24 Ticket 10 : 15 16 25 26 27 28 Ticket 11 : 17 18 21 22 27 28 Ticket 12 : 17 18 23 24 25 26 Ticket 13 : 19 20 21 22 25 26 Ticket 14 : 19 20 23 24 27 28 Ticket 15 : 29 30 31 32 33 34 Ticket 16 : 35 36 37 38 39 40 Ticket 17 : 41 42 43 44 45 46 Ticket 18 : 41 42 43 47 48 49 Ticket 19 : 44 45 46 47 48 49 tip3. choose 6 digits and 40 numbers,when two conditions match,you directly get all 6 in 24 numbers. Strive to predict 8 tickets for 12 numbers with Max ROI for Pick6! Monte Carlo France Member #55589 October 9, 2007 1181 Posts Offline Posted: April 17, 2009, 5:31 pm - IP Logged As for how to make a match5 in 90 lines 100% guaranteed,here is the wheel. Each position cover 3 numbers,like the first position covers 1.2.and 3,etc totally there are 6 lines. then you can use sum to filter to 10 lines and expect also a match5 and some sure 4 numbers. sample 1...1,2,3 2...9,10,11 3...19,20,21 4...29,30,31 5...39,40,41 6...47,48,49 90 lines: 1 9 19 29 39 47 1 9 19 30 40 48 1 9 19 31 41 49 1 9 20 29 40 47 1 9 20 29 40 49 1 9 20 30 41 47 1 9 20 31 39 48 1 9 21 29 39 49 1 9 21 29 40 49 1 9 21 29 41 48 1 9 21 30 39 49 1 9 21 31 40 47 1 10 19 29 39 49 1 10 19 29 40 49 1 10 19 30 41 47 1 10 19 31 39 47 1 10 19 31 39 49 1 10 20 29 41 48 1 10 20 30 39 48 1 10 20 31 39 47 1 10 21 30 40 47 1 10 21 30 40 48 1 10 21 31 41 49 1 11 19 30 41 47 1 11 19 31 39 49 1 11 19 31 40 48 1 11 20 29 41 48 1 11 20 30 40 49 1 11 20 31 39 48 1 11 21 29 39 47 1 11 21 29 39 48 1 11 21 29 41 48 1 11 21 31 39 47 1 11 21 31 41 49 2 9 19 29 40 47 2 9 19 30 39 49 2 9 19 31 40 49 2 9 19 31 41 48 2 9 20 29 39 47 2 9 20 29 39 48 2 9 20 31 41 49 2 9 21 30 40 48 2 9 21 30 41 47 2 10 19 29 39 48 2 10 19 30 40 47 2 10 19 31 41 47 2 10 20 29 41 47 2 10 20 30 39 49 2 10 20 31 40 48 2 10 21 29 40 49 2 10 21 30 41 48 2 10 21 31 39 47 2 11 19 29 41 49 2 11 19 31 39 47 2 11 20 30 39 47 2 11 20 30 40 47 2 11 20 30 40 48 2 11 20 30 40 49 2 11 20 31 40 47 2 11 21 29 40 48 2 11 21 30 41 47 2 11 21 30 41 48 2 11 21 31 39 49 3 9 19 29 41 47 3 9 19 29 41 48 3 9 19 30 40 49 3 9 19 31 39 49 3 9 20 30 39 47 3 9 20 30 41 48 3 9 21 29 41 49 3 9 21 31 39 48 3 9 21 31 40 47 3 9 21 31 40 48 3 10 19 30 40 47 3 10 19 30 41 49 3 10 19 31 41 48 3 10 20 29 40 48 3 10 20 30 41 49 3 10 20 31 40 49 3 10 21 29 39 47 3 10 21 29 41 47 3 10 21 30 39 47 3 10 21 31 39 48 3 11 19 29 40 47 3 11 19 30 39 48 3 11 19 31 39 49 3 11 20 29 39 49 3 11 20 31 41 47 3 11 20 31 41 48 3 11 21 30 40 49 Strive to predict 8 tickets for 12 numbers with Max ROI for Pick6! Monte Carlo France Member #55589 October 9, 2007 1181 Posts Offline Posted: April 17, 2009, 7:40 pm - IP Logged So CD what I can tell you is that I have found out a hidden struture where in 5 numbers per line ,out of 10 lines there hides 3 or 4 lines with 3 winning numbers. The risk is taken by one point ...hard to explain..there are 12 numbers cut from 30 numbers,and when those 12 numbers have first 3 digits number,then i got it. totally I use 30 Numbers to wheel. The problem is that I need 6 last digits.And I cant make sure of it. I am thinking about this,what if I choose 30 numbers and replace each 5 of them with one digits all numbers for example,I choose 01 05 19 28 33,etc GrOUP A I replace them as 01 11 21 31 41,etc GROUP B then I get 6 digits and still cover all 6 numbers. Again after results,I exchange GROUP B those numbers back to those numbers in group A. Strive to predict 8 tickets for 12 numbers with Max ROI for Pick6! Page 7 of 7	4,147	11,425	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50	latest	en	0.917337
https://socratic.org/questions/how-do-you-find-the-next-three-terms-of-the-arithmetic-sequence-19-24-29-34	1,596,668,258,000,000,000	text/html	crawl-data/CC-MAIN-2020-34/segments/1596439735989.10/warc/CC-MAIN-20200805212258-20200806002258-00389.warc.gz	512,727,989	5,747	# How do you find the next three terms of the arithmetic sequence 19, 24, 29, 34,...? Mar 17, 2018 Next three terms are color(brown)(39, 44, 49 #### Explanation: ${a}_{1} = 19$ ${a}_{2} - {a}_{1} = d = 24 - 19 = 5$ ${a}_{3} - {a}_{2} = d = 29 - 24 = 5$ ${a}_{4} - a - 3 = d = 34 - 29 = 5$ Hence ${a}_{1} = 19 , d = 5$ ${a}_{5} = {a}_{4} + d = 34 + 5 = 39$ ${a}_{6} = {a}_{5} + d = 39 + 5 = 44$ ${a}_{7} = {a}_{6} + d = 44 + 5 = 49$ To find the $n \left(t h\right)$ term, ${a}_{n} = {a}_{1} + \left(n - 1\right) \cdot d$	255	531	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "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}	4.34375	4	CC-MAIN-2020-34	latest	en	0.619636
https://www.studystack.com/flashcard-2619291	1,679,493,320,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296943809.76/warc/CC-MAIN-20230322114226-20230322144226-00027.warc.gz	1,141,582,098	18,985	Save or or taken Make sure to remember your password. If you forget it there is no way for StudyStack to send you a reset link. You would need to create a new account. focusNode Didn't know it? click below Knew it? click below Don't Know Remaining cards (0) Know 0:00 Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page. Normal Size Small Size show me how # Work and energy ### Work in Physics In which term do we use work in the daily life? We use the term ''work'' to indicate an activity or an effort. When a force, acting upon an object, does work? A force, acting upon an object, is doing work, when the point of application of the force is displaced. What is the definition of gravitational potential energy? Gravitational potential energy is the work done by the gravity on an object. What is the definition of energy? Energy is a scalar quantity, and its SI unit is the Joule (J). How is 1 J defined? 1 J is defined as 1 N of orce applied through one meter of distance. What is the definition of elastic potential energy? Elastic potential energy is the energy of a stretched or compressed spring and it represents the work done by an elastic force to return the spring to its initial position. What is the definition of Kinetic enegy? Kinetic energy is the energy due to motion. Created by: Roberta_tOR Voices Use these flashcards to help memorize information. Look at the large card and try to recall what is on the other side. Then click the card to flip it. If you knew the answer, click the green Know box. Otherwise, click the red Don't know box. When you've placed seven or more cards in the Don't know box, click "retry" to try those cards again. If you've accidentally put the card in the wrong box, just click on the card to take it out of the box. You can also use your keyboard to move the cards as follows: • SPACEBAR - flip the current card • LEFT ARROW - move card to the Don't know pile • RIGHT ARROW - move card to Know pile • BACKSPACE - undo the previous action If you are logged in to your account, this website will remember which cards you know and don't know so that they are in the same box the next time you log in. When you need a break, try one of the other activities listed below the flashcards like Matching, Snowman, or Hungry Bug. Although it may feel like you're playing a game, your brain is still making more connections with the information to help you out. To see how well you know the information, try the Quiz or Test activity. Pass complete! "Know" box contains: Time elapsed: Retries: restart all cards	616	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}	2.640625	3	CC-MAIN-2023-14	latest	en	0.930841
http://physics.stackexchange.com/questions/35195/what-do-i-call-the-inverse-of-a-propagator?answertab=oldest	1,462,022,939,000,000,000	text/html	crawl-data/CC-MAIN-2016-18/segments/1461860111865.15/warc/CC-MAIN-20160428161511-00173-ip-10-239-7-51.ec2.internal.warc.gz	208,095,787	16,552	# What do I call the inverse of a propagator? Let's suppose I have a theory described by a Lagrangian as follows: $\mathcal{L} = A_\mu \underbrace{\left( \partial^2 g^{\mu\nu} - \partial^\mu \partial^\nu + m^2 g^{\mu \nu} \right)}_{K^{\mu \nu}} A_\nu$ (What's the actual theory does not matter.) I know that the inverse of $K^{\mu \nu}$ is called the propagator, I don't know how to call $K^{\mu \nu}$ itself. I would like to avoid something like ''operator'' because I often switch to the path integral formalism... - ## 1 Answer I) In a free field theory $$S[\phi]~=~ \frac{1}{2} \iint \! d^dx~ d^dy ~\phi^{\alpha}(x) K_{\alpha\beta}(x,y) \phi^{\beta}(y),$$ I would call the integral kernel $$K_{\alpha\beta}(x,y)~=~\frac{\delta^2 S}{\delta\phi^{\alpha}(x)\delta\phi^{\beta}(y)}$$ either the inverse propagator or the Hessian. (Here we assume that $\phi^{\alpha}$ is bosonic (Grassmann-even) in order not to clutter the formulas with sign factors. In case of gauge symmetry, the action should be gauge-fixed to remove zero-modes.) II) In an interacting theory, the free propagator is by definition associated with the free part of the action, i.e. the part which is quadratic in the field variables $\phi$. -	376	1,222	{"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}	2.71875	3	CC-MAIN-2016-18	latest	en	0.833226
https://www.futurelearn.com/info/courses/precalculus/0/steps/32367	1,695,910,704,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00786.warc.gz	876,731,274	35,577	New offer! Get 30% off one whole year of Unlimited learning. Subscribe for just £249.99 £174.99. New subscribers only T&Cs apply # Finding roots in practice Solving exercises about root factorization of polynomials 10.9 Hello. Welcome to “Finding roots in practice” step of week 3. Let us consider the first exercise. We have to find the root factorization of this polynomial. How can we do this? How can we solve this exercise? The idea is the following. We want to find the roots of this polynomial. In general, it’s not so easy. But you have seen in the Francis video a good algorithm to find rational roots of a polynomial, that is, roots of the form p/q, with p and q integers. Then the rational roots of this form of this polynomial 59.8 had to satisfy some conditions, precisely: you have that p has to divide the constant term of our polynomial. Minus or plus 4 is the same. And q has to divide the leading coefficient of our polynomial. Therefore, the possible candidates are the following. We have plus and minus 1/1, because 1 divides 4 and 1 divides 3. Then we have plus and minus 1/3, in that 1 divides 4 and 3 divides 3. Then we have plus and minus 2/1, plus and minus 2/3, plus and minus 4/1, and plus and minus 4/3. These are the possible candidates. Let us check for each one if they are or not roots of this polynomial. And let us start with 1. 147.4 If we substitute to x 1, what do we get? 3 minus 13 plus 16 minus 4. And this is equal– 3 minus 13 is minus 10, plus 16 is 6 minus 4 is 2, which is not 0. Therefore, 1 is not a root of our polynomial. And now we have to check minus 1. But observe, if you substitute to x in this polynomial any negative number, all the terms are negative. And therefore, you cannot get 0. You get a negative number. Therefore, this polynomial has no negative roots. Then it’s not necessary to check the negative candidates. Then let us now consider 1/3. What do we get? 207.1 We get 3 times 1 over 3 cubed minus 13 1 over 3 squared plus 16 times 1/3 minus 4. And this is equal to 1/9 minus 13/9– and it’s a good idea to take 9 as the common denominator now– plus 16 times 3 over 9 and minus 4 times 9 over 9. And what we get is– OK, 9 as common denominator– and then we have 1 minus 13, which is minus 12. Then 16 times 3 is equal 48. 273.3 And then we have minus 4 times 9, minus 36. And you’ll see the numerator is equal to 0, therefore, we get 0. Therefore, 1/3 is a root of our polynomial. And now let us consider the other positive candidate– 2. 298.2 If we substitute 2 to x, we get 3 times 2 cubed minus 13 times 2 squared plus 16 times 2 minus 4, which is equal to 3 times 8, which is 24, minus 13 times 4, which is equal to minus 52, plus 16 times 2, which is plus 32, minus 4. And now you see 24 minus 4 is 20, minus 52 plus 32 is minus 20. And therefore, we get 0. Then also 2 is a root of our polynomial. Observe, we have to check also the other candidates. But observe also now, what do we know? We know that our polynomial, which has degree 3, can be divided by x minus 1/3 times x minus 2. 379.2 This is because these two numbers are roots of this polynomial. And then a good idea to factorize this polynomial is to compute this product, to make the division of this polynomial by this one, and we get the last factor of degree 1. Observe that if this polynomial is divisible by this product, then it is divisible also by any scalar multiple of this polynomial. Therefore, it is divisible also by 3 times this one. I have multiplied by 3 just to eliminate this denominator and to have a simpler computation to do. Then let us compute this product, and we get 3x times x, which is 3x square. 434.5 Then we have 3 times minus 1/3 x, which is minus x minus 6x, therefore is minus 7x. 449.9 And finally, we have the product of these two, which is plus 2. 458.1 Then our polynomial is divisible by– our starting polynomial is divisible by this one. Let us compute the division. Then remember, we have to do the following. We’ll write here 3 x cubed minus 13 x squared plus 16 times x minus 4, and we have to divide by 3 x squared minus 7 times x plus 2. And we get x, 3 x squared times x is 3 x cubed minus 7 x squared plus 2 times x. We consider the subtraction, and we get minus 13 minus minus 7 is like minus 13 plus 7, which is minus 6 x squared, plus 14 times x minus 4. 536.2 Then we have to multiply 3 x squared with minus 2 to get minus 6 x squared. And then we write minus 2, and we consider the multiplication. And we get minus 6 x squared plus 14x minus 4. Consider the subtraction, and you get 0. Clearly, we get 0 as a remainder, because we knew in advance that our polynomial is divisible by this one. Then we can conclude that 3 x cubed minus 13 x squared plus 16x minus 4 is equal to 3 times x minus 1/3 times x minus 2 times, again, another copy of x minus 2. Thank you for your attention. The following exercises are solved in this step. We invite you to try to solve them before watching the video. In any case, you will find below a PDF file with the solutions. ### Exercise 1. Find the root factorization of the following polynomial: [3x^3-13x^2+16x-4.] ### Exercise 2. [Solved only in the PDF file] Find the root factorization of the following polynomial: [3x^5+5x^4+4x^3+4x^2+x-1.] ### Exercise 3. [Solved only in the PDF file] Find the root factorizations of the following polynomials: i) (x^4+2x^3-9x^2-2x+8 ) ii) (x^4-7x^3+17x^2-17x+6) iii) (2x^4+x^3+3x^2+3x-9)	1,562	5,464	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.46875	4	CC-MAIN-2023-40	longest	en	0.895112
https://calculatorzen.com/health-and-fitness-calculators/metric-body-mass-index-bmi-calculator	1,685,299,359,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224644506.21/warc/CC-MAIN-20230528182446-20230528212446-00760.warc.gz	191,192,871	51,537	## Metric Body Mass Index (BMI) Calculator Use our free calculator to determine BMI based on height in centimeters (cm) and weight in kilograms (kg) • What is BMI and how is it measured? The formula is: BMI = (Weight in kilograms) divided by (Height in centimeters squared) An ideal BMI score is one that falls between 18.5 and 24.9, indicating that a person is within the healthy weight range for his or her height. • What is BMI in physical fitness? BMI (body mass index) is a calculated number representing a person's level of fat or obesity level. According to the Centers for Disease Control and Prevention (CDC), a BMI of 30 or above indicates obesity. BMI levels are broken down by weight range and are as follows: BMI < 18.5 is underweight. • How do you calculate BMI? Formula: weight (kg) / [height (m)^2.Calculate BMI by dividing weight in pounds (lbs) by height in inches (in) squared and multiplying by a conversion factor of 703.Example: Weight = 80 kg, Height = 193 cmCalculation: 80 ÷ (1.93 x 1.93) = 21.5 • What are the BMI score ranges and what do they mean? According to the National Institutes of Health (NIH): A BMI of less than 18.5 means that a person is underweight. A BMI of between 18.5 and 24.9 is ideal. A BMI of between 25 and 29.9 is overweight. A BMI over 30 indicates obesity.	337	1,316	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.703125	3	CC-MAIN-2023-23	longest	en	0.946183
https://www.physicsforums.com/threads/two-people-are-pulling-on-a-stubborn-mule-person-one-is-pulling-the.122831/	1,508,795,137,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187826642.70/warc/CC-MAIN-20171023202120-20171023222120-00539.warc.gz	956,480,338	15,904	# Two people are pulling on a stubborn mule. Person one is pulling the 1. Jun 4, 2006 ### oaky6 Two people are pulling on a stubborn mule. Person one is pulling the mule at 125 Newtons (at 60 degrees in the first quadrant) and person 2 is pulling the mule at 65 Newtons (at 75 degrees counterclockwise from the - x - axis in the second quadrant). a)find the single force that is equivalent to the the two forces. i solved for this and got the correct magnitude (177.1 N) and direction (75 degrees counterclockwise from the +x- axis) I'm having problems with part b) whis wants the force and direction that a third person would have to exert on the mule to make the net force equal to zero. any advice? for part a i used vectors to solve the problem, but I'm not certain how to find the force for the third person . #### Attached Files: • ###### p3-18alt.gif File size: 12 KB Views: 95 Last edited: Jun 4, 2006 2. Jun 4, 2006 ### Hootenanny Staff Emeritus HINT: For the net force to be zero the vector sum of the forces must be zero. I.e. if you add the forces together they must equal zero, don't forget direction matters ~H 3. Jun 4, 2006 ### lando45 Find the resultant force of F1 and F2 using trigonometry or otherwise. You then have one force with one magnitude and one direction. For the net force to equal zero, the vector sum of all forces must equal zero, like Hootenanny said. So think about what the third required force would have to be. 4. Jun 4, 2006 ### oaky6 im thinking that the third force would be towards the 3rd or 4th quadrant..also, finding that force would i have to change the direction of f1 and f2 towards the origin? 5. Jun 4, 2006 ### Hootenanny Staff Emeritus	465	1,709	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.609375	4	CC-MAIN-2017-43	longest	en	0.879881
https://dyscalculiaservices.com/?tag=numbers	1,555,649,707,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578527135.18/warc/CC-MAIN-20190419041415-20190419063415-00500.warc.gz	408,818,977	18,115	Children learn by doing Research in education tells us over and over that hands on activities have more impact than verbal or written explanations alone. Simple manipulatives and pictures are used as key elements to illustrate math concepts. Download our free guidelines and templates for games, outright playful and more serious activities to develop the math centers in your children’s brain. There are sample activities to develop number sense, visuo-spatial activities and templates. Have fun! Please register for free, to give us an idea if activities are useful as a guidance to expand the website in the future. We never share your information with any other organization, see our Terms and Conditions – Privacy Policy. Number sense activities Math concept Rules of Thumb General approach for teaching your child Number walk and Bedtime book Learning number words, counting, one to one relationship, starting with the numerals Take a guess before you count Subitizing, small quantity estimation, quantity comparison Learning to count on the kitchen tiles Counting up and counting down, numerals First, Second, and Third place Play a game to learn the cardinal numbers Number cards counting and sorting activity NEW Counting by ones and tens, and sorting by tens Number-line addition and subtraction Number-line, distance, counting, skip counting, addition, subtraction On the Double Introducing the concept of multiplication by doubling Conquer the Multiplication Tables Basic understanding of the multiplication tables till ten Help with fractions Help your child to grasp the concept of fractions Understanding fractions and Fractions Booklet Explanation how to make the most of the Fractions Booklet Fractions Booklet My fractions book: whole and parts, add and subtract like parts Fraction Notation activities and cards NEW Understanding Fraction notation and concepts Cross out cross-multiplication Prevent confusion with cross-multiplication by showing all steps Visuo-spatial activities Math concept Symmetry activity: looking left and right Symmetry drawings: sizes and shapes, line of symmetry, symmetrical shape Ruler number sense activity NEW Estimating size of numbers Thinking about Area by folding Fold to divide an area in similar parts, area is width times length Pegboards Copy patterns and shapes Patterns from concrete to abstract From patterns with concrete objects to number patterns Templates Math concept Counting place-mat Counting to five Number-bonds place-mat Number-bonds to ten Number cards Color coded number cards 1 – 100 Base ten template Ones, tens, hundreds Domino Doubles cards Learning the Doubles from 0 – 9 # Instant Fix Graph paper works wonders both for calculations and understanding concepts Although there is no instant fix for dyscalculia, there is an instant fix to maximize your credits for the math you can do! Using graph paper (quad paper, quadrille paper) will help you both with calculations and understanding concepts! It can benefit applying concepts as well as communicating your thoughts about the math problem. Making your own drawing to represent the question makes it more fun and clarifies all types of math work, you can also add colors to your work! Initially using graph paper might take a few minutes more, but it will save you time in the long run as well as increase your grade. Starting in KG Having a square to write in will make learning to form the numerals easier. KinderGartners and 1st graders need larger squares and should be taught to leave an empty square between each numeral and between lines to prevent the numerals ‘bleed into each other’. Elementary Depending upon fine motor skills / penmanship and the progress in math, somewhat smaller size squares will come in handy now. Using graph paper makes it quick and easy to draw a number-line and to show your additions and subtractions as ‘jumps’, including the multiples of ten numbers as ‘stepping stones’. It can also easily illustrate multiplication as repeated addition and division as repeated subtraction. Extending the number-line to the left from zero later explains negative numbers. Using the sides of the squares as a guide for drawing will enable you to easily visualize and compare areas and perimeters and help you solve problems about area and perimeter. Writing each numeral in it’s own square will help you to keep the digits in multiplications and long divisions aligned, so you do not mix up your units, tens, and hundreds, etc. The grid will help you find lines of symmetry, complete symmetry drawings (like our free download activity) and make artistic patterns. When you cut out shapes or fold and cut the sides of a large square to make a star or snowflake, the printed squares will help you make strait cuts or folds. These activities are enhancing visual-spatial abilities, one of the components that contribute to being a mathematician. Grid paper helps you to make the link between manipulatives and drawing models. By drawing a model the student shows he has internalized the concept that was illustrated by using manipulatives. The model shows his/her thinking and helps the teacher/tutor/parent to see if there are any remaining misconceptions, and if so which help is needed. Middle and High school Together with switching from wide ruled to college ruled paper you will now probably go for the small squares, allowing for more complex algebraic equations and working with geometric shapes. In algebra it will keep those little ‘devilish’ minuses and other small math signs like powers etc. securely locked in your equations. When you use the grid to draw a few points using an x- and y- axis on graph paper it is much easier to immediately ‘see’ the whole line, calculate the slope, and find the x and y intercepts. Drawing your own graph before using your graphic calculator engages the brain more and forms a stronger memory than only looking at a ready made textbook or calculator graph. Translations, rotations, reflections and dilations are easily understood making a drawing on graph paper as well as vectors. Dysgraphia Just getting into the habit of using a separate square for each numeral or sign will land you those extra points you deserve in algebra, in particular when you also have dysgraphia. Do not back off because using graph paper in class or for homework means you need to copy the question on your paper before starting to solve it. The copying (or errors in it) shows your teacher or parent that you have (or have not) read the question correctly. In a nutshell Graph paper can be used in multiple ways, such as for many arithmetic, measurement, algebraic, geometric, and recreational math purposes. It is also beneficial to students without Math problems and should be available in each math class, it should be dubbed ‘Math paper’.	1,345	6,859	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.484375	3	CC-MAIN-2019-18	longest	en	0.885067
https://www.reise-nach-australien.de/mining/calculate-volume-of-a-tapered-cylinder.html	1,620,591,062,000,000,000	text/html	crawl-data/CC-MAIN-2021-21/segments/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00195.warc.gz	1,009,512,819	5,744	# Calculate Volume Of A Tapered Cylinder ### Volume of a circular truncated cone calculator Calculate exact volume of cylindrical tapered fish tank calculates the volume lateral area and surface area of a circular truncated cone given the lower and upper radii and height compute sample volume of cylinder constricted in the center. ### Calculate the volume of a cylinder - dummies 2020-6-10to calculate the volume of a cylinder you need to know its height and the area of its base because a cylinder is a flat-top figure a solid with two congruent parallel bases the base can be either the top or bottom if you know a cylinders. ### Volume calculator This free volume calculator can compute the volumes of common shapes including that of a sphere cone cube cylinder capsule cap conical frustum ellipsoid and square pyramid explore many other math calculators like the area and surface area. . ### Volume of a tapered cylinder formula Volume of a tapered cylinder formula products as a leading global manufacturer of crushing grinding and mining equipments we offer advanced reasonable solutions for any size-reduction requirements including volume of a tapered cylinder formula. ### Volume of a conical cylinder calculator - Volume of a conical cylinder calculator online calculator to find the conical cylinder volume this can be calculated by adding the volume of cylinder and cone you will get the conical cylinder when the edge of the cylinder is cut off the base of the. ### Calculate volume of a truncated cone and its surface Volume to weight weight to volume and cost conversions for refrigerant r-417a liquid r417a with temperature in the range of -30c -22f to 60c 140f weights and measurements millisievert msv is the si unit of any of the quantities expressed as dose. ### Volume of a tapered cylinder yahoo answers 2012-6-4for the example given the original height of the cone would have been 24 inches since the diameter of the top is one-half that of the base you calculate the volume v of the frustum by subtracting the volume of the part removed from the. ### Circular cylinder calculator Calculator online for a circular cylinder calculate the unknown defining surface areas height circumferences volumes and radii of a capsule with any 2 known variables online calculators and formulas for a cylinder and other geometry. ### What is the area of a varying diameter cylinder - quora Rather than calling the 3d object you are asking about a tapering cylinder it is easier to think of it as a truncated cone that is find the surface area of the total cone and then subtract off the surface area of the top part to reveal the. ### Tapered cylinder volume calculator Volume of a tapered cylinder mayukhportfoliocoin volume of a tapered cylinder crusher mill how do you calculate volume of cylinder in taper shape any tapered online conversion - object volumes welcome to onlineconversion object volumes cylinder. ### Volume of a tapered cylinder formula Volume of a tapered cylinder formula products list volume formulas with examples and videos - online math learning cylinder calculate the volume length radius or diameter of a cylinder or cylindrical shaped tank more details get price cone -. ### Volume for a cyliner with tapered sides Jun 14 2013 volume of a tapered cylinder calculator posted atjune 14 2013 47 - 2608 ratings volume of a tapered cylinder calculator volume tapered cylinder the calculations may be slightly off if the sides are tapered or the bottom is. ### Volume calculator - calculate the volume of a cube Volume of a cylinder the volume formula for a cylinder is height x x diameter 2 2 where diameter 2 is the radius of the base d 2 x r so another way to write it is height x x radius 2 visual in the figure below you need two measurements the height of. ### Cone volume formula - omni calculator Cone volume formula a cone is a solid that has a circular base and a single vertex to calculate its volume you need to multiply the base area area of a circle r by height and by. ### Inclined cylinder volume calculation for tanks and pipes The calculated volume for the measurement is a minimum value since it is possible that the tilted cylinder is indeed full the maximum volume is r 2 l equations for inclined cylinder in performing the volume computation a certain group of variables. ### How to calculate the volume of a cylinder in gallons The volume of a cylinder depends on two parameters the radius of its circular cross-section and its length measure these in inches or feet and youll get the volume in cubic inches or cubic feet measure in metric units and youll get the volume in cubic. ### Volume and weight calculator Calculate the volume and weight in english or metric units for over 40 geometric shapes and a variety of materials select from such metals as aluminum cast iron or steel or from such thermoplastics as abs nylon or polycarbonate typical values. ### Volume formulas examples solutions games Related topics more geometry lessons volume games in these lessons we give a table of volume formulas and surface area formulas used to calculate the volume and surface area of three-dimensional geometrical shapes cube cuboid prism solid cylinder. ### Volume calculators - unit conversion calculates 2011-4-6calculate volume of a cylinder enter value and click on calculate result will be displayed enter your values unit radius height results volume of cylinder calculate volume of a square or rectangle room enter value and click on calculate. ### Online conversion - volume of a cylinder 2014-7-22calculate the volume or height of a cylinder or cylindrical tank volume pi radius 2 length enter two of the following the radiusdiameter length. ### Formula volume of cylinder explained with pictures This page examines the properties of a right circular cylindera cylinder has a radius r and a height h see picture below this shape is similar to a soda can the surface area is the area of the top and bottom circles which are the same and the area. ### Surface area calculator - calculate the surface area of Surface area calculator use this surface area calculator to easily calculate the surface area of common 3-dimensional bodies like a cube rectangular box cylinder sphere cone and triangular prism formulas and explanation. ### Tank volume calculator - inch calculator How to calculate the volume of a tank the volume or capacity of a tank can be found in a few easy steps of course the calculator above is the easiest way to calculate tank volume but follow along to learn how to calculate it yourself step one. ### Volume of hollow cylinder equation and calculator Volume of hollow cylinder equation and calculator a cylinder is one of the most basic curvilinear geometric shapes the surface formed by the points at a fixed distance from a given line segment the axis of the cylinder the solid enclosed by this. ### How can the volume of an oval cylinder be If the oval cylinder have a base with major radius a and minor radius b of height h area of the oval base x a x b so volume of oval cylinder x a x b x h hope it. ### Cylinder volume calculator in feet and inches 2020-4-1how to find out the volume of a cylinder work out the volume of a cylinder using feet and inches calculate volume of cylinders or tanks results in either cubic feet cubic inches uk gallons or us gallons cylinder volume formula volume of a. ### How to calculate the volume of a tapered hopper A tapered hopper is generally the shape of a pyramid or cone turned upside down with a large top that tapers down to a smaller bottom when opened gravity causes the material inside the hopper to feed out the bottom the formula for the volume of a. ### Volume of a truncated cone - calculator online Calculate volume of a truncated cone if you know bases radii and height v volume of a truncated cone - calculator online home list of all formulas of the. ### How to calculate drum volume sciencing Multiply the radius squared times pi times the height of the drum pi is approximately 314159265 the final result is the volume of the drum in cubic units the same units you used to make the measurements for example calculate the radius squared 81. ### How to calculate volume of a weird cylinder 2009-9-16any tapered cylinder can be treated as a longer cone minus a smaller cone lets say the cylinder we need the volume for has the following dimensions heighth2 base radiusa top radiusb volume of this cylinder volume of cone with base radius. ### Need to write a program that finds the volume and Need to write a program that finds the volume and surface area of shapes askes a user how many shapes they want to find the v and sa of askes to input values displays the final values and askes.	1,737	8,789	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.234375	3	CC-MAIN-2021-21	latest	en	0.877189
https://madformath.com/calculators/basic-math/base-converters/octal-to-decimal-converter-with-steps/octal-to-decimal-converter-with-steps	1,725,821,585,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651017.43/warc/CC-MAIN-20240908181815-20240908211815-00340.warc.gz	352,922,917	12,050	# ⇩ BASE N TO DECIMAL CONVERTERS ⇩ #### OCTAL TO DECIMAL CONVERTER (WITH STEPS) Enter an octal number. (125)8 = (85)10 SOLUTION We multiply each digit by its place value and add the products. (125)8 = (1 × 82) + (2 × 81) + (5 × 80) = (1 × 64) + (2 × 8) + (5 × 1) = 64 + 16 + 5 = (85)10 OTHER INFORMATION Copied to clipboard Copy Text # ⇩ BASE N TO DECIMAL CONVERTERS ⇩ ## INFORMATION ### HEXADECIMAL AND DECIMAL NUMBERS Binary, octal, decimal and hexadecimal numbering systems are commonly used in mathematics, computer science and electrical engineering. Octal Numbering System (Base 8): In the octal numbering system, numbers are represented by eight digits: 0, 1, 2, 3, 4, 5, 6 and 7. The ones digit has a place value of 80 = 1, the next digit to the left has a place value of 81 = 8, then 82 = 64, and so on. As we move one place to the left, the place value increases by a factor of 8. Decimal Numbering System (Base 10): In the decimal system, numbers are represented using ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Each digit has a place value of 10 raised to a power depending on its position in the number. ### OCTAL TO DECIMAL CONVERSION Each octal number has a unique representation in other numbering systems. Octal to decimal (oct to dec) conversion is the process of computing the equivalent decimal representation of a base 8 (octal) number. To convert an octal number to decimal we multiply each digit by its place value and add the products. Each place value in octal representation corresponds to an exponent of 8. The exponent increases by 1 as we move one digit to the left, starting from zero for the ones digit. ### BINARY TO DECIMAL CONVERSION EXAMPLES #### EXAMPLE: Find the decimal equivalent of 5038. To find the decimal equivalent of this number, we multiply each digit by its place value and then sum up the products. 5038 = 5 · 82 + 0 · 81 + 3 · 80 = 5 · 64 + 0 · 8 + 3 · 1 = 320 + 0 + 3 = 32310 So, the decimal equivalent of 5038 is 32310. #### EXAMPLE: Find the decimal equivalent of the 47.28. The place values of B0.416 are shown below. Multiply those values with the corresponding digits and add the products. ... ... ... ... ... ... ### OCTAL TO DECIMAL CONVERSION TABLE The table below displays the decimal equivalents of the smallest non-negative octal numbers up to 100. 08 = 010 18 = 110 28 = 210 38 = 310 48 = 410 58 = 510 68 = 610 78 = 710 108 = 810 118 = 910 128 = 1010 138 = 1110 148 = 1210 158 = 1310 168 = 1410 178 = 1510 208 = 1610 218 = 1710 228 = 1810 238 = 1910 248 = 2010 258 = 2110 268 = 2210 278 = 2310 308 = 2410 318 = 2510 328 = 2610 338 = 2710 348 = 2810 358 = 2910 368 = 3010 378 = 3110 408 = 3210 418 = 3310 428 = 3410 438 = 3510 448 = 3610 458 = 3710 468 = 3810 478 = 3910 508 = 4010 518 = 4110 528 = 4210 538 = 4310 548 = 4410 558 = 4510 568 = 4610 578 = 4710 608 = 4810 618 = 4910 628 = 5010 638 = 5110 648 = 5210 658 = 5310 668 = 5410 678 = 5510 708 = 5610 718 = 5710 728 = 5810 738 = 5910 748 = 6010 758 = 6110 768 = 6210 778 = 6310 1008 = 6410 1018 = 6510 1028 = 6610 1038 = 6710 1048 = 6810 1058 = 6910 1068 = 7010 1078 = 7110 1108 = 7210 1118 = 7310 1128 = 7410 1138 = 7510 1148 = 7610 1158 = 7710 1168 = 7810 1178 = 7910 1208 = 8010 1218 = 8110 1228 = 8210 1238 = 8310 1248 = 8410 1258 = 8510 1268 = 8610 1278 = 8710 1308 = 8810 1318 = 8910 1328 = 9010 1338 = 9110 1348 = 9210 1358 = 9310 1368 = 9410 1378 = 9510 1408 = 9610 1418 = 9710 1428 = 9810 1438 = 9910 1448 = 10010 ### WHAT IS OCTAL TO DECIMAL CONVERTER? Octal to decimal converter, • Computes the decimal equivalent of the entered octal number, • Describes the solution step by step and • Illustrates the place values. ### HOW TO USE OCTAL TO DECIMAL CONVERTER? You can use octal to decimal converter in two ways. • #### USER INPUTS You can enter an octal number into the input box and then click the 'CONVERT' button. The result and explanations will appear below the calculator. • #### RANDOM INPUTS You can click on the DIE ICON next to the input box to generate a random octal number, which will be automatically entered into the calculator. The result and explanations will then appear below the calculator. You can also create your own examples and practice using this feature. • #### CLEARING THE INPUT BOX To find the decimal equivalent of another hexadecimal number, click on the CLEAR button to clear the input box. • You can copy the generated solution by clicking on the 'Copy Text' link located below the solution panel. • You can also download the solution as an image file with a .jpg extension by clicking on the 'Download Solution' link located at the bottom of the solution panel. You can then share the downloaded image file.	1,559	4,728	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.1875	4	CC-MAIN-2024-38	latest	en	0.796009
https://nl.mathworks.com/matlabcentral/profile/authors/7585805-robert-price?s_tid=cody_local_to_profile	1,579,986,487,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579251681412.74/warc/CC-MAIN-20200125191854-20200125221854-00301.warc.gz	585,270,231	18,458	Community Profile # Robert Price 16 total contributions since 2016 View details... Contributions in View by Solved Finding Perfect Squares Given a vector of numbers, return true if one of the numbers is a square of one of the other numbers. Otherwise return false. E... bijna 4 jaar ago Solved Create times-tables At one time or another, we all had to memorize boring times tables. 5 times 5 is 25. 5 times 6 is 30. 12 times 12 is way more th... bijna 4 jaar ago Solved Fibonacci sequence Calculate the nth Fibonacci number. Given n, return f where f = fib(n) and f(1) = 1, f(2) = 1, f(3) = 2, ... Examples: Inpu... bijna 4 jaar ago Solved Determine whether a vector is monotonically increasing Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return f... bijna 4 jaar ago Solved Swap the first and last columns Flip the outermost columns of matrix A, so that the first column becomes the last and the last column becomes the first. All oth... bijna 4 jaar ago Solved Make a checkerboard matrix Given an integer n, make an n-by-n matrix made up of alternating ones and zeros as shown below. The a(1,1) should be 1. Examp... bijna 4 jaar ago Solved Find all elements less than 0 or greater than 10 and replace them with NaN Given an input vector x, find all elements of x less than 0 or greater than 10 and replace them with NaN. Example: Input ... bijna 4 jaar ago Solved Triangle Numbers Triangle numbers are the sums of successive integers. So 6 is a triangle number because 6 = 1 + 2 + 3 which can be displa... bijna 4 jaar ago Solved Is my wife right? Regardless of input, output the string 'yes'. bijna 4 jaar ago Solved Column Removal Remove the nth column from input matrix A and return the resulting matrix in output B. So if A = [1 2 3; 4 5 6]; and ... bijna 4 jaar ago Solved Select every other element of a vector Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s... bijna 4 jaar ago Solved Determine if input is odd Given the input n, return true if n is odd or false if n is even. bijna 4 jaar ago Solved Given a and b, return the sum a+b in c. bijna 4 jaar ago Solved Find the sum of all the numbers of the input vector Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ... bijna 4 jaar ago Solved Make the vector [1 2 3 4 5 6 7 8 9 10] In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s... bijna 4 jaar ago Solved Times 2 - START HERE Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:... bijna 4 jaar ago	795	2,823	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.21875	3	CC-MAIN-2020-05	latest	en	0.72841
https://lifein19x19.com/viewtopic.php?f=13&t=16420&start=680	1,618,509,619,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038087714.38/warc/CC-MAIN-20210415160727-20210415190727-00548.warc.gz	467,281,099	14,031	It is currently Thu Apr 15, 2021 11:00 am All times are UTC - 8 hours [ DST ] Page 35 of 40 [ 796 posts ] Go to page Previous 1 ... 32, 33, 34, 35, 36, 37, 38 ... 40 Next Print view Previous topic \| Next topic Who will win? EGF pros 69% [ 40 ] AGA pros 22% [ 13 ] Don't know 9% [ 5 ] Author Message Post subject: Re: EGF vs AGA pros win-and-continue match #681 Posted: Fri May 31, 2019 1:30 am Lives in sente Posts: 1224 Liked others: 97 Was liked: 401 Uberdude wrote: Also, the facebook post has a-d not a-c Facebook a, c, d have become a, b, c on the website, while facebook b is in the "requirements" paragraph of the website. This post by jlt was liked by: Javaness2 Top Post subject: Re: EGF vs AGA pros win-and-continue match #682 Posted: Fri May 31, 2019 2:18 am Judan Posts: 6683 Location: Cambridge, UK Liked others: 432 Was liked: 3663 Rank: UK 4 dan KGS: Uberdude 4d OGS: Uberdude 7d OGS also has a rather confusing user interface for the clock in byo-yomi if you are used to other servers. On KGS "0:45 (5)" means you have 45 seconds left and 5 byo-yomi periods remaining, wait a minute and it'll say "0:45 (4)", another minute to "0:45 (3)", then "0:45 (2)", then "0:45 SD", then you lose 45 seconds later if you didn't play in time, or it goes back up to 1:00 SD if you play in time. On OGS the equivalent clock display is "0:45 + 5 x 60s", then "0:45 + 4 x 60s", then "0:45 + 3 x 60s", then "0:45 + 2 x 60s", then "0:45 + 1 x 60s" then 45 seconds later you lose. My initial understanding of that final display is that I have 45 seconds and one more 60 seconds period remaining after that if I take more than 45 seconds (because I read it like maths: 45 + 1 x 60 = 45 + (1x60) = 105 total time remaining, it's the plus sign that's misleading), but you don't, The OGS display is in some ways more logical because the KGS one doesn't distinguish between 45 seconds left of absolute time and 45 seconds left with 1 renewable byo-yomi period, but it's quite a dangerous gotcha that I hope both players are aware of. Is it the player's responsibility or the organisers to ensure this is understood? P.S. I like Fox's display, in final byo-yomi it says "Byo-yomi 1x, 0:45" and then elsewhere on the screen has the game settings "6.5 komi, 1 hour, 5 x 1 min". P.P.S I'm wondering should I send Ali a message about this, or does that actually hurt his winning chances: in the event he times out because he is misled by the confusing clock he could no longer claim ignorance and, echoing the Mateusz ruling, that means he accepts the bad conditions. But if Eric loses on time because of a confusing clock he can claim he didn't know about the bad user interface and therefore should get a rematch. So if I want Ali to win I should tell Eric, ignorance is bliss! Top Post subject: Re: EGF vs AGA pros win-and-continue match #683 Posted: Fri May 31, 2019 4:27 am Honinbo Posts: 10674 Liked others: 3594 Was liked: 3339 Uberdude wrote: OGS also has a rather confusing user interface for the clock in byo-yomi if you are used to other servers. On KGS "0:45 (5)" means you have 45 seconds left and 5 byo-yomi periods remaining, wait a minute and it'll say "0:45 (4)", another minute to "0:45 (3)", then "0:45 (2)", then "0:45 SD", then you lose 45 seconds later if you didn't play in time, or it goes back up to 1:00 SD if you play in time. On OGS the equivalent clock display is "0:45 + 5 x 60s", then "0:45 + 4 x 60s", then "0:45 + 3 x 60s", then "0:45 + 2 x 60s", then "0:45 + 1 x 60s" then 45 seconds later you lose. My initial understanding of that final display is that I have 45 seconds and one more 60 seconds period remaining after that if I take more than 45 seconds (because I read it like maths: 45 + 1 x 60 = 45 + (1x60) = 105 total time remaining, it's the plus sign that's misleading), but you don't, The OGS display is in some ways more logical because the KGS one doesn't distinguish between 45 seconds left of absolute time and 45 seconds left with 1 renewable byo-yomi period, but it's quite a dangerous gotcha that I hope both players are aware of. Is it the player's responsibility or the organisers to ensure this is understood? A competent organizer would make this known to the players. Quote: I would send a message to the organizers or the referee. We do not want another incident, which might sink this tournament. _________________ At some point, doesn't thinking have to go on? Visualize whirled peas. Everything with love. Stay safe. Last edited by Bill Spight on Fri May 31, 2019 5:12 am, edited 1 time in total. Top Post subject: Re: EGF vs AGA pros win-and-continue match #684 Posted: Fri May 31, 2019 4:57 am Dies in gote Posts: 53 Liked others: 3 Was liked: 33 Rank: KGS 2k Bit of an upset in the Amsterdam tournament running at the moment. Ali Jabarin lost to Stephen Hu, 5 dan, in the second round. Lets hope it doesnt affect his confidence for sunday Top Post subject: Re: EGF vs AGA pros win-and-continue match #685 Posted: Fri May 31, 2019 5:25 am Gosei Posts: 1504 Liked others: 815 Was liked: 500 Rank: AGA 3k KGS 1k Fox 2d GD Posts: 61 KGS: dfan Aram wrote: Bit of an upset in the Amsterdam tournament running at the moment. Ali Jabarin lost to Stephen Hu, 5 dan, in the second round. Stephen puts on such a good show of being slightly befuddled during his broadcasts that I forget he is actually a pretty good player. This post by dfan was liked by: Bonobo Top Post subject: Re: EGF vs AGA pros win-and-continue match #686 Posted: Fri May 31, 2019 5:29 am Judan Posts: 6683 Location: Cambridge, UK Liked others: 432 Was liked: 3663 Rank: UK 4 dan KGS: Uberdude 4d OGS: Uberdude 7d Schedule of Amsterdam tournament: Quote: Sunday, June 2 09:30 – 13:00 Round 6 14:00 – 17:30 Round 7 --- Live commentary of the final by Rob van Zeijst 7d 17:45 – 18:15 Prize-giving ceremony If Ali is playing in those 2 rounds he's going to be tired and not in a good condition to play another gruelling game at 8pm Top Post subject: Re: EGF vs AGA pros win-and-continue match #687 Posted: Fri May 31, 2019 5:32 am Gosei Posts: 1504 Liked others: 815 Was liked: 500 Rank: AGA 3k KGS 1k Fox 2d GD Posts: 61 KGS: dfan Uberdude wrote: OGS also has a rather confusing user interface for the clock in byo-yomi if you are used to other servers. I appreciate your description but I'm still confused! What is difference in display between "45 seconds left of main time, followed by 5 byo-yomi periods" and "main time has expired and there are 45 seconds left in your first byo-yomi period" (out of 5)? I agree that "0:45 + 1 x 60s" really should mean that you have one period in reserve after using this one up. Top Post subject: Re: EGF vs AGA pros win-and-continue match #688 Posted: Fri May 31, 2019 5:33 am Dies in gote Posts: 53 Liked others: 3 Was liked: 33 Rank: KGS 2k Uberdude wrote: Schedule of Amsterdam tournament: Quote: Sunday, June 2 09:30 – 13:00 Round 6 14:00 – 17:30 Round 7 --- Live commentary of the final by Rob van Zeijst 7d 17:45 – 18:15 Prize-giving ceremony If Ali is playing in those 2 rounds he's going to be tired and not in a good condition to play another gruelling game at 8pm Indeed.. imagine the transatlantic match ending at around midnight for Ali, and him having played 3 rounds during the day... Top Post subject: Re: EGF vs AGA pros win-and-continue match #689 Posted: Fri May 31, 2019 5:36 am Judan Posts: 6683 Location: Cambridge, UK Liked others: 432 Was liked: 3663 Rank: UK 4 dan KGS: Uberdude 4d OGS: Uberdude 7d dfan wrote: Uberdude wrote: OGS also has a rather confusing user interface for the clock in byo-yomi if you are used to other servers. I appreciate your description but I'm still confused! What is difference in display between "45 seconds left of main time, followed by 5 byo-yomi periods" and "main time has expired and there are 45 seconds left in your first byo-yomi period" (out of 5)? There is none, you cannot distinguish these 2 situations! P.S. I've posted a reply to the official facebook post about the confusing clock and other concerns. This post by Uberdude was liked by 2 people: Bill Spight, gamesorry Top Post subject: Re: EGF vs AGA pros win-and-continue match #690 Posted: Fri May 31, 2019 5:37 am Gosei Posts: 1504 Liked others: 815 Was liked: 500 Rank: AGA 3k KGS 1k Fox 2d GD Posts: 61 KGS: dfan Uberdude wrote: dfan wrote: Uberdude wrote: OGS also has a rather confusing user interface for the clock in byo-yomi if you are used to other servers. I appreciate your description but I'm still confused! What is difference in display between "45 seconds left of main time, followed by 5 byo-yomi periods" and "main time has expired and there are 45 seconds left in your first byo-yomi period" (out of 5)? There is none, you cannot distinguish these 2 situations! Top Post subject: Re: EGF vs AGA pros win-and-continue match #691 Posted: Fri May 31, 2019 6:23 am Honinbo Posts: 9216 Liked others: 1550 Was liked: 1596 KGS: Kirby Tygem: 커비라고해 Uberdude wrote: P.P.S I'm wondering should I send Ali a message about this, or does that actually hurt his winning chances: in the event he times out because he is misled by the confusing clock he could no longer claim ignorance and, echoing the Mateusz ruling, that means he accepts the bad conditions. But if Eric loses on time because of a confusing clock he can claim he didn't know about the bad user interface and therefore should get a rematch. So if I want Ali to win I should tell Eric, ignorance is bliss! It'd be more clear to say, if you message Ali about this, and Ali is still confused about the clock, he should bring it up to the organizers, and they should work out a resolution to resolve the confusion. Why would Ali proceed with the game in a state of confusion, without trying to resolve it? If I tell you to watch out before crossing the road, because a bus is coming, why not try to do something to avoid the bus? _________________ be immersed Top Post subject: Re: EGF vs AGA pros win-and-continue match #692 Posted: Fri May 31, 2019 6:54 am Honinbo Posts: 10674 Liked others: 3594 Was liked: 3339 Kirby wrote: If I tell you to watch out before crossing the road, because a bus is coming, why not try to do something to avoid the bus? OC, if your grandmother is handy, . . . _________________ At some point, doesn't thinking have to go on? Visualize whirled peas. Everything with love. Stay safe. Top Post subject: Re: EGF vs AGA pros win-and-continue match #693 Posted: Sun Jun 02, 2019 11:21 am Oza Posts: 2217 Location: Germany Liked others: 8216 Was liked: 918 Rank: OGS DDK again Universal go server handle: trohde Ali Jabarin (W) vs. Eric Lui (B) https://online-go.com/game/18114039 https://www.twitch.tv/usgoweb _________________ “The only difference between me and a madman is that I’m not mad.” — Salvador Dali ★ Play a slooooow correspondence game with me on OGS? This post by Bonobo was liked by 3 people: Bill Spight, BlindGroup, Joaz Banbeck Top Post subject: Re: EGF vs AGA pros win-and-continue match #694 Posted: Sun Jun 02, 2019 11:59 am Tengen Posts: 5407 Location: Banbeck Vale Liked others: 1032 Was liked: 1424 Rank: 1D AGA GD Posts: 1512 Kaya handle: Test Bonobo wrote: Ali Jabarin (W) vs. Eric Lui (B) https://online-go.com/game/18114039... Lui gave up a huge corner for a wall. EDIT...and won. Top Post subject: Re: EGF vs AGA pros win-and-continue match #695 Posted: Sun Jun 02, 2019 12:39 pm Oza Posts: 2217 Location: Germany Liked others: 8216 Was liked: 918 Rank: OGS DDK again Universal go server handle: trohde French commentary: https://www.twitch.tv/HisokaH _________________ “The only difference between me and a madman is that I’m not mad.” — Salvador Dali ★ Play a slooooow correspondence game with me on OGS? Top Post subject: Re: EGF vs AGA pros win-and-continue match #696 Posted: Sun Jun 02, 2019 1:53 pm Gosei Posts: 1355 Liked others: 110 Was liked: 282 I see Eric won. _________________ Top Post subject: Re: EGF vs AGA pros win-and-continue match #697 Posted: Sun Jun 02, 2019 2:11 pm Dies in gote Posts: 53 Liked others: 3 Was liked: 33 Rank: KGS 2k With what happened in a certain earlier game, i find it rather odd and amusing that some people on OGS and Twitch are suggesting that Ali should resign and not play to the end... You know.. if there is a large 10-ish point difference, how could the game result ever change? Top Post subject: Re: EGF vs AGA pros win-and-continue match #698 Posted: Sun Jun 02, 2019 2:18 pm Lives in gote Posts: 383 Liked others: 385 Was liked: 197 Are the variations from the live commentary being officially recorded somewhere? I missed much of the beginning part of the commentary. I know they may eventually make the video commentary available on youtube, like they did for the first 3 games: https://www.youtube.com/user/USGOWeb but that seems to be done with a great delay (game 4 is not available there yet). _________________ Sorin - 361points.com Top Post subject: Re: EGF vs AGA pros win-and-continue match #699 Posted: Sun Jun 02, 2019 3:09 pm Gosei Posts: 1498 Liked others: 450 Was liked: 420 Rank: 5d GD Posts: 1000 Well, I guess it shouldn't be surprising that some hypocritical commenters insisted that Lui should have resigned when significantly behind during overtime in game 4 but didn't then say that Jabarin should resign when more than 20 points behind in game 5. Or maybe Jabarin was hoping that Lui might lose due to running out of time; it was completely within his right to play to the end. Anyhow, it's good to have the game played without any foolishness. We can look forward to the next game. I remember that decades ago the great Japanese champion Sakata played a tournament game all the way through the counting at the end to lose by more than 10 points. No one suggested that Sakata should have resigned earlier. This post by gowan was liked by: Bonobo Top Post subject: Re: EGF vs AGA pros win-and-continue match #700 Posted: Sun Jun 02, 2019 3:18 pm Oza Posts: 2217 Location: Germany Liked others: 8216 Was liked: 918 Rank: OGS DDK again Universal go server handle: trohde Congrats to Eric Lui! _________________ “The only difference between me and a madman is that I’m not mad.” — Salvador Dali ★ Play a slooooow correspondence game with me on OGS? 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https://oneclass.com/class-notes/us/ucf/mathematics-calculusprecalc.en.html	1,548,045,154,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583755653.69/warc/CC-MAIN-20190121025613-20190121051613-00462.warc.gz	593,035,899	133,125	##### Filter Reset ###### Semester Class Notes (1,031,803) US (404,814) UCF (3,792) MAC (358) # Class Notes for Mathematics Calculus/Precalc at University of Central Florida Sort 358 Results ### Exam Study Guides for Mathematics Calculus/Precalc Courses Covers all exam topics. ##### MAC 2311C Lecture 10: 3.5 :implicit Differentiation 5 3. 1 C OS ( ) = 42 + 2y2 Y ( cos ( )) = 42 + ) y ( zyz product If sin + cos Adat = 8x + ( IF ) y 4g = cos (x ) tgddt ) sx gsin f ( cos( x ) 4g ) + = 8 y + sin ( x ) = 8tys=) Tt COS ( ) 4g 2 = arccos ( e2 ) y = e2 ... 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Physics PHYS 24300 All Professors Notes Practice Me Log In OR Don't have an account? Join OneClass Access over 10 million pages of study documents for 1.3 million courses. Sign up Join to view OR By registering, I agree to the Terms and Privacy Policies Already have an account? Just a few more details So we can recommend you notes for your school. Reset Password Please enter below the email address you registered with and we will send you a link to reset your password. Add your courses Get notes from the top students in your class. Request Course Submit	2,447	8,440	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04	latest	en	0.556002
https://www.scienceforums.net/topic/49769-thrown-at-different-angles-from-same-height/#comment-550253	1,660,824,122,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00328.warc.gz	847,594,955	22,480	# Thrown at different angles, from same height... ## Recommended Posts Treating air resistance as negligible for this case, suppose we have three objects thrown FROM one vertical position TO another vertical position; the initial and final vertical positions being the same for each object. They are thrown at the same speeds, but at different velocities; one is thrown at an upward angle, one is thrown at a downward angle, and one is thrown horizontally. Would they have the same speeds, or different speeds, upon hitting the ground? ##### Share on other sites Try applying conservation of energy and answering this. ##### Share on other sites Try applying conservation of energy and answering this. Well, should I assume that gravitational potential energy and macroscopic kinetic energy are the only forms of energy involved in this case? If so they'd have to land at the same speed, but I'm wondering if there's something I'm missing... ##### Share on other sites No, you are not missing anything... Sounds good to me. ##### Share on other sites Unless I'm thinking about this wrong, they will not have the same final velocities, they will be different. I think the thing here is the fact that they will indeed fall as the same rate ( acceleration = g) , but the direction of their velocity is going to effect their final velocities. For example, Labeling (1) Upward at an angle, (2) downward and (3) horizontally...you want to consider only their vertical velocities, which they are not all the same, (1) and (2) only have the same velocities when heading down if (1) is thrown straight upward and (3) always has a 0 downward velocity. So according to this they will not all have the sames velocities when hitting the ground since $v_f = v_0 + at$. Unless I'm crazy. Edited by darkenlighten ##### Share on other sites The question is not asking for equal velocities upon impact but for equal speeds, i.e. magnitudes of velocities. You are of course be correct about different velocities: the horizontal part of the velocity is differs when thrown and not change during the flight. Note that the question is very homework-like, so I think you should be a bit reluctant to discuss details of a solution until MDJH has presented a solution or at least a week or so has passed. ##### Share on other sites Note that the question is very homework-like, so I think you should be a bit reluctant to discuss details of a solution until MDJH has presented a solution or at least a week or so has passed. Agreed, but I think that MDJH has done so in post 3. Merged post follows: Consecutive posts merged Unless I'm thinking about this wrong, they will not have the same final velocities, they will be different. I think the thing here is the fact that they will indeed fall as the same rate ( acceleration = g) , but the direction of their velocity is going to effect their final velocities. For example, Labeling (1) Upward at an angle, (2) downward and (3) horizontally...you want to consider only their vertical velocities, which they are not all the same, (1) and (2) only have the same velocities when heading down if (1) is thrown straight upward and (3) always has a 0 downward velocity. So according to this they will not all have the sames velocities when hitting the ground since $v_f = v_0 + at$. Unless I'm crazy. You can investigate the 2 vertical cases without worrying about horizontal components. If you send an object vertically at v_0, what is its speed when it hits the ground? ##### Share on other sites The question is not asking for equal velocities upon impact but for equal speeds, i.e. magnitudes of velocities. You are of course be correct about different velocities: the horizontal part of the velocity is differs when thrown and not change during the flight. Note that the question is very homework-like, so I think you should be a bit reluctant to discuss details of a solution until MDJH has presented a solution or at least a week or so has passed. I'm not asking for help with homework, (well, not directly anyway) though I can understand why you got that impression. I did introductory physics courses during my first year of university, (didn't learn that thoroughly though) as well as a few more physics courses since, (again, didn't learn it very thoroughly) but I've been unsure about what to major in for a while and have been meandering with all kinds of introductory courses for subjects other than physics; but I've been considering returning to a physics major, and was just trying to refresh my memory on this kind of stuff, thus I'm going through my first year physics book looking for whatever things I don't get, or am not sure if I get, so as to check my understanding. ##### Share on other sites Their speeds will still be different when hitting the ground, due to only the vertical distances and starting velocities (think of potential and kinetic energies if you like) Edited by darkenlighten ##### Share on other sites Their speeds will still be different when hitting the ground, due to only the vertical distances and starting velocities (think of potential and kinetic energies if you like) What do you mean? ##### Share on other sites I think that's better phrased as: you can only get different final speeds if you have different initial speeds, or different heights. ##### Share on other sites I agree with the assessment that the speeds will be the same and velocities different. If released at the same vertical position (same potential E) with the same speed (same kinetic E) and arrive at the same final vertical position (same final potential E) ... for energy to be conserved the final kinetic Es will need to be the same as well which means the velocity magnitudes (speeds) will be the same...assuming of course they all have the same mass. ##### Share on other sites If it's high enough to reach terminal velocity... they will all approach it. Edit: missed the "negligible" part. ##### Share on other sites I realized I was only considering the vertical velocity components and not the velocity vector...so with that from energy conservation its easy to see $E_o = E_f \Rightarrow -mgh_o + \frac{1}{2}mv{_o}{^2} = -mgh_f + \frac{1}{2}mv{_f}{^2} \Rightarrow v_f = [v{_o}{^2} + 2gh_f]{^ \frac{1}{2}}$ So since $v_o$ and $h_f$ are the same for all so yea that's it. Same speed! And to add to "...", mass of the objects will not affect anything. They could all have different masses and nothing will change. ## Create an account Register a new account	1,434	6,553	{"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}	3.46875	3	CC-MAIN-2022-33	latest	en	0.946851
https://www.ks.uiuc.edu/Research/namd/mailing_list/namd-l.2012-2013/0698.html	1,713,380,506,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817171.53/warc/CC-MAIN-20240417173445-20240417203445-00232.warc.gz	780,923,698	3,143	Re: forces saved in forcedcd From: JC Gumbart (gumbart_at_ks.uiuc.edu) Date: Thu May 03 2012 - 20:11:32 CDT How are you calculating the PMF? Did you try, say, ABF? On May 3, 2012, at 7:51 PM, Xiaohu Li wrote: > Hi, Chris, > Thanks for taking time for pondering on my question. > In my case, I would have to stick with the NVT simulation since I'm using the drude polarizable force field. The reason I ask is because I'm doing a PMF of ions pulled out of bulk(glycerol), which the reaction coordinates is quite simple as the surface normal direction. I'm seeing something quite difficult to explain since from a very naive understanding, the PMF should be flat deep in the bulk(as seen in water, methanol, etc...). I'm getting quite oscillatory pattern in the bulk and the i'm talking about several kcal/mol oscillation(using either wham or the direct forces). That leads me to this question. Something I notice is that the frequency of the Langevin friction affect the self-diffusion. So extrapolate from that, it should affect the forces. > Any more opinions on that? > > > Xiaohu > > On May 3, 2012, at 6:42 PM, Chris Harrison wrote: > >> Yes. However, in the limit of converged sampling the Langevin forces are theoretically canceling out to noise. ...theoretically. In practice, it is entirely possible that when using a Langevin thermostat you might see the periodicity produced from the kicks when analyzing the forces. >> >> "Will the Langevin forces affect PMF calculations?" That entirely depends on the PMF you are performing. Many considerations go into answering that question. Consider: Are the forces being predominately measured in the PMF of a small enough scale that the Langevin kicks can be resolved? If performed correctly, the Langevin forces should look like noise. Another consideration: a purist would argue to perform an NVE simulation, removing as many external forces/perturbations from the system as possible. >> >> Unfortunately the answer to your question is: yes, they can affect PMFs, but a) will it matter for the question you are asking, and b) you will have to test to actually determine for sure; or alternatively pursue an NVE ensemble simulation. >> >> Best, >> Chris >> >> >> >> On Thu, May 3, 2012 at 3:43 PM, Xiaohu Li <xiaohu-li_at_northwestern.edu> wrote: >> Dear NAMD users, >> I have a question regarding the forces saved in forcedcd file. suppose I'm doing NVT using Langevin dynamics, will the forces saved in forcedcd contain the random forces from Langevin dynamics? >> Do those random forces affect, say, PMF calculations if they are included? >> Thanks. >> >> >> >> >> -- >> Chris Harrison, Ph.D. >> NIH Center for Macromolecular Modeling and Bioinformatics >> Theoretical and Computational Biophysics Group >> Beckman Institute for Advanced Science and Technology >> University of Illinois, 405 N. Mathews Ave., Urbana, IL 61801 >> >> http://www.ks.uiuc.edu/Research/namd Voice: 773-570-6078 >> http://www.ks.uiuc.edu/~char Fax: 217-244-6078 >> >> >> > This archive was generated by hypermail 2.1.6 : Tue Dec 31 2013 - 23:21:56 CST	773	3,088	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18	latest	en	0.947562
https://www.reference.com/science-technology/letter-k-used-represent-thousand-6cc818476cf09a87	1,722,723,564,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00603.warc.gz	755,106,404	14,266	# Why Is the Letter “K” Used to Represent a Thousand? The letter “K” is used to represent 1000, because it represents the prefix “kilo,” which means 1000 of something in the metric system. For instance, kilogram means 1000 grams. The prefix “kilo” was taken from the Greek word chilioi or khilioi, which means thousand. This happened when a group of French scientists were commissioned to make the metric system. The metric system was designed to make conversions between units easier by making each unit a factor of 10 larger or smaller than the nearest units. Scientists needed to have standard prefixes that represented how much larger or smaller a specific unit was from the base unit. This makes it easy for a kilometer to be converted to a meter or a kilogram to a gram, because once the prefixes are known all conversions are standard.	183	844	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33	latest	en	0.954144
http://www.math.uwaterloo.ca/~cgodsil/html_notes/topology.html	1,524,579,210,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125946721.87/warc/CC-MAIN-20180424135408-20180424155408-00425.warc.gz	468,958,694	4,108	# Topology This is a plan for a reading course in topology, first some point set topology, then some homotopy. The aims are: 1. to understand how to use Tychonov's theorem to prove the de Bruijn-Erdős theorem 2. to understand the classification of surfaces 3. pick up a working knowledge of covering spaces of graphs and surfaces. A long-term goal is to learn about Lie groups, which requires some minimal understanding of manifolds. ## Part I: James We're using I. M. James "Topological and Uniform Spaces", Springer 1987.(Which I like because it introduces you to ultrafilters, and its short.) 1. Read Chapters 1-5 and do a good fraction of the exercises. This brings you to Tychonov's theorem. 2. Chapters 6, 9. Do a good fraction of the exercises. 3. Write out proofs that the separation conditions collapse on topological groups (i.e., $T_0=T_1=T_2$). Here are some articles that may be useful. 4. Study connectedness (James, ch 9). 5. Arrow's Theorem: This has very little to do with topology, but it's very interesting and has nice proofs using ultrafilters. Some quick finds on the web: There's also a blog article by Tao. (Let me know which of these is most useful.) 6. Look up the Zariski topology on $\fld^d$, for an algebraically closed field $\fld$. Assuming it is compact (true), prove that any ideal in the ring $\fld[\seq x1d]$ is finitely generated. 7. Write up notes on topologies on function spaces, in particular: product, uniform, compact-open. 8. What are the advantages of being Hausdorff? (A short essay, please.) 9. James's terminology is not always standard. Find another reasonable book on topology and prepare a translation table. ## Part II: Homotopy, Covers, Surfaces Any introductory text you find on algebraic topology that starts with homotopy will probably be useful. Two that do work are: • Hatcher "Algebraic Topology" CUP 2002. on-line • Lee "Introduction to Topological Manifolds". (You can get a pdf through the library. This also covers a lot of the stuff in James.) One goal is to be able to explain the sentences "the covering graphs of a finite connected graph correspond to the permutation representations of its fundamental group" and "an embedding of a graph in a surface determines a homomorphism from its fundamental group into the fundamental group of the surface". This is covered in Chapter 0 and in Sections 1.1 and 1.3 of Hatcher. (Note that this is much denser than James's book.) I will give a lecture on free groups at some point. We start by studying fundamental groups and covering spaces. As a group, provide a set of notes dealing with the following. 1. Write down definitions for simplicial complex, cell complex, manifold, smooth manifold. Provide interesting examples. In case, describe the natural maps. 2. Write down definitions for: isotopy, ambient isotopy, homotopy. Knots may prove useful for examples. 3. Let $X$ be a topological space and assume $a\in X$. A loop in $X$ based at $a$ is a map $f:[0,1]\to X$ such that $f(0)=f(1)=a$. The loops in $X$ based at $a$ form a monoid. Homotopy is an equivalence relation on loops (proof), and the equivalence classes of loops at $a$ form a group (proof), denoted $\pi_1(X,a)$ and called the first fundamental group of $X$. [For us, it will be the first and only, so we will usually drop the "first".] If $X$ is connected in some appropriate sense (which you should specify) and $b$ is a second point in $X$, determine the relation between $\pi_1(X,a)$ and $\pi_1(X,b)$. 4. Determine the fundamental group of the circle. Go over a proof of the "fundamental" "theorem of algebra" based on this. 5. Suppose $\cC$ is a cell complex and the graph $X$ be its 1-skeleton. Each walk in the graph determines a path in the complex. When are two walks homotopic? (Here and elsewhere, two paths are homotopic if they have the same end points and there is a homotopy from one to the other that fixes the end points.) 6. Write up a definition of covering space. State the significant results relating paths in $X$ and paths in a covering space $Y$. 7. Assume $X$ is not too evil and let $a$ be a point in $X$. Let $\hat{X}$ denote the set of homotopy classes of paths in $X$ that start at $a$. Show that $\hat{X}$ is a covering space for $X$. Show that $\pi_1(X,a)$ acts freely on $\hat{X}$, and that the orbits of $\pi_1$ are the fibres of the cover. Show that if $X$ is a manifold, so is $\hat X$. 8. (Continuing from the previous.) Show that if $Y$ covers $X$, then $\hat{X}$ covers $Y$ and the composition of the two covering maps is the map with the orbits of $\pi_1(X)$ as its fibres. 9. Determine the fundamental groups of the real projective plane and the torus. 10. Suppose $Y_1$ and $Y_2$ cover $X$. By considering the map from $Y_1\times Y_2$ to $X\times X$, show that there is subspace of $Y_1\times Y_2$ that covers $Y_1$ and $Y_2$. Show that the obvious diagram commutes.	1,280	4,895	{"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}	3.109375	3	CC-MAIN-2018-17	latest	en	0.914427
https://it.mathworks.com/matlabcentral/cody/problems/1974-length-of-a-short-side/solutions/418079	1,586,353,459,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585371813538.73/warc/CC-MAIN-20200408104113-20200408134613-00265.warc.gz	521,721,341	15,563	Cody # Problem 1974. Length of a short side Solution 418079 Submitted on 14 Mar 2014 by Guillaume 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 %% b = 1; c = 2; a_correct = sqrt(3); tolerance = 1e-12; assert(abs(calculate_short_side(b,c)-a_correct)<tolerance); 2 Pass %% b = 4; c = 5; a_correct = 3; tolerance = 1e-12; assert(abs(calculate_short_side(b,c)-a_correct)<tolerance); 3 Pass %% b = 12; c = 13; a_correct = 5; tolerance = 1e-12; assert(abs(calculate_short_side(b,c)-a_correct)<tolerance); 4 Pass %% b = 8; c = 10; a_correct = 6; tolerance = 1e-12; assert(abs(calculate_short_side(b,c)-a_correct)<tolerance);	248	751	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.984375	3	CC-MAIN-2020-16	latest	en	0.624613
https://newonlinecourses.science.psu.edu/stat414/node/179/	1,556,000,562,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578593360.66/warc/CC-MAIN-20190423054942-20190423080942-00213.warc.gz	501,131,679	8,408	# Normal Approximation to Binomial Printer-friendly version As the title of this page suggests, we will now focus on using the normal distribution to approximate binomial probabilities. The Central Limit Theorem is the tool that allows us to do so. As usual, we'll use an example to motivate the material. ### Example Let Xi denote whether or not a randomly selected individual approves of the job the President is doing. More specifically: • Let Xi = 1, if the person approves of the job the President is doing, with probability p • Let Xi = 0, if the person does not approve of the job the President is doing with probability 1 − p Then, recall that Xi is a Bernoulli random variable with mean: $\mu=E(X)=(0)(1-p)+(1)(p)=p$ and variance: $\sigma^2=Var(X)=E[(X-p)^2]=(0-p)^2(1-p)+(1-p)^2(p)=p(1-p)[p+1-p]=p(1-p)$ Now, take a random sample of n people, and let: Y = X1 + X2 + ... + Xn Then Y is a binomial(np) random variable, y = 0, 1, 2, ... , n, with mean: $\mu=np$ and variance: $\sigma^2=np(1-p)$ Now, let n = 10 and p = ½, so that Y is binomial(10, ½). What is the probability that exactly five people approve of the job the President is doing? Solution. There is really nothing new here. We can calculate the exact probability using the binomial table in the back of the book with n = 10 and p = ½. Doing so, we get: \begin{align} P(Y=5)&= P(Y \leq 5)-P(Y \leq 4)\\ &= 0.6230-0.3770\\ &= 0.2460\\ \end{align} That is, there is a 24.6% chance that exactly five of the ten people selected approve of the job the President is doing. Note, however, that Y in the above example is defined as a sum of independent, identically distributed random variables. Therefore, as long as n is sufficiently large, we can use the Central Limit Theorem to calculate probabilities for Y. Specifically, the Central Limit Theorem tells us that: $Z=\dfrac{Y-np}{\sqrt{np(1-p)}}\stackrel {d}{\longrightarrow} N(0,1)$. Let's use the normal distribution then to approximate some probabilities for Y. Again, what is the probability that exactly five people approve of the job the President is doing? Solution. First, recognize in our case that the mean is: $\mu=np=10\left(\dfrac{1}{2}\right)=5$ and the variance is: $\sigma^2=np(1-p)=10\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=2.5$ Now, if we look at a graph of the binomial distribution with the rectangle corresponding to Y = 5 shaded in red: we should see that we would benefit from making some kind of correction for the fact that we are using a continuous distribution to approximate a discrete distribution. Specifically, it seems that the rectangle Y = 5 really includes any Y greater than 4.5 but less than 5.5. That is: $P(Y=5)=P(4.5< Y < 5.5)$ Such an adjustment is called a "continuity correction." Once we've made the continuity correction, the calculation reduces to a normal probability calculation: Now, recall that we previous used the binomial distribution to determine that the probability that Y = 5 is exactly 0.246. Here, we used the normal distribution to determine that the probability that Y = 5 is approximately 0.251. That's not too shabby of an approximation, in light of the fact that we are dealing with a relative small sample size of n = 10! Let's try a few more approximations. What is the probability that more than 7, but at most 9, of the ten people sampled approve of the job the President is doing? Solution. If we look at a graph of the binomial distribution with the area corresponding to 7 < Y ≤ 9 shaded in red: we should see that we'll want to make the following continuity correction: $P(7<Y \leq 9)=P(7.5< Y < 9.5)$ Now again, once we've made the continuity correction, the calculation reduces to a normal probability calculation: By the way, you might find it interesting to note that the approximate normal probability is quite close to the exact binomial probability. We showed that the approximate probability is 0.0549, whereas the following calculation shows that the exact probability (using the binomial table with n = 10 and p = ½) is 0.0537: $P(7<Y \leq 9)=P(Y\leq 9)-P(Y\leq 7)=0.9990-0.9453=0.0537$ Let's try one more approximation. What is the probability that at least 2, but less than 4, of the ten people sampled approve of the job the President is doing? Solution. If we look at a graph of the binomial distribution with the area corresponding to 2 ≤ Y < 4 shaded in red: we should see that we'll want to make the following continuity correction: $P(2 \leq Y <4)=P(1.5< Y < 3.5)$ Again, once we've made the continuity correction, the calculation reduces to a normal probability calculation: \begin{align} P(2 \leq Y <4)=P(1.5< Y < 3.5) &= P(\dfrac{1.5-5}{\sqrt{2.5}}<Z<\dfrac{3.5-5}{\sqrt{2.5}})\\ &= P(-2.21<Z<-0.95)\\ &= P(Z>0.95)-P(Z>2.21)\\ &= 0.1711-0.0136=0.1575\\ \end{align} By the way, the exact binomial probability is 0.1612, as the following calculation illustrates: $P(2 \leq Y <4)=P(Y\leq 3)-P(Y\leq 1)=0.1719-0.0107=0.1612$ Just a couple of comments before we close our discussion of the normal approximation to the binomial. (1) First, we have not yet discussed what "sufficiently large" means in terms of when it is appropriate to use the normal approximation to the binomial. The general rule of thumb is that the sample size n is "sufficiently large" if: np ≥ 5 and n(1 − p) ≥ 5 For example, in the above example, in which p = 0.5, the two conditions are met if: np = n(0.5) ≥ 5 and n(1 − p) = n(0.5) ≥ 5 Now, both conditions are true if: n ≥ 5(10/5) = 10 Because our sample size was at least 10 (well, barely!), we now see why our approximations were quite close to the exact probabilities. In general, the farther is away from 0.5, the larger the sample size n is needed. For example, suppose p = 0.1. Then, the two conditions are met if: npn(0.1) ≥ 5 and n(1 − p) = n(0.9) ≥ 5 Now, the first condition is met if: n ≥ 5(10) = 50 And, the second condition is met if: n ≥ 5(10/9) = 5.5 That is, the only way both conditions are met is if n ≥ 50. So, in summary, when p = 0.5, a sample size of n = 10 is sufficient. But, if p = 0.1, then we need a much larger sample size, namely n = 50. (2) In truth, if you have the available tools, such as a binomial table or a statistical package, you'll probably want to calculate exact probabilities instead of approximate probabilities. Does that mean all of our discussion here is for naught? No, not at all! In reality, we'll most often use the Central Limit Theorem as applied to the sum of independent Bernoulli random variables to help us draw conclusions about a true population proportion p. If we take the Z random variable that we've been dealing with above, and divide the numerator by n and the denominator by n (and thereby not changing the overall quantity), we get the following result: $Z=\dfrac{\sum X_i-np}{\sqrt{np(1-p)}}=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\stackrel {d}{\longrightarrow} N(0,1)$ The quantity: $\hat{p}=\dfrac{\sum\limits_{i=1}^n X_i}{n}$ that appears in the numerator is the "sample proportion," that is, the proportion in the sample meeting the condition of interest (approving of the President's job, for example). In Stat 415, we'll use the sample proportion in conjunction with the above result to draw conclusions about the unknown population proportion p. You'll definitely be seeing much more of this in Stat 415!	2,153	7,416	{"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}	4.5625	5	CC-MAIN-2019-18	latest	en	0.868642
https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Cocurvature	1,618,935,918,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618039476006.77/warc/CC-MAIN-20210420152755-20210420182755-00428.warc.gz	1,188,984,224	4,287	# Cocurvature In mathematics in the branch of differential geometry, the cocurvature of a connection on a manifold is the obstruction to the integrability of the vertical bundle. ## Definition If M is a manifold and P is a connection on M, that is a vector-valued 1-form on M which is a projection on TM such that PabPbc = Pac, then the cocurvature ${\displaystyle {\bar {R}}_{P}}$ is a vector-valued 2-form on M defined by ${\displaystyle {\bar {R}}_{P}(X,Y)=(\operatorname {Id} -P)[PX,PY]}$ where X and Y are vector fields on M.	149	535	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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	0.895113
http://www.tntech.edu/cesr/labgrid/print/?tmpl=component	1,398,365,846,000,000,000	text/html	crawl-data/CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00031-ip-10-147-4-33.ec2.internal.warc.gz	926,442,881	3,227	## LabGrid ### A Laboratory sized power Grid (LabGrid) Electrical engineering programs typically have motors, generators, and power meters as part of their regular laboratory curriculum but an actual multi-generator power grid which can be used to study disturbances in a power grid is rare. Figure 1 shows LabGrid in an early stage of development with control computer and National Instruments data acquisition system. Figure 2 highlights some of the grid components such as the Hampden synchronous machines for generators and a lumped circuit transmission line models. Figure 1 Graduate students Chayanon Sontidpanya during an early stage of LabGrid construction. Figure 2 Various components of LabGrid The Servo Motors are the prime movers of the system, much like a coal fired power plant, or a diesel generator, the electric power is generated from the synchronous generator and fed into the grid consisting of transmission lines, busses, and loads. Computer controlled relays disturbance to be introduced to the grid, or separating of the grid into multiple islands depending on the grid configuration under study. Figure 3 Undergraduate student Steven Corum running LabGrid’s LabView control program ### Areas of Study Many areas of power grid research can benefit from a physical power grid that is independent of the local power utility. Here are a few areas currently under study: • Measurements of frequency in the power grid don’t always correlate well with generator rotational speeds providing incentive for a new laboratory sized power grid to study this phenomenon. • Islanding detection and implementation algorithms can be tested. • Improving simulation models through studying areas where physical measurements and simulation models don’t agree. Figure 4: A Matlab Simulink model of a two generator system. ### Two Generator Speed Demonstration In the real world power grid you could never tell the generators to operate at different frequencies from each other (for any extended period of time) but with the LabGrid there is no such limitation. Figure 5 shows what happens when two different generators are set to two different speeds. The systems are subject to control loops and they fight each other as the two systems get more and more out of phase with one another till they snap away from each other when they are too out of phase to transmit significant power. This behavior cycles over and over as the two systems cycle between being in synch and out of synch with each other. Figure 5 Measured frequencies and generator speeds expressed as electrical frequencies. ### Microgrid Test bed The Smart Grid Group at TTU, recognizing the benefits a physical grid have banded together to create a larger Micro-Grid Test Bed which will incorporate the smaller LabGrid in the near future.	545	2,832	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-15	latest	en	0.884419
http://www.icalculator.info/india/terminology/short-term-capital-gains.html	1,496,115,886,000,000,000	text/html	crawl-data/CC-MAIN-2017-22/segments/1495463613780.89/warc/CC-MAIN-20170530031818-20170530051818-00496.warc.gz	636,536,993	7,814	Free Online Calculators: # What are Short-term Capital Gains? Capital gains are profits made from selling capital assets. Some examples of capital assets are land, property, jewelry, mutual funds, shares, and art & collectibles. As per income tax laws, you need to pay tax on any capital gains. The amount of tax depends on whether you've made short-term capital gains or long-term capital gains. ### Short-term Capital Gains in Detail Short-term capital gains are the profits you make on selling of assets, which you have held for a short term as per income tax laws. The tax on short-term capital gains varies by the asset type. ### Calculation of short-term capital Gains Tax: Based on asset type, there are two types of short-term capital gains: • Short-term capital gains under section 111A • Short-term capital gains outside of section 111A ### Short-term capital gains under section 111A: Section 111A is applicable to any short-term gains made from investments in equity or equity-based mutual funds. For such instruments, short-term is considered as 1 year. As per Section 111A, such short-term capital gains are taxed at a flat rate of 15%. Example 1: Say, you purchased shares of a certain company for ₹100,000 and sold them after 9 months for ₹150,000. Since you sold the shares within 1 year, this gain will qualify as a short-term capital gain under section 111A. In such a case, the short-term gain is ₹50,000 (₹150,000 less ₹100,000), and the short-term capital gain tax will be ₹7,500 (15% of ₹50,000). ## Popular India Tax & Finance Guides and Calculators People who read 'Tax Returns: Understanding Short-term Capital Gains?' also read the following tax and finance guides and calculators:	392	1,720	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22	longest	en	0.937103
https://yutsumura.com/given-the-characteristic-polynomial-find-the-rank-of-the-matrix/	1,586,525,533,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585371896913.98/warc/CC-MAIN-20200410110538-20200410141038-00267.warc.gz	1,221,661,321	27,609	# Given the Characteristic Polynomial, Find the Rank of the Matrix ## Problem 484 Let $A$ be a square matrix and its characteristic polynomial is given by $p(t)=(t-1)^3(t-2)^2(t-3)^4(t-4).$ Find the rank of $A$. (The Ohio State University, Linear Algebra Final Exam Problem) ## Solution. Note that the degree of the characteristic polynomial $p(t)$ is the size of the matrix $A$. Since the degree of $p(t)$ is $3+2+4+1=10$, the size of the matrix $A$ is $10\times 10$. From the characteristic polynomial, we see that the eigenvalues of $A$ are $1,2,3,4$. In particular, $0$ is not an eigenvalue of $A$. Hence the null space of $A$ is zero dimensional, that is, the nullity of $A$ is $0$. By the rank-nullity theorem, we have $\text{rank of A} +\text{ nullity of A}=10.$ Hence the rank of $A$ is $10$. ## Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University) This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568). The other problems can be found from the links below. ### 8 Responses 1. 06/28/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 2. 06/28/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 3. 07/12/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 4. 08/02/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 5. 08/12/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 6. 08/17/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 7. 10/05/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 8. 10/15/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] This site uses Akismet to reduce spam. Learn how your comment data is processed. ##### Diagonalize the 3 by 3 Matrix Whose Entries are All One Diagonalize the matrix $A=\begin{bmatrix} 1 & 1 & 1 \\ 1 &1 &1 \\ 1 & 1 & 1 \end{bmatrix}.$... Close	583	2,048	{"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": 0, "x-ck12": 0, "texerror": 0}	3.921875	4	CC-MAIN-2020-16	latest	en	0.800399
https://www.banhamdark.co.uk/uk/21992-how-many-linear-feet-is-a-fence-panel.html	1,579,716,326,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579250607314.32/warc/CC-MAIN-20200122161553-20200122190553-00104.warc.gz	764,299,114	6,140	## how many linear feet is a fence panel ### Figure fence needed to enclose property - Hoover Fence Co. Vinyl Fence Panels ; Rolled Fence Fabric / Posts ; . How many feet of fencing would it take to accomplish this? . 511 x 4 = 2044 linear feet of fence to enclose 6 . ### 8ft panel is how many linear ft - Composite Decking Price 8ft panel is how many linear ft. Home > review > wpc products. . how many linear feet are in 2 8ft wood fence panels About how many linear feet are . ### how do i measure a 6' x 8' wood panel fence to linear ft . How do i measure a 6' x 8' wood panel fence to linear ft? . if your panel is 6 feet high and eight feet wide, you simple take your linear feet/panel width. ### How to Measure Fencing \| eHow How to Measure Fencing. . Fencing is sold in linear feet. . Any remaining footage smaller than a fence panel determines how many lineal feet of partial panels you . ### how many fence panels make linear feet how many fence panels make linear feet. Products Case. How . Any remaining footage smaller than a fence panel determines how many lineal feet of partial panels you . ### how many linear feet is a fence panel - waterproofboards Home > Products > how many linear feet is a fence panel. Products. Decking Series ; QZ Outdoor Decking (Hollow) QZ Decking (solid) DIY Decking ; Wall Panel Series ; ### Fence Calculator - SASCO . in either linear feet or . Horse Fence; Panels; T-posts; . Dealers; News; Fencing Center . Fence Wires; Fence Coatings; Fence Knots; Fence Calculator; Fence . ### How Many Linear Feet Is A Fence Panel How Many Linear Feet Is A Fence Panel Egypt,1 linear foot equals? how much fencing?, Yahoo Answers 1 linear foot equals? how much fencing? ### How to Calculate the Lumber Requirements for a Fence \| Home . How to Calculate the Lumber Requirements for a Fence. . A 200-foot fence therefore needs 400 linear feet of railing. 5. . If you're installing a panel fence, . ### how many fence panels make linear feet Any remaining footage smaller than a fence panel determines how many lineal feet of partial panels you will need . So you would need 75*3=225 linear feet of fence . ### how many linear feet is a fence panel - Best Terrace Decking how many linear feet is a fence panel . Composite Decking Price how many linear feet are in 2 8ft wood fence panels About how many linear feet are in 2 8ft . ### how many linear feet are in 2 8ft wood fence panels how many linear feet are in 2 8ft wood fence panels Home > Solution > how many linear feet are in 2 8ft wood fence panels ### How Many Linear Feet Is A Fence Panel Egypt how many linear feet is a fence panel . how many linear feet are in 2 8ft wood fence panels how many linear feet are in 2 8ft wood fence panels. ### 8ft panel is how many linear ft - Composite Decking Price Products 1 - 877 . Plank Panel Material, . how many linear feet are in 2 8ft wood fence panels . 8ft panel is how many linear ft ### how many linear feet is a fence panel - Outside Wpc Deck Bekaert high tensile fence with class three coating You should be able to estimate how many linear feet your fence . Million linear feet. Fence/ Wood Panel . ### fence - How can I estimate square footage of fencing for . I have a fence that looks like the . How can I estimate square footage of fencing for . Example 4ft high X 100 feet long X 2 sides = 800 square feet. 800 . ### How to Calculate the Lumber Requirements for a Fence \| Home . How to Calculate the Lumber Requirements for a Fence. . a panel fence, . by the height of the fence to find the number of linear feet of slat or picket . ### 1 linear foot equals? how much fencing? \| Yahoo Answers . A linear foot means the measurment of the fence length. . 1 linear foot equals? how much fencing? . Fence panels are usually sold in 8' lengths. ### How much cedar fence in linear feet will I need? \| The Home . Someone offered me 400 linear feet of cedar fence . And how long are the panels so you can determine how many posts you will need.	976	4,031	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05	latest	en	0.899114
https://physics.stackexchange.com/questions/707377/is-this-a-rigorous-derivation-of-coriolis-effect-it-seems-pretty-simple	1,713,875,118,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00843.warc.gz	408,498,885	44,041	# Is this a rigorous derivation of Coriolis effect? It seems pretty simple Coriolis effect is usually derived in a long twice differentiated equation of motion between a lab frame and rotated frame. Afterwards you group together all the terms and you get all the fictitious forces. However no intuition is gained in it for me. For the coriolis effect, it seems it is the force that appropriately changes the angular momentum of an object, or the friction force in the rotating frame that is needed to keep an object on a straight path w.r.t. the rotating frame. So for an object moving radially outward from the center of the rotating disk, the corrective friction force will be (ignoring the other effects): $$$$\text{Angular momentum of the object is:} \\ L = mr^2 \omega= m(vt)^2 \omega, \ \ \ \ \ [r=vt] \\$$$$ $$$$\text{differentiating w.r.t. time} \\ \frac{dL}{dt} = \frac{d}{dt} (mv^2t^2 \omega) = mv^2 \omega \frac{d (t^2)}{dt} \\$$$$ $$$$\text{expressing it in terms of torque} \\ \frac{dL}{dt} = \tau = F_{\text{cor}}r = 2 mv^2 \omega t \\ F_{\text{cor}} = 2 mv \omega \frac{(vt)}{r} = \boxed{2mv\omega}$$$$ There is a related derivation for centrifugal force as well using conservation of linear momentum. $$\frac{dp}{dt} = \frac{d (p\theta)}{dt} = p\omega = mv \frac{v}{r} = \frac{mv^2}{r}$$ Why aren't these derivations used more often? • your case is just for one dimensional in three dimensional the Coriolis force is $~\vec F_c=2\,m\,(\vec\omega\times \vec v)~$ can you obtain this result ? – Eli May 7, 2022 at 15:31 • Isn't that an arbitrary generalization? I mean if you have an object with a random three dimensional velocity in a rotating frame you could decompose the velocity in radial/tangential velocity and apply conservation of (angular momentum) to get it. But the gist is exactly the same, it just gets a bit messier in 3D I believe. Or I could be mistaken May 7, 2022 at 19:36 • Related : Velocity in a turning reference frame. May 7, 2022 at 21:27 Let me first give a general derivation for the case of an object that is physically going around a center of attraction, with the provided centripetal force dialed in to be equal in magnitude and opposite in direction to the centrifugal effect, at every distance to the center of attraction. That case is circumnavigating motion with a force according to Hooke's law. With a centripetal force according to Hooke's law the solution of the equation of motion can be described with the following parametric equation. $$x = a \cos(\Omega t)$$ $$y = b \sin(\Omega t)$$ $$a$$ half the length of the major axis $$b$$ half the length of the minor axis $$\Omega$$ 360 degrees divided by the duration of one revolution So the motion is a superpositon of two harmonic oscillations, perpendicular to each other. The uploaded animated GIF represents a rotating disk. On the left the rotating disk as viewed from a stationary point of view, on the right the rotating disk as viewed from a co-rotating point of view. The arrow represents the centripetal force. The magnitude of the force increases in proportion to the distance to the center of circumnavigation. Along the rim there is a division in 4 quadrants, the quadrants serve as a reference. The angular velocity is not constant; the angular velocity increases and decreases, cycling twice for every full circumnavigation. Likewise the distance to the center of attraction cycles twice for every full circumnavigation. The parametric equation can be rearranged as follows: $$x = \frac{a+b}{2} \cos(\Omega t) + \frac{a-b}{2} \cos(\Omega t)$$ $$y = \frac{a+b}{2} \sin(\Omega t) - \frac{a-b}{2} \sin(\Omega t)$$ After transformation of the motion to the rotating coordinate system the motion relative to the co-rotating coordinate system is as follows: $$x = \frac{a-b}{2} \cos(2 \Omega t)$$ $$y = - \frac{a-b}{2} \sin(2 \Omega t)$$ Notice that the motion along the small circle ( the motion relative to the co-rotating coordinate system) cycles with a frequency of $$2\Omega$$. I will refer to the motion depicted on the right hand side of the animated GIF as 'motion along the epi-circle' The next step is to obtain an expression for the acceleration of the moving object relative to the co-rotating coordinate system. Let $$a_c$$ be the acceleration relative to the co-rotating coordinate system. The magnitude of the acceleration towards the center of the epi-circle is given by the standard expression for the required centripetal acceleration for sustained circular motion: $$a_c = \omega^2r \tag{1}$$ (I use the lowercase $$\omega$$ here for the motion along the epi-circle to distinguish it from the uppercase $$\Omega$$ that I used for the angular velocity of the system.) This expression gives the magnitude. To sustain uniform circular motion the acceleration must be perpendicular to the instantaneous velocity. That gives the direction of $$a_c$$: perpendicular. As stated earlier, the motion along the epi-circle occurs at twice the frequency of the angular velocity of the system: 2$$\Omega$$ Inserting that into (1): $$a_c = (2\Omega)^2r \tag{2}$$ (2) does not feature the velocity relative to the rotating coordinat system explicitly. Let $$v_c$$ be the velocity of the object relative to the co-rotating coordinate sytem. Then we can use this expression: $$2\Omega r = v_c \tag{3}$$ to modify (2). This modification gives us the following expression for the acceleration of the object with respect to the co-rotating coordinate system: $$a_c = 2\Omega v_c \tag{4}$$ Discussion So we see that if the force of attraction increases linear with distance to the center of attraction, Hooke's law, then the motion with respect to the co-rotating coordinate system is given by the coriolis acceleration $$2\Omega v_c$$ The required centripetal force to sustain circular motion is $$m \omega^2 r$$ So: when the provided centripetal force is according to Hooke's law there is no opportunity for centrifugal effect; a Hooke's law centripetal force counteracts centrifugal effect everywhere. Therefore: when the centripetal force is according to Hooke's law the acceleration with respect to the co-rotating coordinate system is described entirely by the Coriolis term: $$2\Omega v_c$$ The coriolis effect is the same in all directons. The magnitude is always $$2\Omega v_c$$ The direction is always perpendicular to the instantaneous velocity. In many cases you effectively get Hooke's law automatically. For instance, let's say you are on a merry-go-round, and you have been instructed to hold on. When the angular velocity of the merry-go-round increases you hold on more firmly, in response to the need for more centripetal force. In many cases the setup is designed to produce required centripetal force automatically. Example: a liquid mirror telescope Planar motion This derivation is for motion in a plane. The following property is general: when the centripetal force is always towards a single point of attraction the resulting motion is inherently planar motion. Because of that planar nature it is sufficient to derive for the case with two spatial degrees of freedom. indeed you can use it for three dimensional case $$\vec L=\mathbf I\,\vec\omega$$ with the inertia tensor for a mass point $$\mathbf I=-m\,\mathbf{r}_\times\,\mathbf{r}_\times\quad, \mathbf r_\times=\left[ \begin {array}{ccc} 0&-r_{{z}}&r_{{y}} \\ r_{{z}}&0&-r_{{x}}\\ -r_{{y}}&r_{{x}}&0\end {array} \right]$$ you obtain $$\vec L=-m\,\mathbf{r}_\times\,\mathbf{r}_\times\,\vec\omega$$ with $$~\vec r=\vec v\,t$$ $$\vec L=-m\,t^2\,\mathbf{v}_\times\,\mathbf{v}_\times\,\vec\omega\quad\Rightarrow\\ \vec{\dot{ L}}=-2\,m\,t\,(\vec v\times\vec v\times\vec \omega)=\vec r\times\vec F_c =\vec{v}\,t\times \vec F_c\quad\Rightarrow\\ +2\,m\,\vec v\times (\vec \omega\times \vec v)=\vec v\times\vec F_c$$ thus the coriolis force $$\boxed{\vec F_c=2\,m\,(\vec\omega\times\vec v)}$$ because $$~\vec{\dot{r}}=\vec{v}~$$ you should use this derivation $$\vec L=-m\,\mathbf{r}_\times\,\mathbf{r}_\times\,\vec\omega= -m\,\left(\vec{r}\times\vec{r}\times\vec\omega\right)$$ $$\vec{\dot{L}}=-m\,\left(\vec{\dot{r}}\times\vec{r}\times\vec\omega \right)-m\,\left(\vec{r}\times\vec{\dot{r}}\times\vec\omega\right)\\ 2\,m\,\vec r\times\vec\omega\times\vec v=\vec r\times\vec F_c\quad\Rightarrow\\ 2\,m\,(\vec\omega\times\vec v)=\vec F_c$$ remarks $$~\vec\omega~$$ is constant and not depending on time • Thank you. Is there a similar one for linear momentum? May 7, 2022 at 21:32 • @bananenheld The centrifugal force from Newton second law \begin{aligned}m\dfrac{d\overrightarrow{v}}{dt}=\overrightarrow{F}\\ \overrightarrow{v}=\overrightarrow{w}\times \overrightarrow{r}\\ m\left( \overrightarrow{w}\times \overrightarrow{v}\right) =\overrightarrow{F}\\ m\left( \overrightarrow{w}\times \overrightarrow{w}\times \overrightarrow{r}\right) =\overrightarrow{F}\end{aligned} – Eli May 8, 2022 at 11:46 In my earlier answer I derived the coriolis effect for the general case. The general case is that the circumnavigating object can have any direction of velocity with respect to the rotating coordinate system. In this answer I will addres the more limited case that you raised in your question: if a object is moving radially with respect to a rotating disk, what is the magnitude of the tendendy of that object to change angular velocity? In your derivation you used a substitution $$r=vt$$ to introduce a factor $$t$$ for time, but that substitution isn't necessary. The expression for angular momentum: $$m\omega r^2$$ The mass $$m$$ is constant anyway, so the statement of conservation can without loss of validity be narrowed down to: $$\omega r^2 = \text{constant} \tag{1}$$ Hence: $$\frac{d(\omega r^2)}{dt} = 0 \tag{2}$$ Differentiating: $$r^2 \frac{d\omega}{dt} + \omega \frac{d(r^2)}{dt} = 0 \tag{3}$$ Using the chain rule to convert $$\tfrac{d(r^2)}{dt}$$ into $$2r\tfrac{dr}{dt}$$: $$r^2 \frac{d\omega}{dt} + 2 r \omega \frac{dr}{dt} = 0 \tag{4}$$ Divide by $$r$$, and then rearrange: $$r \frac{d\omega}{dt} = - 2 \omega \frac{dr}{dt} \tag{5}$$ I will use $$a_t$$ for acceleration in the direction tangential to a concentric circle (hence perpendicular to the radial direction), and $$v_r$$ for velocity in radial direction. Then $$r\tfrac{dw}{dt} = a_t$$ and $$\tfrac{dr}{dt} = v_r$$ The result: $$a_t = -2\omega v_r \tag{6}$$ Multiplying both sides with $$m$$ for mass gives the amount of force required to keep the object at the same angular velocity when it has a radial velocity $$v_r$$ $$F_t = -2m\omega v_r \tag{7}$$ Not the general coriolis effect This derivation addresses only the case of radial velocity. However, the general coriolis effect is the same for every direction of motion with respect to the rotating coordinate system, so this derivation isn't general enough.	3,026	10,858	{"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": 58, "wp-katex-eq": 0, "align": 0, "equation": 3, "x-ck12": 0, "texerror": 0}	3.875	4	CC-MAIN-2024-18	latest	en	0.855716
https://www.sawaal.com/time-and-distance-questions-and-answers/travelling-at-4-5th-of-his-usual-speed-a-man-is-15-minutes-late-what-is-his-usual-time-to-cover-the-_40077	1,656,108,666,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00137.warc.gz	1,044,045,433	14,506	2 Q: # Travelling at 4/5th of his usual speed, a man is 15 minutes late. What is his usual time to cover the same distance? A) 15 min B) 1 hr C) 75 min D) 45 min Explanation: Q: A car covers 25 km at a uniform speed. If the speed had been 8 knv/h more, it would have taken 10 hours less for the same journey. What is the speed of the car (in km/h)? A) 4 B) 3 C) 2.5 D) 2 Explanation: 0 21 Q: The ratio between the speeds of two trains is 5 : 7. If the first train covers 300 km in 3 hours, then the speed (in km/h) of the second train is: A) 100 B) 150 C) 140 D) 120 Explanation: 0 139 Q: The ratio between the speeds of two trains is 2 : 5. If the first train runs 350 km in 5 h, then the difference between the speeds (in km/h) of both the trains is: A) 350 B) 105 C) 165 D) 180 Explanation: 2 395 Q: Given that the lengths of the paths of a ball thrown with different speeds by two boy sare the same, if they take 0.6 seconds and 1 second respectively to cover the said length, what is the average speed of travel for the first throw, if the same for the second is 96 km/h? A) 160 km/h B) 200 km/h C) 150 km/h D) 100 km/h Explanation: 1 406 Q: A man divided his journey into three parts of distances of 18 km, 20 km and 27 km. He travelled the distances at the speeds of 6 km/h, 5 km/h and 9 kn/h,respectively. What was his average speed during the entire journey? A) 6.5 km/h B) 4.5 km/h C) 7.5 km/h D) 5.5 km/h Explanation: 2 650 Q: The distance between two railway stations is 1176 km. To cover this distance, an express train takes 5 hours less than a passenger train while the average speed of the passenger train is 70 km/h less than that of the express train. The time taken by the passenger train to complete the travel is: A) 12 hours B) 23 hours C) 17 hours D) 18 hours Explanation: 2 742 Q: The distance between the places H and O is D units. The average speed that gets a person from H to O in a stipulated time is S units. He takes 20 minutes more time than usual if he travels at 60 km/h, and reaches 44 minutes early if he travels at 75 km/h. The sum of the numerical values of D and S is: A) 376 B) 358 C) 344 D) 384 Explanation: 0 481 Q: A train that is running at the speed of 72 km/h crossesan electric pole in 36 seconds. The length of the train (in metres) is: A) 720 m B) 460 m C) 360 m D) 620 m	730	2,362	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27	latest	en	0.914217
https://convertoctopus.com/236-years-to-minutes	1,679,921,912,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00092.warc.gz	218,464,346	8,127	## Conversion formula The conversion factor from years to minutes is 525949.2, which means that 1 year is equal to 525949.2 minutes: 1 yr = 525949.2 min To convert 236 years into minutes we have to multiply 236 by the conversion factor in order to get the time amount from years to minutes. We can also form a simple proportion to calculate the result: 1 yr → 525949.2 min 236 yr → T(min) Solve the above proportion to obtain the time T in minutes: T(min) = 236 yr × 525949.2 min T(min) = 124124011.2 min The final result is: 236 yr → 124124011.2 min We conclude that 236 years is equivalent to 124124011.2 minutes: 236 years = 124124011.2 minutes ## Alternative conversion We can also convert by utilizing the inverse value of the conversion factor. In this case 1 minute is equal to 8.0564589424097E-9 × 236 years. Another way is saying that 236 years is equal to 1 ÷ 8.0564589424097E-9 minutes. ## Approximate result For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred thirty-six years is approximately one hundred twenty-four million one hundred twenty-four thousand eleven point two minutes: 236 yr ≅ 124124011.2 min An alternative is also that one minute is approximately zero times two hundred thirty-six years. ## Conversion table ### years to minutes chart For quick reference purposes, below is the conversion table you can use to convert from years to minutes years (yr) minutes (min) 237 years 124649960.4 minutes 238 years 125175909.6 minutes 239 years 125701858.8 minutes 240 years 126227808 minutes 241 years 126753757.2 minutes 242 years 127279706.4 minutes 243 years 127805655.6 minutes 244 years 128331604.8 minutes 245 years 128857554 minutes 246 years 129383503.2 minutes	465	1,778	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.1875	4	CC-MAIN-2023-14	latest	en	0.803377
https://convertilo.com/cubic-centimeters-to-milliliters	1,624,537,030,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00194.warc.gz	176,174,555	5,563	# Cubic Centimeters to Milliliters - cm3 to ml ## How to convert from Cubic Centimeters to Milliliters The conversion factor between Cubic Centimeters and Milliliters is 1. To convert an amount in Cubic Centimeters in Milliliters, multiply the the desired amount in Cubic Centimeters by 1: Amount(Cubic Centimeters) × 1(Conversion Factor) = Result(Milliliters) ## Definition of units Let's see how both units in this conversion are defined, in this case Cubic Centimeters and Milliliters: ### Cubic Centimeter (cm3) A cubic centimeter (SI unit symbol: cm3; non-SI abbreviations: cc and ccm) is a commonly used unit of volume which is derived from SI-unit cubic meter. One cubic centimeter is equal to 1⁄1,000,000 of a cubic meter, or 1⁄1,000 of a liter, or one milliliter; therefore, 1 cm3 ≡ 1 ml. ### Milliliter (ml) A milliliter (also written "millilitre", SI symbol ml) is a non-SI metric system unit of volume which is commonly used as liquid unit. It is equal to 1/1000 liter, or one cubic centimeter, therefore, 1ml = 1/1000 L =1 cm3. ## Cubic Centimeters to Milliliters conversion table Below is the conversion table you can use to convert from Cubic Centimeters to Milliliters Cubic Centimeters (cm3) Milliliters (ml) 1 Cubic Centimeters 1 Milliliters 2 Cubic Centimeters 2 Milliliters 3 Cubic Centimeters 3 Milliliters 4 Cubic Centimeters 4 Milliliters 5 Cubic Centimeters 5 Milliliters 6 Cubic Centimeters 6 Milliliters 7 Cubic Centimeters 7 Milliliters 8 Cubic Centimeters 8 Milliliters 9 Cubic Centimeters 9 Milliliters 10 Cubic Centimeters 10 Milliliters	452	1,579	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2021-25	longest	en	0.749629
https://studydaddy.com/question/math-17	1,611,407,802,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703538082.57/warc/CC-MAIN-20210123125715-20210123155715-00603.warc.gz	592,334,796	8,277	QUESTION # Math There are a total of 108 bikes.if the ratio of blue bikes to black bikes is 4 to 5,how many of the bikes are blue? \$10.00 * * ** blue ****** * ** * black bikesThe * * * * ****** * + *** ** * of * blue *** to * black *** 4 to 5 x * ******************** * or \$10.00 * ****** * blue ***** * black bikegiven ***** * ***** * * : 5a * b = * * ****** * a + * = 180asked ***** bikes * ********* **** **************** ************* * ********* i * ********* * ******* * ***** 45a = * * **** * **** * 80so *** are ** *** attached * **** ** version ** this **** in studydaddy ** ** in ** or • @ • 1 order completed \$10.00 * ** bikes i **** * ** * black bikes **** * ******** **** of * * * * **** of bikes 5/9 bike multiply the fraction bikes * * ** of **** * number of ** * *********** * **** of black * is ********** number ** * 48The **** of * * is or \$10.00 *** total *** * ** * black *** *** * * * 5 ** * = ** 9x * 108 * * 108 * 9 * * 12 ** *** = * * 12 * * *** bikes * * * = * total = ** * ** * 108 ***** or	384	1,398	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.25	3	CC-MAIN-2021-04	latest	en	0.500693
https://mersenneforum.org/showthread.php?s=b2e5f847f5442f349c427cbe6c68dd7f&t=17609&goto=nextnewest	1,603,993,694,000,000,000	text/html	crawl-data/CC-MAIN-2020-45/segments/1603107904834.82/warc/CC-MAIN-20201029154446-20201029184446-00396.warc.gz	420,481,560	10,349	mersenneforum.org Prime Numbers from 103,374,001 to 103,374,989 Register FAQ Search Today's Posts Mark Forums Read 2020-10-06, 07:52 #1 tuckerkao Jan 2020 23·3·5 Posts Prime Numbers from 103,374,001 to 103,374,989 How do you call this specific group of prime numbers? 103,374,163 103,374,361 103,374,613 103,374,631 Also wondering who has "Intel® Core™ Processor i9-10980XE Eighteen-Core 3.0GHz (Turbo 4.8GHz) 24.75MB"? 2020-10-06, 07:55 #2 retina Undefined "The unspeakable one" Jun 2006 My evil lair 5,821 Posts Quote: Originally Posted by tuckerkao How do you call this specific group of prime numbers? 103,374,163 103,374,361 103,374,613 103,374,631 It doesn't matter how you call them, because they still won't come when you call them. 2020-10-06, 18:41 #3 Viliam Furik Jul 2018 Martin, Slovakia 22·3·19 Posts Quote: Originally Posted by tuckerkao How do you call this specific group of prime numbers? 103,374,163 103,374,361 103,374,613 103,374,631 I would say "partially incompletely permutable primes", as only part of numbers digits are permutable, and not all permutations of these digits within the number result in prime. 2020-10-06, 22:10 #4 tuckerkao Jan 2020 23×3×5 Posts Quote: Originally Posted by Viliam Furik I would say "partially incompletely permutable primes", as only part of numbers digits are permutable, and not all permutations of these digits within the number result in prime. I searched https://www.mersenne.org/report_recent_cleared/ and I cannot find any number strings with 103374 in it. Just wondering how many M103,374,xxx have already been identified as non-primes? 2020-10-06, 23:06 #5 S485122 Sep 2006 Brussels, Belgium 2×7×113 Posts Quote: Originally Posted by tuckerkao I searched https://www.mersenne.org/report_recent_cleared/ and I cannot find any number strings with 103374 in it. Just wondering how many M103,374,xxx have already been identified as non-primes? Just go to the Results / Detailed Results / Exponent Status and you will have your answer. Jacob https://www.mersenne.org/report_expo...p_hi=103375000 https://www.mersenne.org/report_expo...p_hi=103375000 2020-10-06, 23:08 #6 petrw1 1976 Toyota Corona years forever! "Wayne" Nov 2006 43·103 Posts Quote: Originally Posted by tuckerkao I searched https://www.mersenne.org/report_recent_cleared/ and I cannot find any number strings with 103374 in it. Just wondering how many M103,374,xxx have already been identified as non-primes? https://www.mersenne.org/report_fact...99&tftobits=72 2020-10-06, 23:23 #7 Dr Sardonicus Feb 2017 Nowhere 2·32·199 Posts Quote: Originally Posted by tuckerkao I searched https://www.mersenne.org/report_recent_cleared/ and I cannot find any number strings with 103374 in it. Just wondering how many M103,374,xxx have already been identified as non-primes? If I count the "factored" entries in the report for that range, 33 out of the 61 Mp have known factors. 2020-10-06, 23:48 #8 tuckerkao Jan 2020 23·3·5 Posts Quote: Originally Posted by Dr Sardonicus If I count the "factored" entries in the report for that range, 33 out of the 61 Mp have known factors. Only some of the prime entries show "No Factors under 2^74". What types of the tests are to check the factors within a specific range? 2020-10-07, 00:18 #9 Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 8,747 Posts Quote: Originally Posted by tuckerkao Only some of the prime entries show "No Factors under 2^74". What types of the tests are to check the factors within a specific range? Use colored balls to do the checking. That should do the job. Oh, wait, maybe find a little bit about the project whose board you are posting on and you will know. https://www.mersenne.org/various/math.php 2020-10-07, 00:26 #10 tuckerkao Jan 2020 23·3·5 Posts I already figured out Factor=103374163,74,75 in worktodo.txt or AdvancedTest=103374163 for the complete test 2020-10-07, 01:45 #11 tuckerkao Jan 2020 23×3×5 Posts Quote: Originally Posted by S485122 Just go to the Results / Detailed Results / Exponent Status and you will have your answer. Jacob https://www.mersenne.org/report_expo...p_hi=103375000 I need some explanations, what are the residue and shift on those tables? Also, if I finish 1 LL sequence, do I post screenshots or something? Similar Threads Thread Thread Starter Forum Replies Last Post Anirban Information & Answers 1 2018-07-02 05:26 VicDiesel Programming 12 2017-04-20 21:16 Arxenar Miscellaneous Math 38 2013-06-28 23:26 Citrix Lounge 5 2010-04-05 21:34 Unregistered Miscellaneous Math 8 2008-11-09 07:45 All times are UTC. The time now is 17:48. Thu Oct 29 17:48:14 UTC 2020 up 49 days, 14:59, 1 user, load averages: 2.16, 2.08, 2.13	1,451	4,745	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45	latest	en	0.878394
http://www.slideserve.com/cyrah/ece-1466-modern-optics-course-notes-part-6	1,493,350,732,000,000,000	text/html	crawl-data/CC-MAIN-2017-17/segments/1492917122726.55/warc/CC-MAIN-20170423031202-00387-ip-10-145-167-34.ec2.internal.warc.gz	692,851,691	18,395	# ECE-1466 Modern Optics Course Notes Part 6 - PowerPoint PPT Presentation 1 / 15 ECE-1466 Modern Optics Course Notes Part 6. Prof. Charles A. DiMarzio Northeastern University Spring 2002. Lecture Overview. Some Radiometry Terminology Equations Relating Radiometric Parameters Photometric Parameters Some Numbers A Little Bit of Scattering Theory I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ECE-1466 Modern Optics Course Notes Part 6 Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - ## ECE-1466Modern OpticsCourse NotesPart 6 Prof. Charles A. DiMarzio Northeastern University Spring 2002 Chuck DiMarzio, Northeastern University ### Lecture Overview • Terminology • Photometric Parameters • Some Numbers • A Little Bit of Scattering Theory • Some Applications in Microscopy Chuck DiMarzio, Northeastern University Chuck DiMarzio, Northeastern University M, Flux/Proj. Area Notes: Spectral x=dx/dn or dx/dl: Add subscript n or w, divide units by Hz or mm. F, Flux Watts Luminous Flux Lumens Watts/m2 Luminous Exitance Lumens/m2=Lux 1 W is 683 L at 555 nm. Watts/m2/sr Luminance Lumens/m2/sr 1 Lambert= (1L/cm2/sr)/p I, Flux/W L,Flux/AW Watts/sr Luminous Intensity Lumens/sr E, Flux/Area Rcd. Watts/m2 Illuminance Lumens/m2=Lux 1 ftLambert= (1L/ft2/sr)/p 1mLambert= (1L/m2/sr)/p 1 Ft Candle=1L/ft2 1 Candela=1cd=1L/sr Chuck DiMarzio, Northeastern University 1.8 This curve shows the relative sensitivity of the eye. To convert to photometric units from radiometric, multiply by 683 Lumens Per Watt y 1 Photopic Sensitivity 0 400 500 600 700 800 Wavelength, nm Chuck DiMarzio, Northeastern University dA2 dA’ dA1 dW2 dW1 z Chuck DiMarzio, Northeastern University Our Example = 0.0037/p W/m2/sr ~ 0.001 W/m2/sr at f/1 Half-Lux Camera = 0.0044 W/m2/sr Chuck DiMarzio, Northeastern University ### Black-Body Equation (1) Chuck DiMarzio, Northeastern University m 10 10 m / 2 5 10 0 10 -5 10 -10 10 l M -1 0 1 2 10 10 10 10 l m , Wavelength, m ### Black Body Equations (2) 10000 5000 2000 500 1000 T=300k Chuck DiMarzio, Northeastern University Data from The Science of Color, Crowell, 1953 3000 Exoatmospheric filename=m1695.m Sea Level 2 5000 K Black Body Normalized to 1000 W/m 2500 2 6000 K Black Body Normalized to 1560 W/m m m / 2 2000 1500 1000 l E 500 0 0 200 400 600 800 1000 1200 1400 1600 1800 2000 l , Wavelength, nm Chuck DiMarzio, Northeastern University ### Tungsten Lamps: Hot is Good! • 3000 K • 20 Lumens per Watt • lpeak=1.22mm • x = .4357y = .4032 z = .1610 • 3400 K note: (3400/3000)4=1.64) • 34 Lumens per Watt note: 20X1.64=33 • lpeak=1.09mm • x = .4112y = .3935 z = .1953 y x Chuck DiMarzio, Northeastern University ### Quartz-Halogen Lights • Tungsten Filament • Higher Temperature = Brighter, Whiter • Requires Quartz Envelope • Tungsten Evaporates More Rapidly • Halogen Catalyst • Prevents Tungsten Deposit on Hot Envelope • Tungsten Redeposits on the Filament • Evaporation and Redeposition Requires Thicker Filament • Lower Resistance Requires Lower Voltage Chuck DiMarzio, Northeastern University Lighting Efficiency 1000000 Fluorescent 94 Lumens/Watt at 7000K (Highest Efficiency Black Body) Hi Pressure Na Metal Halide 100000 Lo Pressure Na Incandescent 10000 Light Output, Lumens Thanks to John Hilliar (NU MS ECE 1999) for finding lighting data from Joseph F. Hetherington at www.hetherington.com. 10 June 1998 1000 20.7 Lumens/Watt at 3000K 100 10000 1 10 100 1000 Power Input, Watts ### Source Intensity 0.142 W • Fraction of Light in Filter Passband • Given by Black-Body Equation • Numerical Calculation is Easiest 100W Black Body Spectral and Integrated Flux Density Rev 2.17 by Chuck DiMarzio, Northeastern University 1992,1993,1995, 1997 .49600 to .50400 micrometers, T = 3000.0 K Maximum Spectral Radiant Exitance = .81762E+06 W/m^2/micron in band Radiant Exitance in Band 6541.5 Watts/m^2 Wide Band Radiant Exitance .45925E+07 Watts/m^2 Fraction of total in band.14244E-02 Spectrum on bbsre.dat **************************************************************************** Photocurrent per Area in Band 2669.5 Amps/m^2 .16663E+23 photons/sec/m^2 Average Responsivity .40808 Amps/Watt .39257E-18 Joules/photon (in band) ... Chuck DiMarzio, Northeastern University • Mostly a Geometric Problem • G describes non-uniformity • Like Antenna Gain Distance =R E = GP/(4pR2) Power =P E = (1?)0.14 W/[4p(0.3)m2] ~ 0.12 W/m2 Comparable to a dark cloud Chuck DiMarzio, Northeastern University	1,663	5,109	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17	longest	en	0.675007
https://pythonhosted.org/nengo/examples/combining.html	1,498,584,243,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128321497.77/warc/CC-MAIN-20170627170831-20170627190831-00077.warc.gz	778,297,823	29,748	Notebook # Nengo Example: Combining This example demonstrates how to create a neuronal ensemble that will combine two 1-D inputs into one 2-D representation. In [1]: import matplotlib.pyplot as plt %matplotlib inline import nengo <IPython.core.display.Javascript at 0x7f21581a4fd0> ## Step 1: Create the neural populations Our model consists of three ensembles, two input ensembles and one 2-D ensemble that will represent the two inputs as one two-dimensional signal. In [2]: model = nengo.Network(label='Combining') with model: # Our input ensembles consist of 100 leaky integrate-and-fire neurons, # representing a one-dimensional signal A = nengo.Ensemble(100, dimensions=1) B = nengo.Ensemble(100, dimensions=1) # The output ensemble consists of 200 leaky integrate-and-fire neurons, # representing a two-dimensional signal output = nengo.Ensemble(200, dimensions=2, label='2D Population') ## Step 2: Create input for the model We will use sine and cosine waves as examples of continuously changing signals. In [3]: import numpy as np with model: # Create input nodes generating the sine and cosine sin = nengo.Node(output=np.sin) cos = nengo.Node(output=np.cos) ## Step 3: Connect the network elements¶ In [4]: with model: nengo.Connection(sin, A) nengo.Connection(cos, B) # The square brackets define which dimension the input will project to nengo.Connection(A, output[1]) nengo.Connection(B, output[0]) ## Step 4: Probe outputs Anything that is probed will collect the data it produces over time, allowing us to analyze and visualize it later. In [5]: with model: sin_probe = nengo.Probe(sin) cos_probe = nengo.Probe(cos) A_probe = nengo.Probe(A, synapse=0.01) # 10ms filter B_probe = nengo.Probe(B, synapse=0.01) # 10ms filter out_probe = nengo.Probe(output, synapse=0.01) # 10ms filter ## Step 5: Run the model¶ In [6]: # Create our simulator with nengo.Simulator(model) as sim: # Run it for 5 seconds sim.run(5) ## Step 6: Plot the results¶ In [7]: # Plot the decoded output of the ensemble plt.figure() plt.plot(sim.trange(), sim.data[out_probe][:, 0], 'b', label="2D output") plt.plot(sim.trange(), sim.data[out_probe][:, 1], 'g', label="2D output") plt.plot(sim.trange(), sim.data[A_probe], 'r', label="A output") plt.plot(sim.trange(), sim.data[sin_probe], 'k', label="Sine") plt.legend() Out[7]: <matplotlib.legend.Legend at 0x7f211c896d50> The graph shows that the input signal (Sine), the output from the 1D population (A output), and the 2D population (green line) are all equal. The other dimension in the 2D population is shown in blue. Download combining as an IPython notebook or Python script.	722	2,653	{"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}	2.859375	3	CC-MAIN-2017-26	latest	en	0.620047
http://www.naclhv.com/2017/01/bayesian-evaluation-for-likelihood-of.html	1,606,415,191,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141188899.42/warc/CC-MAIN-20201126171830-20201126201830-00408.warc.gz	144,583,879	16,401	Blog pages Bayesian evaluation for the likelihood of Christ's resurrection (Part 38) So then, here is the summary of the basic idea: We assume that the "skeptic's distribution" will take the form of a generalized Pareto distribution. We will determine the shape parameter of the distribution by looking at how many "outliers" it has. A person's resurrection report is considered an "outlier" if it has at least 20% of the evidence compared to the non-Christian resurrection report with the maximum evidence. The "non-Christian resurrection report with the maximum evidence" is taken to be Krishna or Aristeas, with Apollonius not too far behind. These are taken as having 1/24 of the evidence for the resurrection of Jesus. Recall that the "some people say... " level of evidence - as per PuhuaOsirisZalmoxis, etc. - corresponded to 1/60th of the evidence for Christ's resurrection. This corresponds to 40% of 1/24, meaning that anyone else with the "some people say..." level of evidence would pass the 20% threshold for an "outlier". We can therefore calculate the relationship between the number of outliers and the likely shape parameters, and thereby calculate the probability for the "skeptic's distribution" naturally generating a Jesus-level of evidence for a resurrection. Here is the plan for the program: We will consider shape parameters from 0.02 to 2.1, in increasing intervals of 0.02. That is to say, we will consider shape parameters 0.02, 0.04, 0.06, etc. all the way up to 2.1. Our region of interest will lie in this range. We will create a generalized Pareto distribution with that shape parameter, then simulate drawing the maximum value of 1e9 samples from that distribution. We will then estimate the number of outliers in that distribution, and the probability of that distribution generating a sample more than 24 times larger than the maximum. We will do this 10000 times for each value of the shape parameter. This gives us a table of more than a million rows, with each row containing the shape parameter, the number of outliers, and the probability of generating a Jesus-level of evidence. If we assume equal prior weights for each of the shape parameters, we can consider the final distribution of the shape parameters to be just its distribution from the subset of the table where the number of outliers is equal to the actual, historical value. That is to say, we just have to look at where the theory fits the data, and consider only those theories. This satisfies Bayes' theorem. Likewise, the probability of the "skeptic's distribution" achieving a Jesus-level of evidence for a resurrection will just be the mean value of that probability from the same subset. Every assumption and choice made above favors the skeptic's case. Therefore, the probability obtained at the end will be a maximum probability; the skeptic cannot hope for more than that in explaining Jesus's resurrection "naturally". We will go over the actual computer program in the next post. You may next want to read: 15 puzzle: a tile sliding game How to think about the future (Part 2)	677	3,109	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.640625	4	CC-MAIN-2020-50	longest	en	0.891869
http://www.halfbakery.com/idea/skydiving_20in_20a_20car_20without_20parachute/addlink	1,582,491,087,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875145839.51/warc/CC-MAIN-20200223185153-20200223215153-00253.warc.gz	196,355,060	8,026	h a l f b a k e r y We are investigating the problem and will update you shortly. meta: account: browse anonymously, or get an account and write. user: pass: register, # skydiving in a car without parachute (+2) [vote for, against] Construct a giant ramp. At the top it will be nearly vertical.Car will land on the ramp at this point. It will be in vertical position. Its tyres will gently brush the ramp. Ramps slope will gradually go on decreasing as car moves along the ramp downward. At the end of ramp slope will be near horizintal, where car will exit the ramp. If constructed properly, skydiver should not feel any impact. Car will have control surfaces like rudder, flaps etc so that it can be precisely navigated to the top edge of ramp. It will be interesting to calulate the required height of ramp.I think max "g" humans can survive is 5. That should be the key. Terminal velocity of car will be 120 mph. In other words cars speed will be 120 mph when it will land on the ramp. — VJW, Jan 30 2012 Alternative_20Tube_...for_20dare_20devils Tangentially related idea [hippo, Jan 30 2012] If you're not logged in, you can see what this page looks like, but you will not be able to add anything. Short name, e.g., Bob's Coffee Destination URL. E.g., https://www.coffee.com/ Description (displayed with the short name and URL.) No matter how gently the ramp curves from vertical to horizontal, the g-forces involved will be severe. It would also have to be a damned accurate skydrive to hit the top of the ramp with the car oriented properly. Any variation that resulted in the car hitting the slope rather than the vertical would lead to a catastrophic impact. You're probably also going to need a specially modified car with incredibly stiff suspension and bolstered steering like a top-fuel dragster. This, in turn, would require a perfectly smooth ramp, else the tiniest bump would send the car flying. Try it out with a Hot Wheels track and you'll see what I mean. It will probably take dozens of attempts before it works the way you want it to; with a real car, carrying real passengers, you only get one try. — Alterother, Jan 30 2012 It's not impossible. However, the terminal velocity of a car, if nose-down, will be well over 120mph - I'd guess 150-200mph, depending on the car. A Volvo might be good (boxy), or better yet a caravan or similar. So, assuming a t.v. of 150mph (70m/s), and a maximum G of 5 (50m/s/s which is very conservative), you'll want to curve out over about a 1.5 second period, or about 50-60m of vertical distance. So, this is by no means un-doable, if you could hit the start of the ramp accurately. Actually, there are several skydivers who have proposed to land parachuteless in a wingsuit, using a similar arrangement. — MaxwellBuchanan, Jan 30 2012 If the car is aerodynamic enough to be able to not tumble madly out of control as it falls, its terminal velocity may be somewhat higher than 120mph. Let's say though that the vertical velocity of the car is 50 metres/s. The velocity of a decellerating body is given by v=u-at. If 5g is allowed and g=10, and we want the car to come to rest then (only looking at the vertical component of the car's motion) this becomes 0=50-50x1 so t=1 and all this happens in 1 second (that seems very quick! - have I got this wrong?). Then putting t=1 into s=ut-0.5at^2 = 50-0.5x50 gives us 25m for the height of the ramp. [EDIT: Evidently [Max-B] types faster than I do] — hippo, Jan 30 2012 //1 second (that seems very quick! - have I got this wrong?)// No, it's right. A regular parachute fully deploys in maybe 2 seconds; a reserve chute will deploy in something like 1 second; in both cases, the skydiver's vertical velocity is near zero by the time the chute's fully deployed. Most of the braking happens during the later stages of deployment, so the peak deceleration is actually a bit higher. A steady 1 second deceleration at 5G would be comfortable. — MaxwellBuchanan, Jan 30 2012 There is definitely a theme park roller-coaster type ride that does exactly this, albeit that the vehicle is attached to rails. The humans seem to survive the G absolutely fine. Forgotten where it is though and damned if I can find it through Google... — goff, Jan 30 2012 Drop Of Doom at West Edmonton Mall is an eight story free-fall with rapid decelleration. <side story> A hotel my mom worked at handed out travel coupon booklets to new guests. when the new shipment of booklets arrived the last page of each one was entirely W.E. Mall free rides, so she and a coworker carefully cut hundreds of these pages out with razors and split them up for their kids. I lost count of how many times I've ridden the drop'o'doom... if you hold a coin in one hand and wait for the drop you can make the coin hover weightless above your hand for the whole decent. It's pretty cool. — 2 fries shy of a happy meal, Jan 30 2012 Not that I'm assuming this uses an actual car, but... A front engine car will tend to fall in roughly the right orientation for this to work. The weight at the nose causes it to punch through the air more readily than the tail, and the shape of the car will favor a slightly angled fall. (wind pressure against the roof vs the relatively flat underside). So it's just down to actually landing on the ramp. — MechE, Jan 30 2012 ^Given that you really really would need small reaction jets to line the car up properly anyways, I think a mid/rear-engined vehicle would serve just as well. Or do it from the top of a cliff to the ramp at the bottom: cliff-diving instead of sky-diving. — FlyingToaster, Jan 30 2012 // with a real car, carrying real passengers, you only get one try. // To be pedantic (and why not ?) you only get one try per car; you should still have the aircraft and most, if not all, of the ramp. // better yet a caravan or similar // If nothing else, you will at least attract the undivided attention of Jezza Clarkson. // He didn't warm me he was doing this and it scared the bejeezus out of me. // It may have scared the bejeezus out of him, too, if it was an unintentional event ... // So it's just down to actually landing on the ramp // Well, there you are. It's easy. What are you waiting for ? — 8th of 7, Jan 30 2012 I like the cliff version. Could be installed on Beachy Head or some other tourist spot, so punters could drive their car over the edge of the cliff and end up on the beach. — pocmloc, Jan 30 2012 ... or indeed, "Oh my God, we're all going to die...." — 8th of 7, Jan 30 2012 Would it be easier to build a really big ramp, and hit it at a precise angle and position, or just make a 100x100 foot soft landing spot out of lots and lots of foam? — DIYMatt, Jan 30 2012 // make a 100x100 foot soft landing spot out of lots and lots of foam// In the case of ramp, car will be able to continue the journey horizontally, without losing much energy as if nothing happened. Humans can survive 5 g without passing out, but they will surely notice it. However, if the ramp is designed for something like 2g or 1 g, it would be a very smooth ride and passengers in the cars may not even notice it. — VJW, Jan 31 2012 Keep a watch for the Chevy Sonic commercial coming up during the Superbowl. — RayfordSteele, Jan 31 2012 back: main index	1,844	7,329	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2020-10	latest	en	0.923977
http://coq.inria.fr/pylons/contribs/files/Continuations/v8.4/Continuations.weight.decl_exc.html	1,406,215,485,000,000,000	text/html	crawl-data/CC-MAIN-2014-23/segments/1405997889379.27/warc/CC-MAIN-20140722025809-00247-ip-10-33-131-23.ec2.internal.warc.gz	45,344,015	2,556	# Library Continuations.weight.decl_exc Inductive Txlev : Set := \| xl1 : Txlev \| xl2 : Txlev. Inductive Txval : TxlevSet := \| xv1 : Txval xl1 \| xv2 : boolTxval xl2. Ltac AutoTx := intro l; elim l; simpl in \|- *; intros v Hv; elim Hv; auto with v62. Section inverse_of_xv2. Theorem xl2_bool : Txval xl2bool. intro x; elim x. exact false. intro b; exact b. Defined. Theorem l_xl2 : l : Txlev, l = xl2Txval lTxval xl2. simple induction l. intro E; discriminate E. intros e v; exact v. Defined. Theorem inv_xv2_xl2_bool_aux : (l : Txlev) (v : Txval l) (e : l = xl2), l_xl2 l e v = xv2 (xl2_bool (l_xl2 l e v)). simple induction v. intros; absurd (xl1 = xl2); [ discriminate \| assumption ]. auto with v62. Qed. Theorem inv_xv2_xl2_bool : v : Txval xl2, v = xv2 (xl2_bool v). intro v. cut (xl2 = xl2); [ intro e \| auto with v62 ]. cut (l_xl2 xl2 e v = v); [ intro e'; elim e' \| auto with v62 ]. apply inv_xv2_xl2_bool_aux. Qed. End inverse_of_xv2. Theorem noteq_xv2_true_false : xv2 true = xv2 falseFalse. intro E; discriminate E. Qed.	373	1,041	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2014-23	longest	en	0.400146
https://testbook.com/question-answer/if-a-full-wave-rectifier-operates-at-a-frequency-o--602e8001b5070611da4ca696	1,725,707,107,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00023.warc.gz	541,556,786	46,984	# If a full-wave rectifier operates at a frequency of x Hertz, then the frequency of the output ripples in Hertz will be 1. 2x 2. x/2 3. √2 4. x2 Option 1 : 2x Free Noun & Pronoun: Fill In The Blanks (Most Important Rules) 1.2 Lakh Users 15 Questions 15 Marks 12 Mins ## Detailed Solution The correct answer is option 1) i.e. 2x CONCEPT: • Full-wave rectifiers: A full-wave rectifier is a rectifier circuit that converts alternating current to direct current, and allows both the positive and negative cycles of the AC cycle to pass through it using diodes. • Full-wave rectifiers use multiple diodes. A diode is used because a diode can conduct current only in one direction at a particular instance of time. • AC current can be represented as a waveform and a wave is said to complete one cycle when the wave pattern repeats. The frequency is associated with one complete cycle of a wave. • From the figure, the output wave completes one cycle in half the time taken by input to complete one cycle. • Therefore, the output frequency in a full-wave rectifier is equal to 2 times the input frequency. EXPLANATION: The given frequency of the input is x Hertz The fundamental frequency of the output ripples = 2 times the input frequency = 2x Hertz.	334	1,258	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.125	3	CC-MAIN-2024-38	latest	en	0.721381
http://www.thefullwiki.org/Continuum_mechanics/Reynolds_transport_theorem	1,548,228,675,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547584203540.82/warc/CC-MAIN-20190123064911-20190123090911-00437.warc.gz	384,842,669	7,792	• Wikis Continuum mechanics/Reynolds transport theorem: Wikis Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles. Study guide Up to date as of January 14, 2010 Reynolds transport theorem Let Ω(t) be a region in Euclidean space with boundary $\partial \Omega (t)$. Let $\mathbf{x}(t)$ be the positions of points in the region and let $\mathbf{v}(\mathbf{x},t)$ be the velocity field in the region. Let $\mathbf{n}(\mathbf{x},t)$ be the outward unit normal to the boundary. Let $\mathbf{f}(\mathbf{x},t)$ be a vector field in the region (it may also be a scalar field). Show that $\cfrac{d}{dt}\left(\int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)} (\mathbf{v}\cdot\mathbf{n})\mathbf{f}~\text{dA} ~.$ This relation is also known as the Reynold's Transport Theorem and is a generalization of the Leibniz rule. Content of example. Proof: Let Ω0 be reference configuration of the region Ω(t). Let the motion and the deformation gradient be given by $\mathbf{x} = \boldsymbol{\varphi}(\mathbf{X}, t)~; \qquad\implies\qquad \boldsymbol{F}(\mathbf{X},t) = \boldsymbol{\nabla}_{\circ} \boldsymbol{\varphi} ~.$ Let $J(\mathbf{X},t) = \det[\boldsymbol{F}(\mathbf{X},t)]$. Then, integrals in the current and the reference configurations are related by $\int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV} = \int_{\Omega_0} \mathbf{f}[\boldsymbol{\varphi}(\mathbf{X},t),t]~J(\mathbf{X},t)~\text{dV}_0 = \int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\text{dV}_0 ~.$ The time derivative of an integral over a volume is defined as $\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \lim_{\Delta t \rightarrow 0} \cfrac{1}{\Delta t} \left(\int_{\Omega(t + \Delta t)} \mathbf{f}(\mathbf{x},t+\Delta t)~\text{dV} - \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) ~.$ Converting into integrals over the reference configuration, we get $\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \lim_{\Delta t \rightarrow 0} \cfrac{1}{\Delta t} \left(\int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t+\Delta t)~J(\mathbf{X},t+\Delta t)~\text{dV}_0 - \int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\text{dV}_0\right) ~.$ Since Ω0 is independent of time, we have \begin{align} \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) & = \int_{\Omega_0} \left[\lim_{\Delta t \rightarrow 0} \cfrac{ \hat{\mathbf{f}}(\mathbf{X},t+\Delta t)~J(\mathbf{X},t+\Delta t) - \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)}{\Delta t} \right]~\text{dV}_0 \ & = \int_{\Omega_0} \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)]~\text{dV}_0 \ & = \int_{\Omega_0} \left( \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]~J(\mathbf{X},t)+ \hat{\mathbf{f}}(\mathbf{X},t)~\frac{\partial }{\partial t}[J(\mathbf{X},t)]\right) ~\text{dV}_0 \end{align} Now, the time derivative of $\det\boldsymbol{F}$ is given by (see Gurtin: 1981, p. 77) $\frac{\partial J(\mathbf{X},t)}{\partial t} = \frac{\partial }{\partial t}(\det\boldsymbol{F}) = (\det\boldsymbol{F})(\boldsymbol{\nabla} \cdot \mathbf{v}) = J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\boldsymbol{\varphi}(\mathbf{X},t),t) = J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t) ~.$ Therefore, \begin{align} \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) & = \int_{\Omega_0} \left( \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]~J(\mathbf{X},t)+ \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right) ~\text{dV}_0 \ & = \int_{\Omega_0} \left(\frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]+ \hat{\mathbf{f}}(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~J(\mathbf{X},t) ~\text{dV}_0 \ & = \int_{\Omega(t)} \left(\dot{\mathbf{f}}(\mathbf{x},t)+ \mathbf{f}(\mathbf{x},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~\text{dV} \end{align} where $\dot{\mathbf{f}}$ is the material time derivative of $\mathbf{f}$. Now, the material derivative is given by $\dot{\mathbf{f}}(\mathbf{x},t) = \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t} + [\boldsymbol{\nabla} \mathbf{f}(\mathbf{x},t)]\cdot\mathbf{v}(\mathbf{x},t) ~.$ Therefore, $\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \int_{\Omega(t)} \left( \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t} + [\boldsymbol{\nabla} \mathbf{f}(\mathbf{x},t)]\cdot\mathbf{v}(\mathbf{x},t) + \mathbf{f}(\mathbf{x},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~\text{dV}$ or, $\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \left( \frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \mathbf{f}\cdot\mathbf{v} + \mathbf{f}~\boldsymbol{\nabla} \cdot \mathbf{v}\right)~\text{dV} ~.$ Using the identity $\boldsymbol{\nabla} \cdot (\mathbf{v}\otimes\mathbf{w}) = \mathbf{v}(\boldsymbol{\nabla} \cdot \mathbf{w}) + \boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{w}$ we then have $\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \left(\frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \cdot (\mathbf{f}\otimes\mathbf{v})\right)~\text{dV} ~.$ Using the divergence theorem and the identity $(\mathbf{a}\otimes\mathbf{b})\cdot\mathbf{n} = (\mathbf{b}\cdot\mathbf{n})\mathbf{a}$ we have ${ \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)}\frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)}(\mathbf{f}\otimes\mathbf{v})\cdot\mathbf{n}~\text{dV} = \int_{\Omega(t)}\frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)}(\mathbf{v}\cdot\mathbf{n})\mathbf{f}~\text{dV} ~. }$ References 1. M.E. Gurtin. An Introduction to Continuum Mechanics. Academic Press, New York, 1981. 2. T. Belytschko, W. K. Liu, and B. Moran. Nonlinear Finite Elements for Continua and Structures. John Wiley and Sons, Ltd., New York, 2000.	2,566	6,278	{"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": 24, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2019-04	latest	en	0.685238
https://www.informit.com/articles/article.aspx?p=2832588&seqNum=10	1,701,743,517,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100540.62/warc/CC-MAIN-20231205010358-20231205040358-00290.warc.gz	936,831,680	45,887	# Dealing with Impairments in Wireless Communication This chapter is from the book ## Problems 1. Re-create the results from Figure 5.5 but with 16-QAM. Also determine the smallest value of Mrx such that 16-QAM has a 1dB loss at a symbol error rate of 10−4. 2. Re-create the results from Example 5.3 assuming training constructed from Golay sequences as [a8; b8], followed by 40 randomly chosen 4-QAM symbols. Explain how to modify the correlator to exploit properties of the Golay complementary pair. Compare your results. 3. Consider the system where s[n] is a zero-mean WSS random process with correlation rss[n], υ[n] is a zero-mean WSS random process with correlation rυυ[n], and s[n] and υ[n] are uncorrelated. Find the linear MMSE equalizer g such that the mean squared error is minimized: 1. Find an equation for the MMSE estimator g. 2. Find an equation for the mean squared error (substitute your estimator in and compute the expectation). 3. Suppose that you know rss[n] and you can estimate ryy[n] from the received data. Show how to find rυυ[n] from rss[n] and ryy[n]. 4. Suppose that you estimate ryy[n] through sample averaging of N samples, exploiting the ergodicity of the process. Rewrite the equation for g using this functional form. 5. Compare the least squares and the MMSE equalizers. 4. Consider a frequency-flat system with frequency offset. Suppose that 16-QAM modulation is used and that the received signal is where = foffsetTs and exp(j2πn) is unknown to the receiver. Suppose that the SNR is 10dB and the packet size is N = 101 symbols. The effect of is to rotate the actual constellation. 1. Consider y[0] and y[1] in the absence of noise. Illustrate the constellation plot for both cases and discuss the impact of on detection. 2. Over the whole packet, where is the worst-case rotation? 3. Suppose that is small enough that the worst-case rotation occurs at symbol 100. What is the value of such that the rotation is greater than π/2? For the rest of this problem assume that is less than this value. 4. Determine the such that the symbol error rate is 10−3. To proceed, first find an expression for the probability of symbol error as a function of . Make sure that is included somewhere in the expression. Set equal to 10−3 and solve. 5. Let T be a Toeplitz matrix and let T* be its Hermitian conjugate. If T is either square or tall and full rank, prove that TT is an invertible square matrix. 6. Consider the training structure in Figure 5.37. Consider the problem of estimating the channel from the first period of training data. Let s[0], s[1], . . . , s[Ntr−1] denote the training symbols, and let s[Ntr], s[Nt + 1], . . . , s[N − 1] denote the unknown QAM data symbols. Suppose that we can model the channel as a frequency-selective channel with coefficients h[0], h[1], . . . , h[ℓ]. The received signal (assuming synchronization has been performed) is 1. Write the solution for the least squares channel estimate from the training data. You can use matrices in your answer. Be sure to label the size of the matrices and their contents very carefully. Also list any critical assumptions necessary for your solution. 2. Now suppose that the estimated channel is used to estimate an equalizer that is applied to y[n], then passed to the detector to generate tentative symbol decisions. We would like to improve the detection process by using a decision-directed channel estimate. Write the solution for the least squares channel estimate from the training data and tentative symbol decisions. 3. Draw a block diagram for the receiver that includes synchronization, channel estimation from training, equalization, reestimation of the channel, reequalization, and the additional detection phase. 4. Intuitively, explain how the decision-directed receiver should perform relative to only training as a function of SNR (low, high) and Ntr (small, large). Is there a benefit to multiple iterations? Please justify your answer. 7. Let be the Toeplitz matrix that is used in computing the least squares equalizer . Prove that if , is full rank. 8. Consider a digital communication system where the same transmitted symbol s[n] is repeated over two different channels. This is called repetition coding. Let h1 and h2 be the channel coefficients, assumed to be constant during each time instance and estimated perfectly. The received signals are corrupted by υ1[n] and υ2[n], which are zero-mean circular symmetric complex AWGN of the same variance σ2, that is, υ1[n], υ2[n] ~ Nc(0, σ2). In addition, s[n] has zero mean and . We assume that s[n], υ1[n], and υ2[n] are uncorrelated with each other. The received signals on the two channels are given by Suppose that we use the equalizers g1 and g2 for the two time instances, where g1 and g2 are complex numbers such that \|g1\|2 + \|g2\|2 = 1. The combined signal, denoted as z[n], is formed by summing the equalized signals in two time instances: If we define the following vectors: then and 1. Write an expression for z[n], first in terms of the vectors h and g, then in terms of g1, g2, h1, and h2. Thus you will have two equations for z[n]. 2. Compute the mean and the variance of the noise component in the combined signal z[n]. Remember that the noise is Gaussian and uncorrelated. 3. Compute the SNR of the combined signal z[n] as a function of σ, h1, h2, g1, and g2 (or σ, h, and g). In this case the SNR is the variance of the combined received signal divided by the variance of the noise term. 4. Determine g1 and g2 as functions of h1 and h2 (or g as a function of h) to maximize the SNR of the combined signal. Hint: Recall the Cauchy-Schwarz inequality for vectors and use the conditions when equality holds. 5. Determine a set of equations for finding the LMMSE equalizers g1 and g2 to minimize the mean squared error, which is defined as follows: Simplify your equation by exploiting the orthogonality principle but do not solve for the unknown coefficients yet. Hints: Using the vector format might be useful, and you can assume you can interchange expectation and differentiation. First, expand the absolute value, then take the derivative with respect to g and set the result equal to zero. Simplify as much as possible to get an expression of g as a function of h1, h2, and σ. 6. The autocorrelation matrix Ryy[0] of y[n] is defined as Compute Ryy[0] as a function of σ, h1, and h2. Hint: and for random processes a[n] and b[n]. 7. Now solve the set of equations you formulated in part (e). 9. Consider an order L-tap frequency-selective channel. Given a symbol period of T and after oversampling with factor Mrx and matched filtering, the received signal is where s is the transmitted symbols and υ is AWGN. Suppose a fractionally spaced equalizer (FSE) is applied to the received signal before downsampling. The FSE f[k] is an FIR filter with tap spacing of T/Mrx and length MN. Figure 5.38 gives the block diagram of the system. 1. Give an expression for w[k], the output of the FSE, and yM [n], the signal after downsampling. 2. We can express the FSE filter coefficients in the following matrix form: We can interpret every (j + 1)th column of F as a T-spaced subequalizer fj[n] where j = 0, 2, ..., M −1. Similarly, we can express the FSE output, the channel coefficients, and the noise as T-spaced subsequences: Using these expressions, show that 3. Now, express yM [n] as a function of s[n], fj[n], hj[n], and υj[n]. This expression is known as the multichannel model of the FSE since the output is the sum of the symbols convolved with the T-spaced subequalizers and subsequences. 4. Draw a block diagram of the multichannel model of the FSE based on part (c). 5. Now consider a noise-free case. Given the channel matrix Hj given by we define where fj is the (jth + 1) column of F. Perfect channel equalization can occur only if the solution to h = Cf, e = ed lies in the column space of H, and the channel H has all linearly independent rows. What are the conditions on the length of the equalizer N given Mrx and L to ensure that the latter condition is met? Hint: What are the dimensions of the matrices/vectors? 6. Given your answer in (e), consider the case of M = 1 (i.e., not using an FSE). Can perfect equalization be performed in this case? 10. Consider an order L-tap frequency-selective channel. After matched filtering, the received signal is where s[n] is the transmitted symbols, h[ℓ] is the channel coefficient, and υ[n] is AWGN. Suppose the received symbols corresponding to a known training sequence {1, −1, −1, 1} are {0.75 + j0.75, −0.75 − j0.25, −0.25 − j0.25, 0.25 + j0.75}. Note that this problem requires numerical solutions; that is, when you are asked to solve the least squares problems, the results must be specific numbers. MATLAB, LabVIEW, or MathScript in LabVIEW might be useful for solving the problem. 1. Formulate the least squares estimator of in the matrix form. Do not solve it yet. Be explicit about your matrices and vectors, and list your assumptions (especially the assumption about the relationship between L and the length of the training sequence). 2. Assuming L = 1, solve for the least squares channel estimate based on part (a). 3. Assume we use a linear equalizer to remove the eff ects of the channel. Let be an FIR equalizer. Let nd be the equalizer delay. Formulate the least squares estimator of given the channel estimate in part (b). Do not solve it yet. 4. Determine the range of values of nd. 5. Determine the nd that minimizes the squared error Jf[nd] and solve for the least squares equalizer corresponding to this nd. Also provide the value of the minimum squared error. 6. With the same assumptions as in parts (a) through (e) and with the value of nd found in part (e), formulate and solve for the direct least squares estimator of the equalizer’s coefficients . Also provide the value of the squared error. 11. Prove the following two properties of the DFT: 1. Let x[n] ↔ X[k] and x1[n] ↔ X1[k]. If X1[k] = ej2πkm/nX[k], we have 2. Let y[n] ↔ Y [k], h[n] ↔ H[k] and s[n] ↔ S[k]. If Y [k] = H[k]S[k], we have . 12. Consider an OFDM system. Suppose that your buddy C.P. uses a cyclic postfix instead of a cyclic prefix. Thus for n = 0, 1, . . . , N + Lc where Lc is the length of the cyclic prefix. The values of w[n] for n < 0 and n > N + Lc are unknown. Let the received signal be 1. Show that you can still recover with a cyclic postfix instead of a cyclic prefix. 2. Draw a block diagram of the system. 3. What are the differences between using a cyclic prefix and a cyclic postfix? 13. Consider an SC-FDE system with equivalent system . The length of the cyclic prefix is Lc. Prove that the cyclic prefix length should satisfy LcL. 14. Consider an OFDM system with N = 256 subcarriers in 5MHz of bandwidth, with a carrier of fc = 2GHz and a length L = 16 cyclic prefix. You can assume sinc pulse shaping. 1. What is the subcarrier bandwidth? 2. What is the length of the guard interval? 3. Suppose you want to make the OFDM symbol periodic including the cyclic prefix. The length of the period will be 16. Which subcarriers do you need to zero in the OFDM symbol? 4. What is the range of frequency offsets that you can correct using this approach? 15. Consider an OFDM system with N subcarriers. 1. Derive a bound on the bit error probability for 4-QAM transmission. Your answer should depend on h[k], No, and dmin. 2. Plot the error rate curves as a function of SNR for N = 1, 2, 4, assuming that h[0] = Ex, h[1] = Ex/2, h[2] = −jEx/2, and h[3] = Exe−j2π/3/3. 16. How many pilot symbols are used in the OFDM symbols during normal data transmission (not the CEF) in IEEE 802.11a? 17. IEEE 802.11ad is a WLAN standard operating at 60GHz. It has much wider bandwidth than previous WLAN standards in lower-frequency bands. Four PHY formats are defined in IEEE 802.11ad, and one of them uses OFDM. The system uses a bandwidth of 1880MHz, with 512 subcarriers and a fixed 25% cyclic prefix. Now compute the following: 1. What is the sample period duration assuming sampling at the Nyquist rate? 2. What is the subcarrier spacing? 3. What is the duration of the guard interval? 4. What is the OFDM symbol period duration? 5. In the standard among the 512 subcarriers, only 336 are used as data subcarriers. Assuming we use code rate 1/2 and QPSK modulation, compute the maximum data rate of the system. 18. Consider an OFDM communication system and a discrete-time channel with taps . Show mathematically why the cyclic prefix length Lc must satisfy LLc. 19. In practice, we may want to use multiple antennas to improve the performance of the received signal. Suppose that the received signal for each antenna can be modeled as Essentially you have two observations of the same signal. Each is convolved by a different discrete-time channel. In this problem we determine the coefficients of a set of equalizers and such that where Δ is a design parameter. 1. Suppose that you send training data . Formulate a least squares estimator for finding the estimated coefficients of the channel and . 2. Formulate the least squares equalizer design problem given your channel estimate. Hint: You need a squared error. Do not solve it yet. Be explicit about your matrices and vectors. 3. Solve for the least squares equalizer estimate using the formulation in part (b). You can use matrices in your answer. List your assumptions about dimensions and so forth. 4. Draw a block diagram of a QAM receiver that includes this channel estimator and equalizer. 5. Now formulate and solve the direct equalizer estimation problem. List your assumptions about dimensions and so forth. 6. Draw a block diagram of a QAM receiver that includes this equalizer. 20. Consider a wireless communication system with a frame structure as illustrated in Figure 5.39. Training is interleaved around (potentially different) bursts of data. The same training sequence is repeated. This structure was proposed relatively recently and is used in the single-carrier mode in several 60GHz wireless communication standards. Let the training sequence be and the data symbols be {s[n]}. Just to make the problem concrete, suppose that Suppose that the channel is linear and time invariant. After matched filtering, synchronization, and sampling, the received signal is given by the usual relationship Assume that N is much greater than Ntr, and that NtrL. 1. Consider the sequence . Show that there exists a cyclic prefix of length Ntr. 2. Derive a single-carrier frequency-domain equalization structure that exploits the cyclic prefix we have created. Essentially, show how we can recover from {y[n]}. 3. Draw a block diagram for the transmitter. You need to be as explicit as possible in indicating how the training sequence gets incorporated. 4. Draw a block diagram for the receiver. Be careful. 5. Suppose that we want to estimate the channel from the training sequence using the training on either side of the data. Derive the least squares channel estimator and determine the minimum value of Ntr required for this estimator to satisfy the conditions required by least squares. 6. Can you use this same trick with OFDM, using training for the cyclic prefix? Explain why or why not. 21. Consider an OFDM communication system. Suppose that the system uses a bandwidth of 40MHz, 128 subcarriers, and a length 32 cyclic prefix. Suppose the carrier frequency is 5.785GHz. 1. What is the sample period duration assuming sampling at the Nyquist rate? 2. What is the OFDM symbol period duration? 3. What is the subcarrier spacing? 4. What is the duration of the guard interval? 5. How much frequency offset can be corrected, in hertz, using the Schmidl-Cox method with all odd subcarriers zeroed, assuming only fine offset correction? Ignore the integer offset. 6. Oscillators are specified in terms of their offset in parts per million. Determine how many parts per million of variation are tolerable given the frequency offset in part (e). 22. Consider an OFDM communication system with cyclic prefix of length Lc. The samples conveyed to the pulse-shaping filter are Recall that is the cyclic prefix. Suppose that the cyclic prefix is designed such that the channel order L satisfies Lc = 2L. In this problem, we use the cyclic prefix to perform carrier frequency offset estimation. Let the receive signal after match filtering, frame offset correction, and downsampling be written as 1. Consider first a single OFDM symbol. Use the redundancy in the cyclic prefix to derive a carrier frequency offset estimator. Hint: Exploit the redundancy in the cyclic prefix but remember that Lc = 2L. You need to use the fact that the cyclic prefix is longer than the channel. 2. What is the correction range of your estimator? 3. Incorporate multiple OFDM symbols into your estimator. Does it work if the channel changes for different OFDM symbols? 23. Synchronization Using Repeated Training Sequences and Sign Flipping Consider the framing structure illustrated in Figure 5.40. This system uses a repetition of four training sequences. The goal of this problem is to explore the impact of multiple repeated training sequences on frame synchronization, frequency offset synchronization, and channel estimation. 1. Suppose that you apply the frame synchronization, frequency offset estimation, and channel estimation algorithms using only two length Ntr training sequences. In other words, ignore the two additional repetitions of the training signal. Please comment on how frame synchronization, frequency offset estimation, and channel estimation work on the aforementioned packet structure. 2. Now focus on frequency offset estimation. Treat the framing structure as a repetition of two length 2Ntr training signals. Present a correlation-based frequency offset estimator that uses two length 2Ntr training signals. 3. Treat the framing structure as a repetition of four length Ntr training signals. Propose a frequency offset estimator that uses correlations of length Ntr and exploits all four training signals. 4. What range of frequency offsets can be corrected in part (b) versus part (c)? Which is better in terms of accuracy versus range? Overall, which approach is better in terms of frame synchronization, frequency offset synchronization, and channel estimation? Please justify your answer. 5. Suppose that we flip the sign of the third training sequence. Thus the training pattern becomes T, T,T, T instead of T, T, T, T. Propose a frequency offset estimator that uses correlations of length Ntr and exploits all four training signals. 6. What range of frequency offsets can be corrected in this case? From a frame synchronization perspective, what is the advantage of this algorithm versus the previoius algorithm you derived? 24. Synchronization in OFDM Systems Suppose that we would like to implement the frequency offset estimation algorithm. Suppose that our OFDM systems operate with N = 128 subcarriers in 2MHz of bandwidth, with a carrier of fc = 1GHz and a length L = 16 cyclic prefix. 1. What is the subcarrier bandwidth? 2. We would like to design a training symbol that has the desirable periodic correlation properties that are useful in the discussed algorithm. What period should you choose and why? Be sure to consider the effect of the cyclic prefix. 3. For the period you suggest in part (b), which subcarriers do you need to zero in the OFDM symbol? 4. What is the range of frequency offsets that you can correct using this approach without requiring the modulo correction? 25. Consider the IEEE 802.11a standard. By thinking about the system design, provide plausible explanations for the following: 1. Determine the amount of carrier frequency offset correction that can be obtained from the short training sequence. 2. Determine the amount of carrier frequency offset correction that can be obtained from the long training sequence. 3. Why do you suppose the long training sequence has a double-length guard interval followed by two repetitions? 4. Why do you suppose that the training comes at the beginning instead of at the middle of the transmission as in the mobile cellular system GSM? 5. Suppose that you use 10MHz instead of 20MHz in IEEE 802.11a. Would you rather change the DFT size or change the number of zero subcarriers? 26. Answer the following questions for problem 17: 1. How much frequency offset (in Hertz) can be corrected using the Schmidl-Cox method with all odd subcarriers zeroed, assuming only fine offset correction? Ignore the integer offset. 2. Oscillator frequency offsets are specified in parts per million (ppm). Determine how many parts per million of variation are tolerable given the result in part (a). 27. GMSK is an example of continuous-phase modulation (CPM). Its continuous-time baseband transmitted signal can be written as where a[n] is a sequence of BPSK symbols and ϕ(t) is the CPM pulse. For GSM, T = 6/1.625e6 ≈ 3.69µs. The BPSK symbol sequence a[n] is generated from a differentially binary (0 or 1) encoded data sequence d[n] where where ⊕ denotes modulo-2 addition. Let us denote the BPSK encoded data sequence as Then (5.398) can be rewritten in a simpler form as Consider the Gaussian pulse response Denote the rectangle function of duration T in the usual way as For GSM, BT = 0.3. This means that B = 81.25kHz. The B here is not the bandwidth of the signal; rather it is the 3dB bandwidth of the pulse g(t). The combined filter response is Then the CPM pulse is given by Because of the choice of BT product, it is recognized that This means that the current GMSK symbol a[n] depends most strongly on the three previous symbols a[n − 1], a[n − 2], and a[n − 3]. Note that the phase output ϕ(t) depends on all previous bits d[n] because of the infinite integral. Consequently, the GSM modulator is initialized by the state give in Figure 5.41. The state is reinitialized for every transmitted burst. The classic references for linearization are [171, 173], based on the pioneering work on linearizing CPM signals in [187]. Here we summarize the linearization approach of [171, 173], which was used in [87] for the purpose of blind channel equalization of GMSK signals. We sketch the idea of the derivation from [87] here. 1. Exploiting the observation in (5.405), argue that for t ∈ [nT, (n+1)T). This makes the dependence of s(t) on the current symbol and past three symbols clearer. 2. Explain how to obtain 3. Using the fact that a[n] are BPSK modulated with values +1 or −1, and the even property of cosine and the odd property of sine, show that 4. Now let Note that for any real number t = sgn(t) \|t\|. Now define It can be shown exploiting the symmetry of ϕ(t) that Substitute in for β to obtain 5. Show that where and for t ∈ [0, 5T]. This is called the one-term approximation. 6. Modify the typical transmit block diagram to implement the transmit waveform in (5.415). 7. Modify the typical receiver block diagram to implement the receiver processing in a frequency-selective channel. 28. Plot the path loss for a distance from 1m to 200m for the following channel models and given parameters. The path loss should be plotted in decibels. 1. Free space assuming Gt = Gr = 3dB, and λ = 0.1m 2. Mean log distance using a 1m reference with free-space parameters as in part (a) with path-loss exponent β = 2 3. Mean log distance using a 1m reference with free-space parameters as in part (a) with path-loss exponent β = 3 4. Mean log distance using a 1m reference with free-space parameters as in part (a) with path-loss exponent β = 4 29. Plot the path loss at a distance of 1km for wavelengths from 10m to 0.01m for the following channel models and given parameters. The path loss should be plotted in decibels and the distance plotted on a log scale. 1. Free space assuming Gt = Gr = 3dB 2. Mean log distance using a 1m reference with free-space parameters as in part (a) with path-loss exponent β = 2 3. Mean log distance using a 1m reference with free-space parameters as in part (a) with path-loss exponent β = 3 4. Mean log distance using a 1m reference with free-space parameters as in part (a) with path-loss exponent β = 4 30. Consider the path loss for the log-distance model with shadowing for σshad = 8dB, η = 4, a reference distance of 1m with reference loss computed from free space, assuming Gr = Gr = 3dB, for a distance from 1m to 200m. Plot the mean path loss. For every value of d you plot, also plot the path loss for ten realizations assuming shadow fading. Explain what you observe. To resolve this issue, some path-loss models have a distance-dependent σshad. 31. Consider the LOS/NLOS path-loss model with Plos(d) = ed/200, free space for the LOS path loss, log distance without shadowing for the NLOS with β = 4, reference distance of 1m, Gt = Gr = 0dB, and λ = 0.1m. Plot the path loss in decibels for distances from 1 to 400m. 1. Plot the LOS path loss. 2. Plot the NLOS path loss. 3. Plot the mean path loss. 4. For each value of distance in your plot, generate ten realizations of the path loss and overlay them on your plot. 5. Explain what happens as distances become larger. 32. Consider the path loss for the log-distance model assuming that β = 3, with reference distance 1m, Gt = Gr = 0dB and λ = 0.1m, and σshad = 6dB. The transmit power is Ptx = 10dBm. Generate your plots from 1 to 500m. 1. Plot the received power based on the mean path-loss equation. 2. Plot the probability that Prx(d) < −110dBm. 3. Determine the maximum value of d such that Prx(d) > −110dBm 90% of the time. 33. Consider the LOS/NLOS path-loss model with Plos(d) = ed/200, free space for the LOS path loss, log distance without shadowing for the NLOS with β = 4, reference distance of 1m, Gt = Gr = 0dB, and λ = 0.1m. The transmit power is Ptx = 10dBm. Generate your plots from 1 to 500m. 1. Plot the received power based on the mean path-loss equation. 2. Plot the probability that Prx(d) < −110dBm. 3. Determine the maximum value of d such that Prx(d) > −110dBm 90% of the time. 34. Consider the same setup as in the previous problem but now with shadowing on the NLOS component with σshad = 6dB. This problem requires some extra work to include the shadowing in the NLOS part. 2. Plot the probability that Prx(d) < −110dBm. 3. Determine the maximum value of d such that Prx(d) > −110dBm 90% of the time. 35. Consider a simple cellular system. Seven base stations are located in a hexagonal cluster, with one in the center and six surrounding. Specifically, there is a base station at the origin and six base stations located 400m away at 0°, 60°, 120°, and so on. Consider the path loss for the log-distance model assuming that β = 4, with reference distance 1m, Gt = Gr = 0dB, and λ = 0.1m. You can use 290K and a bandwidth of B = 10MHz to compute the thermal noise power as kTB. The transmit power is Ptx = 40dBm. 1. Plot the SINR for a user moving along the 0° line from a distance of 1m from the base station to 300m. Explain what happens in this curve. 2. Plot the mean SINR for a user from a distance of 1m from the base station to 300m. Use Monte Carlo simulations. For every distance, generate 100 user locations randomly on a circle of radius d and average the results. How does this compare with the previous curve? 36. Link Budget Calculations In this problem you will compute the acceptable transmission range by going through a link budget calculation. You will need the fact that noise variance is kTB where k is Boltzmann’s constant (i.e., k = 1.381 × 10−23W/Hz/K), T is the effective temperature in kelvins (you can use T = 300K), and B is the signal bandwidth. Suppose that field measurements were made inside a building and that subsequent processing revealed that the data fit the log-normal model. The path-loss exponent was found to be n = 4. Suppose that the transmit power is 50mW, 10mW is measured at a reference distance d0 = 1m from the transmitter, and σ = 8dB for the log-normal path-loss model. The desired signal bandwidth is 1MHz, which is found to be sufficiently less than the coherence bandwidth; thus the channel is well modeled by the Rayleigh fading channels. Use 4-QAM modulation. 1. Calculate the noise power in decibels referenced to 1 milliwatt (dBm). 2. Determine the SNR required for the Gaussian channel to have a symbol error rate of 10−2. Using this value and the noise power, determine the minimum received signal power required for at least 10−2 operation. Let this be Pmin (in decibels as usual). 3. Determine the largest distance such that the received power is above Pmin. 4. Determine the small-scale link margin LRayleigh for the flat Rayleigh fading channel. The fade margin is the difference between the SNR required for the Rayleigh case and the SNR required for the Gaussian case. Let PRayleigh = LRayleigh + Pmin. 5. Determine the large-scale link margin for the log-normal channel with a 90% outage. In other words, find the received power Plarge-scale required such that Plarge-scale is greater than PRayleigh 90% of the time. 6. Using Plarge-scale, determine the largest acceptable distance in meters that the system can support. 7. Ignoring the effect of small-scale fading, what is the largest acceptable distance in meters that the system can support? 8. Describe the potential impact of diversity on the range of the system. 37. Suppose that field measurements were made inside a building and that subsequent processing revealed that the data fit the path-loss model with shadowing. The path-loss exponent was found to be n = 3.5. If 1mW was measured at d0 = 1m from the transmitter and at a distance of d0 = 10m, 10% of the measurements were stronger than −25dBm, find the standard deviation σshad for the log-normal model. 38. Through real measurement, the following three received power measurements were obtained at distances of 100m, 400m, and 1.6km from a transmitter: 1. Suppose that the path loss follows the log-distance model, without shadowing, with reference distance d0 = 100m. Use least squares to estimate the path-loss exponent. 2. Suppose that the noise power is −90dBm and the target SNR is 10dB. What is the minimum received signal power required, also known as the receiver sensitivity? 3. Using the value calculated in part (b), what is the approximate acceptable largest distance in kilometers that the system can support? 39. Classify the following as slow or fast and frequency selective or frequency flat. Justify your answers. In some cases you will have to argue why your answer is correct by making reasonable assumptions based on the application. Assume that the system occupies the full bandwidth listed. 1. A cellular system with carrier frequency of 2GHz, bandwidth of 1.25MHz, that provides service to high-speed trains. The RMS delay spread is 2µs. 2. A vehicle-to-vehicle communication system with carrier frequency of 800MHz and bandwidth of 100kHz. The RMS delay spread is 20ns. 3. A 5G communication system with carrier frequency of 3.7GHz and bandwidth of 200MHz. 4. A 60GHz wireless personal area network with a bandwidth of 2GHz and an RMS delay spread of 40ns. The main application is high-speed multimedia delivery. 5. Police-band radio. Vehicles move at upwards of 100mph and communicate with a base station. The bandwidth is 50kHz at 900MHz carrier. 40. Consider a wireless communication system with a frequency carrier of fc = 1.9GHz and a bandwidth of B = 5MHz. The power delay profile shows the presence of three strong paths and is plotted in Figure 5.42. 1. Compute the RMS delay spread. 2. Consider a single-carrier wireless system. Classify the system as frequency flat or frequency selective and slow or fast. Assume the system provides service to vehicle-to-vehicle communication at the speed of υ = 120km/h and not packet transmission but symbol transmission. You should use the stricter relationship between coherence time Tcoh and maximum Doppler frequency fm by . 3. Consider an OFDM system with IFFT/FFT size of N = 1024. What is the maximum-length cyclic prefix needed to allow for efficient equalization in the frequency domain with OFDM? 4. For the specified OFDM system, which also provides services to vehicle-to-vehicle communication at urban speeds of 30mph, is the channel slow or fast fading? Justify your answer. 41. Suppose that Rdelay(τ) = 1 for τ ∈ [0, 20µs] and is zero otherwise. 1. Compute the RMS delay spread. 2. Compute the spaced-frequency correlation function Sdelaylag). 3. Determine if the channel is frequency selective or frequency flat for a signal with bandwidth B = 1MHz. 4. Find the largest value of bandwidth such that the channel can be assumed to be frequency flat. 42. Suppose that for τ ∈ [0, 20µs]. 1. Compute the RMS delay spread. 2. Compute the spaced-frequency correlation function Sdelaylag). 3. Determine if the channel is frequency selective or frequency flat for a signal with bandwidth B = 1MHz. 4. Find the largest value of bandwidth such that the channel can be assumed to be frequency flat. 43. Sketch the design of an OFDM system with 80MHz channels, supporting pedestrian speeds of 3km/h, an RMS delay spread of 8µs, operating at fc = 800MHz. Explain how you would design the preamble to permit synchronization and channel estimation and how you would pick the parameters such as the number of subcarriers and the cyclic prefix. Justify your choice of parameters and explain how you deal with mobility, delay spread, and how much frequency offset you can tolerate. You do not need to do any simulation for this problem; rather make calculations as required and then justify your answers. 44. Sketch the design of an SC-FDE system with 2GHz channels, supporting pedestrian speeds of 3km/h, and an RMS delay spread of 50ns, operating at fc = 64GHz. Explain how you would design the preamble to permit synchronization and channel estimation and how you would pick the parameters. Justify your choice of parameters and explain how you deal with mobility, delay spread, and how much frequency offset you can tolerate. You do not need to do any simulation for this problem; rather make calculations as required and then justify your answers. 45. Consider an OFDM system with a bandwidth of W = 10MHz. The power delay profile estimates the RMS delay spread to be σRMS,delay = 5µs. Possible IFFT/FFT sizes for the system are N = {128, 256, 512, 1024, 2048}. Possible cyclic prefix sizes (in terms of fractions of N) are {1/4, 1/8, 1/16, 1/32}. 1. What combination of FFT and cyclic prefix sizes provides the minimum amount of overhead? 2. Calculate the fraction of loss due to this overhead. 3. What might be some issues associated with minimizing the amount of overhead in this system? 46. Compute the maximum Doppler shift for the following sets of parameters: 1. 40MHz of bandwidth, carrier of 2.4GHz, and supporting 3km/h speeds 2. 2GHz of bandwidth, carrier of 64GHz, and supporting 3km/h speeds 3. 20MHz of bandwidth, carrier of 2.1GHz, and supporting 300km/h speeds 47. In fixed wireless systems, the Clarke-Jakes spectrum may not be appropriate. A better model is the Bell Doppler spectrum where has been found to fit to data [363] where KB is a constant so that ∫fSDoppler(f)df = 1. In this case, the maximum Doppler shift is determined from the Doppler due to moving objects in the environment. 1. Determine KB for a maximum velocity of 160km/h and a carrier frequency of 1.9GHz. 2. Determine the RMS Doppler spread. 3. Is a signal with bandwidth of B = 20MHz and blocks of length 1000 symbols well modeled as time invariant? 48. Look up the basic parameters of the GSM cellular system. Consider a dual-band phone that supports fc = 900MHz and fc = 1.8GHz. What is the maximum velocity that can be supported on each carrier frequency so that the channel is sufficiently constant during a burst period? Explain your work and list your assumptions. 49. Look up the basic parameters of an IEEE 802.11a system. What is the maximum amount of delay spread that can be tolerated by an IEEE 802.11a system? Suppose that you are in a square room and that the access point is located in the middle of the room. Consider single reflections. How big does the room need to be to give this maximum amount of channel delay? 50. Consider 4-QAM transmission. Plot the following for error rates of 1 to 10−5 and SNRs of 0dB to 30dB. This means that for higher SNRs do not plot values of the error rate below 10−5. • Gaussian channel, exact expression • Gaussian channel, exact expression, neglecting the quadratic term • Gaussian channel, exact expression, neglecting the quadratic term and using a Chernoff upper bound on the Q-function • Rayleigh fading channel, Chernoff upper bound on the Q-function 51. Computer In this problem, you will create a flat-fading channel simulator function. The input to your function is x(nT/Mtx), the sampled pulse-amplitude modulated sequence sampled at T/Mtx where Mtx is the oversampling factor. The output of your function is r(nT/Mrx), which is the sampled receive pulse-amplitude modulated signal (5.7). The parameters of your function should be the channel h (which does not include ), the delay τd, Mtx, Mrx, and the frequency offset . Essentially, you need to determine the discrete-time equivalent channel, convolve it with x(nT/Mtx), resample, incorporate frequency offset, add AWGN, and resample to produce r(nT/Mrx). Be sure that the noise is added correctly so it is correlated. Design some tests and demonstrate that your code works correctly. 52. Computer In this problem, you will create a symbol synchronization function. The input is y(nT/Mrx), which is the matched filtered receive pulse-amplitude modulated signal, Mrx the receiver oversampling factor, and the block length for averaging. You should implement both versions of the MOE algorithm. Demonstrate that your code works. Simulate x(t) obtained from 4-QAM constellations with 100 symbols and raised cosine pulse shaping with α = 0.25. Pass your sampled signal with Mtx ≥ 2 through your channel simulator to produce the sampled output with Mrx samples per symbol. Using a Monte Carlo simulation, with an offset of τd = T/3 and h = 0.3ejπ/3, perform 100 trials of estimating the symbol timing with each symbol synchronization algorithm for SNRs from 0dB to 10dB. Which one gives the best performance? Plot the received signal after correction; that is, plot for the entire frame with real on the x-axis and imaginary on the y-axis for 0dB and 10dB SNR. This should give a rotated set of 4-QAM constellation symbols. Explain your results. 53. Computer In this problem, you will implement the transmitter and receiver for a flat-fading channel. The communication system has a training sequence of length 32 that is composed of length 8 Golay sequences and its complementary pairs as [a8, b8, a8, b8]. The training sequence is followed by 100 M-QAM symbols. No symbols are sent before or after this packet. Suppose that Mrx = 8, τd = 4T +T/3, and h = ejπ/3. There is no frequency offset. Generate the matched filtered receive signal. Pass it through your symbol timing synchronizer and correct for the timing error. Then perform the following: 1. Develop a frame synchronization algorithm using the given training sequence. Demonstrate the performance of your algorithm by performing 100 Monte Carlo simulations for a range of SNR values from 0dB to 10dB. Count the number of frame synchronization errors as a function of SNR. 2. After correction for frame synchronization error, estimate the channel. Plot the estimation error averaged over all the Monte Carlo trials as a function of SNR. 3. Compute the probability of symbol error. Perform enough Monte Carlo trials to estimate the symbol error reliably to 10−4. 54. Computer Consider the same setup as in the previous problem but now with a frequency offset of = 0.03. Devise and implement a frequency offset correction algorithm. 1. Assuming no delay in the channel, demonstrate the performance of your algorithm by performing 100 Monte Carlo simulations for a range of SNR values from 0dB to 10dB. Plot the average frequency offset estimation error. 2. After correction for frequency synchronization error, estimate the channel. Plot the estimation error averaged over all the Monte Carlo trials as a function of SNR. 3. Perform enough Monte Carlo trials to estimate the symbol error reliably to 10−4. 4. Repeat each of the preceding tasks by now including frame synchronization and symbol timing with τd as developed in the previous problem. Explain your results. 55. Computer In this problem, you will create a multipath channel simulator function where the complex baseband equivalent channel has the form and the channel has multiple taps, each consisting of complex amplitude αk and delay τk. The input to your function is x(nT/Mtx), the sampled pulse-amplitude modulated sequence sampled at T/Mtx where Mtx is the oversampling factor. The output of your function is r(nT/Mrx), which is sampled from Your simulator should include the channel parameters Mtx, Mrx, and the frequency offset . Check that your code works correctly by comparing it with the flat-fading channel simulator. Design some other test cases and demonstrate that your code works. ### InformIT Promotional Mailings & Special Offers I would like to receive exclusive offers and hear about products from InformIT and its family of brands. I can unsubscribe at any time. ## Overview Pearson Education, Inc., 221 River Street, Hoboken, New Jersey 07030, (Pearson) presents this site to provide information about products and services that can be purchased through this site. 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https://education.apple.com/story/250010751	1,721,565,903,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00343.warc.gz	199,023,820	25,543	# Using GarageBand to Teach Equivalent Ratios Growing up, music became an important part of my life. I enjoyed the creativity that playing an instrument gave me, and still gives me to this day. But one of the greatest gifts that music gave me personally was a greater insight into Mathematics. I struggled with Math all my life, and it was not until I had been playing an instrument for about a year that I started to see those connections, and suddenly fractions began to make sense to me. I knew that music would be a great tool to use with my students to help reinforce math concepts. With this project, I decided to use the Beat Sequencer in GarageBand as that tool. To start the activity, open up GarageBand and swipe left or right until you get to the Drums, click on the bottom toolbar for the Beat Sequencer. Once it is open, you would just need to click on various boxes to turn on that beat and create a drum sequence. In this activity, I use it a little differently and have the students look at it a little differently. I create couplets (line pairs) for each ratio. The bass drum and snare drum are the 1st ratio. The hand claps and closed hi-hat cymbal are the 2nd ratio. The open high-hat and floor tom are the 3rd ratio. And the 4th ratio is the tom-tom and the drumsticks. I start creating a group beat by starting with ratios equivalent to 1:2. I allow students to turn on whatever beat they want, but it has to be on those lines. So the bass drum will get only one beat on the entire line, and the snare drum will get 2 beats only. Then we move to the second pair, which is the hand clap the closed hi-hat cymbal, and create a ratio of 2:4 And then continue the process to finish the 3rd and 4th ratios. Once you have finished, click the play button and listen to the beat you created. I then have the students create a ratio and the equivalent ratios to random ratios such as 2:3. Below is a sample song. Enjoy music and ratios. 4 replies August 27, 2022 What a creative approached to mathematics! Music does it every time! August 28, 2022 This is wonderful! What a great creative activity! I struggle to find ways to use GarageBand (in general) but also as a math teacher. This is wonderful and the kids would really love it. Thanks for sharing! September 02, 2022 This is a wonderful lesson to help connect math to real-world engagement. What a great lesson to help think fractionally and then be able to apply this new knowledge to other content areas like science! Thanks for sharing! September 08, 2022 This would be a great addition to the Mathematics Area under Teaching and Learning! Maximum file size: 400MB 125: 125 220: 220 ## This post contains content from YouTube. If you choose to view this content, YouTube may collect and process certain personal data. You can view YouTube’s <a href="https://www.youtube.com/t/privacy" target="_blank">privacy policy here<span class="a11y">(opens in new window)</span>.</a> ## This post contains content from YouTube. You have rejected content from YouTube. If you want to change your consent, press the button below.	727	3,120	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.75	4	CC-MAIN-2024-30	latest	en	0.965029
https://oeis.org/A089908	1,575,760,621,000,000,000	text/html	crawl-data/CC-MAIN-2019-51/segments/1575540502120.37/warc/CC-MAIN-20191207210620-20191207234620-00456.warc.gz	484,398,591	4,332	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!) A089908 Sum of digits of numbers between 0 and (7/9)(10^n-1). 6 0, 28, 609, 9555, 130501, 1654947, 20049393, 235493839, 2704938285, 30549382731, 340493827177, 3754938271623, 41049382716069, 445493827160515, 4804938271604961, 51549382716049407, 550493827160493853 (list; graph; refs; listen; history; text; internal format) OFFSET 0,2 COMMENTS From a suggestion of Yalcin Aktar LINKS Index entries for linear recurrences with constant coefficients, signature (22,-141,220,-100). FORMULA a(n) = s(7, n-1) where s(a, k)=a(k+1)+a^2sum(i=0, k, i10^(k-i))+sum(i=0, k, 5a(9(k-i)+a- 1)10^(k-i-1)). a(n) = (7(-22(-1+10^n)+9(4+910^n)n))/162. G.f.: 7x(15x^2-x+4) / ((x-1)^2(10x-1)^2). - Colin Barker, Jun 14 2013 CROSSREFS Cf. A089903, A089904, A089905, A087330, A089906, A089907, A089909, A034967. Sequence in context: A264459 A004371 A283096 * A038121 A240684 A184329 Adjacent sequences: A089905 A089906 A089907 * A089909 A089910 A089911 KEYWORD nonn,base,easy AUTHOR Benoit Cloitre, Nov 14 2003 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 7 18:12 EST 2019. Contains 329847 sequences. (Running on oeis4.)	615	1,841	{"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}	3.265625	3	CC-MAIN-2019-51	latest	en	0.673694
https://nebusresearch.wordpress.com/2017/11/23/as-i-try-to-figure-out-what-wronski-thought-pi-was/	1,638,265,153,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00257.warc.gz	465,697,059	39,996	# As I Try To Figure Out What Wronski Thought ‘Pi’ Was A couple weeks ago I shared a fascinating formula for π. I got it from Carl B Boyer’s The History of Calculus and its Conceptual Development. He got it from Józef Maria Hoëne-Wronski, early 19th-century Polish mathematician. His idea was that an absolute, culturally-independent definition of π would come not from thinking about circles and diameters but rather this formula: $\pi = \frac{4\infty}{\sqrt{-1}}\left\{ \left(1 + \sqrt{-1}\right)^{\frac{1}{\infty}} - \left(1 - \sqrt{-1}\right)^{\frac{1}{\infty}} \right\}$ Now, this formula is beautiful, at least to my eyes. It’s also gibberish. At least it’s ungrammatical. Mathematicians don’t like to write stuff like “four times infinity”, at least not as more than a rough draft on the way to a real thought. What does it mean to multiply four by infinity? Is arithmetic even a thing that can be done on infinitely large quantities? Among Wronski’s problems is that they didn’t have a clear answer to this. We’re a little more advanced in our mathematics now. We’ve had a century and a half of rather sound treatment of infinitely large and infinitely small things. Can we save Wronski’s work? Start with the easiest thing. I’m offended by those $\sqrt{-1}$ bits. Well, no, I’m more unsettled by them. I would rather have $\imath$ in there. The difference? … More taste than anything sound. I prefer, if I can get away with it, using the square root symbol to mean the positive square root of the thing inside. There is no positive square root of -1, so, pfaugh, away with it. Mere style? All right, well, how do you know whether those $\sqrt{-1}$ terms are meant to be $\imath$ or its additive inverse, $-\imath$? How do you know they’re all meant to be the same one? See? … As with all style preferences, it’s impossible to be perfectly consistent. I’m sure there are times I accept a big square root symbol over a negative or a complex-valued quantity. But I’m not forced to have it here so I’d rather not. First step: $\pi = \frac{4\infty}{\imath}\left\{ \left(1 + \imath\right)^{\frac{1}{\infty}} - \left(1 - \imath\right)^{\frac{1}{\infty}} \right\}$ Also dividing by $\imath$ is the same as multiplying by $-\imath$ so the second easy step gives me: $\pi = -4 \imath \infty \left\{ \left(1 + \imath\right)^{\frac{1}{\infty}} - \left(1 - \imath\right)^{\frac{1}{\infty}} \right\}$ Now the hard part. All those infinities. I don’t like multiplying by infinity. I don’t like dividing by infinity. I really, really don’t like raising a quantity to the one-over-infinity power. Most mathematicians don’t. We have a tool for dealing with this sort of thing. It’s called a “limit”. Mathematicians developed the idea of limits over … well, since they started doing mathematics. In the 19th century limits got sound enough that we still trust the idea. Here’s the rough way it works. Suppose we have a function which I’m going to name ‘f’ because I have better things to do than give functions good names. Its domain is the real numbers. Its range is the real numbers. (We can define functions for other domains and ranges, too. Those definitions look like what they do here.) I’m going to use ‘x’ for the independent variable. It’s any number in the domain. I’m going to use ‘a’ for some point. We want to know the limit of the function “at a”. ‘a’ might be in the domain. But — and this is genius — it doesn’t have to be. We can talk sensibly about the limit of a function at some point where the function doesn’t exist. We can say “the limit of f at a is the number L”. I hadn’t introduced ‘L’ into evidence before, but … it’s a number. It has some specific set value. Can’t say which one without knowing what ‘f’ is and what its domain is and what ‘a’ is. But I know this about it. Pick any error margin that you like. Call it ε because mathematicians do. However small this (positive) number is, there’s at least one neighborhood in the domain of ‘f’ that surrounds ‘a’. Check every point in that neighborhood other than ‘a’. The value of ‘f’ at all those points in that neighborhood other than ‘a’ will be larger than L – ε and smaller than L + ε. Yeah, pause a bit there. It’s a tricky definition. It’s a nice common place to crash hard in freshman calculus. Also again in Intro to Real Analysis. It’s not just you. Perhaps it’ll help to think of it as a kind of mutual challenge game. Try this. 1. You draw whatever error bar, as big or as little as you like, around ‘L’. 2. But I always respond by drawing some strip around ‘a’. 3. You then pick absolutely any ‘x’ inside my strip, other than ‘a’. 4. Is f(x) always within the error bar you drew? Suppose f(x) is. Suppose that you can pick any error bar however tiny, and I can answer with a strip however tiny, and every single ‘x’ inside my strip has an f(x) within your error bar … then, L is the limit of f at a. Again, yes, tricky. But mathematicians haven’t found a better definition that doesn’t break something mathematicians need. To write “the limit of f at a is L” we use the notation: $\displaystyle \lim_{x \to a} f(x) = L$ The ‘lim’ part probably makes perfect sense. And you can see where ‘f’ and ‘a’ have to enter into it. ‘x’ here is a “dummy variable”. It’s the falsework of the mathematical expression. We need some name for the independent variable. It’s clumsy to do without. But it doesn’t matter what the name is. It’ll never appear in the answer. If it does then the work went wrong somewhere. What I want to do, then, is turn all those appearances of ‘∞’ in Wronski’s expression into limits of something at infinity. And having just said what a limit is I have to do a patch job. In that talk about the limit at ‘a’ I talked about a neighborhood containing ‘a’. What’s it mean to have a neighborhood “containing ∞”? The answer is exactly what you’d think if you got this question and were eight years old. The “neighborhood of infinity” is “all the big enough numbers”. To make it rigorous, it’s “all the numbers bigger than some finite number that let’s just call N”. So you give me an error bar around ‘L’. I’ll give you back some number ‘N’. Every ‘x’ that’s bigger than ‘N’ has f(x) inside your error bars. And note that I don’t have to say what ‘f(∞)’ is or even commit to the idea that such a thing can be meaningful. I only ever have to think directly about values of ‘f(x)’ where ‘x’ is some real number. So! First, let me rewrite Wronski’s formula as a function, defined on the real numbers. Then I can replace each ∞ with the limit of something at infinity and … oh, wait a minute. There’s three ∞ symbols there. Do I need three limits? Ugh. Yeah. Probably. This can be all right. We can do multiple limits. This can be well-defined. It can also be a right pain. The challenge-and-response game needs a little modifying to work. You still draw error bars. But I have to draw multiple strips. One for each of the variables. And every combination of values inside all those strips has give an ‘f’ that’s inside your error bars. There’s room for great mischief. You can arrange combinations of variables that look likely to break ‘f’ outside the error bars. So. Three independent variables, all taking a limit at ∞? That’s not guaranteed to be trouble, but I’d expect trouble. At least I’d expect something to keep the limit from existing. That is, we could find there’s no number ‘L’ so that this drawing-neighborhoods thing works for all three variables at once. Let’s try. One of the ∞ will be a limit of a variable named ‘x’. One of them a variable named ‘y’. One of them a variable named ‘z’. Then: $f(x, y, z) = -4 \imath x \left\{ \left(1 + \imath\right)^{\frac{1}{y}} - \left(1 - \imath\right)^{\frac{1}{z}} \right\}$ Without doing the work, my hunch is: this is utter madness. I expect it’s probably possible to make this function take on many wildly different values by the judicious choice of ‘x’, ‘y’, and ‘z’. Particularly ‘y’ and ‘z’. You maybe see it already. If you don’t, you maybe see it now that I’ve said you maybe see it. If you don’t, I’ll get there, but not in this essay. But let’s suppose that it’s possible to make f(x, y, z) take on wildly different values like I’m getting at. This implies that there’s not any limit ‘L’, and therefore Wronski’s work is just wrong. Thing is, Wronski wouldn’t have thought that. Deep down, I am certain, he thought the three appearances of ∞ were the same “value”. And that to translate him fairly we’d use the same name for all three appearances. So I am going to do that. I shall use ‘x’ as my variable name, and replace all three appearances of ∞ with the same variable and a common limit. So this gives me the single function: $f(x) = -4 \imath x \left\{ \left(1 + \imath\right)^{\frac{1}{x}} - \left(1 - \imath\right)^{\frac{1}{x}} \right\}$ And then I need to take the limit of this at ∞. If Wronski is right, and if I’ve translated him fairly, it’s going to be π. Or something easy to get π from. I hope to get there next week. ## Author: Joseph Nebus I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him. ## 15 thoughts on “As I Try To Figure Out What Wronski Thought ‘Pi’ Was” 1. You know, when you first posted about the Wronski Pi, I thought of doing something like this. But then I got distracted by shiny things. Still, this looks like a fun diversion! I think I’m gonna try my hand at it, too, from a Non-Standard Analysis direction. Liked by 1 person 1. Certainly understand the distractions. I’ve been wrestling with a lot of them myself lately. I’d be delighted if you did try a nonstandard analysis approach. It would probably be pretty close to what Wronski was trying to get at, if I’m getting from Boyer a fair sense of what Wronski was thinking. (To those who don’t know what I’m talking about: nonstandard analysis is this approach to analysis that’s grown over the last 50 years. It’s built on ideas like what if we suppose there's some number that's larger than zero but still less than the reciprocal of all positive integers?'' You might answer there's no such thing. A nonstandard analyst might respond they said the same things about negative numbers and about imaginary numbers and about different sizes of infinity. If the reasoning is sound and the results are fruitful, who's to say there'sno such thing”? It’s still “nonstandard” because, well, it’s not the way we all grow up learning analysis. And we don’t all grow up learning analysis this way because it’s still a relatively new approach and there’s objections to the kinds of abstraction you need to be comfortable with to get interesting results. Maybe check back in another fifty years and see how opinions on it have changed.) Liked by 1 person 2. This is very interesting! Perhaps Wronski had some method to his madness after all. Like 1. Oh, he certainly had method to this madness at least. I’ve worked through the expressions and I think I know roughly where he was going. It’s just dangerous getting there. Liked by 2 people 1. That is nice to hear. Most biographies I read portray Wronski as a crazy guy. Like 1. I confess I haven’t read a proper biography of Wronski. (Come to it, I’m not sure there is one.) I’ve just had short essays, mostly on the web, that might be factually right but not necessarily fair. It’s hard to judge on just a thousand words about someone. Liked by 1 person 3. Multiplying by infinity seems related to the old joke about asking someone to give more than 100% of effort to something. Something that only exists because you calculated it. Like 1. Well, I’m not sure. I mean, when you’re talking about percentages of something you do get your choice of what the 100 percent'' level is, after all. Like, the Space Shuttle Main Engines routinely ran at 104 percent maximum thrust, and later 109 percent maximum thrust, because themaximum thrust” level was set early in the design and when refinements came in that made the engine more powerful, it was easier to say normal procedure was running at above 100 percent than to go back and change the value of “designed maximum thrust” everywhere. Multiplying by infinity, that gets tricky because you do have to think carefully what you mean by doing it. And you can easily get ridiculous results if you aren’t thinking carefully. Or if you want to play pranks on people. See every humorous mathematical science fiction story ever that isn’t about Moebius strips. I’m not sure there’s a direct link, although there are clever mappings of the real numbers to finite strips that probably make it possible to get from percentages to infinities quickly enough. Liked by 1 person This site uses Akismet to reduce spam. Learn how your comment data is processed.	3,196	12,796	{"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": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.796875	4	CC-MAIN-2021-49	longest	en	0.92472
https://practicaldev-herokuapp-com.global.ssl.fastly.net/erikhei/comment/17bc1	1,618,778,063,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038860318.63/warc/CC-MAIN-20210418194009-20210418224009-00388.warc.gz	561,351,914	24,062	## DEV Community is a community of 605,533 amazing developers We're a place where coders share, stay up-to-date and grow their careers. # Discussion on: What are Linked Lists and How Do I Make One in Python? ## Replies for: Shorter technique: my_list = List() ;) But yes, what you showed is the way you'd create your own linked list in Python! Now I'd be curious why you... Whiteboarding with Erik If you're not worried about performance, then yes, it's easier to use a Python list. The reason for learning how to implement a linked list is along the lines of learning math when you can use a calculator–it's more about learning the concept. It's hotly debated in the developer community whether technical interviews that ask data structures and algorithms problem are really necessary. I don't know the answer to this. This article series is meant to prepare students who only know Python for these kinds of interview questions. One advantage is that this simple linked list class provides a base to make our own specialized list class with custom methods, which I'll do in a future article when we use our linked list class to make a simple blockchain. So stay tuned for the DIY Bitcoin article! Jason C. McDonald • Edited Ah, okay, I was merely under the impression that `List` had been implemented as a linked list. (I already know the difference between fixed-length array, dynamic-length array, and linked list, but your explanation will be helpful to anyone who isn't.) Whiteboarding with Erik I also was unaware how `list` is implemented in Python. So we both learned something :) Jason C. McDonald • Edited On nerd-level discussion of data structures (language agnostic now), another good option for a dynamic contiguous collection is a circular buffer, which is designed such that the head and tail "float". It's a bit hard to explain — basically, you don't require head to be at element 0, but implement logic for "wrapping around" from the last element to the first, as long as you haven't yet reached tail. It makes for a time complexity of O(n/2) on insertion/deletion without reallocation, instead of the O(n) of a typical contiguous collection. Even better, head and tail operations are always Θ(1). While this is still a bit higher than the consistent O(1) of a linked list, you don't have the cache misses. It's also better-suited for random access. No idea how you'd effectively implement this in Python, however. I always use pointer arithmetic to pull it off.	535	2,494	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17	latest	en	0.946783
https://www.coursehero.com/file/143327/f2006lecNotesWeek06/	1,498,214,653,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128320049.84/warc/CC-MAIN-20170623100455-20170623120455-00072.warc.gz	842,912,520	24,523	f2006lecNotesWeek06 # f2006lecNotesWeek06 - 19 Inclusion/Exclusion Principle and... This preview shows pages 1–3. 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: 19 Inclusion/Exclusion Principle and Pigeonhole Principle 19.1 Inclusion/exclusion principle • \| A ∪ B \| = \| A \| + \| B \| - \| A ∩ B \| no proof use diagrams to convince them of validity • \| A ∪ B ∪ C \| = \| A \| + \| B \| + \| C \| - \| A ∩ B \| - \| A ∩ C \| - \| B ∩ C \| + \| A ∩ B ∩ C \| again, use diagrams to convince them • an example 40 people asked about usage of three drugs: A , B , and C for headaches 23 use A 18 use B 31 use C 11 use A and B 19 use A and C 14 use B and C 37 use at least one how many are using none? how many are using all 3? how many are using exactly one? • 4 or more sets show how to generalize 19.2 Basic pigeonhole principle • Given n pigeons placed in k < n pigeonholes, at least one pigeonhole contains more than one pigeon • note that the principle does not say that any particular pigeonhole will contain anything • text proves validity of principle — we will take it as an axiom 19.3 Simple examples of basic principle • there must be at least two people in first year Engineering with the same birthday • at least two students at U of T have the same first and last initials 19.4 Extended principle • suppose we want to put 10 objects in 3 containers by principle, we can say that at least one container must have at least two in fact, we can say something stronger at least one container must have at least 4 objects • Generally, given n pigeons in k pigeonholes, at least one pigeonhole has at least n k pigeons 19.5 Simple examples of extended principle • in a group of 50 people, there must be at least 5 who were born in the same month 40 19.6 More complex examples • many problems can be solved using the principle the trick is usually trying to find out what the pigeons and pigeonholes should be • an example consider a party of n people, all either mutual acquaintances or mutual strangers i.e. never have a situation in which a knows b but b does not know a (if they want something more practical, processors connected directly to each other)... View Full Document ## This note was uploaded on 04/19/2008 for the course ECE 190 taught by Professor Carter during the Fall '06 term at University of Toronto. ### Page1 / 6 f2006lecNotesWeek06 - 19 Inclusion/Exclusion Principle and... This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online	683	2,718	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.84375	4	CC-MAIN-2017-26	longest	en	0.912781
https://brilliant.org/problems/how-did-i-not-see-that/	1,524,818,664,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524127095762.40/warc/CC-MAIN-20180427075937-20180427095937-00476.warc.gz	565,311,376	11,208	# How did I not see that.. Level pending Find the sum of all possible values of $$\theta \in [-\pi, \pi]$$ such that $\sin^3\left(\dfrac{1}{2}\theta\right) + \sin^3\left(\dfrac{1}{3}\theta\right) + \sin\left(\dfrac{1}{6}\theta\right) + \sin\left(\dfrac{1}{12}\theta\right) = 0$ ×	111	282	{"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}	2.609375	3	CC-MAIN-2018-17	latest	en	0.686186
https://cartplayer.com/14-table/	1,674,841,867,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764495001.99/warc/CC-MAIN-20230127164242-20230127194242-00224.warc.gz	167,843,767	18,893	“The Table of 14: A Key to Understanding Multiplication” 14 Table: Multiplication is a fundamental concept in mathematics that is essential in our daily lives. The table of 14 is a crucial part of understanding multiplication and is typically taught in primary and middle school. In this blog post, we will explore the importance of learning the table of 14 and how to easily memorize it. The table of 14 includes the products of 14 multiplied by the numbers 1 through 10. It is a chart that lists the products of 14 and the numbers 1 to 10. The table of 14 is important because it forms the foundation for more advanced mathematical concepts such as fractions, decimals, and algebra. Additionally, understanding the table of 14 can help students develop better problem-solving skills, which can be applied to other areas of their lives. Learning the table of 14 can be done in several ways. The most common method is rote memorization, where students repeat the table over and over again until they have it memorized. This method is effective, but it can be tedious and time-consuming. Another approach is to use flashcards or other visual aids to help students associate the numbers with their corresponding products. One of the ways to make it easier to learn the table of 14 is to relate it to other times tables. For example, 14 x 2 = 28 which is also 7 x 4. This means that if you know the 7 times table, you can find the corresponding number in the 14 times table by simply doubling it. Another way is to use the distributive property, the distributive property states that the product of a sum and a number is the same as the sum of the products. For example, 14 x (3+2) = 14×3 + 14×2 = 42 + 28 = 70. By breaking down the numbers into smaller parts, it can make it easier to memorize the table of 14. In conclusion, table of 14 is an essential part of mathematics education. It provides students with a solid foundation for more advanced mathematical concepts and helps them develop essential problem-solving skills. It is important for students to learn and memorize the table of 14 to improve their mathematical proficiency and achieve success in their studies. The Tables of 14 and 15: A Key to Understanding Multiplication The table of 14 includes the products of 14 multiplied by the numbers 1 through 10, while the table of 15 includes the products of 15 multiplied by the numbers 1 through 10. Here is an example of the table of 14: 14 x 1 = 14 14 x 2 = 28 14 x 3 = 42 14 x 4 = 56 14 x 5 = 70 14 x 6 = 84 14 x 7 = 98 14 x 8 = 112 14 x 9 = 126 14 x 10 = 140 And here is an example of the table of 15: 15 x 1 = 15 15 x 2 = 30 15 x 3 = 45 15 x 4 = 60 15 x 5 = 75 15 x 6 = 90 15 x 7 = 105 15 x 8 = 120 15 x 9 = 135 15 x 10 = 150 It’s important to note that a multiplication table can be extended to any number of rows or columns, it’s a way of showing the product of a number when it’s multiplied by other numbers. These tables of 14 and 15 are common ones that are taught in primary and middle school, but it can be extended to other numbers if needed. It’s also important to notice that the table of 15 has a pattern of adding 15 to each product, while the table of 14 doesn’t have a specific pattern. This can be an additional tool to memorize the tables as you can use the pattern to guess the next product. The 14 Times Table in Word: A Step-by-Step Guide • 14 times 1 is 14 (14×1 =14) • 14 times 2 is 28 (14×2 = 28) • 14 times 3 is 42 (14×3 = 42) • 14 times 4 is 56 (14×4 = 56) • 14 times 5 is 70 (14×5 = 70) • 14 times 6 is 84 (14×6 = 84) • 14 times 7 is 98 (14×7 = 98) • 14 times 8 is 112 (14×8 = 112) • 14 times 9 is 126 (14×9 = 126) • 14 times 10 is 140 (14×10 = 140) Multiply with Ease: Tricks for Learning the 14 Times Table Here are a few multiplication tricks for the 14 times table: 1. Doubling and Halving: To find the product of 14 x an even number, you can double the number and then add a zero to the end. For example, 14 x 8 = 112. You can also find the product of 14 x an even number by halving the number and then multiplying by 28. For example, 14 x 8 = (8/2) x 28 = 4 x 28 = 112. 2. Adding and Subtracting: To find the product of 14 x an odd number, you can add or subtract 13 from the number and then multiply by 14. For example, 14 x 7 = 7 – 13 + 14 = -2 x 14 = -28. 3. Using the commutative property: The commutative property states that the order of the numbers being multiplied does not affect the product. So, for example, 14 x 3 = 3 x 14 = 42. 4. Using the distributive property: The distributive property states that the product of a sum and a number is the same as the sum of the products. For example, 14 x (3+2) = 14×3 + 14×2 = 42 + 28 = 70. 5. Relating to other times tables: Some numbers in the 14 times table can be related to other times tables. For example, 14 x 5 = 70 which is also 7 x 10 and also 2 x 35. 6. Last digit: The last digit of any product in the 14 times table will be always 4. It’s important to note that these tricks are helpful tools for quickly solving problems, but it’s always good to understand the underlying mathematical concepts for a better understanding. The Tables of 14 to 20: A Key to Understanding Multiplication The table of 14 to 20 includes the products of 14, 15, 16, 17, 18, 19, and 20 multiplied by the numbers 1 through 10. Here is an example of the table of 14 to 20: 14 x 1 = 14 14 x 2 = 28 14 x 3 = 42 14 x 4 = 56 14 x 5 = 70 14 x 6 = 84 14 x 7 = 98 14 x 8 = 112 14 x 9 = 126 14 x 10 = 140 15 15 x 1 = 15 15 x 2 = 30 15 x 3 = 45 15 x 4 = 60 15 x 5 = 75 15 x 6 = 90 15 x 7 = 105 15 x 8 = 120 15 x 9 = 135 15 x 10 = 150 16 16 x 1 = 16 16 x 2 = 32 16 x 3 = 48 16 x 4 = 64 16 x 5 = 80 16 x 6 = 96 16 x 7 = 112 16 x 8 = 128 16 x 9 = 144 16 x 10 = 160 17 17 x 1 = 17 17 x 2 = 34 17 x 3 = 51 17 x 4 = 68 17 x 5 = 85 17 x 6 = 102 17 x 7 = 119 17 x 8 = 136 17 x 9 = 153 17 x 10 = 170 18 18 x 1 = 18 18 x 2 = 36 18 x 3 = 54 18 x 4 = 72 18 x 5 = 90 18 x 6 = 108 18 x 7 = 126 18 x 8 = 144 18 x 9 = 162 How to write the table of 14? Here are the steps to write the table of 14: Begin by writing the number 14 at the top of a sheet of paper or in a word processing document. Write the numbers 1 through 10 in a column next to the number 14, with the number 1 at the top and the number 10 at the bottom. Multiply 14 by each of the numbers in the column. This will give you the products of 14 multiplied by each of the numbers 1 through 10. Write the products in a second column, next to the corresponding number. Once you have written the products of 14 multiplied by each of the numbers 1 through 10, you have completed the table of 14. Example: 14 x 1 = 14 14 x 2 = 28 14 x 3 = 42 14 x 4 = 56 14 x 5 = 70 14 x 6 = 84 14 x 7 = 98 14 x 8 = 112 14 x 9 = 126 14 x 10 = 140 It’s important to note that a multiplication table can be extended to any number of rows or columns, it’s a way of showing the product of a number when it’s multiplied by other numbers. This table of 14 is a common one that is taught in primary and middle school, but it can be extended to other numbers if needed. Multiplication Times Tables for Kids: Tables 1 - 40 and Lot More to Learn Rs. 99 Rs. 94 Amazon.in as of January 25, 2023 8:45 am	2,260	7,273	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.8125	5	CC-MAIN-2023-06	latest	en	0.933603
https://economics.stackexchange.com/tags/causal-inference/hot	1,723,081,850,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00162.warc.gz	179,888,718	22,749	# Tag Info Accepted ### Seemingly Unrelated Regression Estimation - Equivalent to OLS Standard errors? Assume that for each observation $i = 1,\ldots, N$, we have $M$ equations: $$y_{i,j} = x_{i,j}\beta_j + \varepsilon_{i,j}$$ Where $i = 1,\ldots, N$ enumerates individuals and and $j = 1,\ldots, M$ ... • 12.5k ### Subsampling vs. m out of n bootstrap You are correct. The bootstrap samples with replacement while subsampling samples without replacement. Theoretically, when $m \ll n$, it does not matter if you sample with or without replacement (if ... • 12.5k ### Identifying assumption meaning "Identification" is the most loaded term in econometrics. There are multiple cheap talk equilibria with regard to its meaning. It is used with different intended (but related and overlapping)... • 2,619 Accepted ### What are the meanings of "difference-in-differences" and "causal estimand"? The difference-in-differences “estimator” is a method used to “estimate” a target parameter of interest. That parameter is also commonly referred to as the causal “estimand” (that which is to be ... Accepted ### Bootstrap always valid under asymptotic Normality? Theorem 2.1 in Horowitz (2019) is what you are looking for • 384 Accepted ### Articles on the definition of causality in Economics As 1muflon1 said, the concept of causality in economics is intertwined with the concept of causality in other sciences, in particular statistic and econometrics, and the concept of probabilistic ... • 4,122 Accepted ### Reverse DiD: or using always treated as control In principle yes, but you would not use 'reverse treatment' but standard terminology. Here just withdrawal of some stimuli is the treatment. That is you would still code the withdrawal as $D=1$. For ... • 57.4k ### Identifying assumption meaning I think the best way how to explain this is to first quickly explain what identification actually is. As mentioned in this thread: For example, in the John Stachurski "A Primer in Econometric ... • 57.4k ### Skepticism about the claims of instrument variable validity/exclusion through a statistical test—the Arellano-Bond Test If yes, then how does this square with the general point that causality/exclusion cannot generally be established with statistical tests... It seems to me that "[exogeneity of IV] cannot ... • 2,619 Accepted ### How does counterfactual for continuous variables work? The Neymen-Rubin potential outcomes terminology is is not typically used in economics outside policy evaluation where your policy will be binary. This being said there are still counterfactuals. For ... • 57.4k Accepted ### Can Difference-in-Difference be used when the treatment effects get smaller with time since treatment? It depends on a few things. First, if you expect treatment effects change over time, then you want to estimate an event-study style DD specification. If you have a single treatment timing (all ... • 384 1 vote • 3,728 1 vote ### Is matching combined with Diff-in-Diff a bad idea? Is matching combined with Diff-in-Diff a bad idea? No. But, as with anything, it depends on the context. One point to consider is that you should match on pre-treatment characteristics, because ... • 6,238 1 vote ### Is it possible to integrate propensity score (weighting or matching) and synthetic control method? You can't integrate the two because there is nothing to integrate. Both approaches are solving similar problems (getting the right counterfactual for causal comparisons) in similar ways (applying ... • 6,238 1 vote ### Diff-in-diff with two control groups: compare parallel trends Background The typical thing to do is visually inspect the pre-treatment trends for the control and treatment groups. Whether you want it or not, you might be biased when looking at the visual ... • 919 1 vote ### Diff-in-diff with two control groups: compare parallel trends I'm not understanding-- could you please talk a little about why both can't be used here? It's been a while since I've really worked with DD, but as far as I'm aware, there's not real metric you can ... 1 vote ### Discussing Difference-in-Difference Assumption of the treatment assigment The first source you found is correct, parallel trend is not sufficient, you can find the same assumption mentioned in multiple places (e.g here). One of the identifying assumptions of DiD is that: ... • 57.4k 1 vote ### Reverse DiD: or using always treated as control For those interested, there is work on this: Kim, K., & Lee, M. J. (2019). Difference in differences in reverse. Empirical Economics, 57(3), 705-725. • 1,843 1 vote ### Identifying assumption meaning "Identification" is the professional jargon in econometrics for "asserting that the outputs from an econometric model do indeed estimate what we want and declare that they estimate"... • 33.9k Only top scored, non community-wiki answers of a minimum length are eligible	1,158	4,939	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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}	2.625	3	CC-MAIN-2024-33	latest	en	0.903483
https://rebelsky.cs.grinnell.edu/Courses/CS151/2003S/EBoards/merge-sort.txt	1,719,355,732,000,000,000	text/plain	crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00633.warc.gz	416,017,668	1,169	CSC151, Class 50: Merge Sort Overview: * Questions * Lab * Reflection Notes: * Good job on presentations yesterday! It's hard to fit stuff in ten minutes. * Read Records in Scheme * Questions on the exam? Question 1: less-than Suppose we have two vectors, a and b. When is a less than b? when element[0] of a is less than element[0] of b or when element[0] of a equals element [0] of b and element[1] of a is less than element [1] of b or when elements[0..1] of a equal elements [0..1] of b and element[2] of a is less than element [2] of b and so on and so forth You may find that you run out of elements in one vector before the other. In that case, the vector that ends first is smaller. Can we assume that element 0 of each vector is a string or number? NO! Part of the fun of this problem is deep recursion ---------------------------------------- Merge sort Do the lab. For problem 1a, I wanted something like (merge (list 1 2 3) (list 1 1.9 2.5) <=) For problem 1d, your solution should look something like (merge cs-faculty young-cs-kids (lambda (person1 person2) ...)) (define foo (time xxx))	300	1,101	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.3125	3	CC-MAIN-2024-26	latest	en	0.877822
https://www.teachexcel.com/excel-tutorial/absolute-and-relative-cell-references-in-excel_1268.html	1,638,960,626,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964363465.47/warc/CC-MAIN-20211208083545-20211208113545-00605.warc.gz	1,062,898,115	12,526	# Absolute and Relative Cell References in Excel Author: In this tutorial I am going to cover the difference between Absolute and Relative Cell References in Excel and show you how to use them and why they are so important. ## Relative Cell References Relative Cell references in a formula change as they are copied or filled down a spreadsheet. What this means is that the cell reference will update relative to the original cell reference as it is copied. To better understand this look at the following example: Here I have a table with 2 columns and 20 rows. I want to add column A and column B together in a Total column. I start by entering the addition formula into cell C2 which references cells A2 and B2. If you think about these cell references as relative to cell C2, I am referencing the 2 cells to the left of the current cell (C2). Now I copy my formula down the Total column: My cell references have updated relative to where I have copied my initial formula. I have cell C21 selected yet the formula now references cells A21 and B21. These 2 cells are the 2 cells to left, relative to the formula cell (C21). ## Absolute Cell References An Absolute cell reference does the opposite of a Relative cell reference. It does not update/change as a formula is copied or filled. It remains the same. This is very useful when you want to use a single value for calculations copied over a range of cells. Looking back to the previous example, I will now add a Tax column: I start by entering my formula and selecting the cell G1 to incorporate my tax rate. Once I have selected the cell reference I wish to change to an Absolute cell I hit F4 on the keyboard to cycle through the Absolute cell reference options. (You can type the \$ symbols that appear by hand but F4 is faster) \$G\$1 is the absolute cell reference. This cell reference wont change at all whether you copy the formula down the rows or across the columns. So if I copy the formula in D2 down to D21: Notice that the Absolute cell reference \$G\$1 doesnt change. The other Absolute cell references you can do are: G\$1 This cell reference wont change if you copy the formula down the rows but will change its column reference as it is copied across the columns. \$G1 This cell reference wont change if you copy the formula across the columns but will change its row reference as it is copied down the rows. The dollar sign \$ is what makes the elements of a cell reference Absolute. Look at how these Absolute cell references change as they are copied down the rows or columns: ## Absolute Cell References Across Different Worksheets in Excel Absolute and Relative cell references also work across different sheets. Just start entering your formula: Then select the next worksheet and select the cell or range you want: Hit F4 to get the Absolute cell reference and then Enter and you will be returned to the original worksheet: You can then copy the formula like normal: Dont be put off by the extra-long cell reference here. It works the exact same way. It is just longer due to the Sheet2! part which is what references the cell on the Sheet2 worksheet. Dont forget about Absolute and Relative cell references! They are very important for making powerful and useful formulas and functions in Excel! #### Question? Ask it in our Excel Forum Our Excel Courses #### Professional Input Forms in Excel • Skill Level: All Levels • \| • 5 hours+ How to make a fully featured professional form in Excel that is unbreakable. This includes how to use the form to store, view, edit, and delete data from a data storage worksheet. #### Send Emails from Excel (Multiple Methods) • Skill Level: All Levels • \| • 2 hours Send Emails from Excel using VBA and Macros. This course starts from the Basics and builds up to more advanced examples with attaching workbooks, worksheets, PDF's, automatically sending emails, including a signature, error handling, increasing speed, and more. Similar Content on TeachExcel Calculate Percent Change in Excel Tutorial: How to calculate the percentage amount that a value has changed in Excel. This includes q... Apply Conditional Formatting to Multiple Cells with a Single Formula Tutorial: How to use a single formula to apply conditional formatting to multiple cells at once in ... 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Tutorial Details	1,040	4,975	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49	longest	en	0.892158
http://www.office-archive.com/4-excel/76fbcd02014d8fba.htm	1,656,841,234,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656104215805.66/warc/CC-MAIN-20220703073750-20220703103750-00003.warc.gz	103,849,712	6,055	Incorrect order of operations in formula Author Message Incorrect order of operations in formula I just noticed that Excel does not use correct order of operations with - and ^. For example, the formula = - 5 ^ 2 incorrectly displays 25 instead of -25. In other words, -x^2 is interpreted as (-x)^2, when it should be interpreted as -(x^2). This is also a problem with multiplication; 1-5^2 also yields 25. On the other hand, addition is fine; if it wasn't I'd be really worried. So 0-5^2 yields -25. If anyone can tell me why this makes sense in somebody's world, I'd like to know. Also, if the problem needs to be reported, how do I do it? Thanks, Jonathan Bishop Wed, 18 Jun 1902 08:00:00 GMT Incorrect order of operations in formula Jonathan, That is indeed the correct order of operation. Negation is the highest (first) priority of standard arithmetic operations. Subtraction, which uses the same symbol as negation, is lower priority than negation or exponentiation, but is a completely different operation. The two formulas =-5^2 and =0-5^2 are NOT the same. In the first, the - is a negation operator. In the second, it is a subtraction operator. This is the way programming languages work. It is not a problem, and does not need to be reported. Cordially, Chip Pearson http://home.gvi.net/~cpearson/excel Quote: >I just noticed that Excel does not use correct order of operations >with - and ^. For example, the formula = - 5 ^ 2 incorrectly displays >25 instead of -25. In other words, -x^2 is interpreted as (-x)^2, when >it should be interpreted as -(x^2). >This is also a problem with multiplication; 1-5^2 also yields 25. >On the other hand, addition is fine; if it wasn't I'd be really worried. >So 0-5^2 yields -25. >If anyone can tell me why this makes sense in somebody's world, >I'd like to know. >Also, if the problem needs to be reported, how do I do it? >Thanks, >Jonathan Bishop Wed, 18 Jun 1902 08:00:00 GMT Incorrect order of operations in formula This issue was raised in the Excel newsgroups about a year or so ago, and I'm still waiting for someone to cite me their persuasive authority for what they deem to be a "correct" order, or that an expression "should be" interpreted a particular way. Many programmers systematically avoid any problem by routinely using appropriate parentheses to force the intended meaning. My own notion of what's going on is that there is a fairly well accepted notion of a negation operator, -, that is different from the subtraction operator, with a different order of precedence in most programming languages, and people try to assert that it should be assigned the same order of precedence as the subtraction operator--but I've never seen a decent discussion of this. It seemed at the time to be an issue that people got emotional over, but nobody wanted to rationally discuss it. People even expressed the view that it was an arrogant Microsoft innovation to "reverse" the "basic rules". I'm curious why the poster here thinks that there is some natural or unnatural law about the order of precedence in this case. Let's get a good, hot (but civil, of course) thread going. Quote: > I just noticed that Excel does not use correct order of operations > with - and ^. For example, the formula = - 5 ^ 2 incorrectly displays > 25 instead of -25. In other words, -x^2 is interpreted as (-x)^2, when > it should be interpreted as -(x^2). > This is also a problem with multiplication; 1-5^2 also yields 25. > On the other hand, addition is fine; if it wasn't I'd be really worried. > So 0-5^2 yields -25. > If anyone can tell me why this makes sense in somebody's world, > I'd like to know. > Also, if the problem needs to be reported, how do I do it? > Thanks, > Jonathan Bishop Wed, 18 Jun 1902 08:00:00 GMT Incorrect order of operations in formula Jonathan, You are comparing negation and subtraction as equals, but they are not: You might want to consult the help before reporting this: The order in which Microsoft Excel performs operations in formulas If you combine several operators in a single formula, Microsoft Excel performs the operations in the order shown in the following table. If a formula contains operators with the same precedence ? for example, if a formula contains both a multiplication and division operator ? Microsoft Excel evaluates the operators from left to right. To change the order of evaluation, enclose the part of the formula to be calculated first in parentheses. Operator Description : (colon) Reference Operators , (comma) (single space) Negation (as in 1) % Percent ^ Exponentiation and / Multiplication and division + and Addition and subtraction & Connects two strings of text (concatenation) = < > <= >= <> Comparison Regards, Tom Ogilvy Quote: >Jonathan Bishop Wed, 18 Jun 1902 08:00:00 GMT Incorrect order of operations in formula Hi Jonathan, Excel is using the order as shown in help under "calculating formulas, calculation order". Negation is figured before Exponentiation. I always use parentheses whether they're needed or not. For example I'd use =(25)+1 even though =25+1 would give the same result. I find it easier to read and less prone to errors. In your case you can force the order you want with = -(5^2). I'll leave it for others to argue whether Excel's order of calculation is the "correct" order. HTH Denny Campbell Grand Rapids, MI USA Quote: >I just noticed that Excel does not use correct order of operations >with - and ^. For example, the formula = - 5 ^ 2 incorrectly displays >25 instead of -25. In other words, -x^2 is interpreted as (-x)^2, when >it should be interpreted as -(x^2). >This is also a problem with multiplication; 1-5^2 also yields 25. >On the other hand, addition is fine; if it wasn't I'd be really worried. >So 0-5^2 yields -25. >If anyone can tell me why this makes sense in somebody's world, >I'd like to know. >Also, if the problem needs to be reported, how do I do it? >Thanks, >Jonathan Bishop Wed, 18 Jun 1902 08:00:00 GMT Incorrect order of operations in formula I realized after writing that my comparison with subtraction was flawed, and that the unary minus is a different operator than the binary minus. So, I didn't help by confusing the issue. but... FORTRAN seems to follow the convention I considered "correct;" a=-52 print,a yields -25.00000 C isn't a good reference since the power operator is a function The IDL (Integrated Data Language) yields -25 for print,-5^2. So, Excel is the only "language" I regularly use that acts other than the way I see as intuitive. I would be interested in knowing why just for my edification. My reasons for thinking that it makes more sense to perform ^ before - is that a) that would be the normally understood order if I were writing formulas by hand, and b) it is more compact. I apologize for reopening this debate and for confusing the issue; I guess I didn't search for the right words on dejanews when I was looking to see if it had been discussed before. Thanks, Jonathan Quote: >Jonathan, >That is indeed the correct order of operation. Negation is the highest >(first) priority of standard arithmetic operations. Subtraction, which >uses the same symbol as negation, is lower priority than negation or >exponentiation, but is a completely different operation. >The two formulas >=-5^2 >and >=0-5^2 >are NOT the same. In the first, the - is a negation operator. In the >second, it is a subtraction operator. >This is the way programming languages work. It is not a problem, and >does not need to be reported. >Cordially, >Chip Pearson >http://home.gvi.net/~cpearson/excel >>I just noticed that Excel does not use correct order of operations >>with - and ^. For example, the formula = - 5 ^ 2 incorrectly >displays >>25 instead of -25. In other words, -x^2 is interpreted as (-x)^2, >when >>it should be interpreted as -(x^2). >>This is also a problem with multiplication; 1-5^2 also yields 25. >>On the other hand, addition is fine; if it wasn't I'd be really >worried. >>So 0-5^2 yields -25. >>If anyone can tell me why this makes sense in somebody's world, >>I'd like to know. >>Also, if the problem needs to be reported, how do I do it? >>Thanks, >>Jonathan Bishop Wed, 18 Jun 1902 08:00:00 GMT Incorrect order of operations in formula Actually, Jonathan, I sorta enjoyed it. As has been pointed out before in this group (I think it was this one, and I think Chip did the pointing), once you add VB or VBA to your set of languages you will get the behavior you expect: -5^2 => -25, as you also get with MatLab, Mathematica and S-Plus. Excel is the only language I know of that gives higher precedence to the unary "-" than to the power operator. But along the lines of what Alan suggested, I use parens liberally just for my own head's sake (it needs a lot of help). Regards to all, TMY Research Quote: > I realized after writing that my comparison with subtraction was > flawed, and that the unary minus is a different operator than the > binary minus. So, I didn't help by confusing the issue. > but... > FORTRAN seems to follow the convention I considered "correct;" > a=-52 > print,a > yields -25.00000 > C isn't a good reference since the power operator is a function > The IDL (Integrated Data Language) yields -25 for > print,-5^2. > So, Excel is the only "language" I regularly use that > acts other than the way I see as intuitive. I would be > interested in knowing why just for my edification. > My reasons for thinking that it makes more sense to > perform ^ before - is that a) that would be the > normally understood order if I were writing formulas > by hand, and b) it is more compact. > I apologize for reopening this debate and for > confusing the issue; I guess I didn't search for the > right words on dejanews when I was looking to see > if it had been discussed before. > Thanks, > Jonathan > >Jonathan, > >That is indeed the correct order of operation. Negation is the highest > >(first) priority of standard arithmetic operations. Subtraction, which > >uses the same symbol as negation, is lower priority than negation or > >exponentiation, but is a completely different operation. > >The two formulas > >=-5^2 > >and > >=0-5^2 > >are NOT the same. In the first, the - is a negation operator. In the > >second, it is a subtraction operator. > >This is the way programming languages work. It is not a problem, and > >does not need to be reported. > >Cordially, > >Chip Pearson > >http://home.gvi.net/~cpearson/excel > >>I just noticed that Excel does not use correct order of operations > >>with - and ^. For example, the formula = - 5 ^ 2 incorrectly > >displays > >>25 instead of -25. In other words, -x^2 is interpreted as (-x)^2, > >when > >>it should be interpreted as -(x^2). > >>This is also a problem with multiplication; 1-5^2 also yields 25. > >>On the other hand, addition is fine; if it wasn't I'd be really > >worried. > >>So 0-5^2 yields -25. > >>If anyone can tell me why this makes sense in somebody's world, > >>I'd like to know. > >>Also, if the problem needs to be reported, how do I do it? > >>Thanks, > >>Jonathan Bishop Wed, 18 Jun 1902 08:00:00 GMT Incorrect order of operations in formula Actually, Jonathan, I sorta enjoyed it. As has been pointed out before in this group (I think it was this one, and I think Chip did the pointing), once you add VB or VBA to your set of languages you will get the behavior you expect: -5^2 => -25, as you also get with MatLab, Mathematica and S-Plus. Excel is the only language I know of that gives higher precedence to the unary "-" than to the power operator. But along the lines of what Alan suggested, I use parens liberally just for my own head's sake (it needs a lot of help). Regards to all, TMY Research Quote: > I realized after writing that my comparison with subtraction was > flawed, and that the unary minus is a different operator than the > binary minus. So, I didn't help by confusing the issue. > but... > FORTRAN seems to follow the convention I considered "correct;" > a=-52 > print,a > yields -25.00000 > C isn't a good reference since the power operator is a function > The IDL (Integrated Data Language) yields -25 for > print,-5^2. > So, Excel is the only "language" I regularly use that > acts other than the way I see as intuitive. I would be > interested in knowing why just for my edification. > My reasons for thinking that it makes more sense to > perform ^ before - is that a) that would be the > normally understood order if I were writing formulas > by hand, and b) it is more compact. > I apologize for reopening this debate and for > confusing the issue; I guess I didn't search for the > right words on dejanews when I was looking to see > if it had been discussed before. > Thanks, > Jonathan > >Jonathan, > >That is indeed the correct order of operation. Negation is the highest > >(first) priority of standard arithmetic operations. Subtraction, which > >uses the same symbol as negation, is lower priority than negation or > >exponentiation, but is a completely different operation. > >The two formulas > >=-5^2 > >and > >=0-5^2 > >are NOT the same. In the first, the - is a negation operator. In the > >second, it is a subtraction operator. > >This is the way programming languages work. It is not a problem, and > >does not need to be reported. > >Cordially, > >Chip Pearson > >http://home.gvi.net/~cpearson/excel > >>I just noticed that Excel does not use correct order of operations > >>with - and ^. For example, the formula = - 5 ^ 2 incorrectly > >displays > >>25 instead of -25. In other words, -x^2 is interpreted as (-x)^2, > >when > >>it should be interpreted as -(x^2). > >>This is also a problem with multiplication; 1-5^2 also yields 25. > >>On the other hand, addition is fine; if it wasn't I'd be really > >worried. > >>So 0-5^2 yields -25. > >>If anyone can tell me why this makes sense in somebody's world, > >>I'd like to know. > >>Also, if the problem needs to be reported, how do I do it? > >>Thanks, > >>Jonathan Bishop Wed, 18 Jun 1902 08:00:00 GMT Incorrect order of operations in formula This isssue is discussed at length in the comp.apps.spreadsheets FAQ, which is maintained by Russell Schulz and is available at: ----- Posted by John Walkenbach of JWalk & Associates ----- ----- Visit "The Spreadsheet Page" ----- ----- http://www.j-walk.com/ss ----- Quote: >I just noticed that Excel does not use correct order of operations >with - and ^. For example, the formula = - 5 ^ 2 incorrectly displays >25 instead of -25. In other words, -x^2 is interpreted as (-x)^2, when >it should be interpreted as -(x^2). >This is also a problem with multiplication; 1-5^2 also yields 25. >On the other hand, addition is fine; if it wasn't I'd be really worried. >So 0-5^2 yields -25. >If anyone can tell me why this makes sense in somebody's world, >I'd like to know. >Also, if the problem needs to be reported, how do I do it? >Thanks, >Jonathan Bishop Wed, 18 Jun 1902 08:00:00 GMT Page 1 of 1 [ 9 post ] Relevant Pages	4,078	15,211	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.296875	3	CC-MAIN-2022-27	latest	en	0.935422
http://stackoverflow.com/questions/15772617/conditional-statement-in-a-one-line-lambda-function-in-python/15772665	1,406,225,425,000,000,000	text/html	crawl-data/CC-MAIN-2014-23/segments/1405997890181.84/warc/CC-MAIN-20140722025810-00168-ip-10-33-131-23.ec2.internal.warc.gz	347,804,856	18,301	# Conditional statement in a one line lambda function in python? Apologies if this has been asked before, but I couldn't see it anywhere. Essentially I've come across a scenario where i need to make use of an if statement inside a lambda function. What makes it difficult is that ideally it needs to be in a single line of code (if thats even possible?) Normally, i would write this: ``````T = 250 if (T > 200): rate = 200exp(-T) else: rate = 400exp(-T) return (rate) `````` However i need it to look like this: ``````rate = lambda(T) : if (T>200): return(200exp(-T)); else: return(400exp(-T)) `````` I realize the easier thing to do would to take the decision making outside of the lambda functions, and then have a separate lambda function for each case, but its not really suitable here. The lambda functions are stored in an array and accessed when needed, with each array element corresponding to a particular "rate" so having two separate rows for the same "rate" would mess things up. Any help would be greatly appreciated, or if its not possible, some confirmation from others would be nice :) - I doubt-squared that you need a one-line lambda function, because I doubt that you need a one-line function, and I doubt that you need a lambda function as opposed to a named function. What's the point of using a `lambda` if you're immediately going to give your function the name `rate` anyhow? – DSM Apr 2 '13 at 19:30 Too bad I can only +1 DSM once… But to make it clear: the easier thing to do is not to "take the decision making outside of the lambda functions", but to just define them as regular functions in the first place. You can store regular functions in an array. There's nothing special about lambdas; they're more limited functions, not more powerful functions. – abarnert Apr 2 '13 at 19:42 Well the function needed to fit in one line, which i didn't think you could do with a named function? Also these functions are stored as strings which are then evaluated using "eval" which i wasn't sure how to do with regular functions. – Nathan Bush Apr 2 '13 at 19:48 First, why does it need to fit in one line? Second, why are they stored as strings and then evaluated using `eval`? Those are both very bad requirements. I suspect there's an XY problem here, and if you told us what you were actually trying to do, we could explain the right way to do it. – abarnert Apr 2 '13 at 20:17 Use the `exp1 if cond else exp2` syntax. ``````rate = lambda T: 200exp(-T) if T>200 else 400exp(-T) `````` Note you don't use `return` in lambda expressions. - Spot on, thank you so much :) – Nathan Bush Apr 2 '13 at 19:49 Yes, you can use the shorthand syntax for `if` statements. ``````rate = lambda(t): (200 * exp(-t)) if t > 200 else (400 * exp(-t)) `````` Note that you don't use explicit `return` statements in`lambda`s either. - Further shortening: `rate = lambda(t): (200 if t > 200 else 400) * exp(-t)` – shahkalpesh Apr 2 '13 at 19:32 @shahkalpesh Quite true, though arguably less explicit in illustrating the answer to the question. – Silas Ray Apr 2 '13 at 19:34 The parens after "lambda" are also superfluous. – gdbdmdb Apr 2 '13 at 19:40 The right way to do this is simple: ``````def rate(T): if (T > 200): return 200exp(-T) else: return 400exp(-T) `````` There is absolutely no advantage to using `lambda` here. The only thing `lambda` is good for is allowing you to create anonymous functions and use them in an expression (as opposed to a statement). If you immediately assign the `lambda` to a variable, it's no longer anonymous, and it's used in a statement, so you're just making your code less readable for no reason. The `rate` function defined this way can be stored in an array, passed around, called, etc. in exactly the same way a lambda function could. It'll be exactly the same (except a bit easier to debug, introspect, etc.). From a comment: Well the function needed to fit in one line, which i didn't think you could do with a named function? I can't imagine any good reason why the function would ever need to fit in one line. But sure, you can do that with a named function. Try this in your interpreter: ``````>>> def foo(x): return x + 1 `````` Also these functions are stored as strings which are then evaluated using "eval" which i wasn't sure how to do with regular functions. Again, while it's hard to be 100% sure without any clue as to why why you're doing this, I'm at least 99% sure that you have no reason or a bad reason for this. Almost any time you think you want to pass Python functions around as strings and call `eval` so you can use them, you actually just want to pass Python functions around as functions and use them as functions. But on the off chance that this really is what you need here: Just use `exec` instead of `eval`. You didn't mention which version of Python you're using. In 3.x, the `exec` function has the exact same signature as the `eval` function: ``````exec(my_function_string, my_globals, my_locals) `````` In 2.7, `exec` is a statement, not a function—but you can still write it in the same syntax as in 3.x (as long as you don't try to assign the return value to anything) and it works. In earlier 2.x (before 2.6, I think?) you have to do it like this instead: ``````exec my_function_string in my_globals, my_locals `````` - By the time you say `rate = lambda whatever...` you've defeated the point of lambda and should just define a function. But, if you want a lambda, you can use 'and' and 'or' ``````lambda(T): (T>200) and (200exp(-T)) or (400exp(-T)) `````` - ``````eval_op = {	1,467	5,615	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23	latest	en	0.944497
https://www.statistics-lab.com/%E6%95%B0%E5%AD%A6%E4%BB%A3%E5%86%99%E9%9A%8F%E6%9C%BA%E8%BF%87%E7%A8%8B%E4%BD%9C%E4%B8%9A%E4%BB%A3%E5%86%99stochastic-processes%E4%BB%A3%E8%80%83conditional-expectation/	1,679,654,235,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00680.warc.gz	1,095,525,906	37,094	### 数学代写\|随机过程作业代写Stochastic Processes代考\|Conditional expectation statistics-lab™ 为您的留学生涯保驾护航在代写随机过程Stochastic Processes方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写随机过程Stochastic Processes代写方面经验极为丰富，各种代写随机过程Stochastic Processes相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 数学代写\|随机过程作业代写Stochastic Processes代考\|Projections in Hilbert spaces 3.1.1 Definition A linear space with the binary operation $X \times X \rightarrow \mathbb{R}$. mapping any pair in $X \times X$ into a scalar denoted $(x, y)$, is called a unitary space or an inner product space iff for all $x, y, z \in \mathbb{X}$, and $\alpha, \beta \in \mathbb{R}$, the following conditions are satisfied: (s1) $(x+y, z)=(x, z)+(y, z)$, (s2) $(\alpha x, y)=\alpha(x, y)$, (s3) $(x, x) \geq 0$, (s4) $(x, x)=0$ iff $x=0$. (s5) $(x, y)=(y, x)$. The number $(x, y)$ is called the scalar product of $x$ and $y$. The vectors $x$ and $y$ in a unitary space are termed orthogonal iff their scalar product is 0 . 3.1.2 Example The space $l^{2}$ of square summable sequences with the scalar product $(x, y)=\sum_{n=1}^{\infty} \xi_{n} \eta_{n}$ is a unitary space; here $x=\left(\xi_{n}\right){n \geq 1}$, $y=\left(\eta{n}\right){n \geq 1}$. The space $C{[0,1]}$ of continuous functions on $[0,1]$ with the scalar product $(x, y)=\int_{0}^{1} x(s) y(s) \mathrm{d} s$ is a unitary space. Another important example is the space $L^{2}(\Omega, \mathcal{F}, \mu)$ where $(\Omega, \mathcal{F}, \mu)$ is a measure space, with $(x, y)=\int_{\Omega} x y \mathrm{~d} \mu$. The reader is encouraged to check conditions (s1)-(s5) of the definition. In particular, if $\mu$ is a probability space, we have $(X, Y)=E X Y$. Note that defining, as customary, the covariance of two square integrable random variables $X$ and $Y$ as $\operatorname{cov}(X, Y)=E\left(X-(E X) 1_{\Omega}\right)\left(Y-(E Y) 1_{\Omega}\right)$ we obtain $\operatorname{cov}(X, Y)=(X, Y)-E X E Y$. 3.1.3 Cauchy-Schwartz-Bunyakovski inequality For any $x$ and $y$ in a unitary space, $$(x, y)^{2} \leq(x, x)(y, y)$$ Proof Define the real function $f(t)=(x+t y, x+t y)$; by (s3) it admits non-negative values. Using (s1)-(s2) and (s5): $$f(t)=(x, x)+2 t(x, y)+t^{2}(y, y) ;$$ so $f(t)$ is a second order polynomial in $t$. Thus, its discriminant must be non-positive, i.e. $4(x, y)^{2}-4(x, x)(y, y) \leq 0$. ## 数学代写\|随机过程作业代写Stochastic Processes代考\|Banach spaces As we have mentioned already, the notion of a Banach space is crucial in functional analysis and in this book. Having covered the algebraic aspects of Banach space in the previous section, we now turn to discussing topological aspects. A natural way of introducing topology in a linear space is by defining a norm. Hence, we begin this section with the definition of a normed space (which is a linear space with a norm) and continue with discussion of Cauchy sequences that leads to the definition of a Banach space, as a normed space “without holes”. Next, we give a number of examples of Banach spaces (mostly those that are important in probability theory) and introduce the notion of isomorphic Banach spaces. Then we show how to immerse a normed space in a Banach space and provide examples of dense algebraic subspaces of Banach spaces. We close by showing how the completeness of a Banach space may be used to prove existence of an element that satisfies some required property. 2.2.1 Normed linear spaces Let $X$ be a linear space. A function $\|\cdot\|$ : $\mathrm{X} \rightarrow \mathbb{R}, x \mapsto\|x\|$ is called a norm, if for all $x, y \in \mathbb{X}$ and $\alpha \in \mathbb{R}$, (n1) $\|x\| \geq 0$, (n2) $\|x\|=0$, iff $x=\Theta$, (n3) $\|\alpha x\|=\|\alpha\|\|x\|$, (n4) $\|x+y\| \leq\|x\|+\|y\|$. If (n2) does not necessarily hold, $\|\cdot\|$ is called a semi-norm. Note that if $\|\cdot\|$ is a semi-norm, then $\|\Theta\|=0$ by (n3) and 2.1.4. A pair (X, $\|\cdot\|)$, where $X$ is a linear space and $\|\cdot\|$ is a norm in $X$ called a normed linear space, and for simplicity we say that $X$ itself is a normed linear space (or just normed space). 2.2.2 Exercise $(n 3)-(n 4)$ imply that for $x, y \in \mathbb{X}$, $$\|\|x\|-\|y\|\| \leq\|x \pm y\| .$$ ## 数学代写\|随机过程作业代写Stochastic Processes代考\|Definition and existence of conditional expectation 3.2.1 Motivation Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. If $B \in \mathcal{F}$ is such that $\mathbb{P}(B)>0$ then for any $A \in \mathcal{F}$ we define conditional probability $\mathbb{P}(A \mid B)$ (probability of $A$ given $B$ ) as $$\mathbb{P}(A \mid B)=\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} .$$ As all basic courses in probability explain, this quantity expresses the fact that a partial knowledge of a random experiment (” $B$ happened”) influences probabilities we assign to events. To take a simple example, in tossing a die, the knowledge that an even number turned up excludes three events, so that we assign to them conditional probability zero, and makes the probabilities of getting 2,4 or 6 twice as big. Or, if three balls are chosen at random from a box containing four red, four white and four blue balls, then the probability of the event $A$ that all three of them are of the same color is $3\left(\begin{array}{l}4 \ 3\end{array}\right) /\left(\begin{array}{c}12 \ 3\end{array}\right)=\frac{3}{55}$. However, if we know that at least one of the balls that were chosen is red, the probability of $A$ decreases and becomes $\left(\begin{array}{l}4 \ 3\end{array}\right)\left[\left(\begin{array}{c}12 \ 3\end{array}\right)-\left(\begin{array}{l}8 \ 3\end{array}\right)\right]^{-1}=\frac{3}{130}$. By the way, if this result does not agree with the reader’s intuition, it may be helpful to remark that the knowledge that there is no red ball among the chosen ones increases the probability of $A$, and that it is precisely the reason why the knowledge that at least one red ball was chosen decreases the probability of $A$. An almost obvious property of $\mathbb{P}(A \mid B)$ is that, as a function of $A$, it constitutes a new probability measure on the measurable space $(\Omega, \mathcal{F})$. It enjoys also other, less obvious, and maybe even somewhat surprising properties. To see that, let $B_{i}, i=1, \ldots, n, n \in \mathbb{N}$ be a collection of mutually disjoint measurable subsets of $\Omega$ such that $\bigcup_{i=1}^{n} B_{i}=\Omega$ and $\mathbb{P}\left(B_{i}\right)>0$. Such collections, not necessarily finite, are often called dissections, or decompositions, of $\Omega$. Also, let $A \in \mathcal{F}$. Consider all functions $Y$ of the form $$Y=\sum_{i=1}^{n} b_{i} 1_{B_{i}}$$ where $b_{i}$ are arbitrary constants. How should the constants $b_{i}, i=1, \ldots, n$ be chosen for $Y$ to be the closest to $X=1_{A}$ ? The answer depends, of course, on the way “closeness” is defined. We consider the distance $$d(Y, X)=\sqrt{\int_{\Omega}(Y-X)^{2} d \mathbb{P}}=\|Y-X\|_{L^{2}(\Omega, \mathcal{F}, \mathbb{P})}$$ In other words, we are looking for constants $b_{i}$ such that the distance $\|Y-X\|_{L^{2}(\Omega, \mathcal{F}, \mathbb{P})}$ is minimal; in terms of $3.1 .12$ we want to find a projection of $X$ onto the linear span of $\left{1_{B_{i}}, i=1, \ldots, n\right}$. Calculations are easy; the expression under the square-root sign in $(3.6)$ is \begin{aligned} \sum_{i=1}^{n} \int_{B_{i}}\left(Y-1_{A}\right)^{2} \mathrm{dP} &=\sum_{i=1}^{n} \int_{B_{i}}\left(b_{i}-1_{A}\right)^{2} \mathrm{dP} \ &=\sum_{i=1}^{n}\left[b_{i}^{2} \mathbb{P}\left(B_{i}\right)-2 b_{i} \mathbb{P}\left(B_{i} \cap A\right)+\mathbb{P}(A)\right], \end{aligned} and its minimum is attained when $b_{i}$ are chosen to be the minima of the binomials $b_{i}^{2} \mathbb{P}\left(B_{i}\right)-2 b_{i} \mathbb{P}\left(B_{i} \cap A\right)+\mathbb{P}(A)$, i.e. if $$b_{i}=\frac{\mathbb{P}\left(A \cap B_{i}\right)}{\mathbb{P}\left(B_{i}\right)}=\mathbb{P}\left(A \mid B_{i}\right) .$$ Now, this is very interesting! Our simple reasoning shows that in order to minimize the distance (3.6), we have to choose $b_{i}$ in (3.5) to be conditional probabilities of $A$ given $B_{i}$. Or: the conditional probabilities $\mathbb{P}\left(A \mid B_{i}\right)$ are the coefficients in the projection of $X$ onto the linear span of $\left{1_{B_{i}}, i=1, \ldots, n\right}$. This is not obvious from the original definition at all. ## 数学代写\|随机过程作业代写Stochastic Processes代考\|Projections in Hilbert spaces 3.1.1 定义二元运算的线性空间X×X→R. 映射任何对X×X成一个标量表示(X,是), 称为酉空间或内积空间 iffX,是,和∈X，和一种,b∈R，满足以下条件： (s1)(X+是,和)=(X,和)+(是,和), (s2)(一种X,是)=一种(X,是), (s3)(X,X)≥0, (s4)(X,X)=0当且当X=0. (s5)(X,是)=(是,X). 3.1.2 示例空间l2具有标量积的平方和序列(X,是)=∑n=1∞Xn这n是一个单一的空间；这里X=(Xn)n≥1, 是=(这n)n≥1. 空间C[0,1]上的连续函数[0,1]和 3.1.3 Cauchy-Schwartz-Bunyakovski 不等式对于任意X和是在一个单一的空间里， (X,是)2≤(X,X)(是,是) F(吨)=(X,X)+2吨(X,是)+吨2(是,是); ## 数学代写\|随机过程作业代写Stochastic Processes代考\|Banach spaces 2.2.1 范数线性空间 LetX是一个线性空间。一个函数\|⋅\| : X→R,X↦\|X\|被称为规范，如果对所有人X,是∈X和一种∈R, (n1)\|X\|≥0, (n2)\|X\|=0, 当且X=θ, (n3)\|一种X\|=\|一种\|\|X\|, (n4)\|X+是\|≤\|X\|+\|是\|. 2.2.2 练习(n3)−(n4)暗示对于X,是∈X, \|\|X\|−\|是\|\|≤\|X±是\|. ## 数学代写\|随机过程作业代写Stochastic Processes代考\|Definition and existence of conditional expectation 3.2.1 动机让(Ω,F,磷)是一个概率空间。如果乙∈F是这样的磷(乙)>0那么对于任何一种∈F我们定义条件概率磷(一种∣乙)（概率一种给定乙）作为 d(是,X)=∫Ω(是−X)2d磷=\|是−X\|大号2(Ω,F,磷) ∑一世=1n∫乙一世(是−1一种)2d磷=∑一世=1n∫乙一世(b一世−1一种)2d磷 =∑一世=1n[b一世2磷(乙一世)−2b一世磷(乙一世∩一种)+磷(一种)], b一世=磷(一种∩乙一世)磷(乙一世)=磷(一种∣乙一世). ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。	4,211	10,605	{"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}	3.984375	4	CC-MAIN-2023-14	latest	en	0.497879
https://mathoverflow.net/questions/165168/japanese-theorem-on-cyclic-polygons-higher-dimensional-generalizations	1,618,927,745,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618039398307.76/warc/CC-MAIN-20210420122023-20210420152023-00002.warc.gz	482,414,548	30,529	# “Japanese Theorem” on cyclic polygons: Higher-dimensional generalizations? A beautiful theorem known as the Japanese Theorem (Wikipedia, MathWorld) says that, no matter how one triangulates a cyclic (inscribed in a circle) polygon, the sum of the radii of the incircles is the same: (Wikipedia image) Q. Does this generalize to higher dimensions? In particular, if one partitions a convex polyhedron in $\mathbb{R}^3$, all of whose vertices lie on a sphere, into tetrahedra, is the sum of the radii of the inspheres independent of the tetrahedralization? A bit of search has not resulted in an answer, which suggests that the answer may well be No... Addendum. Following TMA's suggestion, I computed the radii sum for the five-tetrahedron partition—$1.334$, and the sum for the six-tetrahedron partition—$1.242$. Barring a computation error, this settles the question negatively. Too bad! • I think the answer is no even for the unit cube (5 versus 6 decomposition), but you are in a better position to compute it than I. – The Masked Avenger May 4 '14 at 23:59 • Thanks for checking. I think there are at least two distinct ways to make a 6 part tetrahedral decomposition of the cube. If you take any statistic based on the inscribed spheres that naturally suggests itself as an invariant, say sum of surface areas, you can test it out on those 3 decompositions ( plus more suggested by prisms ) to see if you have an invariant. – The Masked Avenger May 5 '14 at 2:05 • What if you fix a triangulation of the polyhedron, or the number of tetrahedra in the triangulation? After all, in the 2-dimensional case all sides are already simplices. Of course, as The Masked Avenger suggests, another invariant you can consider is the sum of the areas. – Marco Golla May 5 '14 at 9:25 • I checked and $\sum r^2_i$ is not the same for the two partitions of the cube. – Joseph O'Rourke May 5 '14 at 10:29 • @JosephO'Rourke In a different angle, is this result used in Computer Science anywhere? – john mangual May 5 '14 at 11:54	523	2,026	{"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}	3.15625	3	CC-MAIN-2021-17	latest	en	0.917405
https://allaboutpuzzles.blogspot.com/2012/09/	1,696,183,105,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233510924.74/warc/CC-MAIN-20231001173415-20231001203415-00733.warc.gz	106,141,247	33,737	## Aeroplane..(Microsoft puzzle) Riddle: The puzzle question is : On Bagshot Island, there is an airport. The airport is the homebase of an unlimited number of identical airplanes. Each airplane has a fuel capacity to allow it to fly exactly 1/2 way around the world, along a great circle. The planes have the ability to refuel in flight without loss of speed or spillage of fuel. Though the fuel is unlimited, the island is the only source of fuel. What is the fewest number of aircraft necessary to get one plane all the way around the world assuming that all of the aircraft must return safely to the airport? How did you get to your answer? Notes: (a) Each airplane must depart and return to the same airport, and that is the only airport they can land and refuel on ground. (c) The time and fuel consumption of refueling can be ignored. (so we can also assume that one airplane can refuel more than one airplanes in air at the same time.) (d) The amount of fuel airplanes carrying can be zero as long as the other airplane is refueling these airplanes. What is the fewest number of airplanes and number of tanks of fuel needed to accomplish this work? (we only need airplane to go around the world) ## Rectangle 8 part puzzle Question: We have a rectangle It is divided in eight parts by three vertical and one horizontal line so that there are 8 chambers. Now we have numbers from 1-8 to be filled in these chambers. Rule : No two consecutive numbers must be present either side to side or diagonal Invalid situation example Given 5 at position 2 then 4 cannot occur at any of the give position. ## Minimum no. of rats needed for finding poison in bottle....google puzzle Question: There are 8 bottles, one has poison. What's the minimum number of rats you need to find the poison bottle in time T, and how? (You get the rats you need all at once, feed them all at the same time, and poison kills them after time T.) Solution: representation in binary numbers 1-0001 2-0010 3-0011 4-0100 ## prove that n(n2-1) is div by 24...amazon puzzle Question: prove that n(n2-1) is div by 24 Note: n is odd and >=3 Solution: N(N^2-1)=(N-1)N(N+1) means it must be divisible by 2 and 3 since 24=2X2X2X3 ## Solution:Weight difference puzzle....morgan stanley logical puzzle Solution: I believe the solution is as follows (assume initial configuration to be an empty bucket at the top and a bucket with the stone in it at the bottom): 1. Prince goes down, rock goes up. 2. Queen goes down, Prince goes up. 3. (the rock is dropped from the top of the tower) ## Weight difference puzzle....morgan stanley logical puzzle Question: There are 3 people on a tower which may collapse due to fire. King(78 kg),Queen(42kg) and Prince(36kg).There is a pulley on the tower with baskets tied to it on both the sides of rope around the pulley. There is a 30kg stone in one of the baskets. There can be two persons or a person and a stone or a person or a stone in the baskets keeping in mind that the weight difference is not more than 6kg else rope would break. You have to bring all the three on ground safely ## Probability of rolling a 10 and an 11 before rolling a 7? Question: What is the probability of rolling a 10 and an 11 before rolling a 7? Solution: As stated, must roll a 10 and 11 before rolling a 7 P(10)=3/36 P(11)=2/36 P(7)= 6/36 Two options: a) 10 first or b) 11 first a) P that 10 is rolled before 7 is rolled: P(10)/P(7)= 3/6 then 11 rolled before 7 is rolled: P(11)/P(7)= 2/6 ## Solution: Toggling of 100 switches.. Solution: This is a little tricky. Say switches nos are 1 to 100. Here after 1st round, all switches are ON (All were OFF, so toggling will make all ON) after 2nd round, all switches divisible by 2 (switch no divisible by 2) are TOGGLED after 3rd round, all switches divisible by 3 (switch no divisible by 3) are TOGGLED after 4th round, all switches divisible by 4 (switch no divisible by 4) are TOGGLED ....... ## Toggling of 100 switches.. Question: Suppose there are 100 lights, which are all off. First round past them, turn all on; Second round past, turn every other off; Third round, turn every third on; vice verse; Ask the 100 round past, which light will be on? ## Probability of rolling a 10 and an 11 before rolling a 7 Question: What is the probability of rolling a 10 and an 11 before rolling a 7? ## Solution: Maximum no. of matches puzzle Solution: 56 points are distributed to 8 team. In the worst case, team0 loses all the games, he gets 0 point. team1 win two games with team0 and loses all other games, he gets 2 points. In the same way, team2 gets 4 points, team3 gets 6 points. So there are 44 points left which can be distributed to the remaining 4 teams. So the assurance points for a team should be 11 points. ## Maximum no. of matches puzzle Question: Consider a series in which 8 teams are participating. each team plays twice with all other teams. 4 of them will go to the semi final.How many matches should a team win, so that it will ensure that it will go to semi finals.? ## Minimum no. of queens in chess board Question: Imagine there are infinite number of Queens (Chess Game Piece) with u. Find the minimum number of queens required so that every square grid on the chess board is under the attack of a queen. Arrange this minimum no. of Queens on a chess board. ## Solution:Amazon puzzle: find probability of men to go in room Solution: 1 - 1/2 * 1/2 * 1/2 * 1/2 => 15/16 explanation: It is the combined probability of all possible events minus the probability that all four people will NOT open the door (1/2)^4. Another way to look at this problem would be to take the sum of the probabilities of all possible ways that the door will be opened: ## Amazon puzzle: find probability of men to go in room Question: There are 4 people in a closed room and you are waiting outside to enter into the room. You can enter only when one of them opens the door. The probability that somebody will open the door is 1/2. Now what is the probability that the door will be opened so that you can go inside? ## Solution:Find average salary without knowing other`s salary Solution: Lets name these employees A, B, C and D 1. A chooses any random value and whispers it to B privately 2. B chooses any random value and whispers it to C privately 3. C chooses any random value and whispers it to D privately 4. D chooses any random value and whispers to to A privately ## Find average salary without knowing other`s salary Question: How can four employees calculate the average of their salaries without knowing other's salary? ## Solution:Distance between 2 poles...amazon puzzle Solution: The distance between poles is zero. In this case, the cable will be hanging directly down and its center be 8mts down which is half of its length. The height of the poles are 15mts, so 15mts - 8mts = 7mts. POST YOUR OPINION IF YOU HAVE BETTER SOLUTION ## Distance between 2 poles...amazon puzzle Question: There are two poles of equal height 15mts. One cable with length 16mts is hanging between that two poles. The height from center of the cable to earth is 7mts then what is the distance between that two poles. ## Solution: Arrange numbers 1-8 in 2X4 matrix with no consecutive...Adobe puzzle Solution: 1. keep no. in group of two alternate numbers. {1,3} {2,4} {5,7} {6,8} 2.now fill the center 4 block with two group containing first and last number. (1,3) {6,8} - 1 8 - - 3 6 - 3. now fill with other group. 5 1 8 4 7 3 6 2 ## Arrange numbers 1-8 in 2X4 matrix with no consecutive...Adobe puzzle Question: There is a 2X4 matrix, in this you are supposed to arrange the numbers from 1-8, so that no consecutive numbers are adjacent(vertically, horizontally and diagonally) to each other. It is possible to do if one keeps on trying it randomly but it can be done with an intelligent approach too. What would that be? ## Solution: Find no. of acute traingle from cube having 8 vertices Solution: Acute angled triangle can be formed by connecting the diagonal of three adjacent faces of the cube. The number of ways three adjacent faces can be chosen is (6x4x2)/(3x2x1) = 8 So the total number of acute angled triangle is 8. POST YOUR OPINION IF YOU HAVE BETTER SOLUTION ## Find no. of acute traingle from cube having 8 vertices Question: Think of the 8 vertices of a given cube. You are allowed to join three vertices to form a triangle. How many such unique acute triangles can you make ?? ## Solution of 6 litre from 7 & 4 litre Solution: 1. Fill 7 liter tumbler fully: 7 - 0 2. Fill 4 liter tumbler fully by pouring from the 7 liter one: 3 - 4 3. Empty the small one: 3 - 0 ## 6 litre from 7 & 4 litre Question: There are two tumblers one 7 litre tumbler, one 4 litre tumber and ample water. Get 6 liters of water in 7 litre tumbler. Name Email * Message *	2,286	8,860	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.84375	4	CC-MAIN-2023-40	latest	en	0.939728
http://www.math.niu.edu/~rusin/known-math/98/moebius	1,429,598,046,000,000,000	text/plain	crawl-data/CC-MAIN-2015-18/segments/1429246640550.27/warc/CC-MAIN-20150417045720-00004-ip-10-235-10-82.ec2.internal.warc.gz	642,305,016	3,721	From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Moebius tranform Date: 8 Jun 1998 17:58:44 GMT In article <357BEDBB.2E87@univ-tln.fr>, Langevin Philippe wrote: >What is the Moebius transform of a function from E into >Z when E is an ordered set ? > >what about the basic Properties ? I believe you're thinking of the Mobius Inversion function mu on a partially ordered set. I may be messing up some details, but the defining property is: If f: E -> Z is any function and g: E-> Z is then defined by g(e) = Sum( f(e'), e' < e ) then f may be recovered from g by the formula f(e) = Sum( mu(e',e)g(e'), e' < e ). (In order for these expressions to make sense we must assume each element of E has only finitely many predecessors.) Thus for example if E = {1, 2, 3, ...} with the usual ordering, mu(e', e) = 1 if e'=e, and =-1 if e'=e-1. If instead this same set E is partially ordered by divisibility (" e' < e " means e' \| e ) then mu(e', e) = mu(e/e'), where the second mu is the Moebius function on integers: it's 0 for integers divisible by a proper square, and otherwise is (-1)^d where d is the number of distinct prime divisors of e/e'. I seem to recall that the formula mu(e',e)=mu(e/e') is valid in more general contexts in which " e/e' " makes sense, e.g. if E is the lattice of subgroups of a p-group, ordered by inclusion. (Here mu(e', e) = 0 is e' is not normal in e.) You must be working in the context of ordered sets and lattices; see e.g. http://www.math.niu.edu/~rusin/known-math/index/06-XX.html There's a Moebius transform in complex analysis too, but that's different. (It would be a map from the Riemann sphere to itself given by a fractional linear map.) dave ============================================================================== From: hrubin@stat.purdue.edu (Herman Rubin) Newsgroups: sci.math.research Subject: Re: Moebius Inversion Date: 16 Jun 1998 14:45:34 -0500 In article <35865509.7E42@univ-tln.fr>, Langevin Philippe wrote: >Hey ! >Let m be a positive integer and let E={0,1,...,m-1}, I denote >by <= the usual order on E. >Now let n be a positive integer, E^n is a partially ordered set >with the order \| defined by : > x \| y iff x_i<=y_i for all i in [1,n] >What does one know about the Moebius function of ( E^n, \|) ? >References ? This seems too easy to need references. The definition of the Moebius function is enough, and it is a little easier to see it in generality. In fact, the USUAL Moebius function is a slight generalization of the special form you have indicated. Suppose that we have partially ordered sets A_j, all of which are rooted trees with the root designated as 0, and we consider the weak direct product of the A_j, with only a finite number of the coordinates not 0. Then the Moebius function of the product is just the product of the Moebius functions! As an example, consider the positive integers ordered by divisibility. This is the product over all primes p of the powers (starting from the 0's power) of the primes. For A_p = {p^n: n = 0, 1, ... }, we have the linear ordering Moebius function. Combine these by multiplication to get the usual one. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399 hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ============================================================================== From: nikl+sm000079@pchelwig1.mathematik.tu-muenchen.de (Gerhard Niklasch) Newsgroups: sci.math Subject: Re: Moebius tranform Date: 8 Jun 1998 17:36:58 GMT In article <357BEDBB.2E87@univ-tln.fr>, Langevin Philippe writes: \|> Hey ! Bonjour. :) \|> What is the Moebius transform of a function from E into \|> Z when E is an ordered set ? \|> \|> what about the basic Properties ? You don't really give enough context to be sure what is being meant, but here's one possibility. (I don't know it by the name of `Moebius _transform_', which is usually reserved for fractional linear trans- formations of the completed complex plane -- `Moebius inversion' would be more appropriate.) Assume that E is a partially ordered set such that, for every element x of E, the set down(x) =[def] {y in E : y <= x} is finite. The classical example is the set of positive integers, partially ordered by divisibility. Let A be any abelian group, and f any A-valued function on E. Then the following is well-defined: F[f]: E --> A, F[f](x) = sum_{y<=x} f(y) , the `summatory function' of f. The `Moebius inversion problem' is to retrieve the function f, given the function F. A more or less obvious idea is to write f(x) = sum_{y<=x} c(y,x) F(y) and try to deduce the (integer-valued) coefficients c(y,x), acting on the Z-module A in the natural way, by induction on x >= y for each fixed y. If all goes well, one can find a system of coefficients c(y,x) which doesthe trick for all F and depends on E alone (more precisely, on the structure of the interval of elements z of E which satisfy y <= z <= x, with the induced partial order -- the coefficients do not even depend on A). (Semi-technical note: The group A might be `too small' to nail down the coefficients. However, it is obvious that there exists a group which is `large enough', viz. the free abelian group on the set E, and if we can find a coefficient system which works in this case, then it will work in all cases.) Carrying this out for the standard example gives you the usual Moebius function c(y,x) = mu(x/y) -- this is arguably the most natural way to define this function, followed by a proof that it behaves in the well-known way with respect to prime factorization. (Exercise: if f is a multiplicative arithmetic function, then so is F, and vice versa.) In general, it is clear that c(y,y) has to be +1 for all y (consider functions f which are supported on the singleton x), and (for the same reason) that c(y,x) must be -1 whenever x is an immediate successor of y. One proves uniqueness first, and only afterwards shows that the unique solution is in fact a solution. When E is a simplicial set (each down(x) looks like the Boolean algebra of subsets of a finite set, ordered by set inclusion), you recover the classical `inclusion-exclusion' counting principle of combinatorics. I don't remember whether any further conditions need to be imposed on E, and am too lazy to check the details -- you'll discover them when you work out the inductive argument. :^) Enjoy, Gerhard -- Gerhard Niklasch * spam totally unwelcome * http://hasse.mathematik.tu-muenchen.de/~nikl/ ******* all browsers welcome * This .signature now fits into 3 lines and 77 columns * newsreaders welcome ==============================================================================	1,758	6,814	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.234375	3	CC-MAIN-2015-18	latest	en	0.828724
http://mathhelpforum.com/calculus/218269-surface-area-curve-about-y-axis.html	1,527,331,877,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794867416.82/warc/CC-MAIN-20180526092847-20180526112847-00359.warc.gz	184,585,416	10,922	1. ## Surface area of a curve about the y axis the equation is . I found that x=sqrt(1-y) and that x'=-1/(2sqrt(y-1)). I've been using integral(2pixsqrt(1+x'^2)) but i keep just getting mixed up with my calculations. help would be appreciated. 2. ## Re: Surface area of a curve about the y axis If you rotate this function about the y-axis, each cross-section parallel to the x-axis will be a circle. The circumference of each circle is \displaystyle \displaystyle \begin{align} 2\pi r = 2\pi x = 2\pi \, \sqrt{ 1 - y } \end{align}, and if you add up all these circumferences over \displaystyle \displaystyle \begin{align} 0 \leq y \leq 1 \end{align} then you will get the total surface area. So \displaystyle \displaystyle \begin{align} SA &= \int_0^1{2\pi \, \sqrt{ 1 - y} \, dy} \end{align} 3. ## Re: Surface area of a curve about the y axis so i used that equation and my end result was (4pi)/3, but its apparently incorrect. did i do something wrong? 5. ## Re: Surface area of a curve about the y axis I put 2pi out the front and integrated sqrt(1-y) and got -2/3 (1-y)^3/2. when i subbed in 1 I got 0 and when i subbed in 1 i got -2/3, and 0 - -2/3= 2/3. and then multiplied by 2pi 6. ## Re: Surface area of a curve about the y axis I agree with your answer and solution, so I'm not sure what's gone wrong. 7. ## Re: Surface area of a curve about the y axis when you wrote y between 0 and 1, did you actually mean x perhaps cause that is what the question asks. its the only thing i can think of for it not too work	478	1,545	{"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}	4.03125	4	CC-MAIN-2018-22	latest	en	0.891441
https://proofwiki.org/wiki/Inscribing_Regular_Hexagon_in_Circle/Porism	1,679,440,893,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00128.warc.gz	533,632,959	11,508	# Inscribing Regular Hexagon in Circle/Porism ## Porism to Inscribing Regular Hexagon in Circle In the words of Euclid: From this it is manifest that the side of the hexagon is equal to the radius of the circle. And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle an equilateral and equiangular hexagon in conformity with what was explained in the case of the pentagon. And further by means to those explained in the case of the pentagon we can both inscribe a circle in a given hexagon and circumscribe one about it. ## Proof In the same way as for the regular pentagon, we can draw tangents to the circle at the vertices of the regular hexagon. This will draw a regular hexagon which has been circumscribed about the circle. Further, in a similar way to methods used for the regular pentagon, a circle can be inscribed in a regular hexagon and circumscribed about a regular hexagon. $\blacksquare$	239	1,034	{"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}	3.6875	4	CC-MAIN-2023-14	latest	en	0.931049
https://edurev.in/p/123546/NCERT-Solutions-Class-12-Maths-Chapter-2-Inverse-Trigonometric-Functions	1,726,213,766,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00681.warc.gz	211,735,210	65,799	NCERT Solutions Exercise 2.1: Inverse Trigonometric Functions NCERT Solutions Class 12 Maths Chapter 2 - Inverse Trigonometric Functions ``` Page 1 Class XII Chapter 2 – Inverse Trigonometric Functions Maths Page 1 of 38 Exercise 2.1 Question 1: Find the principal value of Let sin -1 Then sin y = We know that the range of the principal value branch of sin -1 is and sin Therefore, the principal value of Question 2: Find the principal value of We know that the range of the principal value branch of cos -1 is . Therefore, the principal value of . Question 3: Find the principal value of cosec -1 (2) www.ncrtsolutions.blogspot.com Page 2 Class XII Chapter 2 – Inverse Trigonometric Functions Maths Page 1 of 38 Exercise 2.1 Question 1: Find the principal value of Let sin -1 Then sin y = We know that the range of the principal value branch of sin -1 is and sin Therefore, the principal value of Question 2: Find the principal value of We know that the range of the principal value branch of cos -1 is . Therefore, the principal value of . Question 3: Find the principal value of cosec -1 (2) www.ncrtsolutions.blogspot.com Class XII Chapter 2 – Inverse Trigonometric Functions Maths Page 2 of 38 Let cosec -1 (2) = y. Then, We know that the range of the principal value branch of cosec -1 is Therefore, the principal value of Question 4: Find the principal value of We know that the range of the principal value branch of tan -1 is Therefore, the principal value of Question 5: Find the principal value of We know that the range of the principal value branch of cos -1 is Therefore, the principal value of ``` Mathematics (Maths) for JEE Main & Advanced 209 videos\|443 docs\|143 tests FAQs on NCERT Solutions Class 12 Maths Chapter 2 - Inverse Trigonometric Functions 1. What are inverse trigonometric functions? Ans. Inverse trigonometric functions are mathematical functions that help find the angle or angles when the trigonometric ratio is known. They are the inverse operations of trigonometric functions and are denoted by sin^(-1), cos^(-1), tan^(-1), etc. 2. How do inverse trigonometric functions work? Ans. Inverse trigonometric functions work by taking the ratio of a side length in a right triangle to its hypotenuse and determining the corresponding angle. They help in finding the angle when the trigonometric ratio is given, making them useful in solving various mathematical problems. 3. What are the ranges and domains of inverse trigonometric functions? Ans. The ranges and domains of inverse trigonometric functions depend on the specific function. For example, the domain of sin^(-1)(x) is [-1, 1], and the range is [-π/2, π/2]. Similarly, the domain of cos^(-1)(x) is [-1, 1], and the range is [0, π]. 4. How are inverse trigonometric functions used in real-life applications? Ans. Inverse trigonometric functions have numerous real-life applications. They are used in fields such as engineering, physics, and navigation to solve problems involving angles and distances. For example, they can be used to calculate the angle of elevation or depression in surveying or to determine the distance between two points using triangulation. 5. What are some important properties of inverse trigonometric functions? Ans. Some important properties of inverse trigonometric functions include their restricted domains and ranges, their periodicity, and their relationships with trigonometric functions. These properties are essential in understanding and applying inverse trigonometric functions in various mathematical contexts. Mathematics (Maths) for JEE Main & Advanced 209 videos\|443 docs\|143 tests Up next Explore Courses for JEE exam Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests. 10M+ students study on EduRev Related Searches , , , , , , , , , , , , , , , , , , , , , ;	967	3,892	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.4375	4	CC-MAIN-2024-38	latest	en	0.82944
http://convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=register+ton&To=section+modulus	1,569,217,117,000,000,000	text/html	crawl-data/CC-MAIN-2019-39/segments/1568514576047.85/warc/CC-MAIN-20190923043830-20190923065830-00114.warc.gz	45,064,490	3,891	Partner with ConvertIt.com New Online Book! Handbook of Mathematical Functions (AMS55) Conversion & Calculation Home >> Measurement Conversion Measurement Converter Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC. Conversion Result: ```register ton = 2.8316846592 volume (volume) ``` Related Measurements: Try converting from "register ton" to acre foot, barrel, cc (cubic centimeters), coomb, cord foot (of wood), cup, dry quart, freight ton, gallon, methuselah, oil arroba (Spanish oil arroba), oz fluid (fluid ounce), petroleum barrel, rehoboam, seam, teaspoon, timber foot, vedro (Russian vedro), wey, wine arroba (Spanish wine arroba), or any combination of units which equate to "length cubed" and represent capacity, section modulus, static moment of area, or volume. Sample Conversions: register ton = .00229568 acre-foot, 1,200 board foot, .78125 cord (of wood), 766,005.19 dram fluid (fluid dram), 83.12 firkin, 935.06 jeroboam, 63,833.77 jigger, 45,960,311.69 minim, 23,937.66 noggin, 321.43 peck (dry peck), 2,992.21 quart (fluid quart), 26.79 sack, 249.35 salmanazar, 2.83 stere, 40.18 strike, 191,501.3 tablespoon, 15,958.44 tea cup, 311.44 UK peck (British peck), 4,983.07 UK pint (British pint), 2,491.53 UK quart (British quart). Feedback, suggestions, or additional measurement definitions? Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks!	476	1,645	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39	longest	en	0.687831
https://m.csmonitor.com/1998/0709/070998.feat.feat.4.html	1,527,304,937,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00618.warc.gz	611,024,587	9,012	Close X # New Number Symbols For a New Millennium Old number challenge (Jim Henry) Come up with a completely new set of symbols for the numbers from 0 to 9. Three winning answers are pictured. Bill Hasek's 1 has one stroke, 2 has two strokes, 3 has three strokes, 4 has four strokes, 5 resembles an F for five, and the rest are curvy, including an upside-down empty bowl for 0. Michael Marcotty's, although similar to our current number symbols, have a neat feature: The symbol for 1 has one angle to it, the symbol for 2 has two angles to it, and so on. Marcotty further proposes replacing the decimal point by the European decimal comma, to avoid confusion with his new raised dot for zero. Dave Rossum's "clock" numerals are original, systematic, and easy to write, although they could be confused, especially on dice or other objects that might be upside down. (Erik Randolph mentions the trouble his young son had confusing M and W, for example.) Incidentally, John Robertson mentions a disadvantage of the English pronunciation of letters as compared with numerals: One often needs to say "b as in boy," but not "9 as in...." New challenge (Aubrey Dunne, Michael Marcotty, and Dave Rossum) Census taker: How old are your three daughters? Mrs. S: The product of their ages is 36, and the sum of their ages is the address on our door here. Census taker: I'm good at math, but I cannot tell. Mrs. S: My eldest daughter has red hair. Census taker: Oh thanks, now I know. Can you figure out how old the three daughters are?	370	1,537	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.265625	3	CC-MAIN-2018-22	latest	en	0.969195
https://www.theengineeringprojects.com/2024/07/total-uncertainty-in-the-measurements.html	1,721,548,374,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763517648.65/warc/CC-MAIN-20240721060614-20240721090614-00576.warc.gz	864,069,860	34,958	Hi friends, I hope you are all well. In this article, we can discuss the uncertainty in the measurements which can be measured. In the era of modern science and technology or modern physics, scientists can measure complex quantities and these measurements are not precise and accurate somehow doubt is present in these measurements, these doubts are suspicious known as uncertainty in the measurements. In physics or other fields of technology and engineering measurement is essential to measure or understand the quantity of a material or an object. Because every measurement is correct there are always some doubts or doubtful digits and they are called uncertainty in the measurements. Now in this article, we can explore the history, definition, quantifying methods, and different techniques that can be used to minimize uncertainty and also explore their applications and significance in different fields of engineering and physics. ## History: All substantial National measurement institutes can research the uncertainty in the measurements and give detailed documents about the measurement which is known as GUM and stands for the “Guide of Uncertainty in the Measurements.” this document gives the details about the uncertainties in the measurements. In metrology measurements when we can take measurements of the object many times it is confirmed that somehow measurements are not correct and precise. Or the doubtful measurements are termed uncertainty. All the measurements are not always correct because the measurement results depend upon the instrument's efficiency and the skilled person who can take measurements That is why uncertainty comes into measurements due to many various factors that may depend upon the environmental factors also. Measurements are essential to determine the quantity of the physical quantity or any objects. Measurements also play a vital role at the economic level. The quality of laboratories can also determined by their calibration results because the exact measurements help to understand the quantity of the object. So that's why the ASME which stands for the American Society of Mechanical Engineers can present different standards for the uncertainty in the measurements. According to their standards, the measurements are done and engineers and scientists in laboratories can also measure the quality of different measurements. ## What is the uncertainty in the measurements: ### Definition: Uncertainty in the measurements is defined as: "The measurement which can we measured have lack of certainty and they have a great difference between the true value and the measurement value which be measured." In simple words, the measurements we can measure have some doubts and their results do not according to the expectations and lack sureness are termed as uncertainty in the measurements. Uncertainty in the measurements is common because not every measurement is accurate or precise. For instance, we can measure the length which is about 6.7cm with a meter ruler but the true value range is about 6.62 or 6.75cm so the uncertainty in this measurement is approximately 0.05cm. Another example is if we measure the height of an object and the measurement which we can measure is 5.5m but the true value range is 5.3m or 5.7m so in this measurement the uncertainty is approximately equal to 0.02m. ## Types of uncertainty in the measurements: There are two major types of uncertainty in measurements which are given below: • Type A uncertainty • Type B uncertainty ## Type A uncertainty in measurements: Type A uncertainty is defined as: "The uncertainty measurement which can be evaluated through the different methods of statical analysis are known as type A uncertainty measurements." Generally, in type A uncertainty measurements we can measure or collect different data about the measurement and then observe the series of collected data, and then evaluate the uncertainty which are present in these measurements. For example, we can take measurements of an object many times or maybe 20 times and then evaluate and observe the results of these measurements and then analyze the uncertainty in these measurements. from empirical data, we can directly identify or determine the type A uncertainty in the measurements. Another example is when we want precise and accurate measurements then we can measure the same measurements many times like if we can measure the length and the measurements range between 5.7cm to 14cm and in between different measurements occur after these measurements we can observe or estimate the average uncertainty. ## Type A uncertainty evaluating methods: Type A uncertainty can be measured by repeated measurements of the measuring object and evaluated through statistical methods or techniques. Some statistical methods that are included in the evaluation of type A uncertainty are given below: • Confidence intervals • Arithmetic means • Standard error of the mean • Degree of freedom • Standard deviation ### Confidence intervals: Confidence interval is defined as: "The true and standard values that can be measured through the measurement of quantity, and confidence interval convey the range of accuracy with a confidence level of the true measured value." ### Formula: This formula can be used for a normal distribution which has a 95% confidence interval. The formula is given there: μ z SEM μ z SEM There, z is for the confidence level which can we desire and it is approximately equal to 1.96 for the normal distribution with a confidence level of 95% ### Arithmetic mean: Arithmetic means is defined as: “The set of measured numbers, the all measured numbers are added and dived by the total numbers which are present in the set and the central number which is present in the set also added with the all measured numbers”. To know an average of all measured measurements we can use the arithmetic formula because, through this statistical technique, we can evaluate the type A uncertainty in the measured sets. ### Formula: The formula which is used to calculate the average of the measured set is given there: x = 1n i=1nxi ### Standard error of the mean: The standard error of the mean is defined as: “ the uncertainty which is present in the average that can be calculated from the set of measurements with no of measurements and the standard deviation, these estimate can be conveyed by the standard error of the mean.” ### Formula: The formulas that are used for calculating the standard error of the mean are given there: SEM= σN there, SEM = standard error of the mean σ= standard deviation N = no of measurements ### Standard deviation: Standard deviation can be defined as: “The average can be measured from the dispersion of the set of collected measurements.” Standard deviation can be used to measure the average variance which can be essential for evaluating the Type A uncertainty. ### Formula: The formula that can be used for the standard deviation or measure the average variances is given there: σ= 1n-1i=1n(xi-x)2 ### Degree of freedom: The degree of freedom can be defined as: After the calculation of the standard deviation and the average the final numbers can be calculated freely for statistical analysis and the final values can help to understand the type A uncertainty and the degree of freedom. ## Formula: The formula that is used for the degree of freedom is given there: v= n - 1 ## Sources of Type A uncertainty in the measurements: The main causes and sources of the Type A uncertainty in the measurements are given there: • Environmental changes • Human factors • Instrumental fluctuations ### Methods used to reduce type A uncertainty: The methods that can reduce the uncertainty in the measurements are given below: • Repeated numbers of measurements • Control environmental factors • Improving measurements techniques ## Type B uncertainty in the measurements: Type b uncertainty can be defined as: "the uncertainty which can be evaluated by using different methods except the statistical analysis of measurements. Type B uncertainty can't be evaluated through statistical analysis they can be evaluated through calibration certificates, scientists' judgment, and through the publishers." Type B uncertainty can be measured differently from Type A uncertainty in the measurements because it is mostly evaluated through the collected information and through the publishers. This type of error is also common the main sources and the causes of these uncertainties can be explained below. ## Type B uncertainty measurement evaluating methods: The methods that can be used to evaluate the type B uncertainties in the measurements are given below: • Expert judgment • Manufacturer specification • Theoretical analysis • Reviews of calibration certificates • Reference material data • Chats of collected data information ### Expert judgment: When we can do measurements but do have not direct measurements and data then the uncertainty measurement is provided to the experts who have experienced and understand the limitations and uncertainties of the measurements. Then the judges understand the uncertainty and then identify the type of uncertainty and try to reduce these uncertainties. Some examples are given below: • The scientists who can do experiments and want to change into theory the council and the judges understand the experiment according to their experience and then allow them. • A well-experienced scientist or metrologist can measure the uncertainty in the measurements through their experience and knowledge. ### Calibration certificates: The type B uncertainty can be evaluated by using the calibration certificate because the calibration certificates convey information and details about the accuracy and precision of the measuring instruments. Calibration certificates also provide information about the correction of uncertainty in the measurements. #### Example: The uncertainty of the voltmeter is 0.05 provided in the calibration certificate and how to recover this uncertainty information is also present in it. ### Reference material data: The uncertain information and documents are provided through the reference material data. The information and values that are provided through these reference data help to improve the uncertainty in the measurements that can be calculated. #### Example: The certified uncertainty of the gas analyzer is approximately about 0.1% and the reference data is provided to minimize the uncertainty in the measurements. ### Manufactures specification: The instrument's specification accuracy, precision, and limits can only defined by the manufacturers because they understand the nature of their instruments and they also determine or estimate the uncertainty that can produced by their instrument during the measurements. #### Example: For the dimensional measurements the instruments we can use a digital micrometer and their accuracy is about 0.002 it can also budget the uncertainty measurement in it. ### Theoretical analysis: There are different theoretical models are present that can convey detailed information about the uncertainties in the measurements. Because these models are based on the assumptions and the practical experiences. By using these models we can also estimate and identify the uncertainties in the measurements which can be measured. #### Example: We can estimate the uncertainty and the precision in the vacuum of the speed of light and we can measure these uncertainties that are based on precision and we can calculate them. ## Sources which can cause type B uncertainty: The sources which can use the uncertainty in the measurements are given below: • Environmental conditions • Previous measurements • Manufactures specification • Theoretical models • Instrumental calibration • Previous measurements ## The combining component of Type B uncertainty: In the combination of different type b components, we can use the root sum square method to estimate and calculate the uncertainty in the measurements. The all components that are combined are independent but we can combine them to estimate the uncertainty in the measurements precisely. ## Formula: uc(y) = i=1nciu(xi)2 ## Methods to minimize the type B uncertainty: We can reduce the type B uncertainty in the measurements if we can follow these given steps. Because it can help to reduce tp understanding of the uncertainties in the measurements. The methods and the steps are given below: • Improved calibration • Used high-quality reference data • Enhanced environmental controls. ## Sources of uncertainty in the measurements: Generally, uncertainty in the measurements occurs due to many sources but the major two are • Random error • Systematic error The sources and details of these errors are given below: ## Random error: Random errors are common because they can caused by many different sources and they may be reduced by doing repeated measurements and by estimating the main cause of error. Some major sources which can cause this error are given there: • Observer variability • Environmental noise • Instrumental fluctuations ## Systematic error: This error occurs due to the imperfect instruments usage and the unskilled persons who can take measurements but the main sources and causes of systematic error are given there: • Methodological error • Instrumental error • Observer error • Environmental factors ## Techniques and methods to reduce uncertainty: The techniques and some methods that are used to reduce the uncertainty are given there: • Replicates and repeat measurements • Randomization • Calibration and standardization • Improved experimental designs • Control variables To reduce the uncertainty in the measurements we can use many different advanced techniques some are given there: • Error-correcting algorithm • Automated data collection • High precision instruments • Skilled persons ### In scientific research the significance of uncertainty: In scientific research the measurement of uncertainty is essential and it is also essential to reduce it because the scientists try to make precise and accurate measurements according to the calibration certificates, the significance of uncertainty in scientific research is given there: • Reproducibility • Peer review • Validation • Transparent reporting ## Application of uncertainty in the management: In the field of modern technology and engineering, in measurements uncertainty and error are common but by using different techniques we can reduce them. Some applications of uncertainty in the management are given there: • Medical and biological research • Pollution monitoring • Quality control • Climate modeling • Safety standards • Drug efficacy • Diagnostic accuracy ## Conclusion: The uncertainty in the measurements is common but in modern science and technology or different fields of science and physics, we can reduce the uncertainty in the measurements using many different techniques because scientists and engineers want to measure the precise and accurate measured values. Because the experts can agree on the measurements which are according to the standard values of the calibration certificates. Because the national measuring institutes and the American Society of Measurements can present the standard suits for measurements that are used to reduce or estimate the uncertainty in the measurements.	2,776	15,549	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.609375	3	CC-MAIN-2024-30	latest	en	0.94756
https://science.howstuffworks.com/nature/climate-weather/atmospheric/farmers-almanac-predict-weather.htm	1,716,680,504,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058858.81/warc/CC-MAIN-20240525222734-20240526012734-00209.warc.gz	433,108,293	47,728	How does the Farmers' Almanac predict weather? By: Kate Kershner \| As anyone who's ever looked at a forecast can tell you, you can't predict the weather too far out. Just ask Punxsutawney Phil, the groundhog that everyone in the United States must blindly salute and take orders from every Feb. 2. All poor Phil has to do is tell us whether there will be six more weeks of winter -- which isn't really even much of a forecast -- and he still manages to get it wrong more than 60 percent of the time (although, to be fair, he is speaking in a complicated groundhogese dialect that a translator might have trouble interpreting) [source: Neuman Punxsutawney]. While major news organizations might cheerfully report a farmer's almanac prediction, their doing so shouldn't lend much weight to the claims. But before we thoroughly confuse you, let's get one thing straight: There is both a Farmers' Almanac and an Old Farmer's Almanac. While they clearly have strong contradictory feelings about apostrophe usage, they can agree on one thing: Weather forecasting should be secret science, emphasis way more on "secret" than "science." The Farmers' Almanac was founded in 1812, and it is indeed the baby compared to the Old Farmer's Almanac and its 1792 origin. They both advertise that they rely on secret formulas to come up with their prognostications. The Farmers' Almanac uses "mathematical and astronomical" formulas, which are passed along from one (anonymous) prognosticator to another -- and only one at a time [source: Farmers' Almanac]. The Old Farmer's Almanac relies on a theory that weather is a result of magnetic storms on the sun's surface, and the forecasts are predicted based on a formula literally locked in a black box at headquarters [source: Old Farmer's Almanac]. Now let's get to the point: While they both claim to have a proprietary pattern that predicts the weather, that's different than actually predicting the weather. And although both claim great results (80 percent accuracy, in fact), meteorologists aren't buying it [source: Neuman]. And that really can't be too surprising, considering they're attempting to predict weather a year out -- if not more, due to publication deadlines. The National Oceanic and Atmospheric Administration's Climate Prediction Center, in fact, will only take a prediction out three months before acknowledging their predictions' lessened reliability [source: Palermo]. So you can't predict the future, or the weather. Predicting the future weather -- as both almanacs promise -- is a losing battle. How accurate is the Farmers Almanac in predicting weather? The Farmers Almanac is about 80% accurate in predicting weather. Sources • Farmers' Almanac. "How Does the Farmers' Almanac Predict the Weather?" 2015. (Jan. 2, 2015) http://farmersalmanac.com/farmers-almanac-forecasts/ • Fine Gardening. "How Do They Create The Farmers' Almanac?" (Jan. 2, 2015) http://www.finegardening.com/how-do-they-create-farmers-almanac • Holthaus, Eric. "Don't Waste Your Money on the Farmers' Almanac." Slate. Sept. 9, 2014. (Jan. 2, 2015) http://www.slate.com/blogs/future_tense/2014/09/09/farmer_s_almanac_old_farmers_almanac_have_terrible_meteorological_track.html • Neuman, Scott. "Decoding the Allure of the Almanac." National Public Radio. March 2, 2012. (Jan. 2, 2015) http://www.npr.org/2012/03/02/147810046/decoding-the-allure-of-the-almanac • Neuman, Scott. "Punxsutawney Phil vs. The Farmers' Almanac." National Public Radio. Feb. 1, 2014. (Jan. 2, 2015) http://www.npr.org/blogs/thetwo-way/2014/02/01/269999572/punxsutawney-phil-vs-the-farmers-almanac-who-do-you-trust • Old Farmer's Almanac. "How We Predict the Weather." 2015. (Jan. 2, 2015). http://www.almanac.com/content/how-we-predict-weather • Palermo, Elizabeth. "Brutal Winter?" Live Science. Aug. 27, 2014. (Jan. 2, 2015) http://www.livescience.com/47575-farmers-almanac-winter-weather-forecast.html	1,014	3,922	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.625	3	CC-MAIN-2024-22	latest	en	0.958297
https://teamtreehouse.com/community/help-please-31	1,722,976,625,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640508059.30/warc/CC-MAIN-20240806192936-20240806222936-00590.warc.gz	463,537,794	36,369	## Welcome to the Treehouse Community Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here. ### Looking to learn something new? Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today. Given the height and width of a wall in inches, calculate the area in feet and assign it to a constant named areaInFeet. // the height and width of a wall let height = 120.0 // in inches let width = 144.0 // in inches let areaInFeet = 22 You need to make sure you divide the length and width by 12 before setting the constant because those are in inches, while the constant is in feet. ```let height = 120.0 / 12 let width = 144.0 / 12 ``` also, ```let areaInFeet = 22 ``` Make sure you use a variable * a variable, not 22. let height = 120.0 / 12 let width = 144.0 / 12 let areaInFeet = height * width works for me This is the best answer, however, statements on one line should be separated by a ;	320	1,308	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33	latest	en	0.901527
http://www.ricardmarxer.com/loudia/doc/classIFFTComplex.html	1,555,634,893,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578526923.39/warc/CC-MAIN-20190419001419-20190419023419-00278.warc.gz	314,363,787	2,841	# IFFTComplex Class Reference Algorithm to perform an Inverse Fast Fourier Transform of a vector of Complex values representing the full FFT. More... `#include <IFFTComplex.h>` List of all members. ## Public Member Functions int fftSize () const void setFftSize (int size, bool callSetup=true) int frameSize () const void setFrameSize (int size, bool callSetup=true) bool zeroPhase () const void setZeroPhase (bool zeroPhase, bool callSetup=true) ## Detailed Description Algorithm to perform an Inverse Fast Fourier Transform of a vector of Complex values representing the full FFT. The IFFT is a fast implementation of an Inverse Discrete Fourier Transform (IDFT). The algorithm takes as input M point vectors of Complex values (M being the FFT size), and returns N point vectors of Real values (N being the frame size). Note that N can be smaller than M. In this case the last ( M - N ) coefficients will be discarded, since it assumes that zero padding has been made at the end of the frame prior to the forward FFT transfor. Alternatively the algorithm can undo the center zeropadding and the N/2 rotation if done durnig the FFT forward transform. This is specified by using the setZeroPhase() method. See also: FFT, FFTComplex, IFFT ## Member Function Documentation int fftSize ( ) const Returns the size of the FFT to be processed. The default is 1024. See also: setFftSize() void setFftSize ( int size, bool callSetup = `true` ) Specifies the size of the FFT to be processed. The given size must be higher than 0. Note that if size is a power of 2 will perform faster. See also: fftSize() int frameSize ( ) const Returns the size of the target frame. The default is 1024. See also: setFrameSize() void setFrameSize ( int size, bool callSetup = `true` ) Specifies the size of the target frame. The given size must be higher than 0. Note that if size is a power of 2 will perform faster. See also: frameSize() bool zeroPhase ( ) const Returns the zero phase setting. The default is true. See also: setZeroPhase() void setZeroPhase ( bool zeroPhase, bool callSetup = `true` ) Specifies the zeroPhase setting. See also: zeroPhase() The documentation for this class was generated from the following files: Generated on Tue Mar 31 20:38:33 2009 for Loudia by 1.5.6	555	2,304	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18	latest	en	0.693735
https://legal-two.com/which-thermodynamic-law-says-that-you-cannot/	1,679,504,927,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296943845.78/warc/CC-MAIN-20230322145537-20230322175537-00189.warc.gz	421,133,895	26,991	Which Thermodynamic Law Says That You Cannot? The second law of thermodynamics says that you cannot get something for nothing. You have to put energy in to get energy out. Checkout this video: What is the thermodynamic law? The thermodynamic law is an important law of physics that says you cannot create or destroy energy. Energy is conserved. This law is also known as the law of conservation of energy. It says that energy can be transformed from one form to another, but it cannot be created or destroyed. The thermodynamic law is a very important law because it helps us understand how the universe works. It helps us understand why things happen the way they do, and how we can use energy to our advantage. For example, the law of conservation of energy says that when you ride a bike, the energy you put into pedaling is converted into kinetic energy, which makes the wheels go around. But the total amount of energy in the universe stays the same. The thermodynamic law also explains why hot things cool down and cold things heat up. It’s because when you transfer heat from a hot object to a cold object, the total amount of energy in the universe stays the same, but now it’s spread out more evenly. The law of conservation of energy is a very important law, and it’s one of the laws that governs how our universe works. What does the law say? The law in question is the second law of thermodynamics, and it says that you cannot create or destroy energy. Energy can be converted from one form to another, but it cannot be created or destroyed. As such, it is impossible to create a perpetual motion machine – a machine that could run forever without any outside input of energy. Why is the law important? Thermodynamic laws are important because they help physicists, engineers, and chemists to understand and predict the behavior of systems that involve energy transfers. The first law of thermodynamics, also known as the law of energy conservation, states that energy can neither be created nor destroyed. In other words, the total amount of energy in the universe is constant. What are some real-world examples of the law in action? One of the most famous examples of the law in action is the so-called “ perpetual motion” machine. This is a device that appears to be able to do work forever without any external energy source. The laws of thermodynamics say that this is impossible, and so any machine that claims to be perpetual motion is always revealed as a fraud. What happens if you violate the law? In thermodynamics, you cannot break the laws. However, if you were to violate one of the laws, the consequences would be dire. For example, if you were to break the law of conservation of energy, you would be violating one of the most fundamental laws of physics. The consequences of this could be catastrophic, and could potentially lead to the destruction of our universe. Are there any exceptions to the law? The laws of thermodynamics are some of the most fundamental and far-reaching laws in all of science. They govern everything from the behavior of subatomic particles to the evolution of the universe as a whole. The first law, also known as the law of energy conservation, states that energy can neither be created nor destroyed. The second law says that entropy (a measure of disorder) always increases over time. And the third law states that it is impossible to reach absolute zero, the coldest possible temperature. In spite of their simplicity and power, the laws of thermodynamics do have a few exceptions. For example, the second law does not apply to closed systems in equilibrium, like a glass of water sitting on a table. In such a system, entropy may remain constant or even decrease over time as molecules move around and find their lowest energy state. The second law also does not apply to isolated systems, like our universe as a whole. If the universe were truly isolated, then its entropy would remain constant forever (or at least for trillions upon quadrillions of years). The second law of thermodynamics says that you cannot. This law is also known as the law of entropy. It is a statement about the universe as a whole, not just about you or your surroundings. The universe is constantly moving from a state of low entropy (order) to high entropy (disorder). You can think of it like a clock ticking. Entropy always increases over time, and it can never be undone. First, by understanding that the universe is constantly moving towards disorder, you can appreciate the importance of order in your own life. Second, by realizing that entropy is inevitable, you can learn to accept change and embrace impermanence. Third, understanding the second law of thermodynamics can help you make better decisions in life. For example, if you want to live a long and healthy life, it makes sense to eat healthy food and exercise regularly – these activities will help reduce your entropy and slow down the aging process. What are some common misconceptions about the law? There are some common misconceptions about the law, which can be summarized as follows: -The law prohibits you from doing work. -The law says that you cannot do work on a system. -The law says that you cannot create energy. -The law applies only to closed systems. What is the history of the law? In the early 1800s, French physicist Nicolas Léonard Sadi Carnot realized that there was a theoretical maximum efficiency that could be achieved by any heat engine. He developed what is now known as the Carnot cycle, which helped to establish the foundation of thermodynamics. In 1824, English physicist and mathematician Peter Guthrie Tait expanded on Carnot’s work and pointed out that there were actually two types of heat: “Orderly” and “Disorderly.” Tait’s work led to the development of the concept of entropy, which is key to understanding the second law of thermodynamics. The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed. The second law of thermodynamics says that entropy ( disorder) will always increase over time. This means that things will tend to move from a state of order to a state of disorder. The second law is often summed up with the phrase “you can’t win,” because it means that it is not possible to convert all of the heat from a system into useful work. Some heat will always be lost to entropy, making it impossible to achieve 100% efficiency. How does the law relate to other laws of thermodynamics? The law of conservation of energy says that you cannot create or destroy energy. The first law of thermodynamics says that you cannot create or destroy matter. The second law of thermodynamics says that you cannot create or destroy entropy. Scroll to Top	1,378	6,810	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.765625	3	CC-MAIN-2023-14	latest	en	0.956986
http://www.ehow.co.uk/how_12266678_use-quadratic-equations-solve-sides-rectangle.html	1,490,614,200,000,000,000	text/html	crawl-data/CC-MAIN-2017-13/segments/1490218189471.55/warc/CC-MAIN-20170322212949-00107-ip-10-233-31-227.ec2.internal.warc.gz	485,450,694	17,239	# How to use quadratic equations to solve for the sides of a rectangle Written by laurel storm • Share • Tweet • Share • Pin • Email If you are given the area of a rectangle, as well as some information about its perimeter or sides, you can derive a simple quadratic equation based on the rectangle's geometric properties. You can then solve this quadratic equation to obtain the rectangle's two sides. The exact process varies depending on whether you are given the rectangle's perimeter, the difference between its length and width, or its length and width expressed as a linear function of x. Skill level: Moderately Easy ## Given the length and width as a linear function of x 1. 1 Call the rectangle's length "l," its width "w." and its area "a." Because the area of a rectangle is equal to its length multiplied by its width, you can create a quadratic equation using the information you are given. For example, if: l = 2x + 4 w = 2x + 1 a = 40 Then: (2x + 4) * (2x + 1) = 40 4x^2 + 8x + 2x + 4 = 40 4x^2 + 10x - 36 = 0 2x^2 + 5x - 18 = 0 2. 2 Solve the equation for x. For example: x = (-5 +/- sqrt(25 +144)) / 4 x = (-5 +/- 13) / 4 x = 8 / 4 OR x = -18 / 4 x = 2 OR x = -4.5 3. 3 Insert the value of x in the original formulas you were given to obtain the length and width of the rectangle. For example: For x = 2, l = (2 * 2) + 4 = 8 and w = (2 * 2) +1 = 5. For x = -4.5, l = (2 * -4.5) + 4 = -5 and w = (2 * -4.5) + 1 = -8. If a particular value of x results in a negative value for l or w, as in this case, discard that solution. Multiply the resulting length by the resulting width to check your work. ## Given either the perimeter or the difference between length and width 1. 1 Call the two sides of the rectangle "x" and "y." If you have been given the value of y in terms of x, skip to the next step. If you have been given the rectangle's perimeter, use the fact that the perimeter of a rectangle is twice the sum of its sides to derive the value of y in terms of x. For example, assuming a rectangle with a perimeter of 26: 2 * (x + y) = 26 x + y = 13 y = 13 - x 2. 2 Replace y with its value in terms of x in the rectangle area formula to obtain the quadratic equation you need to solve. For example, assuming the rectangle in question has an area of 40: x * y = 40 x * (13 - x) = 40 13x - x^2 = 40 13x - x^2 - 40 = 0 x^2 - 13x + 40 = 0 3. 3 Solve the equation for x. The two solutions correspond to the two sides of the rectangle. For example: x = (13 +/- sqrt(13^2 - 4 * 40)) / 2 x = (13 +/- sqrt(169 - 160)) / 2 x = (13 +/- sqrt(9)) / 2 x = (13 +/- 3) / 2 x = (13 - 3) / 2 OR x = (13 + 3) / 2 x = 10 / 2 OR x = 16 / 2 x = 5 OR x = 8 Calculate the rectangle's perimeter and area using the two solutions and compare the results to the values you were given to check your work. ### Don't Miss #### References • All types • Articles • Slideshows • Videos ##### Sort: • Most relevant • Most popular • Most recent	926	2,961	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.65625	5	CC-MAIN-2017-13	latest	en	0.890079
https://de.zxc.wiki/wiki/Quintillion	1,632,312,447,000,000,000	text/html	crawl-data/CC-MAIN-2021-39/segments/1631780057347.80/warc/CC-MAIN-20210922102402-20210922132402-00484.warc.gz	241,479,110	6,045	# Quintillion Quintillion 1000000000000000000000000000000 presentation dual 1100 1001 1111 0010 1100 1001 1100 1101 0000 0100 0110 0111 0101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Octal 14 4762 6234 6404 3165 0000 0000 0000 0000 Duodecimal 7343 1471 0673 65A0 0000 0000 0000 Hexadecimal C 9F2C 9CD0 4675 0000 0000 0000 Morse code - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Mathematical properties sign positive parity straight Factorization ${\ displaystyle 2 ^ {30} \ cdot 5 ^ {30}}$ Divider All numbers that are neither divisible by nor by a prime number other than 2 and 5 (see running text). ${\ displaystyle 2 ^ {31}}$${\ displaystyle 5 ^ {31}}$ A quintillion (formed from French million and Latin quinque , "five") or quinquillion (neo-Latin) is a number that stands for the fifth power of a million . This equates to a million quadrillion, or 10 30 , in digits a 1 with 30 zeros: 1,000,000,000,000,000,000,000,000,000,000 The name is not the same as quintillion in US English , which corresponds to our trillion (10 18 ). In the US, our quintillion equals the nonillion .	542	1,375	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39	latest	en	0.443393
http://www.ma.utexas.edu/pipermail/maxima/2008/011228.html	1,448,618,487,000,000,000	text/html	crawl-data/CC-MAIN-2015-48/segments/1448398448506.69/warc/CC-MAIN-20151124205408-00194-ip-10-71-132-137.ec2.internal.warc.gz	543,597,095	2,603	# [Maxima] floor(log(4)/log(2)) = 1 ?? Stavros Macrakis macrakis at alum.mit.edu Sat Apr 26 23:05:08 CDT 2008 ```On Sat, Apr 26, 2008 at 11:21 PM, Robert Dodier <robert.dodier at gmail.com> wrote: > On 4/26/08, Oliver Kullmann <O.Kullmann at swansea.ac.uk> wrote: > > > floor(log(4) / log(2)); > Yeah. Me neither. This bug has been reported before; > do we know what's going on here? Is someone working on it? > For arguments which aren't obviously rational, floor calculates the argument's value at various precisions, and if the floors agree, it uses that value. This is not a reliable method when the argument is in fact exactly an integer. It would be better if it could calculate the argument using true interval arithmetic (with pessimistic rounding etc.), but Maxima cannot currently perform interval arithmetic. Of course, with that method, many cases would remain unsimplified (e.g. floor(sin(1)^2+cos(1)^2)), since the interval would typically straddle the exact result. Better that than the wrong result, but still frustrating. In general, there is no algorithm that will always give the correct result (a result related to Goedel's theorem). Barton Willis has been working on this. > Furthermore, I couldn't find any simplification rule which would yield log(4) / log(2) = 2 ?? Radcan specializes in algebraic dependencies like this, but doesn't always give the simplest result. Barton and I have talked about using radcan-like techniques for log expressions, but I don't know what the state of the code is. And of course many other cases (trig, nested radicals) wouldn't be handled by this. > What seems to me absolutely indispensable is a general "simplify"-function, which in some reasonable > way applies all simplification rules (without fear that it takes a bit > such errors etc. takes much longer :-(). There are several problems with this. Putting aside the theoretical result that it is in general impossible, there is the practical problem that "simplest" isn't well-defined. We agree that log(4)/log(2) is more complicated then 2, but which is the simplest of the following: sin(%pi/8), sqrt(sqrt(2)-1)/2^(3/4), or sqrt(1-sqrt(2)/2)/sqrt(2) ? On the other hand, as Robert says, there is lots of room for improving our documentation.... -s -------------- next part -------------- An HTML attachment was scrubbed... URL: http://www.math.utexas.edu/pipermail/maxima/attachments/20080427/c3e92530/attachment.htm ```	653	2,461	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48	latest	en	0.949259
http://www.drumtom.com/q/the-kinetic-energy-of-an-object-is-increased-by-a-factor-of-6-5-by-what-factor-is-the-magnitude-of-its-momentum-changed	1,480,832,314,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698541214.23/warc/CC-MAIN-20161202170901-00393-ip-10-31-129-80.ec2.internal.warc.gz	432,551,654	8,246	# The kinetic energy of an object is increased by a factor of 6.5, by what factor is the magnitude of its momentum changed? • The kinetic energy of an object is increased by a factor of 6.5, by what factor is the magnitude of its momentum changed? Answer to The kinetic energy of an object is increased by a factor of 7, ... , by what factor is the magnitude of its momentum changed? Positive: 47 % KE = (P²)/(2m) KE = Kinetic energy. P = Momentium. m = mass. If the momentum of the object is doubled then... KE is directly propotional to Square of the ... Positive: 44 % ### More resources Chapter 6 Work and Kinetic Energy ... then its speed must have changed. (b) ... increases its kinetic energy by a factor of 9 and (d) ... Positive: 47 % D. energy, work, distance. ... B. the kinetic energy of the object. ... C. it is free to move in such a way as to decrease its kinetic energy. Positive: 42 % The momentum of an object is increased by a factor of 4 in magnitude. By what factor is its kinetic energy changed? How it works; Positive: 28 % ... A resultant force changes the velocity of an ... energy (B) force (C) momentum (D) ... By what FACTOR was the kinetic energy increased? Positive: 5 % Anonymous18279 Create new account Discover Questions	311	1,259	{"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}	3.21875	3	CC-MAIN-2016-50	longest	en	0.941154
https://republicofsouthossetia.org/question/576-plants-are-to-be-planted-in-a-garden-in-such-a-way-that-each-row-contains-as-many-plants-as-16143042-33/	1,638,033,159,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00083.warc.gz	574,464,086	13,252	## 576 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows Question 576 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row. in progress 0 3 weeks 2021-11-08T15:06:43+00:00 2 Answers 0 views 0 ## Answers ( ) 1. Let x = number of rows and y = number of plants. (x)(y) = 576 Information provided says x is same as y thus we can state equation as the following (x)(x) = 576 or x^2 = 576 Solve for x, sqrt of 576 which is 24 x = y = 24. Therefore 24 plants in 24 rows	198	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}	3.78125	4	CC-MAIN-2021-49	latest	en	0.945532
https://www.automateexcel.com/functions/percentrank-formula-excel/	1,713,413,512,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817187.10/warc/CC-MAIN-20240418030928-20240418060928-00747.warc.gz	619,118,737	47,950	# PERCENTRANK Function Examples – Excel & Google Sheets Written by Editorial Team Reviewed by Steve Rynearson Last updated on November 9, 2023 This tutorial demonstrates how to use the PERCENTRANK Function in Excel to calculate the rank of a value in a data set as a percentage of the data set. ## What Is the PERCENTRANK Function? The PERCENTRANK Function returns the percentage rank of a value in a given range of data, inclusive of the first and last values. ## PERCENTRANK Is a “Compatibility” Function As of Excel 2010, Microsoft replaced PERCENTRANK with two variations: PERCENTRANK.INC (which is the same as PERCENTRANK) and PERCENTRANK.EXC (which returns the percent rank exclusive of the first and last values). PERCENTRANK still works, so older spreadsheets using it will continue to function as normal. However, Microsoft may discontinue the function at some point in the future, so unless you need to retain compatibility with older versions of Excel, you should use PERCENTRANK.INC or PERCENTRANK.EXC. ## How to Use the PERCENTRANK Function Use PERCENTRANK like this: ``=PERCENTRANK(C4:C13,F3,3)`` The table shows the ratings of a group of chess players. PERCENTRANK can tell us what percentage of the scores in this group are below a given value. Here’s how it works: • The first argument, C4:C13, is the data range • The second argument is the value that we want to rank • The third argument is called “significance”, and it’s simply the number of decimal places we want in the result So, we’ve asked PERCENTRANK what percentage of scores in this group are lower than 2773, and we want the answer to 3 decimal places. Since the value of 2773 actually appears in our data range, the calculation is fairly straightforward. Excel calculates it as follows: ``Percent Rank = Number of values below / (Number of values below + number of values above)`` Plugging our numbers in: ``Percent Rank = 3 / (3 + 6) = 0.333 (or 33.333%)`` ## When Values Don’t Appear in the Data Range What about if we ask for the percent rank of a value that doesn’t appear in the dataset? In such cases, PERCENTRANK first calculates the percent ranks of the two values it lies between, and then calculates an intermediate value. Here’s an example: ``=PERCENTRANK(C4:C13,F3,3)`` Now we want the percent rank of 2774, and we get 0.361, or 36.1%. Why? 2774 sits between Levon’s score of 2773 and Alexander’s score of 2774. In fact, it’s 25% of the way between the two scores, so the percent rank returned is 25% of the distance between the two scores’ percent ranks. The calculation works like this: ``Lower percent rank + (distance(Higher percent rank – lower percent rank))`` We already know Levon’s percent rank is 0.333. Alexander’s is 0.444, so plugging these in: ``0.333 + (0.25 (0.444 – 0.333)) = 0.36075`` Since we’ve set our significance to 3 in the above formula, that rounds to 0.361. ## PERCENTRANK.INC As noted earlier, Microsoft has replaced the PERCENTRANK function, so you should try not to use it wherever possible. One of its replacements, PERCENTRANK.INC, works in exactly the same way: ``=PERCENTRANK.INC(C4:C13,F3,3)`` You define the same arguments in PERCENTRANK.INC that we just saw in PERCENTRANK: • the data range • the value you want to rank • and the significance. As you can see here, it returns the same result. The “INC” part of the function’s name is short for “inclusive”. It means that the function sets the largest value in the set to 100%, the smallest to 0%, and then positions the rest of the scores at intermediate percentages between the two. These intermediate percentages are determined with the calculation: ``1 / (n – 1)`` Where n is the number of data points in the range. Since we have 10 data points, that works out to 1 / (10 – 1) = 0.111 or 11%. So the lowest score’s percent rank is 0, the next one is 11%, then 22%… and so on all the way to 100%. ## PERCENTRANK.EXC PERCENTRANK.EXC is very similar, and you use it in the same way, but it calculates the results a little differently. Use it like this: ``=PERCENTRANK.EXC(C4:C13,F3,3)`` We want the percent rank of 2773, Levon’s score again, but this time we get 0.363, or 36.3% instead of the 33% we saw earlier. Why is that? Well the “EXC” is short for “Exclusive”, meaning PERCENTRANK.EXC excludes the first and last values when calculating the percentage difference between each score. The distances between each score are calculated as follows: ``1 / (n + 1)`` Since we have 10 data points, this works out to 1 / (10 + 1) = 0.0909, just over 9%. So with PERCENTRANK.EXC, the lowest score’s percent rank is 9%, the next 18%, and so on up to 100%. Now we can see where the difference comes from – Levon’s score is the fourth highest, so his percent rank is: ``0.0909 * 4 = 0.363`` Everything else works the same way, including the equation for calculating the percent rank of score that don’t appear in the data set – Excel just uses these 9.09% intervals instead of the 11.111% intervals we saw with PERCENTRANK.INC. ## PERCENTRANK Function in Google Sheets The PERCENTRANK Function works exactly the same in Google Sheets as in Excel:	1,378	5,186	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.734375	4	CC-MAIN-2024-18	latest	en	0.795158
https://journalfeed.org/article-a-day/2019/lessons-from-schrdingers-cat-get-cozy-with-uncertainty/	1,675,057,565,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00565.warc.gz	361,814,024	34,047	# Lessons from Schrödinger’s Cat – Get Cozy with Uncertainty Written by Bo Stubblefield Spoon Feed Statistical methods were developed and applied to clinical trials to quantify uncertainty, not as a decision tool to be used in the face of uncertainty. We should become comfortable with uncertainty, as it is present irrespective of the arbitrary thresholds used for interpretation of data. Why does this matter? p-values are consistently reported in the literature as thresholds for statistical significance. Investigators draw definitive conclusions and make decisions based on arbitrary thresholds. Consequently, there is a not-so-silent revolution occurring which advocates to remove the p-value as our test for statistical significance. This recently published opinion piece uses an analogy to illustrate how we should be conducting and evaluating clinical trials. Be more like a quantum physicist – uncertainty is the world in which we live This is an opinion piece published in Circulation by cardiologist Dr. Milton Packer. Packer uses the difference between Newtonian and quantum physics to draw a parallel to research frameworks that are probabilistic and not deterministic. The thought analogy is erudite, namely Schrödinger’s Cat. Yes, we have included a “cat video” on JF, but the concept is valuable to illustrate randomized clinical trials and their reported p-values. On evaluation of a new drug in a randomized clinical trial, a researcher provides a result in probabilistic terms with a reported confidence interval that estimates the effect size of the drug. Effect size may reach statistical significance as the sample size increases, but the result may still be clinically trivial. The result of a study is only actionable if the researchers determine the degree of uncertainty they are willing to tolerate prior to the start of the study. But, these levels of tolerance are arbitrary. There is no absolute certainty with this result. Furthermore, findings that are statistically significant in one study are frequently unable to be reproduced in a second study (1). This is especially true if there is a large degree of uncertainty (e.g. large confidence intervals) (2). What is the answer? – We must become more comfortable with uncertainty and be careful not to use a p-value to make artificial distinctions between levels of uncertainty that are tolerable and levels that are not. The declaration of a decision (p<0.05 = good) does not resolve the uncertainty. As the p-value falls out of favor in the scientific community, Bayesian analysis has been gaining favor and may be the future of statistical analyses (3,4). Source The Parable of Schrödinger’s Cat and the Illusion of Statistical Significance in Clinical Trials. Circulation. 2019 Sep 9;140(10):799-800. doi: 10.1161/CIRCULATIONAHA.119.041245. Epub 2019 Sep 3. Reviewed by Thomas Davis Works Cited 1. The behavior of the P-value when the alternative hypothesis is true. Biometrics. 1997 Mar;53(1):11-22. 2. Double Vision: Replicating a Trial Showing a Survival Benefit. JACC Heart Fail. 2017 Mar;5(3):232-235. doi: 10.1016/j.jchf.2016.12.017. 3. Scientists rise up against statistical significance. Nature. 2019 Mar;567(7748):305-307. doi: 10.1038/d41586-019-00857-9. 4. The Proposal to Lower P Value Thresholds to .005. JAMA. 2018 Apr 10;319(14):1429-1430. doi: 10.1001/jama.2018.1536. Scroll to Top	779	3,399	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06	longest	en	0.934655
http://mathforum.org/kb/thread.jspa?threadID=2426366&messageID=7956374	1,519,057,413,000,000,000	text/html	crawl-data/CC-MAIN-2018-09/segments/1518891812756.57/warc/CC-MAIN-20180219151705-20180219171705-00137.warc.gz	235,446,107	5,123	Search All of the Math Forum: Views expressed in these public forums are not endorsed by NCTM or The Math Forum. Notice: We are no longer accepting new posts, but the forums will continue to be readable. Topic: Image processing for fluorescence image intensity reading Replies: 1 Last Post: Jan 3, 2013 8:31 PM Search Thread: Advanced Search ImageAnalyst Posts: 13,022 Registered: 12/26/06 Re: Image processing for fluorescence image intensity reading Posted: Jan 3, 2013 8:31 PM Plain Text Reply "Ling Fang Toh" wrote in message <kc3sio\$m0j\$1@newscl01ah.mathworks.com>... > I need to read intensity values for JPEG fluorescence image in gray scale (in black and white), and use the values to calculate temperature readings. > Previously I have tried using imread() and impixelinfo command, as well as intensityValue=Image(X,Y), but i think it only reads rgb values. What I'm looking for are single values in 3 decimal places for precision purpose. I'm wondering if this can be done... Is it also possible to view the values by simply placing the cursor over the image? > > I have posted a similar question earlier but I was unable to comment on the answer to clarify my inquiry further. My apologies that I'm new to this webpage. ============================================= Your jpeg image will be integer - it will have no 3 decimal places after the decimal point, they're all zero because it's integer. You can use impixelinfo() to display rgb or gray level values as you mouse over the image. It works with both, so try it again. intensityValue=Image(X,Y) will only work for grayscale images, not RGB images. You'd need to specify a color channel intensityValue=Image(X,Y, 1) or extract the color channels into separate monochrome images: % Extract the individual red, green, and blue color channels. redChannel = rgbImage(:, :, 1); greenChannel = rgbImage(:, :, 2); blueChannel = rgbImage(:, :, 3); You can also try Answers: http://www.mathworks.com/matlabcentral/answers/ There are a lot of experts over there and you can edit your posts so it's nicer. © The Math Forum at NCTM 1994-2018. All Rights Reserved.	508	2,135	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09	longest	en	0.845926
http://sci.tech-archive.net/Archive/sci.physics.relativity/2005-12/msg01222.html	1,493,584,305,000,000,000	text/html	crawl-data/CC-MAIN-2017-17/segments/1492917125849.25/warc/CC-MAIN-20170423031205-00489-ip-10-145-167-34.ec2.internal.warc.gz	336,994,977	6,338	# Lienard-Wiechert potential • From: D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) • Date: Wed, 14 Dec 2005 14:43:04 +0000 (UTC) There was some confusion a while back about the part that the Lienard- Wiechert potential of an accelerating charge plays, so I thought that I would clarify. Firstly, it should be pointed out that gauge transformations have absolutely no effect on the physics, so any choice of gauge has no effect on the physics. Some people seem to have missed that fact. The Lienard-Wiechert potential is an electromagnetic potential which satisfies the Lorentz condition (that div A + (1/c^2) @\phi/@t = 0) - since the physics is unchanged under gauge transformations, then we are free to take any gauge condition that we like. Maxwell's Equations are inhomogeneous linear equations, and in vacuo, the constitutive equations are linear. Maxwell's Equations are: div B = 0, curl E = - @B/@t, div D = \rho, curl H = J + @D/@t, where \rho is the charge density and J is the current density. Since we are working in a Minkowskian spacetime, which is contractible to a point in the rigorous mathematical definition of that term (see below), then there exist a vector potential A and a scalar potential \phi such that B = curl A, E = - grad \phi - @A/@t. By the statement that Minkowski spacetime (denoted by M) is contractible to a point, I mean that there exists a continuous function f mapping Mx[0,1] to M, and an event x_0 in M, such that f(x,0) = x_0 for all x in M, f(x,1) = x for all x in M. The very fact that such a continuous function exists is enough to guarantee the existence of the required potentials A and \phi. In vacuo, the constitutive equations are D = e_0 E and B = m_0 H, where e_0 is the permittivity of free space and m_0 is the permeability of free space. It follows that div E = \rho/e_0, curl B = m_0 J + (1/c^2) @E/@t, where c^2 = 1/(m_0 e_0), so that c is a speed, equal to 299 792 458 m/s. For a given vector potential A and scalar potential \phi which correspond to the fields E and B, it is possible to solve the equation del^2 \psi - (1/c^2) @^2 \psi/@t^2 = div A + (1/c^2) @\phi/@t for \psi, e.g. \psi(r,t) = (1/(4 \pi)) \int g(r-R,t-\|R\|/c) \|R\| sin(\theta) d\|R\| d\theta d\Phi = (1/(4 \pi)) \int g(r-R,t-\|R\|/c)/\|R\| dV, where the integral with respect to \|R\| is taken from 0 to infinity, the integral with respect to \theta is taken from 0 to \pi, and the integral with respect to \Phi is taken from 0 to 2\pi, and g = div A + (1/c^2) @\phi/@t. When we take the gauge transformation, A' = A - grad \psi, \phi' = \phi + @\psi/@t, then we find that div A' + (1/c^2) @\phi'/@t = 0. Since (A',\phi') is related to (A,\phi) by a gauge transformation, then the physics is unchanged, i.e. A' and \phi' are vector and scalar potentials for the fields E and B. It follows that for any fields E and B, there exist a vector potential A and a scalar potential \phi such that the Lorentz condition, div A + (1/c^2) @\phi/@t = 0, holds. In vacuo, since div E = \rho/e_0, then - del^2 \phi - div @A/@t = \rho/e_0, and so, by the Lorentz condition, - del^2 \phi + (1/c^2) @^2 \phi/@t^2 = \rho/e_0. Since curl B = m_0 J + (1/c^2) @E/@t, then curl curl A + (1/c^2) grad @\phi/@t + (1/c^2) @^2 A/@t^2 = m_0 J. Since curl curl A = grad div A - del^2 A, then grad div A - del^2 A + (1/c^2) grad @\phi/@t + (1/c^2) @^2 A/@t^2 = m_0 J, and so, by the Lorentz condition, - del^2 A + (1/c^2) @^2 A/@t^2 = m_0 J. These equations are linear in A and \phi, as is the Lorentz condition. This means that the most general solution to these equations is equal to one particular solution plus the general solution of the homogeneous equations. The solutions of the homogeneous equations for A and \phi form a vector space over the real numbers. The solutions of the inhomogeneous equation are of the form of a specified solution of the inhomogeneous equations plus an arbitrary solution of the homogeneous equations. The exact solution of the homogeneous equation that you add to the specific solution of the inhomogeneous equation is determined by the boundary conditions and initial conditions There are nonzero solutions of the homogeneous equations, for example, A = a cos(k.r - \omega t), \phi = 0, where \omega = c \|k\|, and where a is a constant vector such that k.a = 0. The corresponding EM field is E = \|k\| c a sin(k.r - \|k\| c t), B = - [k,a] sin(k.r - \|k\| c t). This solution is radiative. In fact, if k and \omega are nonzero, then this is the most general form needed for the term of the type A = b cos(k.r - \omega t), \phi = f cos(k.r - \omega t). If A = b cos(k.r - \omega t), \phi = f cos(k.r - \omega t), for a constant vector b and a constant scalar f, then the Lorentz condition requires that k.b = f \omega/c^2. When we take A' = A - grad \psi, \phi' = \phi + @\psi/@t, where \psi = f/\omega sin(k.r - \omega t), then the map from (A,\phi) to (A',\phi') is a gauge transformation, so that (A',\phi') and (A,\phi) correspond to the same EM field, and A' = (b - f k/\omega) cos (k.r - \omega t), \phi' = 0, and k.(b - f k/\omega) = k.b - f \|k\|^2/\omega = k.b - f \omega/c^2 = 0. It follows that A = a cos(k.r - \omega t), \phi = 0, where \omega = c \|k\|, and k.a = 0, is the most general potential needed of the form, A = b cos(k.r - \omega t), \phi = f cos(k.r - \omega t). Similarly, A = a sin{k.r - \omega t), \phi = 0, where \omega = c \|k\| and k.a = 0, is the most general potential needed of the form, A = b sin(k.r - \omega t), \phi = f sin(k.r - \omega t). Since these are solutions of the homogeneous equation, then they are sourceless solutions, and do not enter into the Lienard-Wiechert potential, which corresponds to the field generated by the charge. The solutions discussed above are radiation solutions. We are specifically interested in the EM field generated by the sources. An additional EM field which is a solution of the homogeneous equations could then be added on afterwards. The EM potential which is generated by the sources is given by A(r,t) = m_0/(4 \pi) \int J(r-R,t-\|R\|/c)/\|R\| dV, \phi(r,t) = 1/(4 \pi e_0) \int \rho(r-R,t-\|R\|/c)/\|R\| dV. This potential obeys the Lorentz condition. Instead of integrating with respect to volume, we could integrate with respect to charge. For an element of charge dq, which is at r-R at time t-\|R\|/c, take an element of volume dV = dS d\|R\|, where dS is an element of area on the sphere of radius R centred on r. As the sphere shrinks through the distance of d\|R\| in time d\|R\|/c, then the contribution to the charge is given by \rho(r-R,t-\|R\|/c) dS d\|R\| - \rho(r-R,t-\|R\|/c) u.R dS dt = \rho(r-R,t-\|R\|/c) (1 - u.R/(\|R\| c)) dS d\|R\| = \rho(r-R,t-\|R\|/c) (1 - u.R/(\|R\| c)) dV, where u is the velocity of the charge dq. It follows that A(r,t) = m_0/(4 \pi) \int u/(\|R\|-u.R/c) dq, \phi(r,t) = 1/(4 \pi e_0) \int 1/(\|R\|-u.R/c) dq, where R is such that the element dq of charge is at r-R at time t-\|R\|/c, and u is the velocity of the element when it is at r-R at time t-\|R\|/c. In the case of a point charge, this reduces to A(r,t) = m_0/(4 \pi) q u/(\|R\|-u.R/c), \phi(r,t) = 1/(4 \pi e_0) q/(\|R\|-u.R/c), where R is such that the charge is at r-R at time t-\|R\|/c (since the charge is travelling at less than the speed of light, then there is at most one value of R which satisfies this condition). If there is no such value of R, then A(r,t) = 0 and \phi(r,t) = 0. Note that u is taken for the charge at r-R at time t-\|R\|/c. In the case where the charge is stationary at 0 (the origin of the spatial coordinate system), R = r and u = 0, so that A(r,t) = 0, \phi(r,t) = q/(4 \pi e_0 \|r\|), which is indeed the electromagnetic potential associated with Coulomb force law. Panofsky and Phillips include a derivation of the EM field corresponding to the Lienard-Wiechert potential, which is therefore the EM field generated by the charge. The EM field is given by E = q/(4 \pi e_0) {(1/s^3) S (1-u^2/c^2) + (1/(c^2 s^3)) [R,[S,du/dt]]}, B = [R,E]/(\|R\| c), where R and u are as before, S = R - \|R\| u/c, s = \|R\| - u.R/c, du/dt is the acceleration of the charge when it is at r-R at time t-\|R\|/c, and for vectors v and w, [v,w] denotes the cross product of v and w. The first term within the braces in the expression for E yields the electric field induced by a uniformly moving charge (and the corresponding term in B in the full expansion yields the magnetic field induced by a uniformly moving charge). The first term in the braces makes no contribution to the radiation. The second term, which depends on the acceleration, Note that the first term in the braces in the expression for E is directed along the line joining where the charge would be at time t if it continued moving at uniform velocity u from r-R at time t-\|R\|/c. Specifically, the electric field at (r,t) generated by a uniformly moving charge is always directed along the line joining the position of the charge at time t with r. If (r,t) is such that no value of R exists for which the charge is at r-R at time t-\|R\|/c, then the contribution to the EM field from the charge is E = 0 and B = 0. In the case of a stationary charge at 0, u = 0, du/dt = 0, R = r, S = r, and s = \|r\|, so that E = q r/(4 \pi e_0 \|r\|^3), B = 0, which is the Coulomb electric field associated with a stationary charge, and a zero magnetic field (which is the magnetic field of a stationary charge). In the case of an instantaneously stationary charge q accelerating with acceleration a, integration of the long-distance Poynting vector over a sphere demonstrates that the charge is losing energy to radiation at the instantaneous rate of q^2 a^2/(6 \pi e_0 c^3) = m_0 q^2 a^2/(6 \pi c), i.e. the instantaneous power being supplied by the charge to the radiation is q^2 a^2/(6 \pi e_0 c^3). ----- .	3,093	9,821	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17	latest	en	0.931499
http://www.mathworks.com/matlabcentral/cody/players/1072042-pedro-villena/solved	1,484,922,314,000,000,000	text/html	crawl-data/CC-MAIN-2017-04/segments/1484560280835.22/warc/CC-MAIN-20170116095120-00281-ip-10-171-10-70.ec2.internal.warc.gz	572,310,756	16,688	Cody # Pedro Villena Rank Score #### Compute a dot product of two vectors x and y Created by Evgeny Problem Group Community #### Determine Whether an array is empty Created by Yudong Zhang Problem Group Community This problem is locked. You need to solve more problems from Cody Challenge group to unlock it. #### Getting the indices from a vector Created by Doug Hull Problem Group Indexing II #### Cosine frequency doubler Created by Richard Moore Problem Group Community This problem is locked. You need to solve more problems from Cody Challenge group to unlock it. This problem is locked. 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You need to solve more problems from Cody Challenge group to unlock it. #### Program an exclusive OR operation with logical operators Created by Tobias Otto Problem Group Community #### Omit columns averages from a matrix Created by Roy Fahn Tags columns, matrix Problem Group Community 1 – 50 of 133	1,012	4,571	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04	latest	en	0.889508
https://ajaytech.co/2019/07/05/what-is-the-difference-between-lapply-vs-sapply/	1,701,232,941,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100056.38/warc/CC-MAIN-20231129041834-20231129071834-00763.warc.gz	119,133,177	17,786	# What is the difference between lapply vs sapply For a more comprehensive list of R Interview questions, visit the link. sapply () is just a wrapper around lapply (). lapply returns a list whereas sapply returns a vector. ```# Here is a small data frame of iris data set. > iris_small Sepal.Length Sepal.Width Petal.Length Petal.Width 1 5.1 3.5 1.4 0.2 2 4.9 3.0 1.4 0.2 3 4.7 3.2 1.3 0.2 51 7.0 3.2 4.7 1.4 52 6.4 3.2 4.5 1.5 53 6.9 3.1 4.9 1.5 101 6.3 3.3 6.0 2.5 102 5.8 2.7 5.1 1.9 103 7.1 3.0 5.9 2.1 ``` Let’s convert this into a List ( which you don’t need to to perform lapply function, but just play along for now ) ```> iris_small_list = as.list(iris_small) \$Sepal.Length [1] 5.1 4.9 4.7 7.0 6.4 6.9 6.3 5.8 7.1 \$Sepal.Width [1] 3.5 3.0 3.2 3.2 3.2 3.1 3.3 2.7 3.0 \$Petal.Length [1] 1.4 1.4 1.3 4.7 4.5 4.9 6.0 5.1 5.9 \$Petal.Width [1] 0.2 0.2 0.2 1.4 1.5 1.5 2.5 1.9 2.1 ``` Let’s calculate the mean across each of the list elements ```> means = lapply(iris_small_list,mean) > means \$Sepal.Length [1] 6.022222 \$Sepal.Width [1] 3.133333 \$Petal.Length [1] 3.911111 \$Petal.Width [1] 1.277778 ``` Each element of the list element gets sent to the mean function as an argument and the result is returned in a list finally. Now, let’s perform the same with sapply ```> means = sapply(iris_small_list,mean) > means Sepal.Length Sepal.Width Petal.Length Petal.Width 6.022222 3.133333 3.911111 1.277778 ``` This time the list elements are converted to a vector. This site uses Akismet to reduce spam. Learn how your comment data is processed.	707	1,896	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.625	3	CC-MAIN-2023-50	latest	en	0.612177

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