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http://vijayschawalla.co.uk/lyu1c5sa/page.php?tag=matrix-multiplication-is-associative-true-or-false-288c6d | 1,618,384,670,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077336.28/warc/CC-MAIN-20210414064832-20210414094832-00416.warc.gz | 98,940,564 | 12,198 | # matrix multiplication is associative true or false
Why? This is a perfectly reasonable assumption and is true, being possible to demonstrate after it has been shown that matrix multiplication is associative. Explain. In particular, matrix multiplication is not "commutative"; you cannot switch the order of the factors and expect to … •Fluently compute a matrix-matrix multiplication. Showing that matrix products are associative. According to me matrix multiplication is not commutative. (c) If A and B are matrices whose product is a zero matrix, then A or B must be the zero matrix… Thus, even though AB = AC and A is not a zero matrix, B does not equal C. Example 13: Although matrix multiplication is not always commutative, it is always associative. 0 6 h. Matrix multiplication is associative. The matrix multiplication is a commutative operation. With this knowledge, we have the following: The statement is false. Let A be an mxn matrix; Let B and C have sizes for which the indicated sums and products are defined: 1) Associative Law of Multiplication: A(BC) = (AB)C 2) Left Distributive Law: A(B+C) = AB+AC 3) Right Distributive Law: (B+C)A = BA+CA 4) r(AB) = (rA)B = A(rB) for any scalar r 5) Identity for Matrix Multiplication: ImA = A = AIn Matrix multiplication shares some properties with usual multiplication. This can be observed from the following examples. Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes!*. Multiply the matrix by the product of transformation matrices; the FIRST transformation should be CLOSEST to the matrix on the LEFT. Want to see this answer and more? True False Equations Video. Ask your question. 1 answer. Answered Both addition and multiplication are associative for whole numbers. Dec 03,2020 - Which of the following property of matrix multiplication is correct:a)Multiplication is not commutative in genralb)Multiplication is associativec)Multiplication is distributive over additiond)All of the mentionedCorrect answer is option 'D'. ... Matrix multiplication is associative. Q: Please show step by step algebra for solving/isolating (h): Step-by-step answers are written by subject experts who are available 24/7. Specifically, the product A~x is a linear combination of the columns of A: Question 2.3. True False Equations Calculator. then both . According to me matrix multiplication is not commutative. Show that X Y e q Y X XY eq YX X Y e q Y X if. This preview shows page 2 out of 2 pages.. b. matrix multiplication? (b) If A is a 3 x 2 matrix and B is a 7 x 3 matrix and C is a 4 x 7 matrix, then the transformation whose standard matrix is CBA is a transformation from R4 to R2. True or false To add or subtract matrices both matrices must have the same dimenson?, When does an addition matrix have no solution?, True or False Dimensions of the resulting matrices=the dimensions of the matrices being added?, 3 0 1 2 8 1 -1 + 3 9 -9 = ? same order same number of columns. Want to see this answer and more? upper triangular matrix diagonal matrix Question No: 13 ( Marks: 1 ) - Please choose one The matrix multiplication is associative True False Question No: 14 ( Marks: 1 ) - Please choose one We can add the matrices of _____. Ask Questions, Get Answers Menu X. home ask tuition questions practice papers mobile tutors pricing How to find the change of coordinates matrix? •Relate composing rotations to matrix-matrix multiplication. {/eq} is the same {eq}4\times 2 False. And what I do in this video you can extend it to really any dimension of matrices for which of the matrix multiplication is actually defined. 10 True or False Quiz Problems about Matrix Operations . Questions are typically answered in as fast as 30 minutes. LOGIN TO VIEW ANSWER. if you are adding or multiplying it does not matter where you put the parenthesis. Join now. Its not okay to arbitrarily reverse the order in which you multiply matrices. d. The following are other important properties of matrix multiplication. We're supposed to use direct matrix multiplication (using ladder operators is in part b of the problem). $$\begin{pmatrix} e & f \\ g & h \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ae + cf & be + df \\ ag + ch & bg + dh \end{pmatrix}$$ False. Mathematics. Best answer. {/eq}. matrix multiplication proceeds exactly as does a regular matrix-matrix multiplication except that individual multiplications of scalars commute while (in general) individual multiplications with matrix blocks (submatrices) do not. | 1,067 | 4,545 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-17 | latest | en | 0.909467 |
https://served-up-fresh.com/qa/question-how-many-hours-are-in-each-month.html | 1,611,193,300,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522150.18/warc/CC-MAIN-20210121004224-20210121034224-00229.warc.gz | 568,282,618 | 8,172 | Question: How Many Hours Are In Each Month?
Is a day exactly 24 hours?
Day Length On Earth, a solar day is around 24 hours.
However, Earth’s orbit is elliptical, meaning it’s not a perfect circle.
That means some solar days on Earth are a few minutes longer than 24 hours and some are a few minutes shorter.
On Earth, a sidereal day is almost exactly 23 hours and 56 minutes..
Why is it 60 minutes in an hour?
Who decided on these time divisions? THE DIVISION of the hour into 60 minutes and of the minute into 60 seconds comes from the Babylonians who used a sexagesimal (counting in 60s) system for mathematics and astronomy. They derived their number system from the Sumerians who were using it as early as 3500 BC.
Who many seconds are in a year?
31,536,000 secondsone year would equal 365 times 24 times 60 times 60 seconds…or 31,536,000 seconds!
What is the maximum working hours per week?
Most workers should not have to work more than an average of 48 hours a week, according to the Working Time Regulations. The Regulations also give you rights to paid holiday, rest breaks and limits on night work. Your average working hours are calculated over a 17-week period.
How many hours are in a month 30 days?
MonthNumber of DaysNumber of HoursApril30720May31744June30720July3174411 more rows
How many months is 2000 work hours?
2.7376 monthsThis conversion of 2,000 hours to months has been calculated by multiplying 2,000 hours by 0.0013 and the result is 2.7376 months.
How many days passed 2020?
366 daysThis page lists all days in 2020 with day and week numbers. The year 2020 has 366 days. This is a leap year.
How many hours in a month and a half?
Months to weeks conversion table1 month730 hours11 months8030 hours12 months8760 hours13 months9490 hours14 months10220 hours16 more rows
How is Manhour calculated?
The total man hours per task is obtained by multiplying the number of people assigned to a task by the total time it takes to complete it. Let’s say, for example, that 15 workers at a metal plant and devote 10 workdays to complete an order of 800 product units.
Can you work 100 hours a week?
An occasional 100-hour week is alright. … The latter figure translated into 13 hours per day, five days a week. Ouch! In a study of high earners, management writers Sylvia Ann Hewlett and Carolyn Buck Luce found that a full 35 percent worked more than 60 hours a week, and 10 percent worked more than 80 hours a week.
How many hours is 2020?
8784 hourshow many hours in 2020? There are 8784 hours in the year 2020.
How many days hours minutes and seconds are in 1 month?
One month, 31 days, 744 hours, 44640 minutes, 2678400 seconds | Let’s Skip to the Good Bit.
Why is a day 24 hours?
Our 24-hour day comes from the ancient Egyptians who divided day-time into 10 hours they measured with devices such as shadow clocks, and added a twilight hour at the beginning and another one at the end of the day-time, says Lomb. “Night-time was divided in 12 hours, based on the observations of stars.
How do you calculate man hours per day?
Multiply the number of workers by the number of hours each one worked to calculate the number of man-hours your business used during that time period. In this example, multiply 10 workers by 160 hours per worker to get 1,600 man-hours. This means that you produced 50,000 pieces using 1,600 man-hours.
What months dont have 30 days?
30 days has September, April, June and November. And 29 in each leap year. A knuckle is “31 days”, and in between each knuckle it isn’t.
How many days are 3 in a week?
Weeks to days conversion table1 week7 days3 weeks21 days4 weeks28 days5 weeks35 days6 weeks42 days16 more rows
How many hours will 2020 end?
When people ask that question, the reply they expect to get is that it’s 261 days until the end of 2020 (as in how many days are left between today, April 15, and the end of the year, December 31). Instead, however, Siri is returning answers such as “It’s 13 hours until then.”
How do I calculate my hours worked per month?
There are basically two ways to calculate the hours per month. With full-time employees, you should assume one employee will work a 40 hour workweek. A quick and easy method of calculating monthly hours is to multiply 40 hours per week by 4 weeks, yielding 160 hours for the month. | 1,062 | 4,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-04 | latest | en | 0.952456 |
https://www.doubtnut.com/question-answer/is-5x2-17x-6-factorisable-if-yes-factorise-it-644442458 | 1,670,532,078,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711360.27/warc/CC-MAIN-20221208183130-20221208213130-00251.warc.gz | 772,199,132 | 33,826 | Home
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# Is 5x^(2) + 17x +6 factorisable ? If yes, factorise it.
Updated On: 27-06-2022
Text Solution
Answer : is factorisable , (x+3)(5x+2) | 96 | 244 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-49 | latest | en | 0.330579 |
https://www.teacherspayteachers.com/Product/Place-Value-and-Patterns-Activity-3494225 | 1,512,957,393,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512054.0/warc/CC-MAIN-20171211014442-20171211034442-00146.warc.gz | 800,698,053 | 24,701 | Total:
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# Place Value and Patterns - Activity
Common Core Standards
Product Rating
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PDF (Acrobat) Document File
103 KB|11 pages
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Product Description
Standard: 5.NBT.1
Corresponding GoMath Lesson: 1.1 Game: Placing Value on Patterns
Materials:
- “Placing Value on Patterns” Sheets
- Game board (whole numbers; multiple of 10)
- Movement Cards or Dice
- Game pieces
Directions:
- Students are put into groups of 2 or 3
- Each group is given a game board and a deck of movement cards or dice
- Each student is given a “Placing Value on Patterns” sheet
- Student uses movement card or dice to determine the space they move to
- Student writes down the number they land on in the gray shaded box
- Then, they fill in the rest of the row/the empty boxes based on the number the
landed on
- If students complete the game board, they may start again
- Students work at their own pace
Sheet Breakdown:
- Sheet 1: Typical (shaded column alternates; challenging)
- Sheet 2: Shaded column is always original number; pattern is consistent
- Sheet 3: Shaded column is always the original number but in the middle rather than
the side; easier to see pattern
- Sheet 4: For students using a calculator; multiplies original number by 10
Total Pages
11 pages
N/A
Teaching Duration
N/A
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 391 | 1,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-51 | longest | en | 0.839341 |
http://betterlesson.com/lesson/resource/2587995/tasks-purrrrrfect-cat-toy-company-docx | 1,477,107,997,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718423.65/warc/CC-MAIN-20161020183838-00347-ip-10-171-6-4.ec2.internal.warc.gz | 28,344,581 | 21,326 | Tasks - Purrrrrfect Cat Toy Company.docx - Section 1: Launch and Explore
Tasks - Purrrrrfect Cat Toy Company.docx
The Purrrrrfect Cat Toy Company
Unit 8: Matrices and Systems
Lesson 1 of 16
Big Idea: Matrices are perfect for large data sets - monthly sales for cat toys are analyzed and matrix operations are introduced.
Print Lesson
10 teachers like this lesson
Standards:
Subject(s):
50 minutes
Tim Marley
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Nothing to upload details close | 301 | 1,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | longest | en | 0.882938 |
https://discuss.leetcode.com/topic/99925/shortest-4-lines-single-pass-o-k-python-solution | 1,513,226,200,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948539745.30/warc/CC-MAIN-20171214035620-20171214055620-00581.warc.gz | 567,306,769 | 8,245 | Shortest (4 Lines) single pass O(k) python solution
• I realized that the 2nd element was always k, 3rd element was increased by the factor of 0.5(1/2) each time k increased by 1 (compared to the previous element), 4th element increased by the factor of 0.33..(1/3) each time k increased by 1 (also compared to the previous element).
So after some time I figured out that the number at answer[i] = answer[i - 1] * (k + 1 - i) / i
I would love to know the formal mathematics proof behind this.
Anyways, here is my code:
``````
class Solution(object):
def getRow(self, rowIndex): | 163 | 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} | 2.765625 | 3 | CC-MAIN-2017-51 | latest | en | 0.966969 |
https://topic.alibabacloud.com/zqpop/radian-value_92093.html | 1,709,491,896,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00713.warc.gz | 560,904,322 | 17,374 | Want to know radian value? we have a huge selection of radian value information on alibabacloud.com
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 915 | 4,205 | {"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} | 3.109375 | 3 | CC-MAIN-2024-10 | latest | en | 0.824156 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1710/2/l/ | 1,620,996,326,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989526.42/warc/CC-MAIN-20210514121902-20210514151902-00565.warc.gz | 907,447,071 | 51,627 | # Properties
Label 1710.2.l Level $1710$ Weight $2$ Character orbit 1710.l Rep. character $\chi_{1710}(1261,\cdot)$ Character field $\Q(\zeta_{3})$ Dimension $72$ Newform subspaces $19$ Sturm bound $720$ Trace bound $11$
# Related objects
## Defining parameters
Level: $$N$$ $$=$$ $$1710 = 2 \cdot 3^{2} \cdot 5 \cdot 19$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 1710.l (of order $$3$$ and degree $$2$$) Character conductor: $$\operatorname{cond}(\chi)$$ $$=$$ $$19$$ Character field: $$\Q(\zeta_{3})$$ Newform subspaces: $$19$$ Sturm bound: $$720$$ Trace bound: $$11$$ Distinguishing $$T_p$$: $$7$$, $$11$$
## Dimensions
The following table gives the dimensions of various subspaces of $$M_{2}(1710, [\chi])$$.
Total New Old
Modular forms 752 72 680
Cusp forms 688 72 616
Eisenstein series 64 0 64
## Trace form
$$72q - 2q^{2} - 36q^{4} + 4q^{8} + O(q^{10})$$ $$72q - 2q^{2} - 36q^{4} + 4q^{8} + 2q^{10} + 4q^{13} + 6q^{14} - 36q^{16} - 12q^{17} + 12q^{19} - 6q^{22} + 4q^{23} - 36q^{25} + 16q^{29} - 16q^{31} - 2q^{32} + 20q^{34} - 2q^{35} + 16q^{37} + 4q^{38} + 2q^{40} - 20q^{41} + 20q^{43} + 12q^{46} - 8q^{47} + 76q^{49} + 4q^{50} + 4q^{52} + 8q^{53} - 12q^{56} + 8q^{58} + 30q^{59} - 24q^{61} + 16q^{62} + 72q^{64} + 8q^{65} - 6q^{67} + 24q^{68} + 4q^{70} + 16q^{71} - 22q^{73} - 34q^{74} + 6q^{76} - 72q^{77} + 36q^{79} - 22q^{82} + 12q^{83} - 16q^{86} + 12q^{88} + 2q^{89} + 28q^{91} + 4q^{92} - 16q^{94} + 4q^{95} - 38q^{97} - 26q^{98} + O(q^{100})$$
## Decomposition of $$S_{2}^{\mathrm{new}}(1710, [\chi])$$ into newform subspaces
Label Dim. $$A$$ Field CM Traces $q$-expansion
$$a_2$$ $$a_3$$ $$a_5$$ $$a_7$$
1710.2.l.a $$2$$ $$13.654$$ $$\Q(\sqrt{-3})$$ None $$-1$$ $$0$$ $$-1$$ $$-2$$ $$q+(-1+\zeta_{6})q^{2}-\zeta_{6}q^{4}+(-1+\zeta_{6})q^{5}+\cdots$$
1710.2.l.b $$2$$ $$13.654$$ $$\Q(\sqrt{-3})$$ None $$-1$$ $$0$$ $$-1$$ $$4$$ $$q+(-1+\zeta_{6})q^{2}-\zeta_{6}q^{4}+(-1+\zeta_{6})q^{5}+\cdots$$
1710.2.l.c $$2$$ $$13.654$$ $$\Q(\sqrt{-3})$$ None $$-1$$ $$0$$ $$1$$ $$-10$$ $$q+(-1+\zeta_{6})q^{2}-\zeta_{6}q^{4}+(1-\zeta_{6})q^{5}+\cdots$$
1710.2.l.d $$2$$ $$13.654$$ $$\Q(\sqrt{-3})$$ None $$-1$$ $$0$$ $$1$$ $$-10$$ $$q+(-1+\zeta_{6})q^{2}-\zeta_{6}q^{4}+(1-\zeta_{6})q^{5}+\cdots$$
1710.2.l.e $$2$$ $$13.654$$ $$\Q(\sqrt{-3})$$ None $$1$$ $$0$$ $$-1$$ $$-10$$ $$q+(1-\zeta_{6})q^{2}-\zeta_{6}q^{4}+(-1+\zeta_{6})q^{5}+\cdots$$
1710.2.l.f $$2$$ $$13.654$$ $$\Q(\sqrt{-3})$$ None $$1$$ $$0$$ $$-1$$ $$-2$$ $$q+(1-\zeta_{6})q^{2}-\zeta_{6}q^{4}+(-1+\zeta_{6})q^{5}+\cdots$$
1710.2.l.g $$2$$ $$13.654$$ $$\Q(\sqrt{-3})$$ None $$1$$ $$0$$ $$-1$$ $$-2$$ $$q+(1-\zeta_{6})q^{2}-\zeta_{6}q^{4}+(-1+\zeta_{6})q^{5}+\cdots$$
1710.2.l.h $$2$$ $$13.654$$ $$\Q(\sqrt{-3})$$ None $$1$$ $$0$$ $$-1$$ $$2$$ $$q+(1-\zeta_{6})q^{2}-\zeta_{6}q^{4}+(-1+\zeta_{6})q^{5}+\cdots$$
1710.2.l.i $$2$$ $$13.654$$ $$\Q(\sqrt{-3})$$ None $$1$$ $$0$$ $$-1$$ $$6$$ $$q+(1-\zeta_{6})q^{2}-\zeta_{6}q^{4}+(-1+\zeta_{6})q^{5}+\cdots$$
1710.2.l.j $$4$$ $$13.654$$ $$\Q(\sqrt{-3}, \sqrt{7})$$ None $$-2$$ $$0$$ $$-2$$ $$0$$ $$q+(-1-\beta _{2})q^{2}+\beta _{2}q^{4}+(-1-\beta _{2}+\cdots)q^{5}+\cdots$$
1710.2.l.k $$4$$ $$13.654$$ $$\Q(\sqrt{-3}, \sqrt{19})$$ None $$-2$$ $$0$$ $$-2$$ $$0$$ $$q+(-1-\beta _{2})q^{2}+\beta _{2}q^{4}+(-1-\beta _{2}+\cdots)q^{5}+\cdots$$
1710.2.l.l $$4$$ $$13.654$$ $$\Q(\sqrt{-3}, \sqrt{73})$$ None $$-2$$ $$0$$ $$2$$ $$-4$$ $$q+(-1+\beta _{2})q^{2}-\beta _{2}q^{4}+(1-\beta _{2})q^{5}+\cdots$$
1710.2.l.m $$4$$ $$13.654$$ $$\Q(\sqrt{-3}, \sqrt{17})$$ None $$-2$$ $$0$$ $$2$$ $$10$$ $$q-\beta _{2}q^{2}+(-1+\beta _{2})q^{4}+\beta _{2}q^{5}+(2+\cdots)q^{7}+\cdots$$
1710.2.l.n $$4$$ $$13.654$$ $$\Q(\sqrt{-2}, \sqrt{-3})$$ None $$2$$ $$0$$ $$2$$ $$4$$ $$q+(1-\beta _{1})q^{2}-\beta _{1}q^{4}+(1-\beta _{1})q^{5}+\cdots$$
1710.2.l.o $$6$$ $$13.654$$ 6.0.29654208.1 None $$-3$$ $$0$$ $$-3$$ $$-6$$ $$q+(-1+\beta _{1})q^{2}-\beta _{1}q^{4}+(-1+\beta _{1}+\cdots)q^{5}+\cdots$$
1710.2.l.p $$6$$ $$13.654$$ 6.0.29654208.1 None $$3$$ $$0$$ $$3$$ $$-6$$ $$q+(1-\beta _{1})q^{2}-\beta _{1}q^{4}+(1-\beta _{1})q^{5}+\cdots$$
1710.2.l.q $$6$$ $$13.654$$ 6.0.29654208.1 None $$3$$ $$0$$ $$3$$ $$2$$ $$q+(1-\beta _{1})q^{2}-\beta _{1}q^{4}+(1-\beta _{1})q^{5}+\cdots$$
1710.2.l.r $$8$$ $$13.654$$ 8.0.4678560000.4 None $$-4$$ $$0$$ $$4$$ $$12$$ $$q-\beta _{1}q^{2}+(-1+\beta _{1})q^{4}+\beta _{1}q^{5}+(1+\cdots)q^{7}+\cdots$$
1710.2.l.s $$8$$ $$13.654$$ 8.0.4678560000.4 None $$4$$ $$0$$ $$-4$$ $$12$$ $$q+\beta _{1}q^{2}+(-1+\beta _{1})q^{4}-\beta _{1}q^{5}+(1+\cdots)q^{7}+\cdots$$
## Decomposition of $$S_{2}^{\mathrm{old}}(1710, [\chi])$$ into lower level spaces
$$S_{2}^{\mathrm{old}}(1710, [\chi]) \cong$$ $$S_{2}^{\mathrm{new}}(38, [\chi])$$$$^{\oplus 6}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(57, [\chi])$$$$^{\oplus 8}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(95, [\chi])$$$$^{\oplus 6}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(114, [\chi])$$$$^{\oplus 4}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(171, [\chi])$$$$^{\oplus 4}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(190, [\chi])$$$$^{\oplus 3}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(285, [\chi])$$$$^{\oplus 4}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(342, [\chi])$$$$^{\oplus 2}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(570, [\chi])$$$$^{\oplus 2}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(855, [\chi])$$$$^{\oplus 2}$$ | 2,669 | 5,259 | {"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-2021-21 | latest | en | 0.279807 |
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Seq.(Seq.(Seq.)) seeking sod (Posted on 2009-03-01)
A sequence {S(x)}, each of whose terms is a positive integer, is defined as:
S(x) = x/2, if x is even, and:
S(x) = x+3, if x is odd
Given that n is an odd positive integer and, S(S(S(n))) = 27, determine sod(n).
Note: sod(n) denotes the sum of the base-10 digits of n.
No Solution Yet Submitted by K Sengupta Rating: 1.0000 (1 votes)
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True but unprovable | Comment 2 of 3 |
A little background information on this sequence, none of which is necessary to solve the problem:
No matter what the value of x, the sequence seems to wind up repeating (cycling) through the digits 4,2,1. For instance, if x = 7, the sequence is
7,10,5,8,4,2,1,4,2,1,4,2,1,4,...
Last time I checked (years ago) it was generally believed that this was true for all x, but that it could not be proved that it was true for all x.
(Godel has proved that there are some theorems that are true but unprovable in any sufficiently rich mathematical system, and many believed that this was one of them.)
Of course, it was also widely believed that the 4-color map theorem was unprovable, until that theorem was proved.
Posted by Steve Herman on 2009-03-02 11:05:47
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https://community.smartsheet.com/discussion/44156/help-ryb-color-change-based-on-dates | 1,721,865,506,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00489.warc.gz | 146,925,377 | 108,713 | # HELP! RYB Color Change Based on Dates
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edited 12/09/19
Hello everyone,
I am currently using this:
=IF(NOT(ISBLANK(Completion4)), IF([Estimated End]4 >= Completion4, "Green", IF(NOT(ISBLANK(Completion4)), IF([Estimated End]4 < Completion4, "Red"))))
Which is working but I try to add this:
IF(ISBLANK(Completion3), IF(NOT(ISBLANK(Supervisor4)), "Yellow"
And it doesn't work, I'm not sure what I am doing wrong. Any ideas?
• ✭✭✭✭✭✭
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How are you trying to add it, what is the issue/error once it is added, and what is the desired outcome?
• edited 03/22/19
Options
Hi Paul,
The first two parts for the green and red work fine but nothing happens when it should turn yellow. I don't get any errors. When I try the whole statement together when it should be turning yellow it doesn't. I'm not sure why because if I use that statement alone it works.
• ✭✭✭✭✭✭
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you are referencing row 3 instead of 4 with the yellow statement which might be contributing to your issue.
=IF(NOT(ISBLANK(Completion4)), IF([Estimated End]4 >= Completion4, "Green", IF(NOT(ISBLANK(Completion4)), IF([Estimated End]4 < Completion4, "Red",IF(ISBLANK(Completion4), IF(NOT(ISBLANK(Supervisor4)), "Yellow"
the above is what I believe you are trying to use. you need to look at the heirarchy of your statements. You never give a false criteria, that is to say you never tell the formula what to do when the opposite of your criteria
• Options
Hi, thank you.
But no that's not it, I changed that already and it's still not working. It's pretty strange.
• ✭✭✭✭✭✭
Options
1. =IF(NOT(ISBLANK(Completion4)),
Return if above statement is true
IF([Estimated End]4 >= Completion4, "Green", IF(NOT(ISBLANK(Completion4)), IF([Estimated End]4 < Completion4, "Red",IF(ISBLANK(Completion4), IF(NOT(ISBLANK(Supervisor4)), "Yellow"
Return if above statement is false
""
2. IF([Estimated End]4 >= Completion4
Return if above statement is true
"Green"
Return if above statement is false
IF(NOT(ISBLANK(Completion4)), IF([Estimated End]4 < Completion4, "Red",IF(ISBLANK(Completion4), IF(NOT(ISBLANK(Supervisor4)), "Yellow"
3. IF(NOT(ISBLANK(Completion4))
Return if above statement is true
IF([Estimated End]4 < Completion4, "Red",IF(ISBLANK(Completion4), IF(NOT(ISBLANK(Supervisor4)), "Yellow"
Return if above statement is false
""
4. IF([Estimated End]4 < Completion4
Return if above statement is true
"Red"
Return if above statement is false
IF(ISBLANK(Completion4), IF(NOT(ISBLANK(Supervisor4)), "Yellow"
5. IF(ISBLANK(Completion4)
Return if above statement is true
IF(NOT(ISBLANK(Supervisor4)), "Yellow"
Return if above statement is false
""
6. IF(NOT(ISBLANK(Supervisor4)), "Yellow"
Return if above statement is true
"Yellow"
Return if above statement is false
""
You have 4 if statements that don't return anything if false. Your formula is linear and does not effectively utilize both criteria of the statements. If statements are composed of 3 parts (Criteria, true, false) and all need to be considered when stacking them. I'll clean up the formula and post it in a response to this so you can see what I am talking about.
• ✭✭✭✭✭✭
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Original
=IF(NOT(ISBLANK(Completion4)), IF([Estimated End]4 >= Completion4, "Green", IF(NOT(ISBLANK(Completion4)), IF([Estimated End]4 < Completion4, "Red",IF(ISBLANK(Completion4), IF(NOT(ISBLANK(Supervisor4)), "Yellow"
1. if C is not blank
2. if EE is greater than or equal to C then "green"
else
3. if C is not blank then if EE is less than C "Red"
else
4. if C is blank then if S is not blank then "Yellow"
Every single statement is reliant on the criteria of the previous, so there are several redundant statements in this. An example would be line 3, we already checked if C is blank in line 1, why check again? Also we already checked if EE is greater than or equal to in line 2, the only other option is less than, so we don't need to check it again in line 3.
For line 4 we introduce a new column (supervisor) but we never tell the program what to do if S is blank, only what to do when the is a value in the cell.
• ✭✭✭✭✭✭
Options
What you can do instead is something like:
=if(not(isblank(Completion4)),if([Estimated End]4 >= Completion4,"Green","Red"),if(isblank(Supervisor4),"","Yellow"
I recommend pasting your solution into notepad and breaking it down into parts to be able to focus on a single part at a time. Trying to understand something like this in a single piece is very difficult, when you break it into individual parts it becomes much easier.
• ✭✭✭✭✭✭
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I also find it helps to list out each scenario (requirement/outcome).
.
If this equals that, then I want it to do this.
If this equals something else, then I want it to do something else.
.
List EVERYTHING out. Don't worry about listing things multiple times. Don't worry about having 5 different things equaling the same thing. Just list it all out. | 1,369 | 4,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.833042 |
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MBD Chapter 5
MBD Chapter 5 - Uniaxial stress = = = = = =-σ1 Eϵ1 σ2...
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Chapter 5: Elastic Strain, Deflection and Stability 5.3 Analysis of Strain – Equiangular Rosettes , = + + ±( - - ) +( + ) ϵ1 2 ϵ0 ϵ120 ϵ2403 2ϵ0 ϵ120 ϵ240 29 ϵ120 ϵ240 23 = ( - ) - - tan2α 3 ϵ120 ϵ240 2ϵ0 ϵ120 ϵ240 5.4 Analysis of Strain – Rectangular Rosettes , = + ±( - ) +( - ) ϵ1 2 ϵ0 ϵ902 ϵ0 ϵ45 2 ϵ45 ϵ90 22 = tan2α - + - ϵ0 2ϵ45 ϵ90ϵ0 ϵ90 5.5 Elastic Stress-Strain Relationships and Three-Dimensional Mohr Circles = - + = - + = σ1 E1 ν2ϵ1 νϵ2 σ2 E1 ν2ϵ2 νϵ1 σ3 0 = - = - =- ( + ) ϵ1 1Eσ1 νσ2 ϵ2 1Eσ2 νσ1 ϵ3 νE σ1 σ2 Uniaxial stress:
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Unformatted text preview: Uniaxial stress: = = = = = =-σ1 Eϵ1 σ2 σ3 0 ϵ1 σ1E ϵ2 ϵ3 νσ1E Look at Tabbed Sections 5.6, 5.7, 5.8 5.10 Column Buckling – Elastic Instability = = = ( ) = ( ) Pcr π2EILe2 Scr PcrA π2E Leρ 2 or ScrE π2 Leρ 2 5.12 Column Design Equations – J.B. Johnson Parabola = =-( ) ≫ = Scr PcrA Sy Sy24π2E Leρ 2 Scr Sy2 = Leρ 2π2ESy 5.14 Equivalent Column Stresses = ≫ = = ( ) Scr Syα α SyScr Sy Leρ 2π2E Euler = =-( ) α SyScr 4π2E4π2E Sy Leρ 2 Johnson...
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https://electronicsforu.com/electronics-projects/prototypes/make-joystick-rf-controlled-robot | 1,579,287,163,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250590107.3/warc/CC-MAIN-20200117180950-20200117204950-00458.warc.gz | 426,047,589 | 40,512 | # Make Your Own Joystick for RF Controlled Robot
Sanjay Kumar Gupta is interested in circuit designing and is an Electronics Enthusiasts
3320
In the DPDT (double pole double throw) based joystick or the RF transmitter module switch based joystick, you must press two buttons simultaneously for a robot to move forward, backward or in circle. In this one, you can press one button for one motion very similar to a normal joystick. You need not practice controlling the robot as you do when you use a DPDT based joystick. This joystick circuit is quite cheaper than the joysticks available in the market. You only have to connect it to a RF transmitter module and make the supply common.
## Components Required
S.NO. COMPONENT NAME QUANTITY 1 ZERO PCB 1 2 7432(OR GATE IC) 2 3 7404(NOT GATE IC) 1 4 PUSH BUTTON 6 5 1K (RESISTOR) 6 6 LED 6 7 330 OHM 1 8 HEADER PIN MALE 6
## Setting up the Circuit
I noted down the logic needed for each move like forward, backward, left, right, clockwise and anticlockwise rotation and made a truth table based on that truth table I designed the circuit which is nothing but a circuit of 6*4 Encoder circuit, on pressing any of six button a BCD code is generated which is for certain move. You may think why I have not used 16*4 encoder IC or 3 input OR gate IC. It’s simply because it is difficult to find in the market.
S.NO. MOTION A B C D 1 FORWARD 0 1 0 1 2 BACKWARD 1 0 1 0 3 RIGHT 0 1 1 1 4 LEFT 1 1 0 1 5 CLOCKWISE ROTATION 1 0 0 1 6 ANTICLOCKWISE ROTATION 0 1 1 0
## Circuit operation
Whenever any input of OR gate goes HIGH its output will be HIGH and NOT gate output will be HIGH when input is LOW, and LOW when input is HIGH. Initially every gate used in circuit is pulled down to zero by pulldown resistor 1K. Suppose forward button is pressed as shown in circuit diagram it will make U1:A (OR gate) go HIGH which will make U1:B(OR gate) go HIGH also it will make U1:C (OR gate) go HIGH which will make U2:A(OR gate) go HIGH before passing not gate code generated is 1010 but after passing through not gates it become 0101 which is required logic to move forward as you can see in the truth table.
Similarly when we press backward button as in circuit diagram it will make U1:D and U2:B (OR gates) go HIGH, here before passing not gates code is 0101 after passing not gate it become 1010 which is required condition for moving backward, led will glow according to the code it will glow indicating HIGH else LOW. Match the each output with truth table and then connect the RF transmitter module with it. In RF transmitter module 4 switches are given connect the output of circuit A, B, C, D to that switch and make the supply common.
## Tutorial Video
If you have any doubt you can watch a video at: | 706 | 2,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-05 | latest | en | 0.881908 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=112967 | 1,369,524,759,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706470784/warc/CC-MAIN-20130516121430-00054-ip-10-60-113-184.ec2.internal.warc.gz | 314,997,304 | 747 | ```Question 153456
See the pattern below
In 1 quarter, there are 25 cents
In 2 quarters, there are 2*25 cents
In 3 quarters, there are 3*25 cents
...................................
In q quarters, there are q*25 cents
Write q*25 as 25q.
``` | 72 | 240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.869169 |
https://www.doubtnut.com/question-answer/in-the-following-cases-find-the-coordinates-of-the-foot-of-the-perpendicular-drawn-from-the-origin-a-2609 | 1,631,894,580,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055684.76/warc/CC-MAIN-20210917151054-20210917181054-00694.warc.gz | 790,441,756 | 75,918 | Home
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# In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.(a) 2x + 3y + 4z 12 = 0 (b) 3y + 4z - 6 = 0(c) x + y + z = 1 (d) 5y + 8 = 0
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1:33 | 443 | 1,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-39 | latest | en | 0.489571 |
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# First Order Kinetics Reactions
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## Jennifer Kordash
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#### Transcript of First Order Kinetics Reactions
Are you scared for your life?
Do you fear your children will have birth defects?
Are you afraid to go outside?
Have no fear! FONBs is here! Nuclear Bombs: They'll blow your mind A reaction where the concentration of one reactant determines the rate of the reaction. The reactant and rate are proportional; every time the reactant doubles, so does the rate. Compared to a second order reaction, when the reactant is doubled, rate is quadrupled. Rate Law What is a first order reaction? Rate=K [A] [B] a rate law for a first order reaction, would just contain one of the variables because the other would be 0 order.
Example: A is the first order because when the concentration doubles, so does the rate. B is 0 order, because when the concentration doubles, the rate remains constant. Rate Constant This is 'k' in the rate law. The rate constant varies in each reaction. This is to adjust for error in the reaction. To solve for 'k' use substitution with data given in the table. Reaction Mechanism The reaction mechanism is made up of at least two separate reactions, one is slow, and one is fast. The two reactions must add together and create the original reaction. The stoichiometry between the two reactions must agree. First Order in Real Life Airbags are a good example of first order kinetics in action. Airbags inflate when an NaN3 pellet is burned and nitrogen gas fills the nylon airbag. The airbag inflates at a rate of 150 to 250 mph in 40 milliseconds. First order kinetics would be the better choice in this case because it allows more control over the reaction. This is because the rate of reaction is directly proportional to the amount of NaN3. A second order reaction would cause the rate of inflation to quadruple causing the bag to inflate too fast and possibly burst. First Order Graphs The Obvious Choice After a crash course in first order reaction, it is clear that first order kinetics is the obvious choice when it comes to nuclear bombs. rate= k [A] [B] 1=k[.1]
1/.1=10
k=10/Ms rate=1
k=?
[A]=.1
[B]=1 so plug in the values... example: A+E-->C+D slow
D+B-->E fast | 625 | 2,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-39 | longest | en | 0.928932 |
http://turan.no/exercise/27853/15.10.15 | 1,669,921,863,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710869.86/warc/CC-MAIN-20221201185801-20221201215801-00545.warc.gz | 56,148,582 | 16,222 | 145 BPM , 73.2%
178 BPM , 89.9%
15.10.15
27.2 km
###### Ascent
383 m (14.1 m/km)
378 m
89 m
185 m
3:10 p.m.
Oct. 15, 2015
67.5 km/h
###### Calories
644 kcal (0.1 kg bodyfat)
75 RPM
67 RPM
111 RPM
• ###### Avg Pedaling Power
214 W 62.9% FTP
183 W
232 W
0.682
45
632 W
###### xPower
221 W
xPower is an alternative to normalized power.
It's using a 25 sec Exponentially Weighted Moving Average instead of 30 second moving average to Address Variability:
A 30 second moving average (30s MA) does not account for the fact
that physiologic response to stress decays with an exponential rather than linear time
course. A logical / intuitive solution would be to address this with an algorithm that
provided an appropriate decay pattern. Theoretically, an exponentially weighted moving
average should provide a more realistic model. (This was in fact the initial basis for the
development of the xPower and BikeScore algorithms.) We selected a 25 second
exponentially weighted moving average (25s EWMA) based upon the fact that this should
track consistently with oxygen kinetics.
A comprehensive explanation of BikeScore/xPower can be found here: http://www.physfarm.com/bikescore.pdf
BikeScore is a trademark of Dr. Philip Friere Skiba, PhysFarm Training Systems LLC
0.7
###### BikeScoreBikeScore is a proprietary unit designed to measure the overall training stress incurred by the athlete during any training period. A 30 second moving average (30s MA) does not account for the fact that physiologic response to stress decays with an exponential rather than linear time course. A logical / intuitive solution would be to address this with an algorithm that provided an appropriate decay pattern. Theoretically, an exponentially weighted moving average should provide a more realistic model. (This was in fact the initial basis for the development of the xPower and BikeScore algorithms.) We selected a 25 second exponentially weighted moving average (25s EWMA) based upon the fact that this should track consistently with oxygen kinetics. It was was developed by Dr. Phil Skiba. A comprehensive explanation of BikeScore can be found here: http://www.physfarm.com/bikescore.pdf BikeScore is a trademark of Dr. Philip Friere Skiba, PhysFarm Training Systems LLC
41
• m
• m
• BPM
• BPM
• BPM
• deg
• VAM
• km/h
• km/h
• km/h
• RPM
• RPM
• RPM
• W
• W
• W
• W
• W
Zoom Start Tid Lengd Elevation Fart Gj. effekt FTP % Avg power/kg Maks effekt Puls Avg cadence Makskadens
15:10, 0 km 58:03 27.2 km +383/-376 m 28.1 km/h 184 W 54.1% 2.1 W/kg 632 W 73.7% 75 RPM 111 RPM
### Time in power zones
Zone Duration %
1 (0% - 55%) 27 minutes 59 seconds 48.2%
2 (55% - 75%) 14 minutes 58 seconds 25.8%
3 (75% - 90%) 6 minutes 24 seconds 11.0%
4 (90% - 105%) 5 minutes 3 seconds 8.7%
5 (105% - 121%) 2 minutes 5 seconds 3.6%
6 (121% - 150%) 1 minute 35 seconds 2.7%
7 (150% - 13 seconds 0.4%
### Time in HR zones
Zone Duration %
0 (0% - 60%) 6 minutes 42 seconds 11.5%
1 (60% - 72%) 18 minutes 49 seconds 32.4%
2 (72% - 82%) 17 minutes 8 seconds 29.5%
3 (82% - 87%) 13 minutes 37 seconds 23.5%
4 (87% - 92%) 2 minutes 1 second 3.5%
### Speed distribution
Name Start Length Ascent % Duration Speed Avg power Est power W/kg VAM Category Avg HR
Utleira - Stein... 18.0 km 1.8 km 56 m 3.0 % 05:52 18.8 km/h 177 W 210.0 W 2.1 W/kg 573 70.7%
18.1 km 3.1 km 65 m 2.2 % 09:21 19.6 km/h 160.3 W 175.7 W 1.9 W/kg 68.7%
Bakke ved fortu... 23.6 km 0.7 km 39 m 6.0 % 03:19 11.9 km/h 194 W 211.6 W 2.3 W/kg 706 68.7%
23.7 km 1.3 km 48 m 3.6 % 06:54 11.6 km/h 147.1 W 134.9 W 1.7 W/kg 65.2%
Zoom Duration Power Avg power/kg Start Length Elevation
5 seconds 575 W 6.7 W/kg 10.1 km 50 m +2 / -0 m
10 seconds 547 W 6.4 W/kg 10.1 km 97 m +4 / -0 m
30 seconds 443 W 5.2 W/kg 10.0 km 277 m +10 / -0 m
1 minute 367 W 4.3 W/kg 10.0 km 546 m +15 / -1 m
4 minutes 282 W 3.3 W/kg 0.5 km 2058 m +42 / -25 m
5 minutes 287 W 3.3 W/kg 0.5 km 2582 m +56 / -28 m
10 minutes 268 W 3.1 W/kg 0.5 km 5612 m +88 / -64 m
20 minutes 259 W 3.0 W/kg 0.4 km 11.2 km +173 / -139 m
30 minutes 231 W 2.7 W/kg 0.1 km 17.0 km +218 / -253 m
Zoom Duration Speed Start Length Elevation
5 seconds 68.7 km/h 25.7 km 92 m +0 / -3 m
10 seconds 67.4 km/h 25.6 km 185 m +0 / -7 m
30 seconds 61.8 km/h 25.5 km 514 m +0 / -19 m
1 minute 55.0 km/h 25.1 km 916 m +0 / -41 m
4 minutes 39.3 km/h 3.3 km 2620 m +19 / -35 m
5 minutes 37.0 km/h 8.4 km 3085 m +30 / -51 m
10 minutes 36.3 km/h 8.6 km 6057 m +55 / -121 m
20 minutes 35.1 km/h 2.6 km 11.7 km +149 / -173 m
30 minutes 34.2 km/h 0.3 km 17.1 km +215 / -254 m | 1,640 | 4,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-49 | latest | en | 0.896992 |
https://numberworld.info/245076569657920 | 1,603,813,527,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894203.73/warc/CC-MAIN-20201027140911-20201027170911-00415.warc.gz | 447,586,115 | 25,127 | # Number 245076569657920
### Properties of number 245076569657920
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 2 * 2 * 5 * 7 * 7 * 7 * 11 * 11 * 13 * 127 * 11177
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
dee555e09240
Base 32:
6uslau14i0
sin(245076569657920)
0.90684028150799
cos(245076569657920)
0.42147444030986
tan(245076569657920)
2.151590214679
ln(245076569657920)
33.132591806862
lg(245076569657920)
14.389301792873
sqrt(245076569657920)
15654921.579424
Square(245076569657920)
6.0062524995293E+28
### Number Look Up
245076569657920 (two hundred forty-five trillion seventy-six billion five hundred sixty-nine million six hundred fifty-seven thousand nine hundred twenty) is a great figure. The cross sum of 245076569657920 is 73. If you factorisate the figure 245076569657920 you will get these result 2 * 2 * 2 * 2 * 2 * 2 * 5 * 7 * 7 * 7 * 11 * 11 * 13 * 127 * 11177. The number 245076569657920 has 1344 divisors ( 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 20, 22, 26, 28, 32, 35, 40, 44, 49, 52, 55, 56, 64, 65, 70, 77, 80, 88, 91, 98, 104, 110, 112, 121, 127, 130, 140, 143, 154, 160, 176, 182, 196, 208, 220, 224, 242, 245, 254, 260, 280, 286, 308, 320, 343, 352, 364, 385, 392, 416, 440, 448, 455, 484, 490, 508, 520, 539, 560, 572, 605, 616, 635, 637, 686, 704, 715, 728, 770, 784, 832, 847, 880, 889, 910, 968, 980, 1001, 1016, 1040, 1078, 1120, 1144, 1210, 1232, 1270, 1274, 1372, 1397, 1430, 1456, 1540, 1568, 1573, 1651, 1694, 1715, 1760, 1778, 1820, 1936, 1960, 2002, 2032, 2080, 2156, 2240, 2288, 2420, 2464, 2540, 2548, 2695, 2744, 2794, 2860, 2912, 3080, 3136, 3146, 3185, 3302, 3388, 3430, 3520, 3556, 3640, 3773, 3872, 3920, 4004, 4064, 4160, 4235, 4312, 4445, 4459, 4576, 4840, 4928, 5005, 5080, 5096, 5390, 5488, 5588, 5720, 5824, 5929, 6160, 6223, 6292, 6370, 6604, 6776, 6860, 6985, 7007, 7112, 7280, 7546, 7744, 7840, 7865, 8008, 8128, 8255, 8470, 8624, 8890, 8918, 9152, 9680, 9779, 10010, 10160, 10192, 10780, 10976, 11011, 11176, 11177, 11440, 11557, 11858, 12320, 12446, 12584, 12740, 13208, 13552, 13720, 13970, 14014, 14224, 14560, 15092, 15367, 15680, 15730, 16016, 16510, 16940, 17248, 17780, 17836, 18161, 18865, 19360, 19558, 20020, 20320, 20384, 21560, 21952, 22022, 22295, 22352, 22354, 22880, 23114, 23716, 24640, 24892, 25168, 25480, 26416, 27104, 27440, 27940, 28028, 28448, 29120, 29645, 30184, 30734, 31115, 31460, 32032, 33020, 33880, 34496, 35035, 35560, 35672, 36322, 37730, 38720, 39116, 40040, 40640, 40768, 41503, 43120, 43561, 44044, 44590, 44704, 44708, 45760, 46228, 47432, 48895, 49049, 49784, 50336, 50960, 52832, 54208, 54880, 55055, 55880, 55885, 56056, 56896, 57785, 59290, 60368, 61468, 62230, 62920, 64064, 66040, 67760, 68453, 70070, 71120, 71344, 72644, 75460, 76835, 77077, 78232, 78239, 80080, 80899, 83006, 86240, 87122, 88088, 89180, 89408, 89416, 90805, 92456, 94864, 97790, 98098, 99568, 100672, 101920, 105664, 107569, 109760, 110110, 111760, 111770, 112112, 115570, 118580, 120736, 122936, 122947, 124460, 125840, 127127, 132080, 135520, 136906, 140140, 142240, 142688, 145288, 145301, 150920, 153670, 154154, 156464, 156478, 160160, 161798, 166012, 172480, 174244, 176176, 178360, 178832, 181610, 184912, 189728, 195580, 196196, 199136, 199771, 203840, 207515, 215138, 217805, 220220, 223520, 223540, 224224, 231140, 237160, 241472, 245245, 245872, 245894, 248920, 251680, 254254, 264160, 271040, 273812, 280280, 284480, 285376, 290576, 290602, 301840, 307340, 308308, 312928, 312956, 320320, 323596, 332024, 342265, 348488, 352352, 356720, 357664, 363220, 369824, 379456, 385385, 391160, 391195, 392392, 398272, 399542, 404495, 415030, 430276, 435610, 440440, 447040, 447080, 448448, 462280, 474320, 479171, 490490, 491744, 491788, 497840, 503360, 508508, 528320, 537845, 539539, 547624, 547673, 560560, 566293, 581152, 581204, 603680, 614680, 614735, 616616, 625856, 625912, 635635, 647192, 664048, 684530, 696976, 704704, 713440, 715328, 726440, 726505, 739648, 752983, 770770, 782320, 782390, 784784, 799084, 808990, 830060, 860552, 860629, 871220, 880880, 889889, 894160, 924560, 948640, 958342, 980980, 983488, 983576, 995680, 998855, 1017016, 1017107, 1075690, 1079078, 1095248, 1095346, 1121120, 1132586, 1162304, 1162408, 1207360, 1229360, 1229470, 1233232, 1251824, 1271270, 1294384, 1328096, 1352417, 1369060, 1393952, 1398397, 1419479, 1426880, 1452880, 1453010, 1505966, 1541540, 1564640, 1564780, 1569568, 1598168, 1598311, 1617980, 1660120, 1721104, 1721258, 1742440, 1761760, 1779778, 1788320, 1849120, 1897280, 1916684, 1961960, 1967152, 1991360, 1997710, 2034032, 2034214, 2151380, 2158156, 2190496, 2190692, 2242240, 2265172, 2324816, 2395855, 2458720, 2458940, 2466464, 2503648, 2542540, 2588768, 2656192, 2697695, 2704834, 2738120, 2738365, 2787904, 2796794, 2831465, 2838958, 2905760, 2906020, 3011932, 3083080, 3129280, 3129560, 3139136, 3196336, 3196622, 3235960, 3320240, 3442208, 3442516, 3484880, 3523520, 3559556, 3576640, 3698240, 3764915, 3833368, 3833711, 3923920, 3934304, 3995420, 4068064, 4068428, 4302760, 4303145, 4316312, 4380992, 4381384, 4449445, 4530344, 4649632, 4791710, 4917440, 4917880, 4932928, 5007296, 5085080, 5085535, 5177536, 5270881, 5395390, 5409668, 5476240, 5476730, 5593588, 5662930, 5677916, 5811520, 5812040, 6023864, 6024403, 6166160, 6229223, 6259120, 6392672, 6393244, 6471920, 6640480, 6762085, 6884416, 6885032, 6969760, 6991985, 7097395, 7119112, 7119749, 7529830, 7666736, 7667422, 7847840, 7868608, 7990840, 7991555, 8136128, 8136856, 8605520, 8606290, 8632624, 8762768, 8898890, 9060688, 9299264, 9466919, 9583420, 9788779, 9835760, 9936353, 10170160, 10171070, 10541762, 10790780, 10819336, 10952480, 10953460, 11187176, 11188177, 11325860, 11355832, 11624080, 12047728, 12048806, 12332320, 12458446, 12518240, 12785344, 12786488, 12943840, 13280960, 13524170, 13770064, 13939520, 13983970, 14194790, 14238224, 14239498, 15059660, 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843340960, 843416420, 858784795, 861489629, 874399064, 895054160, 904208123, 927758062, 963904480, 973762594, 984559576, 996675680, 996764860, 999313216, 1014927485, 1033380712, 1060294928, 1092998830, 1096343248, 1096441346, 1112871536, 1125210944, 1139159840, 1181006528, 1202298713, 1226787520, 1230699470, 1249141520, 1253075824, 1291725890, 1325368660, 1349466272, 1370429060, 1374055672, 1391089420, 1406513680, 1420898479, 1476258160, 1514707040, 1530198362, 1566204640, 1566344780, 1589816480, 1594823776, 1623883976, 1686681920, 1686832840, 1717569590, 1722979258, 1748798128, 1790108320, 1808416246, 1855516124, 1927808960, 1947525188, 1969119152, 1993351360, 1993529720, 2029854970, 2066761424, 2120589856, 2185997660, 2192686496, 2192882692, 2225743072, 2232840467, 2278319680, 2319395155, 2404597426, 2434406485, 2461398940, 2498283040, 2506151648, 2583451780, 2650737320, 2698932544, 2740858120, 2741103365, 2748111344, 2782178840, 2813027360, 2841796958, 2952516320, 3029414080, 3060396724, 3132409280, 3132689560, 3179632960, 3189647552, 3247767952, 3373665680, 3435139180, 3445958516, 3497596256, 3580216640, 3616832492, 3711032248, 3825495905, 3895050376, 3938238304, 3987059440, 4059709940, 4133522848, 4241179712, 4307448145, 4371995320, 4385372992, 4385765384, 4451486144, 4465680934, 4521040615, 4638790310, 4809194852, 4868812970, 4922797880, 4996566080, 5012303296, 5166903560, 5301474640, 5355694267, 5481716240, 5482206730, 5496222688, 5564357680, 5626054720, 5683593916, 5905032640, 6011493565, 6030427403, 6120793448, 6265379120, 6329456861, 6495535904, 6747331360, 6870278360, 6891917032, 6995192512, 7104492395, 7233664984, 7422064496, 7650991810, 7790100752, 7876476608, 7974118880, 8119419880, 8267045696, 8416090991, 8614896290, 8743990640, 8771530768, 8931361868, 9042081230, 9277580620, 9618389704, 9737625940, 9845595760, 9946289353, 10333807120, 10602949280, 10711388534, 10963432480, 10964413460, 10992445376, 11128715360, 11164202335, 11367187832, 12022987130, 12060854806, 12241586896, 12530758240, 12658913722, 12991071808, 13494662720, 13740556720, 13783834064, 14208984790, 14467329968, 14844128992, 15301983620, 15580201504, 15629883269, 15948237760, 16238839760, 16832181982, 17229792580, 17487981280, 17543061536, 17862723736, 18084162460, 18555161240, 19236779408, 19475251880, 19691191520, 19892578706, 20667614240, 21205898560, 21422777068, 21926864960, 21928826920, 22257430720, 22328404670, 22734375664, 24045974260, 24121709612, 24483173792, 25061516480, 25317827444, 26778471335, 27481113440, 27567668128, 28417969580, 28934659936, 29688257984, 30152137015, 30603967240, 31160403008, 31259766538, 31647284305, 32477679520, 33664363964, 34459585160, 34975962560, 35086123072, 35725447472, 36168324920, 37110322480, 38473558816, 38950503760, 39382383040, 39785157412, 41335228480, 42080454955, 42845554136, 43857653840, 44656809340, 45468751328, 48091948520, 48243419224, 48966347584, 49731446765, 50635654888, 53556942670, 54962226880, 55135336256, 56835939160, 57869319872, 58912636937, 60304274030, 61207934480, 62519533076, 63294568610, 64955359040, 67328727928, 68919170320, 69624025471, 71450894944, 72336649840, 74220644960, 76947117632, 77901007520, 78149416345, 79570314824, 84160909910, 85691108272, 87715307680, 89313618680, 90937502656, 96183897040, 96486838448, 99462893530, 101271309776, 107113885340, 109409182883, 113671878320, 117825273874, 120608548060, 122415868960, 125039066152, 126589137220, 134657455856, 137838340640, 139248050942, 142901789888, 144673299680, 148441289920, 155802015040, 156298832690, 159140629648, 168321819820, 171382216544, 175430615360, 178627237360, 192367794080, 192973676896, 198925787060, 202542619552, 214227770680, 218818365766, 227343756640, 235650547748, 241217096120, 244831737920, 250078132304, 253178274440, 269314911712, 275676681280, 278496101884, 289346599360, 294563184685, 312597665380, 318281259296, 336643639640, 342764433088, 348120127355, 357254474720, 384735588160, 385947353792, 397851574120, 405085239104, 428455541360, 437636731532, 454687513280, 471301095496, 482434192240, 500156264608, 506356548880, 538629823424, 547045914415, 556992203768, 589126369370, 625195330760, 636562518592, 673287279280, 696240254710, 714508949440, 765864280181, 795703148240, 856911082720, 875273463064, 942602190992, 964868384480, 1000312529216, 1012713097760, 1094091828830, 1113984407536, 1178252738740, 1250390661520, 1346574558560, 1392480509420, 1531728560362, 1591406296480, 1713822165440, 1750546926128, 1885204381984, 1929736768960, 2025426195520, 2188183657660, 2227968815072, 2356505477480, 2500781323040, 2693149117120, 2784961018840, 3063457120724, 3182812592960, 3501093852256, 3770408763968, 3829321400905, 4376367315320, 4455937630144, 4713010954960, 5001562646080, 5569922037680, 6126914241448, 7002187704512, 7658642801810, 8752734630640, 9426021909920, 11139844075360, 12253828482896, 15317285603620, 17505469261280, 18852043819840, 22279688150720, 24507656965792, 30634571207240, 35010938522560, 49015313931584, 61269142414480, 122538284828960, 245076569657920 ) whith a sum of 812023717478400. The figure 245076569657920 is not a prime number. The figure 245076569657920 is not a fibonacci number. The number 245076569657920 is not a Bell Number. 245076569657920 is not a Catalan Number. The convertion of 245076569657920 to base 2 (Binary) is 110111101110010101010101111000001001001001000000. The convertion of 245076569657920 to base 3 (Ternary) is 1012010202002110111222211121101. The convertion of 245076569657920 to base 4 (Quaternary) is 313232111111320021021000. The convertion of 245076569657920 to base 5 (Quintal) is 224110313303313023140. The convertion of 245076569657920 to base 8 (Octal) is 6756252570111100. The convertion of 245076569657920 to base 16 (Hexadecimal) is dee555e09240. The convertion of 245076569657920 to base 32 is 6uslau14i0. The sine of the figure 245076569657920 is 0.90684028150799. The cosine of the number 245076569657920 is 0.42147444030986. The tangent of the number 245076569657920 is 2.151590214679. The square root of 245076569657920 is 15654921.579424.
If you square 245076569657920 you will get the following result 6.0062524995293E+28. The natural logarithm of 245076569657920 is 33.132591806862 and the decimal logarithm is 14.389301792873. I hope that you now know that 245076569657920 is very amazing figure! | 7,401 | 15,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-45 | latest | en | 0.446728 |
https://www.hand2mind.com/item/handson-standards-common-core-fractions-teachers-resource-guide-grade-5/1790 | 1,604,010,915,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00616.warc.gz | 732,962,726 | 41,184 | # Hands-On Standards®, Common Core Fractions Teacher's Resource Guide, Grade 5
Enrich differentiated fractions instruction with this Common Core teacher guide for the 5th grade
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Hands-On Standards Fractions Common Core Grade 5 Teacher Resource Guide develops students’ conceptual understanding of fractions with targeted, hands-on instruction. Includes 15 lessons focusing on Adding and Subtracting Fractions, Multiplying and Dividing Fractions, and much more.Each hands-on, standards-focused lesson uses one or more of the following most common manipulatives: Color Tiles, Deluxe Rainbow Fraction® Circles, Deluxe Rainbow Fraction® Squares, Fraction Dice, Fraction SAFE-T® Rulers, Fraction Tower® Equivalency Cubes, and Fraction Number Lines.
Hands-On Standards Fractions Common Core Grade 5 Teacher Resource Guide uses engaging hands-on activities to build fluency with fractions. Common Core standards-based lessons include warm-up activities, foundational skill practice, guided and independent practice, remediation, and enrichment opportunities. Differentiation options outline vocabulary use and support, making them perfect for ELL learners.
TOPICS
• Add and subtract unlike fractions
• Add and subtract mixed numbers
Multiply and Divide Fractions
• Fractions as division
• Fraction tower times
• Fraction multiplication
• Division with unit fractions
RESOURCES
Hands-On Standards Fractions Teacher Guide View Sample Lesson
INCLUDES
• Teacher Resource Guide
Brand Hands-On Standards® Fractions Math Fractions, Decimals and Percents(Math) Fractions, Decimals and Percents(Math) 10 5 Each IN79545 In Stock | 359 | 1,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-45 | latest | en | 0.785984 |
http://trilema.com/2014/all-you-may-ever-need-to-know-of-financial-planning/?b=And+your+life&e=game | 1,600,613,613,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198213.25/warc/CC-MAIN-20200920125718-20200920155718-00735.warc.gz | 138,996,834 | 9,559 | # All you may ever need to know of financial planning
Saturday, 01 February, Year 6 d.Tr. | Author: Mircea Popescu
Suppose one day you have sold all that you own, and it is in your pocket as you walk down the street, ten Bitcoin or ten million or whatever it may be.
Suppose that day walking down the street you run into a game, which in a provably fair fashioni offers you 2 in 3 odds to win double your wager back.
The question before you is, what should you do ? Should you press on or stop and play ?
The answer before you is that you should indeed stop and play.ii This notwithstanding any nonsense you may misrepresent as "morals" (such as "don't gamble") and so forth. In point of fact any game with a positive expected value is to be played, period.
So now, that question answered, you stop and play, which sprouts another question : how much should you wager ?
The answer is known as the Kelly criterion, which simply says that you should take your return, multiply it by the odds to win, subtract the odds to lose, and divide the lot by the return. That's the fraction of your money you should wager at any point on any +EV propositioniii. Numerically in our case, (.2/3 * 1 - 1/3)/1= 1/3. We have 66.(6)% odds to win, and 33.(3)% odds to lose, and this offers us a round and thus very convenient answer : at every juncture bet a third of your dough.
Why not bet it all ? Well, what happens if you do bet it all and lose ? You're screwed. And if you win... then you should bet again, right ? What if then you lose ? Still screwed. And if you win... it'll just take a little longer, but overall you're definitely screwed, because sooner or later that 1/3 chance will actualise and that's the end of you. So no, all in is not really a sane strategy, inasmuch as the only guaranteed result is you being screwed.
The Kelly criterion optimises long term gains, which is to say it guarantees that out of a flock of people in which you are the only one playing this way, you will actually have more money at the end of any number of rounds than the average of the rest of the flock. No matter how large the flock, and no matter what other strategies they employ, you'll be on average better off.
This has however sharply discounted that aforementioned cost of opportunity. Suppose you are fifty, and that fortune you're carrying with you is all your life's work, all you've ever made. All of it. The Kelly criterion may be a little too aggressive for you, seeing how it still makes it perfectly possible to lose ten years' worth of hard work in just a few quick seconds.
But what, on the contrary, what if you're seventeen, and that ten is the result of an hour's work mowing a lawn ? Well in that case... Kelly's too slow for you, because even if you lose a lot of your bankroll, you've really not lost so very much in real terms.
That's it : when you're young you want to take bigger risks, because you can easily afford to cover unfortunate turns of events. When you're old you want to take smaller risks, because you ain't covering anything anymore. In any case you wish to strictly play +EV games.
And your life, all of it, from end to end, is nothing but you walking down the street and ending in front of a proposed game.iv
———
1. This "provably fair" is a term of art in the Bitcoin space, and the 2nd most important fundamental reason why fiat gambling has absolutely no prayer in the future, irrespective of what anyone may decree or clamor on the topic. The 1st most important fundamental reason is that Bitcoin gambling is much cheaper to do, and so even without that 2nd substantial jump in quality it'd still run fiat out of the marketplace. In any case : you're well advised to form a good understanding of what "provably fair" actually means, as a more easily digestible and accessible step to actually understanding Bitcoin whole. []
2. A counterargument could be constructed on the cost of opportunity - if you're on your way to getting married perhaps you'd be better advised to carry on, especially if the houri scorned is liable to send you off to meet the maker. []
3. If you're confronting a -EV proposition, the Kelly criterion comes out with a negative fraction, indicating you should bet the other way, which perhaps might mean you should just start your own stand next to the one you've encountered. []
4. And for what it's worth, Wall Street does no better than this, nor does it do anything else, for all the pretense. []
Category: Actiuni si Optiuni
One Response
1. [...] for that potential are proportionally just as handsome if it hits. Without diving into the Kelley Criterion and questions of optimization the first question you should ask of any bet is whether the best outcome adequately rewards you for [...]
»
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https://byjus.com/question-answer/a-compass-has-a-small-balanced-pointer-that-always-points-north-south-this-is-because-2/ | 1,642,926,641,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304217.55/warc/CC-MAIN-20220123081226-20220123111226-00606.warc.gz | 229,001,441 | 16,526 | Question
# A compass has a small balanced pointer that always points North-South. This is because:
A
The stars exert a force on the needle
B
The earth has a magnetic field and the pointer is magnetic
C
Gravity makes the needle point this way
D
The compass needle points to cold places
Solution
## The correct option is D The earth has a magnetic field and the pointer is magnetic The magnetic compass is the oldest instrument for navigation and has been a vital tool for navigators at sea for centuries. The Earth has a substantial magnetic field. The poles of the earth act as magnetic poles, as such a bar magnet, when suspended freely gets attracted towards the poles of Earth. So, it aligns itself in a North-South direction. Physics
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Thread starter 1 week ago
#21
(Original post by old_engineer)
The most direct way of finding n, the direction vector perpendicular to the plane, is to use the direction vectors of L1 and L2. n must be perpendicular to both the line direction vectors.
(Original post by ghostwalker)
In red is a vector parallel to the plane.
The rest of the line doesn't make sense in relation to that.
What do you mean by "the vector to the plane".
The dot product of n with any vector in the plane or parallel to a vector in the plane will be zero.
First one defines a point on l1 and second one is a point on l2.
Please think about what you're trying to say - I am really struggling to make sense of your posts relating to this question.
(Original post by DFranklin)
Yes (noting I haven't actually checked those are the correct direction vectors). And you need to find a solution *other* than x=y=z=0.
Do you know what the vector cross-product is?
thank all 3 of you ever so much! I got 6x + 2y 3z - 14 = 0, as my final answer and it's correct!!!!!!!
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1 week ago
#22
(Original post by Maths&physics)
yes, I know and thats where I got the dot product from.
If you know what a vector cross product is, the standard method is to use it to get a vector perpendicular to both direction vectors.
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Thread starter 1 week ago
#23
(Original post by DFranklin)
If you know what a vector cross product is, the standard method is to use it to get a vector perpendicular to both direction vectors.
yes, that's what I did but I was confused whether the known points of the lines where on the plane - which they are, but not in the way I thought because I drew the lines crossing through the planes, rather than on the plane.
Last edited by Maths&physics; 1 week ago
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1 week ago
#24
so how did you get on with part c of this question?
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Thread starter 1 week ago
#25
(Original post by begbie68)
so how did you get on with part c of this question?
I got the right answer, thanks for asking.
.....are you offering help or would you like some?
Last edited by Maths&physics; 1 week ago
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# How to tally number of audits?
Hello, I am new to working with excel and I was given a task to create a new Daily Report for the facility that I work at. So far I have completed 90% of what I was asked to do but I cannot figure out how to tally the number of audits/reports each employee does in a daily basis. I have included a picture of a piece of the daily report that I have been working on.
The yellow boxes are drop down boxes that the supervisor has the option of typing/clicking in the drop down list (combo box macro) to show who was at work for that day. As you can see you can have any order of employees in one day where Bill may be the first to clock in or maybe even last to clock in. I want to be able to tally however many times he completes an audit or report everyday into a conjunctive monthly tally chart, and yearly.
Is there a function that can accomplish this or perhaps a macro? Thank you in advance!
-C
2 Replies
best response confirmed by CRick1997 (New Contributor)
Solution
# Re: How to tally number of audits?
@CRick1997 Not convinced that the way you collect your data is the most effective, but from judging the screenshot you should be able to use SUMIF.
Personally, I would collect all data in a single table (date, employee, #audits,#reports, etc. ) and create summaries from that table. Per day, per employee, whatever, for example, by using a pivot table.
The attached file contains a mock-up of your schedule with a summary using SUMIF. From column L and onwards, I demonstrated the use of a single structured table, summarising it using a pivot table. You say you are new to Excel, so it may be a bit overwhelming to begin with. But I believe it's better to learn some of the basic features first and do thing right from the start.
# Re: How to tally number of audits?
@Riny_van_Eekelen Thank you! This actually helped me a lot! I see why you said my chart wasn't the best for this task, so I changed it up a little bit so that I can just input the entire column into the function.
-C | 477 | 2,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.963526 |
https://www.jovoto.com/projects/300house/ideas/12619 | 1,685,479,323,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646144.69/warc/CC-MAIN-20230530194919-20230530224919-00097.warc.gz | 886,216,688 | 64,716 | "Tubular" Concrete House
Simple 8ft by 8ft cement home made from sewn tarps and concrete. The tarps act as forms when concrete is pour providing a simple cost effective \$300 house.
Solution
Before I get too far into the description of the design, I think this proposed design is an excellent design or at least a design that should be given to Engineering Universities in India or other countries to give to their engineering students to continue to work on the concept. I think improving and optimizing my proposed design would be interesting design project for engineering students. If you have connections to these universities please forward my design information.
I believe I can produce this house for under \$300. I would have to spend some time doing an in depth structural analysis to verify optimal dimensions to get exact dimensions numbers, but I think my cost estimates are close. Rebar is sold by the pound (\$740 per ton) or kilogram so I was able to get reasonable costs estimates for the rebar. I used the actual cost I paid for the tarp material to figure the cost per square foot. Concrete is made from cement, sand and gravel. I found the cost of cement in India and estimated the cost of sand and gravel to be approximately \$10 per yard. I have estimated the cost of concrete per yard is \$28 per yard.
My vision in creating this \$300 house design was to build 1,000�s of houses simultaneously without costly concrete forms. The proposed design starts with building four corner posts approximately 6 ft high. This is done by supporting a tarp post bag using the internal rebar structure and pouring only a foot of concrete at a time and allowing it to harden before another foot is poured. The tarp bag is not lifting the concrete it is only holding from flowing outward. I estimate the tarp I used could hold 1 vertical foot of concrete without damage in the corner post tarp. This might seem like a slow process, it would take six days to pour the post but I this processes does not require forms. Now imagine creating a small village and in approximately six days I would have all the corner posts created for every house in that village. Think of this design concept of building houses in parallel and if you designed using forms you build in series. The more houses you build simultaneously the more you want to build in parallel.
When the concrete in the corner post reaches the bottom of the header beam rebar, the concrete should be allowed to cure until desired strength compressive strength in the concrete is achieved. Now it is time to hang the wall tarp onto the header rebar beam. (The wall tarp could easily be mass produced.) The upper edges of the wall tarp should have a reinforced hem that a 12mm diameter rebar slid into the hem loop. Brackets made from rest on the header rebar beam and raise the top of the hem(rebar) above the header rebar beam. Rebar would be inserted at the footer location thru the wall tarp to tie the bottom of the posts together with rebar. Remember to put duct tape or something at the end of the rebar so you do not damage the tarp during inserting. Rebar would be inserted at the window location and where needed in the wall for additional strength. Concrete can now be poured into the wall tarp. The length of the wall tarp is designed so the footer rests on the ground so the wall tarp does not hold up the concrete but only keeps concrete from traveling outward. There are some issues that the footer does not shrink in size as the tube part of the wall, but I believe this can easily be solved by sewing the footer to a smaller length. Because the diameter of the wall tubes are smaller than the diameter of the corner posts the stresses in the wall tarp are less and higher vertical amount of concrete can be poured. I was able to pour six feet at one time on my prototype. I would use strong enough tarp material to pour to the height of the bottom of the header rebar beam. Fill window tubes with wet sand and gravel mixture at the window location. I would then allow the concrete in the wall tube to harden. This would then allow you to readjust the wall tarp at the header beam to fill concrete completely around the header rebar beam to the top of the wall tarp. The roof could be finished a variety of ways. I have proposed creating concrete tubes on the ground and then lifting them onto the roof and concreting the tubes together. The house is now livable after cutting openings in the wall tarp for the window and installing the door. The tarp can be left on until if the tarp falls apart or the tarp can be removed and the joints between the tubes filled with mortar.
Update... The 2nd generation header design will increase the estimated cost to \$332 due to the more rebar. However if I use a cheaper door, eliminate foundation and either leave the tarp or mud the wall joints I estimate the cost to be \$296.
I performed a simple test using to simulate how the post will fill with concrete using water (See video below). Polyethylene bag or sheet like material could also be used with heat welded seams if ithe material can be designed to withstand the applied forces and induced stress.
Other entries in this project | 1,132 | 5,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-23 | latest | en | 0.932803 |
https://stats.stackexchange.com/questions/434885/akaike-information-criterion-aic-derivation/435217 | 1,580,185,470,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251773463.72/warc/CC-MAIN-20200128030221-20200128060221-00407.warc.gz | 660,904,175 | 31,237 | # Akaike Information Criterion (AIC) derivation
I am trying to understand the Akaike Information Criterion (AIC) derivation and this resource explains quite well although there are some mysteries for me.
First of all it considers $$\hat{\theta}$$ as the parameters resulting from Maximum Likelihood Estimation (MLE) and it says the difference form the true model can be computed using the Kullback-Leibler distance:
$$\int p(y) \log p(y) dy - \int p(y) \log \hat{p}_j(y) dy$$
Minimizing such a distance is equivalent to maximize the second term referred to as $$K$$. One trivial estimation of $$K$$ is
$$\bar{K} = \frac{1}{N} \sum_{i=1}^N \log p(Y_i, \hat{\theta}) = \frac{\ell_j(\hat{\theta})}{N}$$
Suppose $$\theta_0$$ minimize $$K$$ and let
$$s(y,\theta) = \frac{\partial \log p (y, \theta)}{\partial \theta}$$
be the score and $$H(y,\theta)$$ the matrix of second derivatives.
1. The author later in the proof uses the fact the score has $$0$$ mean: based on what?
Then it says: let $$Z_n = \sqrt{n} (\hat{\theta} - \theta_0)$$
and recall that $$Z_n\rightarrow \mathcal{N}(0, J^{-1}VJ^{-1})$$
where $$J = -E[H(Y,\theta_0)]$$
and $$V= Var(s(Y, \theta_0)$$.
1. Why $$Z_n = \sqrt{n} (\hat{\theta} - \theta_0)$$? Where does it come from?
Then let
$$S_n = \frac{1}{n} \sum_{i=1}^Ns(Y_i, \theta_0)$$
It says that by the Central limit theorem $$\sqrt{n}S_n \rightarrow \mathcal{N}(0,V)$$
1. $$V$$ comes from the definition but why $$0$$ mean? Where does it come from?
2. At some point it says: $$J_n = -\frac{1}{n}\sum_{i=1}^NH(Y_i, \theta_0) - \xrightarrow{P} J$$ What's the meaning of $$- \xrightarrow{P} J$$?
EDIT
Additional question. Defining $$K_0 = \int p(y) \log p(y, \theta_0) dy$$
and $$A_N = \frac{1}{N} \sum_{i=1}^N(\ell(Y_i,\theta_0)-K_0)$$ Why $$E[A_N] =0$$?
• yes you are right, I was reasoning about something else and got confused. – Francesco Boi Nov 7 '19 at 11:38
Consider scalar parameters $$\theta_0$$ and the corresponding scalar estimate $$\hat \theta$$ for simplicity.
I will answer Q1 and Q3 which are essentially asking why is the mean of the score function $$\Bbb{E}_{\theta}(s(\theta)) =0$$. This is a widely known result.. To put it simply, Notice that score function $$s(\theta)$$ depends of the random observations $$X$$. We can take its expectation as follows:
\begin{align} \Bbb{E}_{\theta}(s) & = \int_x f(x;\theta) \frac{\partial \log f(x;\theta)}{\partial \theta} dx \\ &=\int_x \frac{\partial f(x;\theta)}{\partial \theta} dx = 0 \qquad \text{(exchanging integral and derivative)} \end{align}
Now, notice that $$S_n$$ is nothing but averaged-sum of score functions based on independent observations. Hence, its expectation will also be zero.
For Q2) the motivation is to find study the asymptotic properties of our estimator wrt to the true parameter. Let $$\hat{\theta}$$ be the maximizer of $$L_{n}(\theta)=\frac{1}{n} \sum_{i=1}^{n} \log f\left(X_{i} | \theta\right)$$. Now, by meanvalue theorem \begin{align} 0=L_{n}^{\prime}(\hat{\theta}) & =L_{n}^{\prime}\left(\theta_{0}\right)+L_{n}^{\prime \prime}\left(\hat{\theta}_{1}\right)\left(\hat{\theta}-\theta_{0}\right) \quad \text{(for some \theta_1 \in [\hat\theta,\theta_0])}\\ \implies & \left(\hat{\theta}-\theta_{0}\right) = \frac{L_{n}^{\prime}\left(\theta_{0}\right)}{L_{n}^{\prime \prime}\left(\hat{\theta}_{1}\right)} \end{align}
Consider the numerator: \begin{align} \sqrt{n}\left(\frac{1}{n} \sum_{i=1}^{n} l^{\prime}\left(X_{i} | \theta_{0}\right)-\mathbb{E}_{\theta_{0}} l^{\prime}\left(X_{1} | \theta_{0}\right)\right) & = \sqrt{n}(S_n - \Bbb{E}(S_n)) \\ & \rightarrow N\left(0, \operatorname{Var}_{\theta_{0}}\left(l^{\prime}\left(X_{1} | \theta_{0}\right)\right)\right) = N(0,V) \end{align}
Now, the denominator $$L^{''}_n$$ coverges to the Fisher's information $$(J)$$ by LLN. Therefore, for the scalar paramters case, we can see that $$\sqrt{n}(\hat \theta - \theta_0) \rightarrow N(0,\frac{V}{J^2})$$ | 1,327 | 3,935 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 38, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-05 | latest | en | 0.807003 |
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# CS201 Asig 2 Sol
Assalam-o-Alikum
ye solution sahi ha yar?
agar sahi hai tau asaan tareqe se bana de koi :P
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### Replies to This Discussion
Aslam-e-alaikum bhai is mein kya masla mein ne ik are upload with comment ho sakta hai is bar ap ko smajh per jaey.
sath mein ye idea soulution tha. asli wali mein ne abhi upload ki hai.changes zaror karna.
#include<iostream.h>
#include<conio.h>
float avg(int [],int);
float low(int [],int);
float high(int [],int);
int main()
{
int n;
cout"Enter the number of consecutive days to read their temperature";
cin>>n;
coutendl;
int arr[10];
float l,h,a;
for(int i=0;i<n;i++)
{
cout"Enter the temperature for day "i+1" : ";
cin>>arr[i];
coutendl;
}
coutendl;
a=avg(arr,n);
cout"The Average Temperature is "aendl;
h=high(arr,n);
cout"The Highest temperature is "hendl;
l=low(arr,n);
cout"The Lowest temperature is "l;
getch();
}
float avg(int arr[],int n)
{
float sum=0.0,a;
for(int i=0;i<n;i++)
{
sum=sum+arr[i];
}
a=sum/5.0;
return (a);
}
float low(int arr[],int n)
{
float l=100.0;
for(int i=0;i<n;i++)
{
if(arr[i]<l)
l=arr[i];
}
return (l);
}
float high(int arr[],int n)
{
float h=0.0;
for(int i=0;i<n;i++)
{
if(arr[i]>h)
h=arr[i];
}
return (h);
}
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# Sum rows that are not children
Options
edited 12/09/19
I am creating a SS that contains a row for each event and below children for event expenses. The top/parent row calculates the sum of the children beneath it.
I would like to create a Totals row that calculates the sum of all the top/parent rows. Is this something SS does?
Tags:
• ✭✭✭✭✭✭
Options
Just indent all of the events under another parent.
The sum(children()) will only sum up the first level of children.
I hope this helps.
Craig
• ✭✭✭✭✭✭
Options
In situations like this I simply sum the entire column, including children, and divide the total by two since the parent rows already sum the children. That way I'm not sensitive to the parent/children structure.
• ✭✭✭✭✭✭
Options
Jim,
Except if there are more than two levels, then you'd need to know to divide by 3.
I have many sheets indented more than 2 times, but not all of them.
Craig
• edited 06/01/16
Options
Craig, that worked perfectly (your first post) ! I just had to change my totals to the top of my sheet vs. the bottom, but that actually works better anyhow.
I'm not sure I understand how dividing by 2 or 3 does the trick, but I'm good to go.
• ✭✭✭✭✭✭
Options
Craig and Jim,
Yes, I have project budget (plan, commitments, disbursements) in sheets that sometimes have 4 or 5 levels, like this one:
This specific sheet has 220 rows, with lots of children, grandchildren, great-grandchildren...
I would be lost without the sum(children()) formula.
Atus
• edited 02/20/17
Options
Dear All,
How do we count the rows, after applying a filter?
Any idea?
-
Chander.
• ✭✭✭✭✭✭
Options
Amanda,
In Jim's suggestion, he ends up counting everything twice so he divides by 2.
I was thinking that if you had three levels, you would count everything three times, but I might be wrong.
Craig
This discussion has been closed. | 566 | 2,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.941918 |
https://blogs.oracle.com/jag/entry/cosine_redux | 1,433,018,934,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207932705.91/warc/CC-MAIN-20150521113212-00049-ip-10-180-206-219.ec2.internal.warc.gz | 861,733,792 | 9,819 | ### Cosine redux
It's been quite a while since my last blog entry: I had a bit of a technology meltdown that I'm not quite done with yet :-(
I was doing some recreational hacking over the holidays that involved evaluating cosines. I ended up doing (once again!) an implementation of cosine (don't ask why). Given the confused flamage that my previous posts on cosine generated, I figure that showing some code would be useful. I would never have thought that cosine would generate so much email traffic. Yes, I know about Taylor series. No I wasn't trying to insult centuries of standard mathematical practice. Performance is often the art of cheating carefully.
So, here's my implementation of cosine, with its argument in turns (1 turn == 360 degrees):
```public static float cost(float f) {
int bits = Float.floatToRawIntBits(f);
int mantissa = (bits&0x7FFFFF)|(1<<23);
int shift = (bits<<1>>>24)-127+9; // exponent, unbiased, with shift
if(shift>=32 || shift<=-32) return 1;
int fractionBits = shift>=0 ? mantissa<<shift : mantissa>>-shift;
switch(fractionBits>>>30) { // Quadrant is top two bits
case 0: return cosTable[tableIndex];
case 2: return -cosTable[tableIndex];
}
}```
Lets go through this slowly:
1. int bits = Float.floatToRawIntBits(f);
Get the IEEE 754 bits
2. int mantissa = (bits&0x7FFFFF)|(1<<23);
The mantissa is the bottom 23 bits - to which the hidden bit must be prepended.
3. int shift = (bits<<1>>>24)-127+9;
Extract the exponent, correct for the exponent bias, then add a bias to move the binary point to the top of the word.
4. if(shift>=32 || shift<=-32) return 1;
If the shift is too large, the fraction bits would be zero, therefore the result is 1.
5. int fractionBits = shift>=0 ? mantissa<<shirt : mantissa>>-shift;
Shift the mantissa so that it's a fixed point number with the binary point at the top of the 32 bit int. The magic is in what's not here: because the argument is in turns, I get to ignore all of the integer bits (range reduction made trivial); and because it's the cosine function, which is symmetric about the origin, I get to ignore the sign.
The top two bits are the quadrant... extract the bits below that to derive a table index.
7. switch(fractionBits>>>30) {
One case for each quadrant. This switch could be eliminated by making the table 4 times larger.
8. case 0: return cosTable[tableIndex];
Yes! It's just a table lookup! Truly trivial.
...
Since this is just a table lookup, the resulting approximation can be pretty jagged if the table is small. But it's easy to tune the table size depending on accuracy needs. The smaller the table is, the higher the cache hit rate will be, and the more likely it is that the whole table will fit in cache.
Table lookups are a very common way to implement mathematical functions, particularly the periodic ones like cosine. There are all kinds of elaborations. One of the most common for improving accuracy is to do some sort of interpolation between table elements (linear or cubic, usually).
Jeff, I think the problem that your solutions are not helping to solve here is the performance, i.e. the one that Java is having in calling trig functions. Specifically I cannot see how CORDIC will eliminate the runtime overhead that argument reduction brings about in Java when calling x86 hardware trig instructions.
Posted by Alex Lam on January 21, 2006 at 08:08 AM PST #
It's a tradeoff. The better you understand the precision requirements of your application, the better you can tune it. There is a spectacular amount of literature on different tricks on evaluating functions like cosine. The less precision you need, the faster you can make it go. The problem with a general-purpose library like Java's is that it has no information about the precision requirements of the application invoking it. So the safe thing to do is to produce the most accurate answer possible. The difficulty is that in this case, the x86 hardware doesn't do a great job, so we correct for it. The grotesque solution would be to define functions that had extra parameters for (at least) the number of result bits that were needed to be accurate, and the range over which the argument might vary: if these were constants, then a good optimizer could substitute appropriatly tricked-up code fragments.
Posted by James Gosling on January 21, 2006 at 02:02 PM PST #
It is inconvenient that radians depend on PI and PI is an irrational number. At the moment I am writing an arbitrary-precision arithmetics package and is working on implementing the non-linear (trig, exp, log, pow etc.) functions. Apart from revising my maths knowledge I have also looked up (seemingly) similar projects on the Internet (JUMP, JScience, JCM etc.) for possible inspirations. Now I think I have a better sense of direction on how to work things out; I sincerely hope that after all the pain-staking processes I'd come up with an almighty useful Java library ;)
Posted by Alex Lam on January 22, 2006 at 01:04 AM PST #
Hi Alex
I'm searching for a good arbitrary-precision arithmetics package for numerical computations with support for trigonometric functions.
Can you suggest one?
Posted by Axel Kramer on January 22, 2006 at 07:59 PM PST #
Posted by nile black on January 22, 2006 at 10:32 PM PST #
What ever happened to the Java Grande project? Among many other interesting things they were working on - I thought one of the proposals was for a math library that could be processor dependent. In cases in which it was more important to have performance than to have identical answers on all platforms there could be specific optimizations. These would be contained in libraries where it would be clear that this was the trade off you were making.
Posted by Daniel Steinberg on January 22, 2006 at 11:38 PM PST #
"The grotesque solution would be to define functions that had extra parameters for (at least) the number of result bits that were needed to be accurate, and the range over which the argument might vary: if these were constants, then a good optimizer could substitute appropriatly tricked-up code fragments."
What's so grotesque about that? If every function requires information about the error bounds, then enforcing the desired error limits becomes quite complicated. This is doubly true, while there is no standard way to track and describe error bounds. Perhaps JSR 275 will address that. http://www.jcp.org/en/jsr/detail?id=275
I would think that "The Jackpot Way" to solve this problem would be to provide an editor for complex functions and equations. Then a function evaluation engine can evaluate the entire function, tracking the error along the way. Radically different evaluation methods could be chosen, simply by changing the acceptable error.
Total error is a pain to track, but it is almost always the only important thing. The error involved in any single stage of a calculation is only important to the extent that it impacts total error.
If you think this approach requires too much intelligence from the function evaluation engine to ever compete in terms of performance, would you say the same thing about Java itself?
Posted by Curt Cox on January 23, 2006 at 12:04 AM PST #
you sir, are a hacker. I challenge you to pistols at dawn!
Posted by wes on January 23, 2006 at 07:06 PM PST #
Hi Axel, AFAIK, there isn't one (except that there would be one once I've done with it :-D) It's not like extension of non-linear functions from decimal to arbitrary-precision is that straight-forward, I suppose~
Posted by Alex Lam on January 23, 2006 at 09:20 PM PST #
Comments are closed for this entry.
jag
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### 5 Activities for Teaching Angles
5 Activities for Teaching Angles
I love teaching angles - it's short and sweet, and the students always have a lot of success with it - which makes it all the better. :)
1. Interactive Math Journal Entry - this is one of my all-time favorite math journal entries. I use this as a full-class introduction to angles - not as a station. That way, I can gauge what knowledge they come to me with, and we can talk about what we will learn while studying angles. This entry is included in my Interactive Math Journal Resource. If you download the free preview, you can grab this lesson in its entirety. You can also read about this activity in a blog post I have HERE.
2. Tables, Whiteboards and Washi Tape -
Get the tape out and get ready for some hands-on fun! (You don't have to use washi tape - masking tape works perfectly fine, too). To introduce angles at the beginning of our unit, we did a full group lesson on classifying angles. I taped up a table, armed students with a whiteboard marker, and let them classify (acute, right, obtuse, and straight) as many angles as they could. They LOVED this activity! So everyone could fit around the table, I paired up the students so they could take turns marking angles on the table.
Once we were masters at classifying angles, we could move on to measuring angles. This is such a fun activity - perfect for practice at math centers! We just used some fun washi tape to "draw" lines on our whiteboards. We used 5 pieces of tape - and I told them the tape had to be straight and go across the board from one side to the other. From there, students measured the angles made by the tape. You could just leave it at this, but I turned this activity into a game by pairing up the students. Each student had a different color of whiteboard marker. The first student measured an angle (any angle of their choice) and wrote down the measurement using their color. Then, the second student checked the answer. If it was correct, they left it alone. If it was incorrect, they erased the first answer and wrote the correct answer using their color (the second student). The second student then got a chance to pick and measure an angle of their choice - recording the answer in their color. At the end of the activity, the student with the most answers in their color (most correct answers) wins!
3. TIME for Angles - This is a quick activity I do at various points throughout the day while we are studying angles. (I also throw in a few days here and there just to review the concept). At any given time, I ask the students what time it is on our analog clock (that is good practice just in itself). I record the time and ask the students what kind of angle it is - acute, right, obtuse, straight. I also ask them to estimate the size of the angle. When doing this, the students realized that each 5 minute interval was 30 degrees - so helpful for estimating.I just keep a running record on the board by the clock and try to get in at least 5 questions throughout the day.
4. What's in a Name? - This one is another great station activity - and will also make a great math bulletin board display when students are complete. I like to use grid (graph) paper for this activity, but it's not necessary. Have the students write their names in block letters (all straight lines) on the paper. Then, using a protractor, have the students measure all the angles they can find in their names. This is also a great way to start to introduce adjacent angles (and complementary and supplementary angles).
5. Angle Task Cards - This is another great station activity. These Angle Task Cards contain a Minds-On Task which I like to use to introduce the concept to the whole class, with each student completing the challenge task on a whiteboard or paper. The twelve task cards can then be completed independently as part of a station, with each student completing the recording sheet to hand in for a formative assessment. The best part about the recording sheet is that it is perfectly matched up to the answer sheet, so you only need to place the recording sheet over the answer sheet in a light area (like a window) so you can see the answers through the page and easily mark without having to measure each angle. These task cards also make great exit slips after a lesson. You can take a peek at them on TpT by clicking HERE.
I also have a more advanced set of task cards for angle relationships - finding the unknown angles with the properties of complementary and supplementary angles, opposite angles, and interior and exterior angles. You can take a peek at them by clicking on the picture to the right or HERE.
What are some of your best teaching ideas or activities for angles? I'd love for you to leave a comment below to share with us.
1. I LOVE #2...how cute!! Thanks for all of the fabulous ideas :)
Julie
The Techie Teacher
2. These ideas are perfect! So easy and effective. The kids are sure to learn angles in no time with these ideas!
3. What GREAT ideas!! Thank you so much!
4. These are FABULOUS activities. Thanks :)
5. I am moving up to 5th this year, and I am getting so many great ideas from your blog. Thanks!!
6. Love the Time for Angles activity. Kills two birds with one stone; learn to read time and angles at the same time. What a great idea!!
7. I did these activities in my class today, the children loved them. Thanks for sharing! :)
8. #2 was a great idea but the masking tape peeled the surface off my whiteboards. Be careful!
9. Wonderful ideas! I love how you connect angles to the clock and provide layers of activities. | 1,327 | 5,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-22 | latest | en | 0.956601 |
https://en.wikipedia.org/wiki/Wheeler-deWitt_equation | 1,726,839,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00087.warc.gz | 199,263,582 | 34,411 | # Wheeler–DeWitt equation
(Redirected from Wheeler-deWitt equation)
The Wheeler–DeWitt equation[1] for theoretical physics and applied mathematics, is a field equation attributed to John Archibald Wheeler and Bryce DeWitt. The equation attempts to mathematically combine the ideas of quantum mechanics and general relativity, a step towards a theory of quantum gravity.
In this approach, time plays a role different from what it does in non-relativistic quantum mechanics, leading to the so-called "problem of time".[2] More specifically, the equation describes the quantum version of the Hamiltonian constraint using metric variables. Its commutation relations with the diffeomorphism constraints generate the Bergman–Komar "group" (which is the diffeomorphism group on-shell).
## Motivation and background
In canonical gravity, spacetime is foliated into spacelike submanifolds. The three-metric (i.e., metric on the hypersurface) is ${\displaystyle \gamma _{ij}}$ and given by ${\displaystyle g_{\mu \nu }\,\mathrm {d} x^{\mu }\,\mathrm {d} x^{\nu }=(-N^{2}+\beta _{k}\beta ^{k})\,\mathrm {d} t^{2}+2\beta _{k}\,\mathrm {d} x^{k}\,\mathrm {d} t+\gamma _{ij}\,\mathrm {d} x^{i}\,\mathrm {d} x^{j}.}$ In that equation the Latin indices run over the values 1, 2, 3, and the Greek indices run over the values 1, 2, 3, 4. The three-metric ${\displaystyle \gamma _{ij}}$ is the field, and we denote its conjugate momenta as ${\displaystyle \pi ^{ij}}$. The Hamiltonian is a constraint (characteristic of most relativistic systems) ${\displaystyle {\mathcal {H}}={\frac {1}{2{\sqrt {\gamma }}}}G_{ijkl}\pi ^{ij}\pi ^{kl}-{\sqrt {\gamma }}\,{}^{(3)}\!R=0,}$ where ${\displaystyle \gamma =\det(\gamma _{ij})}$, and ${\displaystyle G_{ijkl}=(\gamma _{ik}\gamma _{jl}+\gamma _{il}\gamma _{jk}-\gamma _{ij}\gamma _{kl})}$ is the Wheeler–DeWitt metric. In index-free notation, the Wheeler–DeWitt metric on the space of positive definite quadratic forms g in three dimensions is ${\displaystyle \operatorname {tr} ((g^{-1}dg)^{2})-(\operatorname {tr} (g^{-1}dg))^{2}.}$
Quantization "puts hats" on the momenta and field variables; that is, the functions of numbers in the classical case become operators that modify the state function in the quantum case. Thus we obtain the operator ${\displaystyle {\hat {\mathcal {H}}}={\frac {1}{2{\sqrt {\gamma }}}}{\hat {G}}_{ijkl}{\hat {\pi }}^{ij}{\hat {\pi }}^{kl}-{\sqrt {\gamma }}\,{}^{(3)}\!{\hat {R}}.}$ Working in "position space", these operators are {\displaystyle {\begin{aligned}{\hat {\gamma }}_{ij}(t,x^{k})&\to \gamma _{ij}(t,x^{k}),\\{\hat {\pi }}^{ij}(t,x^{k})&\to -i{\frac {\delta }{\delta \gamma _{ij}(t,x^{k})}}.\end{aligned}}}
One can apply the operator to a general wave functional of the metric ${\displaystyle {\hat {\mathcal {H}}}\Psi [\gamma ]=0,}$ where ${\displaystyle \Psi [\gamma ]=a+\int \psi (x)\gamma (x)\,dx^{3}+\iint \psi (x,y)\gamma (x)\gamma (y)\,dx^{3}\,dy^{3}+\dots ,}$ which would give a set of constraints amongst the coefficients ${\displaystyle \psi (x,y,\dots )}$. This means that the amplitudes for ${\displaystyle N}$ gravitons at certain positions are related to the amplitudes for a different number of gravitons at different positions. Or, one could use the two-field formalism, treating ${\displaystyle \omega (g)}$ as an independent field, so that the wave function is ${\displaystyle \Psi [\gamma ,\omega ]}$.
## Mathematical formalism
The Wheeler–DeWitt equation[1] is a functional differential equation. It is ill-defined in the general case, but very important in theoretical physics, especially in quantum gravity. It is a functional differential equation on the space of three-dimensional spatial metrics. The Wheeler–DeWitt equation has the form of an operator acting on a wave functional; the functional reduces to a function in cosmology. Contrary to the general case, the Wheeler–DeWitt equation is well defined in minisuperspaces like the configuration space of cosmological theories. An example of such a wave function is the Hartle–Hawking state. Bryce DeWitt first published this equation in 1967 under the name "Einstein–Schrödinger equation"; it was later renamed the "Wheeler–DeWitt equation".[3]
### Hamiltonian constraint
Simply speaking, the Wheeler–DeWitt equation says
${\displaystyle {\hat {H}}(x)|\psi \rangle =0,}$
where ${\displaystyle {\hat {H}}(x)}$ is the Hamiltonian constraint in quantized general relativity, and ${\displaystyle |\psi \rangle }$ stands for the wave function of the universe. Unlike ordinary quantum field theory or quantum mechanics, the Hamiltonian is a first-class constraint on physical states. We also have an independent constraint for each point in space.
Although the symbols ${\displaystyle {\hat {H}}}$ and ${\displaystyle |\psi \rangle }$ may appear familiar, their interpretation in the Wheeler–DeWitt equation is substantially different from non-relativistic quantum mechanics. ${\displaystyle |\psi \rangle }$ is no longer a spatial wave function in the traditional sense of a complex-valued function that is defined on a 3-dimensional space-like surface and normalized to unity. Instead it is a functional of field configurations on all of spacetime. This wave function contains all of the information about the geometry and matter content of the universe. ${\displaystyle {\hat {H}}}$ is still an operator that acts on the Hilbert space of wave functions, but it is not the same Hilbert space as in the nonrelativistic case, and the Hamiltonian no longer determines the evolution of the system, so the Schrödinger equation ${\displaystyle {\hat {H}}|\psi \rangle =i\hbar \partial /\partial t|\psi \rangle }$ no longer applies. This property is known as timelessness. Various attempts to incorporate time in a fully quantum framework have been made, starting with the "Page and Wootters mechanism" and other subsequent proposals.[4][5] The reemergence of time was also proposed as arising from quantum correlations between an evolving system and a reference quantum clock system, the concept of system-time entanglement is introduced as a quantifier of the actual distinguishable evolution undergone by the system.[6][7]
### Momentum constraint
We also need to augment the Hamiltonian constraint with momentum constraints
${\displaystyle {\vec {\mathcal {P}}}(x)|\psi \rangle =0}$
associated with spatial diffeomorphism invariance.
In minisuperspace approximations, we only have one Hamiltonian constraint (instead of infinitely many of them).
In fact, the principle of general covariance in general relativity implies that global evolution per se does not exist; the time ${\displaystyle t}$ is just a label we assign to one of the coordinate axes. Thus, what we think about as time evolution of any physical system is just a gauge transformation, similar to that of QED induced by U(1) local gauge transformation ${\displaystyle \psi \to e^{i\theta ({\vec {r}})}\psi ,}$ where ${\displaystyle \theta ({\vec {r}})}$ plays the role of local time. The role of a Hamiltonian is simply to restrict the space of the "kinematic" states of the Universe to that of "physical" states—the ones that follow gauge orbits. For this reason we call it a "Hamiltonian constraint". Upon quantization, physical states become wave functions that lie in the kernel of the Hamiltonian operator.
In general, the Hamiltonian[clarification needed] vanishes for a theory with general covariance or time-scaling invariance.
1. ^ a b DeWitt, Bryce S. (1967-08-25). "Quantum Theory of Gravity. I. The Canonical Theory". Physical Review. 160 (5): 1113–1148. Bibcode:1967PhRv..160.1113D. doi:10.1103/PhysRev.160.1113. ISSN 0031-899X.{{cite journal}}: CS1 maint: date and year (link)
3. ^ Rovelli, Carlo (2001-01-23). Notes for a brief history of quantum gravity. Presented at the 9th Marcel Grossmann Meeting in Roma, July 2000. arXiv:gr-qc/0006061.{{cite book}}: CS1 maint: location (link) CS1 maint: location missing publisher (link) | 2,079 | 7,970 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 28, "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} | 2.921875 | 3 | CC-MAIN-2024-38 | latest | en | 0.786691 |
https://testbook.com/question-answer/if-top-of-earth-retained-is-horizontal-the-coeffi--607d9adcb3792905a82de50d | 1,638,124,740,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00521.warc.gz | 639,491,133 | 31,562 | # If top of earth retained is horizontal, the coefficient of passive earth pressure for retaining wall become:
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1. $$\frac{{1 - \sin \left( \phi \right)}}{{1 + \sin \left( \phi \right)}}$$
2. $$\frac{{1 + \sin \left( \phi \right)}}{{1 - \sin \left( \phi \right)}}$$
3. $$\frac{{\sin \left( \phi \right)}\ -\ 1}{{\sin \left( \phi \right)}\ +\ 1}$$
4. $$\frac{{\sin \left( \phi \right)}\ +\ 1}{{\sin \left( \phi \right)}\ -\ 1}$$
Option 2 : $$\frac{{1 + \sin \left( \phi \right)}}{{1 - \sin \left( \phi \right)}}$$
Free
Building Material & Concrete Technology
15793
20 Questions 20 Marks 25 Mins
## Detailed Solution
Concept:
Coefficient of Active Earth Pressure Coefficient:
$${K_a} = \frac{{1 - \sin \left( \phi \right)}}{{1 + \sin \left( \phi \right)}} = {\tan ^2}\left( {45 - \frac{\phi }{2}} \right)\;$$
Coefficient of Passive Earth Pressure Coefficient:
$${K_p} = \frac{{1 + \sin \left( \phi \right)}}{{1 - \sin \left( \phi \right)}} = {\tan ^2}\left( {45 + \frac{\phi }{2}} \right)\;$$
Point to be remembered:
For any value of shearing angle, the product of active earth pressure coefficient and passive earth pressure coefficient will always equal 1.
$${K_a} \times {K_p} = \frac{{1 - \sin \left( \phi \right)}}{{1 + \sin \left( \phi \right)}} \times \frac{{1 + \sin \left( \phi \right)}}{{1 - \sin \left( \phi \right)}} = 1$$ | 499 | 1,447 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-49 | latest | en | 0.502644 |
https://philosophy.stackexchange.com/questions/48145/a-concept-in-which-an-infinite-force-is-also-limited | 1,620,295,146,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.91/warc/CC-MAIN-20210506083716-20210506113716-00310.warc.gz | 476,058,045 | 38,845 | # A concept in which an infinite force is also limited
Edit (complete rewrite):
OK so I'm completely rephrasing the question.
Let's begin with declaring the concept in hand - A is an infinite, willing, creation force. A "wants" to create all the natural numbers. I'd like here to use the concept of ordinal numbers suggested in the comments, just to deny its use here - A is one complete set, it cannot "move" to a different set (by saying this I admit that I am not sure about the validness of this sentence, and I have zero experience with the set theory, only general guidelines).
Just for clarification first: the infinity of A is it's "will" to create, not the fact that it "wants" to create exactly the idea of all numbers. But, it cannot "change" its will.
Now, let's assume A was able to create all natural numbers (this, might be the invalid assumption I thought of at first. But, as I said, I think this won't exactly be assuming A reached what it "cannot" [according to the definition of infinity] reach, as I've said the infinity in A is its creation force and not the exact creation of numbers). If A created all numbers it can, the very creation force that moves it, the very will that enlivens it, won't have what to "want" anymore. Hence, it's will will be limited.
• To be honest now that I think about it, I'm probably misusing the idea "infinity" by giving it a possibility to have an end. Just wanted to know other people's thoughts about it. – Yechiam Weiss Jan 3 '18 at 15:17
• There seems to be two problems with this. First, you seem to be confusing infinite quantities with absolute infinity (there are infinite sets of integers, for example, that nevertheless do not contain all integers; in fact, you could exclude an infinite number of them... for example, there's an infinite number of odd numbers). The second issue is that you're trying to propose that infinite sets are limited, but the thing you show actually is limited is not an infinite set (suppose you just take counting numbers up to some n; no matter what your n is, you have a finite set). – H Walters Jan 3 '18 at 15:31
• With "A being the array..." what do you mean ? A sequence of number ? The collection of all numbers ? In the first case, it makes sense to say that "it goes to infinity"; in the second case no: it is an infinite "complete" collection (if we agree with Cantor's point of view) taht does not "go" nowhere. – Mauro ALLEGRANZA Jan 3 '18 at 16:50
• What is this "striving?" Does the sequence 1, 2, 3, 4, 5, ... "strive" to be 6? Or perhaps it strives to be 7 but fails. What do you mean by strive? – user4894 Jan 3 '18 at 21:33
• "A is an infinite, willing, creation force" -- WTF does that mean? – user4894 Jan 6 '18 at 0:48
I think what you may be after is modeled by the class of all ordinals, which does exist, in traditional mathematical logic, but cannot be a set.
One definition of the ordinals is the model called L.
L starts by defining the integers in terms of sets:
0 = {}
1 = {0}
2 = {0, 1}
3 = {0, 1, 2}
...
omega is the union of 0, 1, 2, 3, etc.
Then every whole number is in omega, and omega is infinite. But omega is only 'infinite' it is not 'infinity'. We can easily create the object which is "omega union {omega}", and that object is larger than omega itself, the first 'transfinite successor'.
We can continue building from there, and get the class Omega (capital) of all 'ordinals'. Each of these is an infinity that is also limited. So there is no inherent conflict to that concept.
You can prove that every potential model of inclusion is represented somewhere in that class. And any real network of sets is equivalent to a combination of ordinals. So 'L' is a model of all of set theory.
The problem is that if Omega is a set, we could cram it into another set. That would allow us to construct "{Omega} union Omega". Unfortunately that would be another model of inclusion, not isomorphic to any of the ordinals. The way modern set theory avoids this is by declaring Omega not to be a set, only a 'proper' class of sets.
But that is not losing anything from most naive notions of infinity, because one of the standard properties of a realized infinity is already that it would not fit inside anything else.
You can do a very similar and only sightly more confusing construction with pairs of sets, if you want a broader sense of 'all the numbers', which includes models of everything most folks can consider a number. The result is the class of J. H. Conway's 'Hyperreals' as elaborated in Donald Knuth's "Surreal Numbers".
• I assume this is set theory? Please review my edit, as I am aware of these concepts (in other terms), but it isn't what I meant to ask. – Yechiam Weiss Jan 4 '18 at 6:29
• That we have this math is still an indication that many humans see no conflict between the idea of infinity and being limited. Which is at least part of what you asked. And the notion of non-set classes does apply to this odd way you want A = infinity to work. If you can't be clear, you should learn to interpret answers as analogies. You need to consider this, even if it is not directly in the right domain. – user9166 Jan 4 '18 at 22:58 | 1,277 | 5,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-21 | latest | en | 0.970548 |
https://stats.stackexchange.com/questions/142516/principal-component-score-scaling-issue | 1,716,298,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00247.warc.gz | 471,646,062 | 39,300 | # Principal Component Score scaling issue
I am trying to replicate PCA results from an external report, being a paid report I cant see exactly how the numbers are being calculated but I do have a good high level understanding. After a lot of hit and trial the below approach has brought me closest to matching the results:
#small subset from the data (interest rates)
A1 <-
c(1.618,1.711,1.665,1.707,2.178,2.047,1.925,1.936,1.997,1.862)
A2 <-
c(2.342,2.408,2.345,2.397,2.818,2.731,2.637,2.648,2.723,2.59)
A3 <-
c(3.197,3.258,3.191,3.223,3.634,3.587,3.488,3.51,3.611,3.473)
A4 <-
c(3.646,3.707,3.653,3.695,4.104,4.032,3.926,3.956,4.044,3.914)
A5 <-
c(4.428,4.495,4.446,4.49,4.897,4.811,4.734,4.79,4.88,4.738)
A6 <-
c(4.6,4.688,4.649,4.699,4.965,4.898,4.815,4.82,4.898,4.806)
A7 <-
c(5.086,5.154,5.13,5.197,5.413,5.397,5.333,5.345,5.415,5.351)
data <- cbind(A1,A2,A3,A4,A5,A6,A7)
# do PCA on corr because this gives the best match of PC scores plot!
corr <- cor(data)
eg <- eigen(corr)
evectors <- eg\$vectors
scores <- -1*(data %*% evectors) #flip sign to match report
# below check does return approx zeros
summary(data - (scores %*% t(evectors)))
The shape of plot of PC1, PC2 and PC3 scores from above matches uncannily with the ones in the report but the scale is completely different. By the way, the report calls them "PC Values", I am assuming it is same as PC Scores. The PC1 scores from above are around 10 (even though none of the inputs > 5.5), while the report has scores that are much closer to the input numbers. Given that first PC on interest rates represents the level of rates, the PC Scores will be much easier to interpret if I could scale down my PC scores to a level more in line with the input numbers? For example in the above subset, having PC score around 3.8 would be easier to understand and more intuitive.
Could someone please give me some pointers about what this transformation might be? I greatly appreciate any help/suggestions/comments.
• You should not project original data onto the eigenvectors, this does not make a lot of sense to me. Usually in PCA you project the same data that you computed the covariance matrix of, i.e. you project centered and standardized data. Regarding the difference in values: perhaps the report did not use standardized data, i.e. did not divide by the standard deviations? Consider posting the original figures and your replication attempt. Mar 19, 2015 at 21:02
• @amoeba Thank you for your comment, I have made some progress and accordingly updated the question. If you could please spare some time and share your thoughts, that'd be very helpful. Mar 21, 2015 at 16:48
• I don't know -- it is weird. PCA is almost always done on the centered data, meaning that principal components are centered as well, i.e. have mean zero. If "the report has all values greater than zero", it is strange. I have no idea why this could be the case. Does this report perhaps have a Methods section or provide any formulas? Mar 21, 2015 at 16:54
• In addition to what @amoeba said: Have you tried to use an SVD on your original data? (So omit the construction of the correlation/covariance matrix altogether) Maybe you accidentally include some bias in your sample. You mention that "It looks like the output is: constant+lambda*(MyScores). *" are you sure about the *constant part? PCA scores should be zero-meaned. Mar 21, 2015 at 19:29
• As I said above, the algorithm that you now follow in your code: projecting raw uncentered and unscaled data onto the eigenvectors of the correlation matrix -- does not make much sense. If this gives a close match to whatever is displayed in the report, then so much worse for this report. Mar 25, 2015 at 22:46 | 1,078 | 3,715 | {"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} | 2.875 | 3 | CC-MAIN-2024-22 | latest | en | 0.848102 |
https://modelassist.epixanalytics.com/display/EA/An+example+of+using+the+estimate+of+binomial+probabilities+in+risk+analysis | 1,702,016,164,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100724.48/warc/CC-MAIN-20231208045320-20231208075320-00177.warc.gz | 461,355,201 | 16,602 | # An example of using the estimate of binomial probabilities in risk analysis
A (perhaps rather cynical) performance measure for lawyers is the percentage of cases he or she has won. Let's say a person is looking for a lawyer to defend his case and would like to choose the one with the highest performance. There are three lawyers in the area who are knowledgeable in the field and their past performances are displayed in the table below. Assuming that each lawyer's cases were random samples from the same population of cases (so we can assume a Binomial process) since all three lawyers work in the same field, whom should the person choose as his lawyer?
Lawyer Number of trials done Number of trials won % of trials won "Gary" 230 215 93.5% "Jon" 34 32 94.1% "Frank" 17 16 94.1%
The first step is to determine the probability r each lawyer would win a random case. We assume a binomial process for the trials which is probably reasonable if each trial's result is unrelated to any other trial, but will not be entirely correct if random effects change the probability for all trials, in which case it would be a mixture process. Given the binomial assumption, the number of trials n and the number of successes s allow us to estimate the probability p as Beta(s+1,n-s+1,1, using the Bayesian estimate with an uninformed prior (the technique is exactly the same if you use a classically derived uncertainty distribution for r).
A Beta distribution estimate of r is constructed for each lawyer. The three distributions are plotted below:
The density plot on the left shows that there is considerable overlap between the three distributions, which means that we cannot know for sure which lawyer is the best, although the cumulative plot shows that 'Gary' is second order stochastic dominant and should therefore be preferred by the defendant. This may well be sufficient for the defendant's needs, but with simulation we can also describe how confident we are that 'Gary' is actually better than the other two...
The model Lawyers provides the confidence we should have that each lawyer is the best. A random sample from these distributions represent possible true performance rates (probability of success): if a value for 'Gary' is greater than for the other two lawyers in an iteration, then this scenario gives 'Gary' as being the better lawyer.
##### Numerical integration
This model is an example of Numerical Integration: a broad technique that has many applications in risk analysis. The confidence value we are looking for is, in words, the confidence of rate(Gary)=x AND rate(Jon)<x AND rate(Frank)<x, summed over all value of x. In equation terms this is:
where f(x) is the Beta pdf and F(x) is the Beta cdf.
There is also a more efficient approach to the above method which is described separately in the software specific model sections. If you run a simulation you will see the results from both approaches are the same. However, the second approach is more efficient: meaning that it reaches the required value with a specified tolerance with fewer iterations than the original approach.
#### Exercise
Try repeating the exercise using a classical statistic estimate of the "lawyers" rates. The answers will be a little different: why? How would you present the results to the client (the defendant)?
The links to the Lawyers software specific models are provided here:
In the model, an IF function for each lawyer compares the generated values from the three rates and returns a 1 if that lawyer's success rate is the largest and a value of 0 otherwise. Simulation results for this IF function would be equivalent to a Bernoulli(p) distribution, where p is the confidence the analysis gives us that each lawyer is better than the others. The mean of that Bernoulli is just p: the value we are interested in. We therefore use the CB.GetForeStatFN() function to report the p value directly into Excel at the end of a simulation. In this case, we have an 38.7% confidence that 'Gary' is the best, a 33.9% confidence that Jon' is the best, and therefore a 27.5% confidence that 'Frank' is the best.
Moreover, in the Crystal Ball model we are effectively using frequencies of the value that are generated as a replacement for an integration: frGary(x) is replaced by the frequency with which the Gary Beta distribution generates a value x, and FrJon(x) is replaced by the frequency with which the Jon Beta distribution generates a value less than x, and therefore produces a 1 from the IF function.
##### A more efficient approach
Bearing in mind the numerical integration equation equivalent of this analysis, we could make a model that is much more efficient:
Create a Beta(215+1,230-215+1,1) =Beta(216,16,1) distribution for Gary's success rate.
Then calculate the confidence that both Jon's and Frank's rate would be lower:
Cell X: =Beta(216,16,1)
Cell Z: =CB.GetForeStatFN(Cell Y)
The BETADIST() function returns the cdf for the Beta distribution for the other two lawyers' rates.
Running a simulation will give the confidence that Gary is better in Cell Z.
In the model, an IF function for each lawyer compares the generated values from the three rates and returns a 1 if that lawyer's success rate is the largest and a value of 0 otherwise. Simulation results for this IF function would be equivalent to a Bernoulli(p) distribution, where p is the confidence the analysis gives us that each lawyer is better than the others. The mean of that Bernoulli is just p: the value we are interested in. We therefore use the RiskMean() function to report the p value directly into Excel at the end of a simulation. In this case, we have an 38.7% confidence that 'Gary' is the best, a 33.9% confidence that Jon' is the best, and therefore a 27.5% confidence that 'Frank' is the best.
Moreover, in the @RISK model we are effectively using frequencies of the value that are generated as a replacement for an integration: frGary(x) is replaced by the frequency with which the Gary Beta distribution generates a value x, and FrJon(x) is replaced by the frequency with which the Jon Beta distribution generates a value less than x, and therefore produces a 1 from the IF function.
##### A more efficient approach
Bearing in mind the numerical integration equation equivalent of this analysis, we could make a model that is much more efficient:
Create a Beta(215+1,230-215+1) =Beta(216,16) distribution for Gary's success rate.
Then calculate the confidence that both Jon's and Frank's rate would be lower:
Cell X: =RiskBeta(216,16) | 1,436 | 6,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-50 | latest | en | 0.968806 |
https://goodriddlesnow.com/riddles/view/4273 | 1,627,706,807,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154053.17/warc/CC-MAIN-20210731043043-20210731073043-00684.warc.gz | 282,226,349 | 9,854 | # Even it out
Question: How do you make seven an even number?
Riddle Discussion
### Similar Riddles
##### What month do people sleep the least (easy)
Question: During what month do people sleep the least?
##### When is 99 more than 100 (hard)
Question: When is 99 more than 100?
##### Riddle #3363 (medium)
Question: When you're given one, you'll have either two or none?
##### Tricky horse (medium)
Question: A horse jumps over a castle and lands on a man, then the man disappears. How was this possible?
##### The Famous Ship Puzzle (medium)
Question: A Japanese ship was sailing in the Pacific Ocean. The Japanese captain of the ship put his diamond chain and Rolex watch on a shelf, went to get a shower and returned ten minutes later. Now listen carefully, as I will only tell it once: When he returned, both the chain and the watch were missing!! He called the crew of his ship together. There were four of them. A British guy was the cook of the ship. The captain asked him: "Where were you the last ten minutes?" And the cook answered "I was in the cold storage room to select the meat for lunch". A Sri Lankan was the house keeping guy. The captain repeated his question to him, and learnt that the Sri Lankan was at the top of the ship correcting the flag which had been put upside down. An Indian guy was the engineer maintaining the ship. Same question, and the Indian told that the he was in the generator room checking the generator. A French guy also served on the house keeping crew. Same question, and the French told that he was sleeping after the night shift. Within ten seconds the smart captain caught the thief. Who was the thief? How did the captain find him? Source: Puzzlevilla | 384 | 1,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-31 | longest | en | 0.985555 |
https://www.expertsmind.com/library/evaluate-the-magnification-of-the-lens-5544907.aspx | 1,718,439,449,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861584.65/warc/CC-MAIN-20240615062230-20240615092230-00179.warc.gz | 675,324,852 | 13,949 | ### Evaluate the magnification of the lens
Assignment Help Physics
##### Reference no: EM13544907
An object is placed 55.5 cm from a screen.
(a) Where should a converging lens of focal length 8.0 cm be placed to form an image on the screen?
shorter distance ___ cm from the screen
farther distance ___ cm from the screen
(b) Find the magnification of the lens.
magnification if placed at the shorter distance
magnification if placed at the farther distance
#### Questions Cloud
Define the freezing point of solution of glucose : What is the freezing point of a 0.19m solution of glucose, C6H12O6, in water? Kf for water is 1.858 C/m How to calculate the lattice energy for lif : Calculate the lattice energy for LiF(s) given the following; sublimation energy for Li(s) +166KJ/mol, ?Hf for F(g) +77KJ/mol, first ionization energy of Li(g) +520KJ/mol, electron affinity of F(g) -328KJ/mol, enthalpy of formation of LiF(s) -617KJ.. Determine what is the speed of the cars : Two cars traveling with the same speed move directly away from one another. What is the speed of the cars Define 3-phenylpropanal is allowed to mix : Show the product that forms when 3-phenylpropanal is allowed to mix with sodium hydroxide catalyst. Evaluate the magnification of the lens : An object is placed 55.5 cm from a screen. Where should a converging lens of focal length 8.0 cm be placed to form an image on the screen, Find the magnification of the lens Compute the equivalency point and the ph : A 15 mL sample of 0.100 M Ba(OH)2 is titrated with 0.12M HCl. Calculated the equivalency point, and the pH after 5mL, 10 mL, 15 mL, and 20 mL of HCl are added. Evaluate the position and height of the final image : A 1.00-cm-high object is placed 3.80 cm to the left of a converging lens of focal length 7.50 cm. Find the position and height of the final image How much naoh are needed to neutralize solution of h2so4 : How many mL of 0.150 M NaOH are needed to neutralize 50.00 mL of a 0.120 M solution of H2SO4 What is the speed of the box at the equilibrium position : A 5 kg mass is connected to a spring (1000 N/m) platform which undergoes simple harmonic motion in the vertical direction with two complete oscillations each second. What is the speed of the box at the equilibrium position
### Write a Review
#### What is the length of the solenoid
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In proton-beam therapy, a high-energy beam of protons is fired at a tumor. What is the total charge of the protons that must be fired at the tumor to deposit the required energy
#### How long is the nose of the train in station
An airplane is flying with a velocity of 81.0 m/s at an angle of 24.0^circ above the horizontal. When the plane is a distance 108m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment.
#### Calculate how many electrons hit the screen
The current in an electron beam in a cathode-ray tube is measured to be 20 uA. How many electrons hit the screen in 10 seconds
#### What is the force on the particle
The potential energy of a particle moving in the x-y plane is U=a/(x^2+y^2)^1/2, where a is a constant. What is the force on the particle
#### What was the magnitude of the average acceleration
A car travelling at 120 strikes a tree. The front end of the car compresses and driver comes to rest after travelling 0.900.
#### What are two ways to affect the speed of a wave
if you had a small diameter spring and made pulses, then got a different diameter spring, with the same tension and same length of the spring, What are two ways to affect the speed of a wave
#### Find the buoyant force acting on the ice sheet
An ice sheet with area 10:0 m x 10:0 m and uniform thickness of 1.20 m floats in the Arctic Ocean. Assume that the density of ice is 917 kg/m3, Find the buoyant force acting on the ice sheet
#### How much current will flow through the silver wire
When a voltage difference is applied to a piece of copper wire, a 5.0 mA current flows. If the copper wire is replaced with a silver wire with twice the diameter of the copper wire
#### Calculate the amplitude of the electric field
A microwave oven operates with sinusoidal microwaves at a frequency of 2400 MHz. The height of the oven cavity is 25 cm, What is the amplitude of the electric field
#### Compute how far from its base did the diver hit the water
A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and 2.7 s later reaches the water below. How far from its base did the diver hit the water | 1,221 | 4,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.832117 |
http://elmerfem.org/elmerice/wiki/doku.php?id=solvers:forcetostress&rev=1355422870 | 1,582,227,353,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145282.57/warc/CC-MAIN-20200220193228-20200220223228-00374.warc.gz | 43,796,662 | 5,066 | This is an old revision of the document!
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## ForceToStress Solver
### General Informations
• Solver Fortran File: `ForceToStress.f90`
• Solver Name: `ForceToStress`
• Required Output Variable(s): `Stress` (user defined)
• Required Input Variable(s): `Force`
• Optional Output Variable(s): None
• Optional Input Variable(s): None
### General Description
For a given boundary, this solver computes the nodal normal stress equivalent to a given nodal normal force. This can be used also to infer the flux from the knowledge of a debit (should work also to infer a tangential force from the knowledge of a tangential force).
### SIF contents
In the SIF example below, the normal stress on a boundary is inferred from the 3rd component of the Stokes residual.
```Solver 1
Equation = "Navier-Stokes"
Stabilization Method = String Stabilized
Flow Model = Stokes
...
Exported Variable 1 = Flow Solution Loads[Fx:1 Fy:1 Force:1 CEQ Residual:1 ]
End
Solver 2
Equation = "ForceToStress"
Procedure = File "ElmerIceSolvers" "ForceToStress"
Variable = String "Stress"
Variable DOFs = 1
Linear System Solver = Direct
Linear System Direct Method = umfpack
End
! Solve this for body Id 2 (=boundary 3 here)
Equation 2
Active Solvers(1) = 2
End
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Boundary Condition 3
Target Boundaries = 5
Body Id = 2
...
End```
### Examples
In this example, a pressure applied on a boundary is first integrated to get nodal force using the GetHydrostaticLoad Solver, and then the pressure is recovered using the ForceToStress Solver. | 542 | 2,111 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.724914 |
https://subjectcoach.com/tutorials/math/topic/math-definitions-letter-c/chapter/circumference | 1,726,410,115,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00113.warc.gz | 522,507,974 | 29,973 | # Definition of Circumference
The circumference of a circle is its perimeter.
It measures the distance around the curvy edge of the circle.
The circumference of a circle depends on its radius $r$. It is found using the formula
$\text{circumference} = 2 \pi \text{(radius)}$
In this formula, $\pi$ is a constant that is equal to about $3.14$.
### Description
The aim of this dictionary is to provide definitions to common mathematical terms. Students learn a new math skill every week at school, sometimes just before they start a new skill, if they want to look at what a specific term means, this is where this dictionary will become handy and a go-to guide for a student.
### Audience
Year 1 to Year 12 students
### Learning Objectives
Learn common math terms starting with letter C
Author: Subject Coach
You must be logged in as Student to ask a Question. | 197 | 869 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.915003 |
http://www.science-mathematics.com/Mathematics/2019/0615/8958.html | 1,563,773,629,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527531.84/warc/CC-MAIN-20190722051628-20190722073628-00169.warc.gz | 276,728,918 | 4,863 | 3) Eight men can chop down 18 trees in a day. How many trees - science mathematics
3) Eight men can chop down 18 trees in a day. How many trees
## 3) Eight men can chop down 18 trees in a day. How many trees
[From: Mathematics] [author: ] [Date: 06-15] [Hit: ]
3)Eight men can chop down 18 trees in a day. How many trees can 20 men chop down in one day?......
3) Eight men can chop down 18 trees in a day. How many trees can 20 men chop down in one day?
-------------------------------------------------------
Como say: --
8 men_________18 trees
20 men______20/8 x 18 = 5/2 x 18 = 45 trees
-
RR say: 8 men chop 18 trees
1 man shops 1/8 x 18 = 18/8 trees
20 men chop 20 x 18/8 = 360/8 = 45 trees in a day
-
Yas say: 20/8 times as many:
20/8 * 18
45
-
DWRead say: rate = (8 man-days)/(18 trees) = (4 man-days)/(9 trees)
20 man-days × (9 trees)/(4 man-days) = 45 trees
-
Captain Matticus, LandPiratesInc say: 18 trees/day / 8 men = x trees/day / 20 men
20 men * 18 tree/day / 8 men = x trees/day
360 trees/day / 8 = x trees/day
45 trees/day = x trees/day
45 trees
-
Puzzling say: You have 18 trees chopped down by 8 men.
If you divide this by 8, you get the amount of trees chopped by 1 man.
Then multiply by 20 to get the number of trees for 20 men.
18 * 20/8
= 18 * 5/2
= 90/2
= 45 trees
-
llaffer say: work = rate * time
Presuming all men work at the same rate, if the rate of one man is "r", then the rate of 8 men is "8r". Work here is the number of trees, and time is in days (1).
We are given:
w = rt
18 = (8r) * 1
Solve for "r" to find the rate of one man:
18 = 8r
r = 18/8
r = 9/4
Now that we have this, we're asked how many trees 20 men can chop down in one day? We have rate, we have time, solve for work:
w = rt
Rate here will be "20r" since there are 20 people, then we can substitute the value for r that we have from above:
w = 20r * 1
w = 20(9/4)
w = 5(9)
w = 45 trees
-
Mike G say: 18 trees need 8 man-days
1 tree need 8/18 = 4/9 man days
20*9/4 = 45 trees
-
mizoo say: one man in one day: 1/8 * 18 trees
20 men in one day:
20 * (1/8 * 18) = 45 trees
-
Outlier say: 8 men = 18 trees in 1 day
1 man = 18/8 trees in 1 day
So 20 men = 20 * 18/8 = 45 trees in 1 day.
-
keywords: ,3) Eight men can chop down 18 trees in a day. How many trees
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© 2008-2010 science mathematics . Program by zplan cms. Theme by wukong . | 825 | 2,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2019-30 | latest | en | 0.927868 |
http://kidscancode.org/godot_recipes/physics/godot3_kyn_rigidbody1/ | 1,585,506,781,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370495413.19/warc/CC-MAIN-20200329171027-20200329201027-00073.warc.gz | 105,885,687 | 12,433 | # Using Rigid Bodies
In this tutorial, I’ll explain when (and when not) to use rigid bodies, how they work, and demonstrate a few handy tricks to bend them to your will. The examples will use RigidBody2D, but the lessons apply equally to 3D.
## Introduction
RigidBody2D is the physics body in Godot that provides simulated physics. This means that you don’t control a RigidBody2D directly. Instead you apply forces to it (gravity, impulses, etc.) and Godot’s built-in physics engine calculates the resulting movement, including collisions, bouncing, rotating, etc.
You can modify a RigidBody2D’s behavior via properties such as “Mass”, “Friction”, or “Bounce”, which can be set in the Inspector:
The body’s behavior is also affected by the world, via the Project Settings -> Physics properties, or by entering an Area2D that is overriding the global physics properties.
## Using RigidBody2D
One of the benefits of using a rigid body is that a lot of behavior can be gotten “for free” without writing any code. For example, let’s look at making a rudimentary “Angry Birds”-style game with falling blocks. You only need to create RigidBody2Ds for the blocks and projectile, and set their properties. Stacking, falling, and bouncing will automatically be handled by the physics engine.
### Stacking blocks
Start by making a RigidBody2D for the block and adding Sprite and CollisionShape2D children:
Add a texture to the Sprite and a rectangular collision shape. IMPORTANT: Do not change the Scale of the collision shape! In general this is a bad idea, and will result in unexpected collision behavior. Always use the shape’s inner size handles and not the outer Node2D-derived scale handles.
NOTE: For the textures in this example, I’m using the Physics Asset pack from Kenney.nl. It contains a wide variety of blocks in different shapes and materials.
Press “Play” and you’ll see the block fall slowly downward. This is due to the default global gravity. You can find this setting in “Project Settings” under Physics -> 2d. You can also try changing the Block’s Gravity Scale property in the Inspector. I’m using a value of 3.
Now create a Main scene (I usually use a Node) for the root). Add a few StaticBody2D nodes with rectangular collision shapes to serve as your “ground” and walls.
Instance a Block, and then duplicate it (Ctrl-D on Windows and Cmd-D on MacOS) so you can make a nice stack. Something like this:
### Projectile
Create another scene with the same node setup as your Block, but name this one “Ball”. Use one of the round textures and a circular collision shape. Instance this in your Main scene and place it somewhere to the side of the stack of blocks.
To cause a rigid body to move, it must have some velocity. You can give the body an initial velocity using the Linear -> Velocity property. Try setting this to (500, 0).
You can also tinker with the ball’s Friction and Bounce properties. Both of these properties can range from zero to one. I like a bounce of around 0.5.
IMPORTANT: NEVER scale a physics body! If you try, a warning will appear, and when the scene runs, the physics engine will automatically set the scale back to (1, 1).
### Forces
Reset the linear velocity to (0, 0). Now what if you want to be able to toss the ball? You should never set a rigid body’s velocity or position manually - remember, these are simulating “real-world” style physics. In the real world, objects can’t instantly jump from place to place or from a standstill to a high speed. If you try and do so, the physics engine will resist it, and unexpected movement can occur. Instead, we must apply forces which create an acceleration in a certain direction (also known as Newton’s Second Law). Godot physics objects work in the same way.
To add force to a rigid body, you have two functions to choose from:
Adds a continuous force to the body. Imagine a rocket’s thrust, steadily pushing it faster and faster. Note that this adds to any already existing forces. The force continues to be applied until removed.
• apply_impulse()
Adds an instantaneous “kick” to the body. Imagine hitting a baseball with a bat.
We’ll use apply_impulse() to kick the ball when we click, drag, and release the mouse button.
Open “Project Settings” and in the “Input Map” tab, add a new action called “click”. Connect it to the left mouse button.
Next, add a script to the Ball, and add the following code:
extends RigidBody2D
var dragging
var drag_start = Vector2()
func _input(event):
if event.is_action_pressed("click") and not dragging:
dragging = true
drag_start = get_global_mouse_position()
if event.is_action_released("click") and dragging:
dragging = false
var drag_end = get_global_mouse_position()
var dir = drag_start - drag_end
apply_impulse(Vector2(), dir * 5)
This script toggles dragging on when the mouse button is pressed and records the location of the click. When the button is released, we find the vector from the click point to the release point and use that to apply the impulse (multiplying by 5 to scale it up). apply_impulse() also takes an offset as its first parameter. This lets you “hit” the body off center, if you wish. For instance, try setting it to Vector2(25, 0) and you’ll add some spin to the ball when it’s launched.
## Controlling Rigid Bodies
There are cases where you need more direct control of a rigid body. For example, imagine you’re trying to make a version of the classic game “Asteroids”. The player’s spaceship needs to rotate using the left/right arrow keys, and to move forward when the up arrow is pressed.
Here’s the image I’m using for my ship:
I recommend you also go to OpenGameArt and search for a nice space background (but this is totally optional).
Create a new scene for the ship as we did above with the following node structure:
• RigidBody2D
• Sprite
• CollisionShape2D
Note: In Godot 3.0, 0 degrees points to the right (along the x axis). This means you need to add a Rotation of 90 to the Sprite so it will match the body’s direction.
By default, the physics settings provide some damping, which reduces a body’s velocity and spin. In space, there’s no friction, so there shouldn’t be any damping at all. However, for the “Asteroids” feel, we want the ship to stop rotating when we let go of the keys, so set the ship’s Angular -> Damp to 5.
extends RigidBody2D
export (int) var engine_thrust
export (int) var spin_thrust
var thrust = Vector2()
var rotation_dir = 0
var screensize
screensize = get_viewport().get_visible_rect().size
func get_input():
if Input.is_action_pressed("ui_up"):
thrust = Vector2(engine_thrust, 0)
else:
thrust = Vector2()
rotation_dir = 0
if Input.is_action_pressed("ui_right"):
rotation_dir += 1
if Input.is_action_pressed("ui_left"):
rotation_dir -= 1
func _process(delta):
get_input()
func _physics_process(delta):
set_applied_force(thrust.rotated(rotation))
set_applied_torque(rotation_dir * spin_thrust)
Let’s walk through what this script is doing. The two variables, engine_thrust and spin_thrust control how fast the ship can accelerate and turn. In the Inspector, set them to 500 and 25000 respectively (the units of torque make for large numbers). thrust will represent the ship’s engine state: (0, 0) when coasting, or a vector with the length of engine_thrust when powered on. rotation_dir will represent what direction the ship is turning. The screensize variable will capture the size of the screen, which we’ll be using later.
Next, the input() function captures the keystates and sets the ship’s thrust on or off, and the rotation direction (rotation_dir) positive or negative. This function is called every frame in _process().
Finally, physics-related functions should be called in _physics_process(). Here we use set_applied_force() to apply the thrust in whatever direction the ship is facing. Then we use set_applied_torque() to cause the ship to rotate.
Play the scene - you should be able to fly around freely.
## The Position Problem
Another feature of “Asteroids” is that the screen “wraps around”. If the player goes off one side, it teleports to the other side. But we already talked above about how you can’t change a rigid body’s position without breaking the physics engine. This presents a huge problem when working with rigid bodies.
A common mistake is to try adding something like this to _physics_process():
func _physics_process(delta):
if position.x > screensize.x:
position.x = 0
if position.x < 0:
position.x = screensize.x
if position.y > screensize.y:
position.y = 0
if position.y < 0:
position.y = screensize.y
set_applied_force(thrust.rotated(rotation))
set_applied_torque(rotation_dir * spin_thrust)
This fails spectacularly, trapping the player on the edge of the screen (with occasional glitches). So why doesn’t this work? The docs say _physics_process() is for physics-related stuff, right?
Not exactly. _physics_process() is synced to the physics timestep, but that doesn’t make it OK to use for just anything. Hope is not lost, however, the answer is in the docs.
To quote the RigidBody2D docs:
You should not change a RigidBody2D’s position or linear_velocity every frame or even very often. If you need to directly affect the body’s state, use _integrate_forces, which allows you to directly access the physics state.
Allows you to read and safely modify the simulation state for the object. Use this instead of _physics_process if you need to directly change the body’s position or other physics properties.
So there’s our answer. Instead of using _physics_process() we need to use _integrate_forces(), which gives us access to the Physics2DDirectBodyState. I highly recommend you take a look at the linked document, there is a lot of really useful data provided in the physics state object. For our purposes, the key piece of information is the body’s Transform2D. (Explaining transforms is beyond the scope of this document - see Matrices and transforms for more information.)
The body’s position is contained in the transform’s origin. Change _physics_process() to _integrate_forces() and add the following code:
func _integrate_forces(state):
set_applied_force(thrust.rotated(rotation))
set_applied_torque(rotation_dir * spin_thrust)
var xform = state.get_transform()
if xform.origin.x > screensize.x:
xform.origin.x = 0
if xform.origin.x < 0:
xform.origin.x = screensize.x
if xform.origin.y > screensize.y:
xform.origin.y = 0
if xform.origin.y < 0:
xform.origin.y = screensize.y
state.set_transform(xform)
We grab the current transform, change it as necessary, and set it back as the new transform. The physics engine stays happy, and everything works as expected:
## Conclusion
When used properly, rigid bodies are a powerful tool in your Godot toolkit. Many users get in trouble, however, when they use them for the wrong purposes, or fail to understand exactly how they work. | 2,499 | 10,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-16 | longest | en | 0.871346 |
https://mybrowserappstore.com/pint-quart/ | 1,717,019,991,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059408.76/warc/CC-MAIN-20240529200239-20240529230239-00470.warc.gz | 361,682,863 | 28,847 | # Pints to Quarts - Quarts to Pints Converter
## Find out how many Pints in Quarts and Quarts in Pints
Home Converter
Pint - Quart Converter
You can input in any of the fields and get equivalent values.
=
## Pint to Quart and Quart to Pint Conversion
### How many Pints in a Quart (pt to qt)
The formula to convert pint to quart is:
Quarts = Pints / 2 = 0.5 * Pints
For example, to convert 10 pints to quarts, you would divide by 2:
Quarts = 10 pints / 2 = 5 quarts
### How many Quarts in a Pint (qt to pt)
The formula to convert quart to pint is:
Pints = Quarts * 2
To convert 5 quarts to pints, you would multiply by 2:
Pints = 5 quarts * 2 = 10 pints
## What are Pints and Quarts?
• The pint is a unit of volume used in the imperial and US customary systems. It is equal to 16 fluid ounces or 0.56826125 liters.
• The quart is a unit of volume used in the imperial and US customary systems. It is equal to 2 pints or 32 fluid ounces or 0.946352946 liters.
• The pint and quart are related to the gallon, which is a larger unit of volume. One gallon is equal to 8 pints or 4 quarts.
• The pint and quart were originally defined as the volume of a certain amount of wheat or grain. The latest definition of the pint and quart is based on the weight of waterThe current definition of the pint is as follows:
• Imperial pint: 20 fluid ounces (568.26125 milliliters).
• US customary pint: 16 fluid ounces (473.176 milliliters).
The current definition of the quart is as follows:
• Imperial quart: 2 imperial pints (1136.5225 milliliters)
• US customary quart: 2 US customary pints (946.352946 milliliters)
• The pint and quart are still widely used in the United States and other countries that use the imperial system. However, they are becoming less common in other countries, which are increasingly using the metric system.
• The word "pint" comes from the Old French word "pinte," which means "a fifth part." This is because the pint was originally defined as one-fifth of a gallon.
• The word "quart" comes from the Old French word "quartier," which means "a quarter." This is because the quart was originally defined as one-quarter of a gallon.
• The pint and quart are used in a variety of contexts, including cooking, baking, and brewing. They are also used to measure the volume of liquids such as milk, beer, and gasoline.
• The pint and quart are also used in some idiomatic expressions, such as "a pint of bitter" (a glass of beer) and "a quart of ale" (a large quantity of beer).
CharacteristicPintQuart
Unit of volumeYesYes
System of measurementImperial and US customaryImperial and US customary
Definition16 fluid ounces (473.176 milliliters)2 pints (946.352946 milliliters)
Relationship to other units1/8 gallon1/4 gallon
Common usesCooking, baking, brewing, measuring the volume of liquids such as milk, beer, and gasolineCooking, baking, brewing, measuring the volume of liquids such as milk, beer, and gasoline
Symbol ptqt
## When to Use Pints vs. Quarts: A Quick Guide
Pints and quarts are both units of volume, but they are used for different purposes.
Pints are commonly used for:
• Measuring smaller quantities of liquids, such as milk, beer, and juice.
• Measuring the volume of ingredients in recipes.
• Measuring the volume of containers, such as ice cream cartons and yogurt cups.
Quarts are commonly used for:
• Measuring larger quantities of liquids, such as water, milk, and gasoline.
• Measuring the volume of containers, such as milk jugs and oil bottles.
Here are some specific examples of when to use pints and quarts:
• Use a pint to measure the milk of cereal.
• Use a quart to measure the water a pasta pot.
• Use a pint to measure the olive oil for salad dressing.
• Use a quart to measure the milk for weekly baking.
• Use a pint to measure the ice cream for sundae.
• Use a quart to measure the gasoline for lawnmower.
## Pints to Quarts Conversion (pint qt) Chart
Pints (Pt) Quart (qt)
0.25 pt 0.125 qt
0.5 pt 0.25 qt
0.75 pt 0.375 qt
1 pt 0.5 qt
1.5 pt 0.75 qt
2 pt 1 qt
2.5 pt 1.25 qt
3 pt 1.5 qt
3.5 pt 1.75 qt
4 pt 2 qt
4.5 pt 2.25 qt
5 pt 2.5 qt
5.5 pt 2.75 qt
6 pt 3 qt
6.5 pt 3.25 qt
7 pt 3.5 qt
7.5 pt 3.75 qt
8 pt 4 qt
8.5 pt 4.25 qt
9 pt 4.5 qt
9.5 pt 4.75 qt
10 pt 5 qt
12.5 pt 6.25 qt
15 pt 7.5 qt
20 pt 10 qt
25 pt 12.5 qt
50 pt 25 qt
100 pt 50 qt
## Quarts to Pints Conversion (qt pint) Chart
Quart (qt) Pints (Pt)
0.25 qt 0.5 pt
0.5 qt 1 pt
0.75 qt 1.5 pt
1 qt 2 pt
2 qt 4 pt
3 qt 6 pt
4 qt 8 pt
5 qt 10 pt
6 qt 12 pt
7 qt 14 pt
8 qt 16 pt
9 qt 18 pt
10 qt 20 pt
11 qt 22 pt
12 qt 24 pt
13 qt 26 pt
14 qt 28 pt
15 qt 30 pt
20 qt 40 pt
25 qt 50 pt
50 qt 100 pt
75 qt 150 pt
100 qt 200 pt
200 qt 400 pt
## Frequently Asked Questions on Pint v Quart
• Pint and Quart are imperial units of measurement of volume/capacity.
• The difference is in the sizes. A Quart is twice the size of a pint and 1/4th of a gallon
A Pint is 1/8th of a gallon and 2 times of a cup.
1 Pint = 0.5 Quarts, 2 Cups and 16 Fluid Ounces
1 Pint = 1/8th of Gallon and 473.8 ml.
1 Quart = 2 Pints, 4 Cups, 32 Fluid Ounces,
1 Quart = 1/4th of Gallon and 946.3 ml
• 1 Pint = 0.5 Quarts, 2 Cups and 16 Fluid Ounces 1 Pint = 1/8th of Gallon and 473.8 ml. 1 Quart = 2 Pints, 4 Cups, 32 Fluid Ounces, 1 Quart = 1/4th of Gallon and 946.3 ml
• An Imperial or UK Quart is 2 pints (UK) and 2.4 US Pint
• An Imperial or UK Quart is 1.2 times larger than US Quart
• An Imperial or UK Pint is 1.2 times that of a US Pint
• 1 Imperial or UK Pint is 0.6 Quarts
An Imperial Pint is 1.2 times that of US Pint. | 1,738 | 5,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-22 | longest | en | 0.905546 |
http://mathtricks.org/tag/kalendar/ | 1,511,469,278,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806939.98/warc/CC-MAIN-20171123195711-20171123215711-00676.warc.gz | 194,533,825 | 8,908 | ## 2012 Dodecahedron Calendar
January 2nd, 2012 by Math Tricks | 1 Comment | Filed in Polyhedron Nets, Regular Polyhedra
### Dodecahedron Calendar
Happy New Year! And with the new year, I have for you a neat little project that you can complete in a short time. It is a 12 month dodecahedron calendar! It is a perfect project that can be performed by students as an introduction to Platonic Solids, or by anybody who just wants a really cool calendar. If you have very young children, I am sure they will find this a fascinating object – our two oldest boys (5 and 3) really enjoyed it, and now they can say “dodecahedron”!
Simply print the dodecahedron net below, cut it out on the black lines, fold the grey lines, and tape in the tabs such that the dodecahedron is formed.
To print the dodecahedron, click on the image to get the larger file in your browser. You can then right-click the file to save it to your computer, or print it directly.
Click on the image to open the larger file.
A bit of patience is required when assembling the calendar, but I think you will be pleased with the results if you stick with it. Below are some pictures I took while assembling one. I used scissors to cut the dodecahedron, and they worked well enough. If you have an x-acto knife, I would suggest using that instead.
Here is the dodecahedron net cut out and ready to be assembled:
Taping the tabs in:
More folding and taping:
More folding and taping:
Slow and steady, carefully folding and taping. Did I mention that patience is required? Almost done:
Smoothing out the faces with a chop stick:
The finished product:
Our two oldest kids, stunned with the awesomeness of the dodecahedron calendar:
Again, just a bit of patience, and you will have a great calendar in the form of my favorite Platonic Solid. I got a good result on my first attempt. With a little practice, I am sure I can get better results in a shorted time – and so can you! | 488 | 1,961 | {"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} | 2.640625 | 3 | CC-MAIN-2017-47 | latest | en | 0.93717 |
http://intelliseeds.com/topictags.php?top_id=10 | 1,596,952,442,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738425.43/warc/CC-MAIN-20200809043422-20200809073422-00454.warc.gz | 56,145,008 | 13,096 | Math Skills
Practice Skills: Time Bound Skills:
Algebra
Calendar
Comparison
Comparison And Estimation
Counting
Data Handling
Decimals
Division
Estimation
Exponents
Factors
Fractions
Geometry
Graphs
Integers
Location Vocabulary
Measurement
Mixed Operations
Money
Multiplication
Number Sense
Ordering
Patterns
Percentages
Place Value
Placement Assessments
Probability
Ratios And Proportions
Roman Numerals
Roots
Rounding
Shapes
Standardized Mock Tests
Statistics
Subtraction
Time
Counting
A.01: Counting review up to 50
A.02.01: Skip counting by 2 and 3 up to 100
A.03: Skip counting by 5 and 10 up to 100
A.04: Skip counting by 5 and 10 up to 100
A.05: Counting review exercise
C.08: Counting by 10s and 1s
A.01: Counting objects up to 20
A.02: Counting objects up to 30
A.03: Counting objects up to 40
A.04: Counting objects up to 50
A.05: Counting objects up to 70
A.06: Counting objects up to 100
A.07: Number sense using the keys - up to 5
A.08: Number sense using the keys - up to 10
A.09: Number sense using the keys - up to 15
A.10: Number sense using the keys - up to 20
Pre-K:
B.01: Counting objects up to 5 - Assessment 1
B.02: Counting objects up to 5 - Assessment 2
B.03: Counting objects up to 10 - Assessment 1
B.04: Counting objects up to 10 - Assessment 2
B.05: Counting objects up to 15 - Test 1
B.06: Counting objects up to 15 - Test 2
B.07: Counting objects up to 20 - Test 1
B.08: Counting objects up to 20 - Test 2
B.09: Matching Objects Assessment 1
B.10: Matching Objects Assessment 2
B.11: Matching Objects Assessment 3
Total Skills: 27
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Welcome to Intelliseeds Learning Smart learning for your child Intelliseeds learning provides practice in several subjects including Math, Reading Comprehension, Language Arts, Logical Reasoning, Mental Ability, Analytical Ability and Critical Thinking, which are essential to the success of every student. Skills are organized by grades, topics, practice tests and time bound tests. The Math section covers skills such as counting, number operations, algebra, story problems and many more. The IntelliAbility section focuses on thinking skills, patterns, sequences, problem analysis, analogies and over 50 other topics to help develop your child's brain. Click here to learn more about IntelliAbility. The Language Arts section includes reading comprehension, nouns, verbs, adverbs, adjectives, predicates, vowel sounds and many more grammar skills.
Privacy | © 2020 Intelliseeds Learning™ | 751 | 2,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-34 | latest | en | 0.766756 |
http://data.allenai.org/tqa/wave_frequency_L_1066/ | 1,542,626,765,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745761.75/warc/CC-MAIN-20181119105556-20181119131556-00553.warc.gz | 84,569,329 | 25,669 | wave frequency
what is wave frequency
The number of waves that pass a fixed point in a given amount of time is wave frequency. Wave frequency can be measured by counting the number of crests (high points) of waves that pass the fixed point in 1 second or some other time period. The higher the number is, the greater the frequency of the waves. The SI unit for wave frequency is the hertz (Hz), where 1 hertz equals 1 wave passing a fixed point in 1 second. The Figure 1.1 shows high-frequency and low-frequency transverse waves. Q: The wavelength of a wave is the distance between corresponding points on adjacent waves. For example, it is the distance between two adjacent crests in the transverse waves in the diagram. Infer how wave frequency is related to wavelength.
wave frequency and energy
The frequency of a wave is the same as the frequency of the vibrations that caused the wave. For example, to generate a higher-frequency wave in a rope, you must move the rope up and down more quickly. This takes more energy, so a higher-frequency wave has more energy than a lower-frequency wave with the same amplitude. You can see examples of different frequencies in the Figure 1.2 (Amplitude is the distance that particles of the medium move when the energy of a wave passes through them.)
instructional diagrams
No diagram descriptions associated with this lesson
questions
a wave with a higher frequency has a longer wavelength.
``````a. true
--> b. false
``````
the frequency of a wave is the same as the frequency of vibrations that caused the wave.
``````--> a. true
b. false
``````
for waves of the same amplitude, a higher frequency wave has less energy than a shorter frequency wave.
``````a. true
--> b. false
``````
if 20 waves pass a fixed point in 10 seconds, the frequency of the waves is
``````a) 200 hz.
b) 100 hz.
c) 20 hz.
--> d) 2 hz.
``````
the frequency of four different waves is listed below. which wave has the most energy?
``````--> a) 2000 hz.
b) 1000 hz.
c) 200 hz.
d) 20 hz.
``````
diagram questions
No diagram questions associated with this lesson | 514 | 2,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.91222 |
https://yourgametips.com/miscellaneous/what-is-the-value-of-an-ace/ | 1,721,053,552,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00408.warc.gz | 953,331,532 | 38,500 | # What is the value of an ace?
## What is the value of an ace?
Ace is 1 point. Face cards Jack, Queen and King's value is 10 points. Number cards are worth their spot (index) value.
Is an Ace 1 or 11?
Number cards count as their natural value; the jack, queen, and king (also known as "face cards" or "pictures") count as 10; aces are valued as either 1 or 11 according to the player's choice. If the hand value exceeds 21 points, it busts, and all bets on it are immediately forfeit.
How much is an ace worth in Rummy?
Aces are worth 1 point each. Number Cards are worth their face value – for example a six is worth 6 points, a four is 4 points, and so on.
### Is an ace high or low?
The cards are ranked thus, from low to high: 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace. An ace is the highest card, but it can also function as the lowest in completing a straight. The two is usually called a "deuce", and the three is sometimes called a "trey". | 275 | 964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.972981 |
http://web2.0calc.com/questions/solve-the-equation-x-3-1-4 | 1,502,926,682,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102757.45/warc/CC-MAIN-20170816231829-20170817011829-00298.warc.gz | 447,720,553 | 5,719 | +0
# solve the equation x-3=-1+4
0
101
2
solve the equation x-3=-1+4
Guest Feb 13, 2017
Sort:
#1
+75287
0
x - 3 = -1+ 4
x - 3 = 3 add 3 to both sides
x = 6
CPhill Feb 14, 2017
#2
0
Guest Feb 28, 2017
edited by Guest Feb 28, 2017
### 20 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 178 | 500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-34 | longest | en | 0.860848 |
pdg2.lbl.gov | 1,432,387,251,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207927634.1/warc/CC-MAIN-20150521113207-00168-ip-10-180-206-219.ec2.internal.warc.gz | 190,631,385 | 15,870 | From 0-60 in 10 million seconds! – Part 2
This is continuing from the previous post (http://pdg3.lbl.gov/atlasblog/?p=1071), where I discussed how we convert data collected by ATLAS into usable objects. Here I explain the steps to get a Physics result.
I can now use our data sample to prove/disprove the predictions of Supersymmetry (SUSY), string theory or what have you. What steps do I follow? Well, I have to understand the predictions of this theory; is it saying that there will be multiple muons in an event or there will be only one very energetic jet in the event, etc? For instance, the accompanying figure shows the production and decay of SUSY particles, which lead to events with many energetic jets, a muon, and particles that escape the detector without leaving a trace (missing energy), like X1.
Cartoon of the production and decay of SUSY particles
If the signature is unique, then my life is considerably simpler; essentially, I will write some software to go through each event and pick out those that match the prediction (you can think of this as finding the proverbial (metal) needle in a haystack). If the signal I am searching for is not very unique, then I have to be much cleverer (think of this as looking for a fat, wooden needle in a haystack).
First, I have to decide the selection criteria, e.g., I want one muon with momentum greater than, say, 100 GeV/c, or one electron and exactly two jets, etc. Once I’ve decided the selection criteria, I cannot change them, and have to accept the results, whatever they may be. Otherwise, there is a very real danger of biasing the result. To decide these selection criteria, I may look at simulation, i.e., fake data, and/or sacrifice a small portion of real data to do my studies on. With these criteria, I could have a non-zero number of candidate events, or zero events.
In either case, I have to estimate how many events I expect to see due to garden-variety physics effects, which can occur as much as a million or a billion times more frequently, and may produce a similar signature; this is called background. This can happen because our reconstruction software could mis-identify a pion as a muon, or make a wrong measurement of an electron’s energy, or if we produce enough of these garden-variety events a few of them (out in the “tails”) may look like new physics. So I have to think of all the standard processes that can mimic what I am searching for. One way to do this is to run my analysis software on simulated events; since we know what a garden-variety process looks like, we generate tons of fake data and see if some events look like the new effect that I am looking for. I can also use “real” data, and by applying a different set of selection criteria, come up with what we call “data driven background estimate”. If the background estimate is much less than the number of candidate signal events, excitement mounts, and the result pops up on the collaboration’s radar screen.
There is usually a trade-off between increasing the efficiency of finding signal events and reducing background. If you use loose selection criteria, you expect to find more signal events, i.e., increase in efficiency, but also more background. Since the background can overwhelm the signal, one has to be careful. Conversely, if you choose very strict criteria, you could have zero background, but also zero signal efficiency – not very useful!!
There is one more thing that I need to do, which sometimes can take a while, and for which there is definitely no standard prescription. I need to determine systematic uncertainties, i.e., an error estimate for my methodology, on both the signal efficiency, and on the background estimate. For instance, if I use a meter-scale to measure the length of a table, how do I know the meter-scale is correct? I have to quantify the correctness of the meter-scale. A result in our field has to have systematic uncertainties otherwise it is meaningless. This step is usually a source of lot of arguments. For instance, in the paper mentioned in Part 1 (http://arxiv.org/pdf/1110.6191v2.pdf), we say that there is a systematic uncertainty of 6.6% (see section 6). Depending on whether this is smaller (larger) than the statistical uncertainty, we say that the result is statistics (systematics) limited. In the first case, adding more data is necessary, and in the second case, a better understanding is needed. At times, one can have a statistical fluctuation that disappears by adding more data; conversely, many results go by the wayside because of people not understanding systematic effects.
Since there is no fixed recipe to do analysis, I can sometimes run into obstacles, or my results may look “strange”; I then have to step back and think about what is going on. After I get some preliminary results I have to convince my colleagues that they are valid; this involves giving regular progress reports within the analysis group. This is followed by a detailed note, which is reviewed by an internal committee appointed by the experiment’s Publication Committee and/or the Physics Coordinator. If I pass this hurdle, the note is released to the entire collaboration for further review. All along this process, people ask me to do all sorts of checks, or tell me that I am completely wrong, or whatever. Given that every physicist thinks that he/she is smarter than the next, this process can be cantankerous at times, since I have to respond to and satisfy each and every comment. Once the experiment’s leader signs off on the paper, we submit it to a peer-reviewed journal, where the external referee(s) can make you jump through hoops; sometimes their objections are valid, sometimes not. I have been on both sides of this process. Needless to say, as a referee my objections are always valid!!
Depending on the complexity of the analysis, the time from the start to finish can be anywhere from a few months to a year or more (causing a few more grey hair, or in my case a few less hair). The two papers that I mentioned at the start of part 1 took about 1-2 years each. Luckily, I had collaborators and we divided up the work among ourselves, so I could work on both of them in parallel.
Vivek Jain is a Scientist at Indiana University, Bloomington. His current interests range from understanding various aspects of tracking to R-parity violating Supersymmetry. More information about his interests can be found at http://www.indiana.edu/~iubphys/faculty/jain2.shtml
From 0-60 in 10 million seconds! – Part 1
OK, so I’ll try to give a flavour of how the data that we collect gets turned into a published result. As the title indicates, it takes a while! The post got very long, so I have split it in two parts. The first will talk about reconstructing data, and the second will explain the analysis stage.
I just finished working on two papers, which have now been published, one in the Journal of Instrumentation, and the other in Physics Letters B. You can see them here (http://arxiv.org/abs/1110.6191 and http://arxiv.org/abs/1109.2242). By the way, some of the posts I am linking to are from two to three years ago, so the wording may be dated, but the explanations are still correct.
When an experiment first turns on this process is longer than when it has been running for a while, since it takes time to understand how the detector is behaving. It also depends on the complexity of the analysis one is doing. To be familiar with some of the terms I mention below, you should take the online tour of the ATLAS experiment at http://atlas.ch/etours_exper/index.html; slides 7 and 8 will give you an overview of how different particle species are detected and what the various sub-systems look like. For more details you should go take the whole tour; it is meant for non-scientists.
For each event, data recorded by ATLAS is basically a stream of bytes indicating whether a particular sensor was hit in the tracking detectors or the amount of energy deposited in the calorimeter or the location of a hit in the muon system, etc. Each event is then processed through the reconstruction software. This figure gives you an idea of how different particle species leave a signal in ATLAS.
Signals left behind by different particle species
For instance, the part of the software that deals with the tracking detectors will find hits that could be due to a charged particle like a pion or a muon or an electron; in a typical event there may be 100 or more such particles, mostly pions. By looking at the curvature of the trajectory of a particle as it bends in the magnetic field, we determine its momentum (see Seth Zenz’s post on tracking – http://blogs.uslhc.us/?p=481). Similarly, the software dealing with the calorimeter will look at the energy deposits and try to identify clusters that could be due to a single electron or to a spray of particles (referred to as a “jet”), and so on. I believe the ATLAS reconstruction software runs to more than 1 million lines of code! It is very modular, with different parts written by different physicists (graduate students, post-docs, more senior people, etc.).
However, before the reconstruction software can do its magic, a lot of other things need to be done. All the sub-detectors have to be calibrated. What this means is that we need to know how to convert, say, the size of the electronic signal left behind in the calorimeter into energy units such as MeV (million electron volts – the mass of the electron is 0.5 MeV). This work is done using data that we are collecting now (we also rely on old data from test beams, simulation (http://blogs.uslhc.us/?p=843)), and cosmic rays (http://blogs.uslhc.us/?p=1591).
Similarly, we have to know the location of the individual elements of the tracking detectors as precisely as possible. For instance, by looking at the path of an individual track we can figure out precisely where detector elements are relative to one another; this step is known as alignment. Remember, the Pixel detector (http://blogs.uslhc.us/?p=277) can measure distances of the order of 1/10th the thickness of human hair, so knowing its position is critical.
Periodically, we re-reconstruct the data to take advantage of improved in algorithms, calibration and/or alignment and also to have all of the collected data processed with the same version of the software (see Jamie’s post – http://pdg3.lbl.gov/atlasblog/?p=816).
In the next post, I will take you through the analysis stage.
Vivek Jain is a Scientist at Indiana University, Bloomington. His current interests range from understanding various aspects of tracking to R-parity violating Supersymmetry. More information about his interests can be found at http://www.indiana.edu/~iubphys/faculty/jain2.shtml
7 or 8 TeV, a thousand terabyte question!
Event Pile-Up in the New Year?
A very happy new year to the readers of this blog. As we start 2012, hoping to finally find the elusive Higgs boson and other signatures of new physics, an important question needs to be answered first – are we going to have collisions at a center of mass energy of 7 or 8 TeV. While that may not feel like a such a drastic step up, certainly not like going to the full design collision energy of 14 TeV, it does bring its own sets of challenges for ATLAS. Understanding the detector performance is crucial for doing physics with our data, and we will have to make sure all the good work done during 7 TeV collisions can be extended if we run at 8 TeV. More collision energy means more pileup interactions; these occur when our detector can not distinguish between two separate collision events and thus considers them part of the same collision. We need to disentangle the pileup contribution to look at the real single collision event, and while a lot of work has been done in this direction, an increase in pileup is always a cause for concern. However, as someone working closely with Monte-Carlo tuning and production, I know firsthand how big of an issue this is going to be for us.
We need Monte-Carlo samples, or simulated data sets for every analysis, to calculate detector efficiency, backgrounds and what not. Also, a lot of times these Monte-Carlo event generators reflect our best understanding of certain processes, and we want to make sure they are predicting the behavior of real data closely. At times when they do not, we turn the knobs in the Monte-Carlo generators and tune them to match the data. Up until now, this tuning has been done mostly with 7 TeV collision data – although we tried to get the energy extrapolation right by using lower energy Tevatron and ATLAS data. We believe the simulation will do a good job at describing 8 TeV collision data – but we can’t be sure unless we actually compare, and most analysis groups will already want the latest and the best Monte-Carlo samples by the time the data starts coming in!
Then there is the question of size. The combined size of ATLAS 7 TeV Monte-Carlo samples is at least a few thousand terabytes. A very conservative estimate suggests we will need a few of months to produce the 8 TeV samples – the caveat is we can’t start producing them until the decision is actually made to switch to 8 TeV. This will happen immediately after the annual Chamonix meeting in the beginning of February, when the CERN management, engineers and experiment representatives meet to decide. As ramping up the energy results in higher cross-sections for the rare process we want to look at, from a physics perspective it is definitely beneficial, but we have to be ready to utilize this if and when it happens.
With input from Borut Kersevan.
Deepak Kar is a postdoctoral research fellow with University of Glasgow. His physics interest is soft-quantum chromodynamics, and he is currently involved in underlying event analysis activities and Monte Carlo tuning in ATLAS.
Tweeting live #Higgs boson updates from #CERN
My view of CERN's auditorium, 2:15h before seminars began.
“If it’s just a fluctuation of background, it will take a lot of data to kill.”
Dr. Fabiola Gianotti, spokesperson for the ATLAS collaboration, made this statement on Dec. 13, 2011 during a special seminar I attended at CERN. Within the minute that followed, I hurriedly concocted a tweet, tacked on #Higgs and #CERN hashtags, and sent Fabiola’s weighty comment out onto the WWW.
CERN, where the WWW was invented, is the main European particle physics laboratory. I was at the lab for a week to discuss physics and the performance of the ATLAS detector, a 7000-ton apparatus used to examine remnants of high-energy proton collisions delivered by CERN’s 27-km Large Hadron Collider (LHC), straddling the Franco-Swiss border.
This turned out to be no ordinary week. The 2011 LHC program had yielded a fecund data sample, and we needed to take stock of our most promising new-phenomena searches. By far the most anticipated were those of the Higgs boson, hypothetical pieces in the emerging puzzle of the tiniest known subatomic particles. Signal rumours had been swirling around the planet in blogs and other media. I was asked by the media relations department at my home institute, McGill University, to live tweet the Higgs update event. I already knew our ATLAS measurements, but was keen to see results by our competitors, the CMS collaboration, running a complementary detector on the opposite side of the LHC. Exciting times!
We owe much of this excitement to Ernest Rutherford who, while a McGill professor of Experimental Physics early last century, unwittingly helped to kick off the Higgs hunt through his Nobel Prize work on radioactivity. Modern theories that posit the existence of one or more types of Higgs particles seek to unify – into a more symmetric and fundamental theory – two basic forces: 19th-century electromagnetism and Rutherford’s 20th-century radioactivity. As if that weren’t enough, observing Higgs particles would also help to reveal a mechanism by which various fundamental particles are endowed with their non-zero mass values. This gets at the very essence of the physical universe.
More recently, my McGill colleagues and I have taken part in the search for Higgs bosons using the Fermilab Tevatron matter-antimatter collider near Chicago. Just last summer at McGill University, Dr. Adrian Buzatu defended a PhD thesis using Tevatron data to set the world’s best limits on Higgs boson production in the low-mass region that is now revealing hints at the LHC. Adrian recently took up a postdoctoral position in our ATLAS collaboration, working with the University of Glasgow group.
In December, I entered CERN’s auditorium three hours before the seminar was to begin. Within about 30 minutes, all available seats and aisles were jammed. A mob formed outside the auditorium door, but security guards were able to maintain control. Unable to work in all the nervous energy and jostling, I tweeted, “You’d think it was John Lennon coming to CERN today.”
At long last the ATLAS and CMS talks began. Both collaborations had searched extensively for several different signatures, scenarios by which a Higgs particle could be created from LHC collision energy before disintegrating into lighter particles.
Given our detectors’ sensitivities, and the colossal Higgs-mimicking backgrounds, we knew in advance that our samples wouldn’t suffice for a statistically robust 2011 discovery. Nevertheless, both ATLAS and CMS showed suggestive traces in a variety of Higgs signatures. Enticingly, both collaborations ruled out overlapping mass ranges and recorded hints at similar masses.
These indications are thrilling. In this kind of science, discoveries take time and are often preceded by whiffs. We’re also cautious. Our 2011 results could be chance background fluctuations, tantamount to tossing six coins and getting tails every time. Only when our observations are flukier than tossing 20 tails in a row will we claim a discovery.
The 2012 data will likely enable us to observe or rule out the Higgs boson. Either of these outcomes would constitute exhilarating, 21st-century science. I look forward to tweeting about it. | 3,930 | 18,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | longest | en | 0.936339 |
https://walkccc.me/LeetCode/problems/2778/ | 1,718,803,331,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00181.warc.gz | 530,477,323 | 8,523 | # 2778. Sum of Squares of Special Elements ¶
• Time: $O(n)$
• Space: $O(1)$
1 2 3 4 5 6 7 8 9 10 11 12 13 class Solution { public: int sumOfSquares(vector& nums) { const int n = nums.size(); int ans = 0; for (int i = 0; i < n; ++i) if (n % (i + 1) == 0) ans += nums[i] * nums[i]; return ans; } };
1 2 3 4 5 6 7 8 9 10 11 12 class Solution { public int sumOfSquares(int[] nums) { final int n = nums.length; int ans = 0; for (int i = 0; i < n; ++i) if (n % (i + 1) == 0) ans += nums[i] * nums[i]; return ans; } }
1 2 3 4 class Solution: def sumOfSquares(self, nums: List[int]) -> int: return sum(num**2 for i, num in enumerate(nums) if len(nums) % (i + 1) == 0) | 281 | 667 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.158956 |
https://bookdown.org/josephs_david11/tsReview/arp.html | 1,623,875,769,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626008.14/warc/CC-MAIN-20210616190205-20210616220205-00126.warc.gz | 160,864,266 | 7,970 | # Unit 13 AR(p)
This is basically the same thing, with the same conclusion, except instead of having clearly defined behavior based on the sine of the roots, the closer the absolute value of the roots is to one the more dominant that behavior is in the realization. To solve these, we will use this brilliant factor.wge function:
library(tswge)
factor.wge(phi = c(0.7, 0.4, 0.3, -0.4, -0.7, 0.1, -0.1))
##
## Coefficients of Original polynomial:
## 0.7000 0.4000 0.3000 -0.4000 -0.7000 0.1000 -0.1000
##
## Factor Roots Abs Recip System Freq
## 1-2.0903B+1.2507B^2 0.8357+-0.3182i 1.1183 0.0579
## 1+0.7598B+0.8037B^2 -0.4727+-1.0104i 0.8965 0.3196
## 1+0.8141B -1.2283 0.8141 0.5000
## 1-0.1836B+0.1222B^2 0.7512+-2.7602i 0.3496 0.2077
##
##
We can use the absolute reciporacal (speling) of the root to tell if it is stationary easily. If it is less than one, we have stationarity. It has the same property in that if it is closer to one, that is a stronger characteristic in the realization. | 392 | 1,104 | {"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} | 2.96875 | 3 | CC-MAIN-2021-25 | latest | en | 0.800664 |
http://www.mathscore.com/math/practice/Object%20Picking%20Probability/ | 1,369,353,972,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704117624/warc/CC-MAIN-20130516113517-00097-ip-10-60-113-184.ec2.internal.warc.gz | 589,334,563 | 3,932 | ## Object Picking Probability - Sample Math Practice Problems
The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. In the main program, all problems are automatically graded and the difficulty adapts dynamically based on performance. Answers to these sample questions appear at the bottom of the page. This page does not grade your responses.
See some of our other supported math practice problems.
### Complexity=10, Mode=replace
1 A bag contains 6 marbles: 1 red marble, 3 yellow marbles, and 2 blue marbles. If you take one marble out, put it back, and take another marble out, what is the probability that you'll get 1 blue marble followed by 1 red marble? 2 A bag contains 6 balls: 2 red balls, 1 yellow ball, and 3 blue balls. If you take one ball out, put it back, and take another ball out, what is the probability that you'll get 1 yellow ball followed by 1 blue ball?
### Complexity=10, Mode=no replace
1 A bag contains 6 marbles: 1 red marble, 4 yellow marbles, and 1 blue marble. If you take one marble out, don't put it back, and take another marble out, what is the probability that you'll get 1 red marble followed by 1 yellow marble? 2 A bag contains 10 marbles: 2 red marbles, 7 yellow marbles, and 1 blue marble. If you take one marble out, don't put it back, and take another marble out, what is the probability that you'll get 1 blue marble followed by 1 yellow marble?
### Complexity=10
1 A bag contains 8 marbles: 2 red marbles, 2 yellow marbles, and 4 blue marbles. If you take two marbles at the same time, what is the probability that you'll get 1 yellow marble and 1 blue marble? 2 A bag contains 9 balls: 4 red balls, 4 yellow balls, and 1 blue ball. If you take two balls at the same time, what is the probability that you'll get 1 blue ball and 1 yellow ball?
### Complexity=10, Mode=replace
1A bag contains 6 marbles: 1 red marble, 3 yellow marbles, and 2 blue marbles. If you take one marble out, put it back, and take another marble out, what is the probability that you'll get 1 blue marble followed by 1 red marble?
Solution
Begin by noting that since the marbles are replaced, each draw does not depend on previous draws and thus the draws are independent.
P(1 blue marble on first pick) = 2/6 = 1/3.
P(1 red marble on second pick) = 1/6.
P(picking 1 blue marble then picking 1 red marble) = (1/3) × (1/6) = 1/18
2A bag contains 6 balls: 2 red balls, 1 yellow ball, and 3 blue balls. If you take one ball out, put it back, and take another ball out, what is the probability that you'll get 1 yellow ball followed by 1 blue ball?
Solution
Begin by noting that since the balls are replaced, each draw does not depend on previous draws and thus the draws are independent.
P(1 yellow ball on first pick) = 1/6.
P(1 blue ball on second pick) = 3/6 = 1/2.
P(picking 1 yellow ball then picking 1 blue ball) = (1/6) × (1/2) = 1/12
### Complexity=10, Mode=no replace
1A bag contains 6 marbles: 1 red marble, 4 yellow marbles, and 1 blue marble. If you take one marble out, don't put it back, and take another marble out, what is the probability that you'll get 1 red marble followed by 1 yellow marble?
Solution
Begin by noting that the marbles are not replaced and thus each draw depends on previous draws. Thus the draws are dependent.
P(1 red marble on first pick) = 1/6.
P(1 yellow marble on second pick) = 4/(total marbles - 1) = 4/5.
P(picking 1 red marble then picking 1 yellow marble) = (1/6) × (4/5) = 2/15
2A bag contains 10 marbles: 2 red marbles, 7 yellow marbles, and 1 blue marble. If you take one marble out, don't put it back, and take another marble out, what is the probability that you'll get 1 blue marble followed by 1 yellow marble?
Solution
Begin by noting that the marbles are not replaced and thus each draw depends on previous draws. Thus the draws are dependent.
P(1 blue marble on first pick) = 1/10.
P(1 yellow marble on second pick) = 7/(total marbles - 1) = 7/9.
P(picking 1 blue marble then picking 1 yellow marble) = (1/10) × (7/9) = 7/90
### Complexity=10
1A bag contains 8 marbles: 2 red marbles, 2 yellow marbles, and 4 blue marbles. If you take two marbles at the same time, what is the probability that you'll get 1 yellow marble and 1 blue marble?
Solution
Ways of choosing 1 yellow marble and 1 blue marble = yellow marble # × blue marble # = 2 × 4 = 8.
Ways of choosing any 2 marbles = 8 × 7 ÷ 2 = 28.
P(choosing 1 yellow marble and 1 blue marble simultaneously) =
(Ways of choosing 1 yellow marble and 1 blue marble) / (Ways of choosing any 2 marbles) = 8/28 = 2/7. | 1,326 | 4,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2013-20 | longest | en | 0.929097 |
https://www.stockpricetrends.com/company/FHB | 1,726,602,362,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00634.warc.gz | 930,968,303 | 41,621 | # First Hawaiian Inc (FHB)
First Hawaiian, Inc. is a banking holding company for First Hawaiian Bank providing a range of banking services to consumers and business customers in the United States.
## Stock Price Trends
Stock price trends estimated using linear regression.
## Paying users area
The data is hidden behind and trends are not shown in the charts.
Unhide data and trends.
This is a one-time payment. There is no automatic renewal.
#### Key facts
• The primary trend is decreasing.
• The decline rate of the primary trend is 12.26% per annum.
• FHB price at the close of September 16, 2024 was \$23.46 and was higher than the top border of the primary price channel by \$3.33 (16.55%). This indicates a possible reversal in the primary trend direction.
• The secondary trend is increasing.
• The growth rate of the secondary trend is 28.24% per annum.
• FHB price at the close of September 16, 2024 was inside the secondary price channel.
• The direction of the secondary trend is opposite to the direction of the primary trend. This indicates a possible reversal in the direction of the primary trend.
### Linear Regression Model
Model equation:
Yi = α + β × Xi + εi
Top border of price channel:
Exp(Yi) = Exp(a + b × Xi + 2 × s)
Bottom border of price channel:
Exp(Yi) = Exp(a + b × Xi – 2 × s)
where:
i - observation number
Yi - natural logarithm of FHB price
Xi - time index, 1 day interval
σ - standard deviation of εi
a - estimator of α
b - estimator of β
s - estimator of σ
Exp() - calculates the exponent of e
### Primary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### February 16, 2021 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### December 11, 2023 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The primary trend is decreasing.
• The decline rate of the primary trend is 12.26% per annum.
• FHB price at the close of September 16, 2024 was \$23.46 and was higher than the top border of the primary price channel by \$3.33 (16.55%). This indicates a possible reversal in the primary trend direction.
### Secondary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### April 25, 2023 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### September 16, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The secondary trend is increasing.
• The growth rate of the secondary trend is 28.24% per annum.
• FHB price at the close of September 16, 2024 was inside the secondary price channel. | 1,070 | 3,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-38 | latest | en | 0.852893 |
https://espanol.answers.yahoo.com/question/index?qid=20200703190954AAEZAOB | 1,596,839,024,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737225.57/warc/CC-MAIN-20200807202502-20200807232502-00409.warc.gz | 304,024,891 | 31,798 | casey preguntado en Science & MathematicsMathematics · hace 1 mes
# THICC BRAINS NEEDED?
this is the math problem, and i need help solving it step by step please i’m gonna cry
### 9 respuestas
Relevancia
• hace 1 mes
Respuesta preferida
The triangles (ABC and DEC) are similar, so all measurements will be in the same proportion.
In the small triangle (DEC), the longest side is 20 units.
In the big triangle (ABC), the longest side is 30 units (10+20 = 30)
So the proportion of the small triangle compared to the big triangle is 20/30 = 2/3.
To find DE, take 2/3 of AB
2/3 * 17 = 34/3 = 11 1/3
For DC, the proportion of x to x+6 is also 2/3:
x / (x+6) = 2/3
Cross multiply:
3x = 2(x + 6)
3x = 2x + 12
3x - 2x = 12
x = 12
Another way to see that is to notice that CE (20) is twice the length of EB (10). So it makes sense that DC will be double DA (6).
2 * 6 = 12
DC = 12
DE = 11 1/3
P.S. Don't assume this is a right triangle, by the way. It may look like that in the diagram but it is clearly not a right triangle if you check.
• hace 1 mes
The little triangle (CDE)
CD ?
DE ?
CE = 20
The big triangle (ABC)
CD becomes CA
DE becomes AB
CE becomes CB
Given that the angle_1 and the angle_2 are similar, you can deduce that:
CD * ratio = CA ← equation (1)
DE * ratio = AB ← equation (2)
CE * ratio = CB ← equation (3)
The equation (3) gives us:
CE * ratio = CB
20 * ratio = 20 + 10
ratio = 30/20
ratio = 3/2
The equation (1) gives us:
CD * ratio = CA
CD = CA/ratio → where: CA = CD + DA = CD + 6
CD = (CD + 6)/ratio → recall: ratio = 3/2
CD = (CD + 6)/(3/2)
CD = (CD + 6) * (2/3)
3.CD = 2.(CD + 6)
3.CD = 2.CD + 12
→ CD = 12
AC = 6 + 12
→ CA = 18
• hace 1 mes
solve for DC
DC/CE = AC/BC
x/20 = (x + 6)/30
30x = 20(x + 6)
30x = 20x + 120
30x - 20x = 120
10x = 120
DC = 12
solving for DE
CD/DE = AC/AB
12/DE = 18/17
204 = 18DE
• hace 1 mes
As the triangles are in proportion and similar, we know interior angles are identical and the side lengths are in proportion.
So, no trigonometry is required.
Now, as the lengths are in propotion we have:
CD/CA = CE/CBso, x/(x + 6) = 20/(20 + 10)i.e. x/(x + 6) = 2/3so, 3x = 2(x + 6)=> 3x = 2x + 12Hence, x = CD = 12Then, DE/AB = 2/3so, DE/17 = 2/3=> DE = 34/3Hence, DE = 11 1/3:)>
• ¿Qué te parecieron las respuestas? Puedes iniciar sesión para votar por la respuesta.
• hace 1 mes
By Pythagorean Theorem
AC = sqr( 30 ² - 17 ² )
AC = 24.7184
X = 24.7184 - 6
X = 18.7184
• hace 1 mes
ΔABC ~ ΔDEC
θ1 (angle 1) = θ2 (angle 2)
DC = x
AC = x + 6
cos θ1 = 17 / (10 + 20)
cos θ1 = 17 / 30
cos θ1 = 0.566666667
arccos (0.566667) = 55.48189216 ° ≈ 55.5 °
θ1 = 55.5 °
θ2 = 55.5 °
sin θ2 = x / 20
sin 55.5° = x / 20
20(sin 55.5°) = x
x = 20(sin 55.5°)
x = 20(0.823947)
x = 16.47894 ≈ 16.5
x = DC = 16.5 ANSWER
cos θ2 = DE / 20
20(cos θ2) = DE
DE = 20(cos θ2)
DE = 20(0.56667)
DE = 11.33334 ≈ 11.3 ANSWER
• hace 1 mes
△ ABC ~ △DEC
∠1 and ∠2 have the same measure
Find DC and DE
AB = 17
Let DC = x and AC = x + 6
CD/CA = DE/AB = 20/30
30x = 20(x + 6)
x = 12
DC = 12
DE = 11 1/3
• hace 1 mes
CB=30 and CE=20 and EB=10
so
(EB/CB)=(1/3) and (CE/CB)=(2/3)
x=(2/3)6=4
• Anónimo
hace 1 mes
When God was handin' out thicc brains, I musta been standin' in the wrong line, 'cause what I got was a dicc brain.
¿Aún tienes preguntas? Pregunta ahora para obtener respuestas. | 1,371 | 3,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2020-34 | latest | en | 0.828248 |
https://support.nag.com/numeric/mb/nagdoc_mb/manual_25_1/html/s/s30naf.html | 1,713,486,726,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00471.warc.gz | 492,652,873 | 7,848 | Integer type: int32 int64 nag_int show int32 show int32 show int64 show int64 show nag_int show nag_int
Chapter Contents
Chapter Introduction
NAG Toolbox
NAG Toolbox: nag_specfun_opt_heston_price (s30na)
Purpose
nag_specfun_opt_heston_price (s30na) computes the European option price given by Heston's stochastic volatility model.
Syntax
[p, ifail] = s30na(calput, x, s, t, sigmav, kappa, corr, var0, eta, grisk, r, q, 'm', m, 'n', n)
[p, ifail] = nag_specfun_opt_heston_price(calput, x, s, t, sigmav, kappa, corr, var0, eta, grisk, r, q, 'm', m, 'n', n)
Description
nag_specfun_opt_heston_price (s30na) computes the price of a European option using Heston's stochastic volatility model. The return on the asset price, $S$, is
$dS S = r-q dt + vt d W t 1$
and the instantaneous variance, ${v}_{t}$, is defined by a mean-reverting square root stochastic process,
$dvt = κ η-vt dt + σv vt d W t 2 ,$
where $r$ is the risk free annual interest rate; $q$ is the annual dividend rate; ${v}_{t}$ is the variance of the asset price; ${\sigma }_{v}$ is the volatility of the volatility, $\sqrt{{v}_{t}}$; $\kappa$ is the mean reversion rate; $\eta$ is the long term variance. $d{W}_{t}^{\left(\mathit{i}\right)}$, for $\mathit{i}=1,2$, denotes two correlated standard Brownian motions with
$ℂov d W t 1 , d W t 2 = ρ d t .$
The option price is computed by evaluating the integral transform given by Lewis (2000) using the form of the characteristic function discussed by Albrecher et al. (2007), see also Kilin (2006).
$Pcall = S e-qT - X e-rT 1π Re ∫ 0+i/2 ∞+i/2 e-ikX- H^ k,v,T k2 - ik d k ,$ (1)
where $\stackrel{-}{X}=\mathrm{ln}\left(S/X\right)+\left(r-q\right)T$ and
$H^ k,v,T = exp 2κη σv2 tg - ln 1-he-ξt 1-h + vt g 1-e-ξt 1-he-ξt ,$
$g = 12 b-ξ , h = b-ξ b+ξ , t = σv2 T/2 ,$
$ξ = b2 + 4 k2-ik σv2 12 ,$
$b = 2 σv2 1-γ+ik ρσv + κ2 - γ1-γ σv2$
with $t={\sigma }_{v}^{2}T/2$. Here $\gamma$ is the risk aversion parameter of the representative agent with $0\le \gamma \le 1$ and $\gamma \left(1-\gamma \right){\sigma }_{v}^{2}\le {\kappa }^{2}$. The value $\gamma =1$ corresponds to $\lambda =0$, where $\lambda$ is the market price of risk in Heston (1993) (see Lewis (2000) and Rouah and Vainberg (2007)).
The price of a put option is obtained by put-call parity.
The option price ${P}_{ij}=P\left(X={X}_{i},T={T}_{j}\right)$ is computed for each strike price in a set ${X}_{i}$, $i=1,2,\dots ,m$, and for each expiry time in a set ${T}_{j}$, $j=1,2,\dots ,n$.
References
Albrecher H, Mayer P, Schoutens W and Tistaert J (2007) The little Heston trap Wilmott Magazine January 2007 83–92
Heston S (1993) A closed-form solution for options with stochastic volatility with applications to bond and currency options Review of Financial Studies 6 327–343
Kilin F (2006) Accelerating the calibration of stochastic volatility models MPRA Paper No. 2975 http://mpra.ub.uni-muenchen.de/2975/
Lewis A L (2000) Option valuation under stochastic volatility Finance Press, USA
Rouah F D and Vainberg G (2007) Option Pricing Models and Volatility using Excel-VBA John Wiley and Sons, Inc
Parameters
Compulsory Input Parameters
1: $\mathrm{calput}$ – string (length ≥ 1)
Determines whether the option is a call or a put.
${\mathbf{calput}}=\text{'C'}$
A call; the holder has a right to buy.
${\mathbf{calput}}=\text{'P'}$
A put; the holder has a right to sell.
Constraint: ${\mathbf{calput}}=\text{'C'}$ or $\text{'P'}$.
2: $\mathrm{x}\left({\mathbf{m}}\right)$ – double array
${\mathbf{x}}\left(i\right)$ must contain ${X}_{\mathit{i}}$, the $\mathit{i}$th strike price, for $\mathit{i}=1,2,\dots ,{\mathbf{m}}$.
Constraint: ${\mathbf{x}}\left(\mathit{i}\right)\ge z\text{ and }{\mathbf{x}}\left(\mathit{i}\right)\le 1/z$, where $z={\mathbf{x02am}}\left(\right)$, the safe range parameter, for $\mathit{i}=1,2,\dots ,{\mathbf{m}}$.
3: $\mathrm{s}$ – double scalar
$S$, the price of the underlying asset.
Constraint: ${\mathbf{s}}\ge z\text{ and }{\mathbf{s}}\le 1.0/z$, where $z={\mathbf{x02am}}\left(\right)$, the safe range parameter.
4: $\mathrm{t}\left({\mathbf{n}}\right)$ – double array
${\mathbf{t}}\left(i\right)$ must contain ${T}_{\mathit{i}}$, the $\mathit{i}$th time, in years, to expiry, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$.
Constraint: ${\mathbf{t}}\left(\mathit{i}\right)\ge z$, where $z={\mathbf{x02am}}\left(\right)$, the safe range parameter, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$.
5: $\mathrm{sigmav}$ – double scalar
The volatility, ${\sigma }_{v}$, of the volatility process, $\sqrt{{v}_{t}}$. Note that a rate of 20% should be entered as $0.2$.
Constraint: ${\mathbf{sigmav}}>0.0$.
6: $\mathrm{kappa}$ – double scalar
$\kappa$, the long term mean reversion rate of the volatility.
Constraint: ${\mathbf{kappa}}>0.0$.
7: $\mathrm{corr}$ – double scalar
The correlation between the two standard Brownian motions for the asset price and the volatility.
Constraint: $-1.0\le {\mathbf{corr}}\le 1.0$.
8: $\mathrm{var0}$ – double scalar
The initial value of the variance, ${v}_{t}$, of the asset price.
Constraint: ${\mathbf{var0}}\ge 0.0$.
9: $\mathrm{eta}$ – double scalar
$\eta$, the long term mean of the variance of the asset price.
Constraint: ${\mathbf{eta}}>0.0$.
10: $\mathrm{grisk}$ – double scalar
The risk aversion parameter, $\gamma$, of the representative agent.
Constraint: $0.0\le {\mathbf{grisk}}\le 1.0$ and ${\mathbf{grisk}}×\left(1.0-{\mathbf{grisk}}\right)×{\mathbf{sigmav}}×{\mathbf{sigmav}}\le {\mathbf{kappa}}×{\mathbf{kappa}}$.
11: $\mathrm{r}$ – double scalar
$r$, the annual risk-free interest rate, continuously compounded. Note that a rate of 5% should be entered as 0.05.
Constraint: ${\mathbf{r}}\ge 0.0$.
12: $\mathrm{q}$ – double scalar
$q$, the annual continuous yield rate. Note that a rate of 8% should be entered as 0.08.
Constraint: ${\mathbf{q}}\ge 0.0$.
Optional Input Parameters
1: $\mathrm{m}$int64int32nag_int scalar
Default: the dimension of the array x.
The number of strike prices to be used.
Constraint: ${\mathbf{m}}\ge 1$.
2: $\mathrm{n}$int64int32nag_int scalar
Default: the dimension of the array t.
The number of times to expiry to be used.
Constraint: ${\mathbf{n}}\ge 1$.
Output Parameters
1: $\mathrm{p}\left(\mathit{ldp},{\mathbf{n}}\right)$ – double array
$\mathit{ldp}={\mathbf{m}}$.
${\mathbf{p}}\left(i,j\right)$ contains ${P}_{ij}$, the option price evaluated for the strike price ${{\mathbf{x}}}_{i}$ at expiry ${{\mathbf{t}}}_{j}$ for $i=1,2,\dots ,{\mathbf{m}}$ and $j=1,2,\dots ,{\mathbf{n}}$.
2: $\mathrm{ifail}$int64int32nag_int scalar
${\mathbf{ifail}}={\mathbf{0}}$ unless the function detects an error (see Error Indicators and Warnings).
Error Indicators and Warnings
Errors or warnings detected by the function:
Cases prefixed with W are classified as warnings and do not generate an error of type NAG:error_n. See nag_issue_warnings.
${\mathbf{ifail}}=1$
On entry, ${\mathbf{calput}}=_$ was an illegal value.
${\mathbf{ifail}}=2$
Constraint: ${\mathbf{m}}\ge 1$.
${\mathbf{ifail}}=3$
Constraint: ${\mathbf{n}}\ge 1$.
${\mathbf{ifail}}=4$
Constraint: ${\mathbf{x}}\left(i\right)\ge _$ and ${\mathbf{x}}\left(i\right)\le _$.
${\mathbf{ifail}}=5$
Constraint: ${\mathbf{s}}\ge _$ and ${\mathbf{s}}\le _$.
${\mathbf{ifail}}=6$
Constraint: ${\mathbf{t}}\left(i\right)\ge _$.
${\mathbf{ifail}}=7$
Constraint: ${\mathbf{sigmav}}>0.0$.
${\mathbf{ifail}}=8$
Constraint: ${\mathbf{kappa}}>0.0$.
${\mathbf{ifail}}=9$
Constraint: $\left|{\mathbf{corr}}\right|\le 1.0$.
${\mathbf{ifail}}=10$
Constraint: ${\mathbf{var0}}\ge 0.0$.
${\mathbf{ifail}}=11$
Constraint: ${\mathbf{eta}}>0.0$.
${\mathbf{ifail}}=12$
Constraint: $0.0\le {\mathbf{grisk}}\le 1.0$ and ${\mathbf{grisk}}×\left(1.0-{\mathbf{grisk}}\right)×{{\mathbf{sigmav}}}^{2}\le {{\mathbf{kappa}}}^{2}$.
${\mathbf{ifail}}=13$
Constraint: ${\mathbf{r}}\ge 0.0$.
${\mathbf{ifail}}=14$
Constraint: ${\mathbf{q}}\ge 0.0$.
${\mathbf{ifail}}=16$
Constraint: $\mathit{ldp}\ge {\mathbf{m}}$.
W ${\mathbf{ifail}}=17$
Quadrature has not converged to the specified accuracy. However, the result should be a reasonable approximation.
W ${\mathbf{ifail}}=18$
Solution cannot be computed accurately. Check values of input arguments.
${\mathbf{ifail}}=-99$
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
Accuracy
The accuracy of the output is determined by the accuracy of the numerical quadrature used to evaluate the integral in (1). An adaptive method is used which evaluates the integral to within a tolerance of $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left({10}^{-8},{10}^{-10}×\left|I\right|\right)$, where $\left|I\right|$ is the absolute value of the integral.
None.
Example
This example computes the price of a European call using Heston's stochastic volatility model. The time to expiry is $6$ months, the stock price is $100$ and the strike price is $100$. The risk-free interest rate is $5%$ per year, the volatility of the variance, ${\sigma }_{v}$, is $22.5%$ per year, the mean reversion parameter, $\kappa$, is $2.0$, the long term mean of the variance, $\eta$, is $0.01$ and the correlation between the volatility process and the stock price process, $\rho$, is $0.0$. The risk aversion parameter, $\gamma$, is $1.0$ and the initial value of the variance, var0, is $0.01$.
```function s30na_example
fprintf('s30na example results\n\n');
put = 'C';
s = 100.0;
r = 0.05;
q = 0.0;
gamma=1.0;
kappa = 2.0;
eta = 0.01;
var0 = 0.01;
sigmav = 0.225;
corr = 0.0;
x = [100.0];
t = [0.5];
[p, ifail] = s30na( ...
put, x, s, t, sigmav, kappa, corr, var0, eta, gamma, r, q);
fprintf('\nHeston''s Stochastic Volatility Model\n European Call :\n');
fprintf(' Spot = %9.4f\n', s);
fprintf(' Volatility of vol = %9.4f\n', sigmav);
fprintf(' Mean reversion = %9.4f\n', kappa);
fprintf(' Correlation = %9.4f\n', corr);
fprintf(' Variance = %9.4f\n', eta);
fprintf(' Mean of variance = %9.4f\n', var0);
fprintf(' Risk aversion = %9.4f\n', gamma);
fprintf(' Rate = %9.4f\n', r);
fprintf(' Dividend = %9.4f\n\n', 1);
fprintf(' Strike Expiry Option Price\n');
for i=1:1
for j=1:1
fprintf('%9.4f %9.4f %9.4f\n', x(i), t(j), p(i,j));
end
end
```
```s30na example results
Heston's Stochastic Volatility Model
European Call :
Spot = 100.0000
Volatility of vol = 0.2250
Mean reversion = 2.0000
Correlation = 0.0000
Variance = 0.0100
Mean of variance = 0.0100
Risk aversion = 1.0000
Rate = 0.0500
Dividend = 1.0000
Strike Expiry Option Price
100.0000 0.5000 4.0851
``` | 3,787 | 10,864 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 152, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-18 | latest | en | 0.475663 |
https://oeis.org/A074789 | 1,708,793,529,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474541.96/warc/CC-MAIN-20240224144416-20240224174416-00707.warc.gz | 430,714,601 | 4,263 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A074789 Partial sums of usigma(n)^2: square of the sum of unitary divisors of n. 4
1, 10, 26, 51, 87, 231, 295, 376, 476, 800, 944, 1344, 1540, 2116, 2692, 2981, 3305, 4205, 4605, 5505, 6529, 7825, 8401, 9697, 10373, 12137, 12921, 14521, 15421, 20605, 21629, 22718, 25022, 27938, 30242, 32742, 34186, 37786, 40922, 43838 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 REFERENCES L. Toth, An asymptotic formula concerning the unitary divisor sum function, Stud. Univ. Babes-Bolyai, Math. 34, No. 2, 3-10 (1989). LINKS Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 Vaclav Kotesovec, Graph - the asymptotic ratio FORMULA a(n) = Sum_{k=1..n} usigma(k)^2 = Sum_{k=1..n} A034448(k)^2. Asymptotic expression: a(n) = Sum_{k<=n} usigma(k)^2 = (Zeta(2)*Zeta(3)*alpha_1/3)*n^3 + O(n^2*log(n)^4), alpha_1= Product_{p prime} (1+1/p^2-2/p^3-1/p^4-2/p^5+3/p^6), Zeta(2)=A013661 and Zeta(3)=A002117. MATHEMATICA Accumulate[Table[DivisorSum[n, # &, CoprimeQ[#, n/#] &]^2, {n, 1, 50}]] (* Vaclav Kotesovec, Feb 04 2019 *) CROSSREFS Cf. A034448, A064609. Cf. A057434, A061502, A072379. Sequence in context: A099978 A242719 A242489 * A229308 A125075 A209983 Adjacent sequences: A074786 A074787 A074788 * A074790 A074791 A074792 KEYWORD nonn AUTHOR Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Sep 07 2002 STATUS approved
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Last modified February 24 11:15 EST 2024. Contains 370303 sequences. (Running on oeis4.) | 703 | 1,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-10 | latest | en | 0.535406 |
http://stackoverflow.com/questions/tagged/asymptotic-complexity?page=9&sort=newest&pagesize=15 | 1,455,477,871,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454702018134.95/warc/CC-MAIN-20160205195338-00225-ip-10-236-182-209.ec2.internal.warc.gz | 210,765,205 | 28,099 | # Tagged Questions
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Each insertion in a stack is O(1) so is the time taken to insert 'n' elements O(n) ? Can we speak similarly for a hash-table as well ? In average case the time taken to insert 'n' elements in a hash ... | 3,232 | 12,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2016-07 | latest | en | 0.882318 |
http://mathhelpforum.com/advanced-algebra/219303-specific-example-eisenstein-s-theorem-using-r-z.html | 1,481,207,666,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541529.0/warc/CC-MAIN-20161202170901-00128-ip-10-31-129-80.ec2.internal.warc.gz | 173,846,001 | 10,660 | # Thread: Specific example of Eisenstein's Theorem using R = Z
1. ## Specific example of Eisenstein's Theorem using R = Z
Eisenstein's Criterion is stated in Dummit and Foote as follows: (see attachment)
-----------------------------------------------------------------------------------
Proposition 13 (Eisenstein's Criterion) Let P be a prime ideal of the integral domain R and let
$f(x) = x^n + a_{n-1}x^{n-1} + ... ... + a_1x + a_0$
be a polynomial in R[x] (here $n \ge 1$ )
Suppose $a_{n-1}, ... ... a_1, a_0$ are all elements of P and suppose $a_0$ is not an element of $P^2$.
Then f(x) is irreducible in R[x]
------------------------------------------------------------------------------------
The beginning of the proof reads as follows:
Proof: Suppose f(x) were reducible, say f(x) = a(x)b(x) in R[x] where a(x) and b(x) are nonconstant polynomials.
Reducing the equation modulo P and using the assumptions on the coefficients of f(x) we obtain the equation $x^n = \overline{a(x)b(x)}$ in (R/P)[x] where the bar denotes polynomials with coefficients reduced modulo P... .,.. etc. etc.
-----------------------------------------------------------------------------------
I will now take a specific example with R= Z as the integral domain concerned and P = (3) as the prime ideal in Z.
Also take (for example) $f(x) = x^3 + 9x^2 + 21x + 9 = (x+3) (x^2 +6x + 3)$
Now as the proof requires, reduce f(x) mod P
Now using D&F Proposition 2 (see attached) - namely $R[x]/(I) \cong R/I)[x]$ we have
$Z[x]/(3) \cong (Z/(3))[x]$
and so we to obtain $\overline{f(x)}$ we simply reduce the coefficients of f(x) by mod 3
Since $9, 21 \in \overline{0}$
we have $\overline{f(x)} = \overline{x^3}$
The coset TEX] \overline{f(x)} [/TEX] would include elements such as $x^3 + 3, x^3 + 6x^2 + 24x - 3, ... ...$ and so on.
Can someone please confirm my working in this particular case of the Eisenstein proof is correct?
Peter | 562 | 1,937 | {"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": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2016-50 | longest | en | 0.80725 |
http://mathhelpforum.com/statistics/102093-drawing-cards-out-deck-print.html | 1,498,623,246,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322320.8/warc/CC-MAIN-20170628032529-20170628052529-00153.warc.gz | 253,054,460 | 2,984 | # Drawing cards out of deck
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• Sep 13th 2009, 12:51 PM
dude15129
Drawing cards out of deck
So, I am having trouble with this problem....
Emily draws a card from a standard 52-deck. Let A, B, C be the events:
A: Emily draws a club.
B: Emily draws a red card.
C: Emily draws a picture card (Jack, Queen, or King).
What is the probability of at least two of the three events occur?
I have tried but wasn't very successful....i don't think. I have done a problem similar i think and i had to do the probability it wouldn't happen and then just subtract it from 1. I tried it but don't think i did it correctly.
Any help is very appreciated!
• Sep 13th 2009, 01:06 PM
Plato
Quote:
Originally Posted by dude15129
Emily draws a card from a standard 52-deck. Let A, B, C be the events:
A: Emily draws a club.
B: Emily draws a red card.
C: Emily draws a picture card (Jack, Queen, or King).
What is the probability of at least two of the three events occur?
Notation: $P(ABC)$ stands for the probability that all three events occur.
Of course $P(ABC)=0$ because a club is not red.
$P(AC)=\frac{3}{52}$ because there are three clubs face cards.
"What is the probability of at least two of the three events occur?"
$P(AB)+P(AC)+P(BC)+P(ABC)=?$ | 353 | 1,255 | {"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": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.960967 |
https://www.pythonstudio.us/language-processing/wellformed-substring-tables.html | 1,561,593,335,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000609.90/warc/CC-MAIN-20190626234958-20190627020958-00447.warc.gz | 846,847,148 | 8,343 | ## Well Formed Substring Tables
The simple parsers discussed in the previous sections suffer from limitations in both completeness and efficiency. In order to remedy these, we will apply the algorithm design technique of dynamic programming to the parsing problem. As we saw in Section 4.7, dynamic programming stores intermediate results and reuses them when appropriate, achieving significant efficiency gains. This technique can be applied to syntactic parsing, allowing us to store partial solutions to the parsing task and then look them up as necessary in order to efficiently arrive at a complete solution. This approach to parsing is known as chart parsing. We introduce the main idea in this section; see the online materials available for this chapter for more implementation details.
Dynamic programming allows us to build the PP in my pajamas just once. The first time we build it we save it in a table, then we look it up when we need to use it as a subconstituent of either the object NP or the higher VP. This table is known as a well-formed substring table, or WFST for short. (The term "substring" refers to a contiguous sequence of words within a sentence.) We will show how to construct the WFST bottom-up so as to systematically record what syntactic constituents have been found.
Let's set our input to be the sentence in (2). The numerically specified spans of the WFST are reminiscent of Python's slice notation (Section 3.2). Another way to think about the data structure is shown in Figure 8-6, a data structure known as a chart.
I shot an elephant in my pajamas
I shot an elephant in my pajamas
Figure 8-6. The chart data structure: Words are the edge labels of a linear graph structure.
Figure 8-6. The chart data structure: Words are the edge labels of a linear graph structure.
In a WFST, we record the position of the words by filling in cells in a triangular matrix: the vertical axis will denote the start position of a substring, while the horizontal axis will denote the end position (thus shot will appear in the cell with coordinates (1, 2)). To simplify this presentation, we will assume each word has a unique lexical category, and we will store this (not the word) in the matrix. So cell (1, 2) will contain the entry V. More generally, if our input string is a\ai ... an, and our grammar contains a production of the form A ^ ai, then we add A to the cell (i-1, i).
So, for every word in text, we can look up in our grammar what category it belongs to.
>>> text = ['I', 'shot', 'an', 'elephant', 'in', 'my', 'pajamas'] [V -> 'shot']
For our WFST, we create an (n-1) x (n-1) matrix as a list of lists in Python, and initialize it with the lexical categories of each token in the init_wfst() function in Example 8-3. We also define a utility function display() to pretty-print the WFST for us. As expected, there is a V in cell (1, 2).
Example 8-3. Acceptor using well-formed substring table.
def init_wfst(tokens, grammar): numtokens = len(tokens)
wfst = [[None for i in range(numtokens+1)] for j in range(numtokens+1)] for i in range(numtokens):
productions = grammar.productions(rhs=tokens[i]) wfst[i][i+1] = productions[0].lhs() return wfst def complete_wfst(wfst, tokens, grammar, trace=False):
index = dict((p.rhs(), p.lhs()) for p in grammar.productions())
numtokens = len(tokens)
for span in range(2, numtokens+1):
for start in range(numtokens+1-span): end = start + span for mid in range(start+1, end):
nt1, nt2 = wfst[start][mid], wfst[mid][end] if nt1 and nt2 and (nt1,nt2) in index: wfst[start][end] = index[(nt1,nt2)] if trace:
print "[%s] %3s [%s] %3s [%s] ==> [%s] %3s [%s]" % \ (start, nt1, mid, nt2, end, start, index[(nt1,nt2)], end)
return wfst def display(wfst, tokens):
print '\nWFST ' + ' '.join([("%-4d" % i) for i in range(1, len(wfst))]) for i in range(len(wfst)-1): print "%d " % i, for j in range(1, len(wfst)):
print "%-4s" % (wfst[i][j] or '.'), print
>>> tokens = "I shot an elephant in my pajamas".split()
>>> wfst0 = init_wfst(tokens, groucho_grammar)
>>> wfstl = complete_wfst(wfst0, tokens, groucho_grammar)
Returning to our tabular representation, given that we have Det in cell (2, 3) for the word an, and N in cell (3, 4) for the word elephant, what should we put into cell (2, 4) for an elephant? We need to find a production of the form A ^ Det N. Consulting the grammar, we know that we can enter NP in cell (0, 2).
More generally, we can enter A in (i, j) if there is a production A ^ B C, and we find non-terminal B in (i, k) and C in (k, j). The program in Example 8-3 uses this rule to complete the WFST. By setting trace to True when calling the function complete_wfst(), we see tracing output that shows the WFST being constructed:
>>> wfstl = complete_wfst(wfst0, tokens, groucho_grammar, trace=True)
[1] VP [4] PP [7] ==> [1] VP [7] [0] NP [1] VP [7] ==> [0] S [7]
For example, this says that since we found Det at wfst[0][1] and N at wfst[1][2], we can add NP to wfst[0][2].
To help us easily retrieve productions by their righthand sides, we create an index for the grammar. This is an example of a space-time trade-off: we do a reverse lookup on the grammar, instead of having to check through entire list of productions each time we want to look up via the righthand side.
We conclude that there is a parse for the whole input string once we have constructed an S node in cell (0, 7), showing that we have found a sentence that covers the whole input. The final state of the WFST is depicted in Figure 8-7.
Notice that we have not used any built-in parsing functions here. We've implemented a complete primitive chart parser from the ground up!
WFSTs have several shortcomings. First, as you can see, the WFST is not itself a parse tree, so the technique is strictly speaking recognizing that a sentence is admitted by a i^i1-
Figure 8-7. The chart data structure: Non-terminals are represented as extra edges in the chart.
grammar, rather than parsing it. Second, it requires every non-lexical grammar production to be binary. Although it is possible to convert an arbitrary CFG into this form, we would prefer to use an approach without such a requirement. Third, as a bottom-up approach it is potentially wasteful, being able to propose constituents in locations that would not be licensed by the grammar.
Finally, the WFST did not represent the structural ambiguity in the sentence (i.e., the two verb phrase readings). The VP in cell (2,8) was actually entered twice, once for a V NP reading, and once for a VP PP reading. These are different hypotheses, and the second overwrote the first (as it happens, this didn't matter since the lefthand side was the same). Chart parsers use a slightly richer data structure and some interesting algorithms to solve these problems (see Section 8.8).
Your Turn: Try out the interactive chart parser application nltk.app.chartparser().
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#1
07-17-2017, 08:39 PM
Peanuthead Guest Join Date: May 2002 Location: Chicago Posts: 2,056
What is the 2nd fastest thing?
We all know that light is the fastest thing in the universe. (And please don't start an argument about that. Take it as a given.) So, what is the 2nd fastest thing? I'm thinking electricity. Has the speed of electricity been measured? How fast is it? Must be at least close to light, don't you think? Any way to give it a boost so that it equals (or exceeds) light? That would be cool.
#2
07-17-2017, 08:54 PM
boffking Guest Join Date: Nov 2012 Location: New England Posts: 2,391
All electromagnetic radiation travels at the same speed, not just visible light.
#3
07-17-2017, 08:55 PM
Bill Door Charter Member Join Date: Nov 2003 Posts: 4,725
The speed of gravity is the same as the speed of light, so tied for first, or tied for second, take your pick. The wave propagation speed of electricity varies with the medium, but can get up to 0.99c, so a close second. If you're talking about the speed of an actual electron in a wire, that's a lot slower.
#4
07-17-2017, 08:59 PM
Palooka Member Join Date: Jul 2005 Location: Eastern Ontario Posts: 2,469
Electrons in a wire are going to be way slower than random things flying around in outer space.
My WAG is some space radiation is the fastest non-light thing and it's basically moving just short of the speed of light.
#5
07-17-2017, 09:01 PM
Peanuthead Guest Join Date: May 2002 Location: Chicago Posts: 2,056
Quote:
Originally Posted by boffking All electromagnetic radiation travels at the same speed, not just visible light.
Does that mean radio waves and television transmissions?
#6
07-17-2017, 09:02 PM
snfaulkner Guest Join Date: May 2015 Location: 123 Fake Street Posts: 5,088
Quote:
Originally Posted by Peanuthead Does that mean radio waves and television transmissions?
yep
#7
07-17-2017, 09:06 PM
yellowjacketcoder Member Join Date: Jan 2013 Location: Atlanta Posts: 3,086
I'm going to put in for Neutrinos, which are so fast they are sometimes erroneously measured to go faster than light.
#8
07-17-2017, 09:10 PM
Peanuthead Guest Join Date: May 2002 Location: Chicago Posts: 2,056
I'm pretty sure this thread is going to go way beyond my comprehension but I'll try to hang in there. I've already learned something. Yay! Let's hear it for fighting ignorance!
#9
07-17-2017, 09:13 PM
Channing Idaho Banks Guest Join Date: Mar 2015 Location: beautiful Idaho Posts: 2,062
A comet has some good speed as it passes its sun.
#10
07-17-2017, 09:16 PM
Dr. Strangelove Guest Join Date: Dec 2010 Posts: 6,003
The Oh-My-God particle, at 99.999999999999999999999510% of the speed of light.
#11
07-17-2017, 09:28 PM
LSLGuy Charter Member Join Date: Sep 2003 Location: Southeast Florida USA Posts: 20,228
Quote:
Originally Posted by Channing Idaho Banks A comet has some good speed as it passes its sun.
Not even remotely in the same league.
The speed of time/gravity/light/EMR in vacuum is 3E8 meters/sec or 10,000 kilometers per second.
The famous Halley's comet is going about 55 kilometers per second. About 1/200th the speed of time/gravity/etc.
Comets with longer orbits and longer periods can go much faster. So instead of 1/200th the speed of time/gravity/etc they're going 1/50 or 1/30th the speed.
Still nowhere close.
Back to the OP: The magic about the speed of time/gravty/etc is that it is a fixed number. There's nothing else in the universe like that. Any other physical motion or speed of propagation happens at a speed that depends to circumstances. So it isn't sensible to talk about the "second-fastest thing".
It might be possible to talk about the "second fastest combination of thing & circumstances". But even then we get into questions about measurement accuracy, etc.
E.g. the speed of light propagation through fiberoptic glass is measurably obviously significantly slower than the speed of light propagation through a true idealized vacuum. But what if we built slightly better glass that halved the difference? And improved it again to halve the difference again? How fast is "as fast as light in vacuum (the "official c"), less some teeny tiny amount we can barely detect?
Last edited by LSLGuy; 07-17-2017 at 09:29 PM.
#12
07-17-2017, 09:30 PM
wolfpup Guest Join Date: Jan 2014 Posts: 7,496
All those other things are physically constrained. The speed of light (in a vacuum) is not. That's a really fundamental difference. The speed of light isn't some arbitrary "speed", it's the rate at which causality propagates in spacetime, and as such is one of the constants of the universe. Lots of things travel slower than light, but those are limited by physical properties and not universal constants. For example, the propagation speed of light in various media other than a vacuum, or the speeds to which subatomic particles can be accelerated in particle colliders. But those aren't particularly interesting limits. They depend on arbitrary material properties or available energies.
Last edited by wolfpup; 07-17-2017 at 09:33 PM.
#13
07-17-2017, 09:49 PM
Darren Garrison Guest Join Date: Oct 2016 Posts: 4,658
Quote:
Originally Posted by LSLGuy Not even remotely in the same league. The speed of time/gravity/light/EMR in vacuum is 3E8 meters/sec or 10,000 kilometers per second.
I'll give you a moment while you facepalm.
Meanwhile, I also immediately thought "neutrinos" when I saw the thread title. Yes, you get oh my god particles under very extreme circumstances, but your garden variety neutrinos travel at just a whisker below the speed of light.
#14
07-17-2017, 10:00 PM
LSLGuy Charter Member Join Date: Sep 2003 Location: Southeast Florida USA Posts: 20,228
Nice catch. Thanks.
3E8 meters/sec = 3E5 km/sec = 30,000 kilometers/sec. So the rest of my numbers are mis-stated by a factor of 3.
Last edited by LSLGuy; 07-17-2017 at 10:03 PM.
#15
07-17-2017, 10:06 PM
LSLGuy Charter Member Join Date: Sep 2003 Location: Southeast Florida USA Posts: 20,228
One more time: Edit upon edit upon timed out edit:
Nice catch. Thanks.
3E8 meters/sec = 3E5 km/sec = 300,000 kilometers/sec. So the rest of my numbers are mis-stated by a factor of 30.
I had some alarm bells in my head as I was typing. That 1/200 number felt too big. But like an idiot I just ignored the alarm bells & hit [submit].
Halley's apehelion isn't 1/200th of c. It's 1/6000th.
Must be time for my sabbatical from the 'Dope. That's 4 dumb mistakes in utterly unrelated posts in the last 4 days.
#16
07-17-2017, 11:18 PM
Francis Vaughan Member Join Date: Sep 2009 Location: Adelaide, Australia Posts: 4,562
I would vote for neutrinos as well. Indeed they would seem to fit the OP's question as perfectly as we know.
OTOH there has been some suggestion that due to the nature of the vacuum, there is an intrinsic very slight roughness to the path that photons take through space, so even though they travel at c, they do so on a very slightly longer path, and if measured at macroscopic scales the apparent speed is very slightly less than c. I'm not sure that this could not however be more akin to suggesting that the vacuum has a refractive index ever so slightly larger than one.
#17
07-18-2017, 08:10 AM
Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 72,722
We can't actually compare the speed of neutrinos to the speed of the Oh-my-god particle. In both cases, what we actually measure directly is the energy. If you know the mass of the particle and its energy, then it's straightforward to calculate its speed. We're pretty sure that OMG was a proton, so we can use the mass of a proton (which we know very well) to calculate its speed. But we don't actually know the mass of the neutrino. We have an upper bound for the mass, so we can calculate a lower bound for the speed, but they might be much, much lighter, and so must be going even faster to reach that energy.
One can in principle turn that around, by taking neutrinos that have traveled a very long distance, and use the difference in their arrival time, combined with their energies, to determine their mass. This has been done with the neutrinos from supernova 1987a. But the error bars in the measurements were too large, and so you can get better results from other experiments, and we haven't been able to reproduce the measurement, since we haven't had any sufficiently distant sources of neutrinos since 1987.
#18
07-18-2017, 09:43 AM
Quartz Charter Member Join Date: Jan 2003 Location: Home of the haggis Posts: 27,331
Quote:
Originally Posted by Peanuthead We all know that light is the fastest thing in the universe.
How about the expansion of the universe itself?
#19
07-18-2017, 09:44 AM
Darren Garrison Guest Join Date: Oct 2016 Posts: 4,658
But whatever the speed of a neutrino is, it is fast, and it is innate. To get an "oh my god" particle you have to take a common slow particle and stick it in a gigantic natural accelerator. So can you say that anything other than a neutrino travels at closer to the speed of light without outside help?
#20
07-18-2017, 09:53 AM
Pleonast Charter Member Join Date: Aug 1999 Location: Los 'Kamala'ngeles Posts: 6,346
Quote:
Originally Posted by wolfpup All those other things are physically constrained. The speed of light (in a vacuum) is not. That's a really fundamental difference. The speed of light isn't some arbitrary "speed", it's the rate at which causality propagates in spacetime, and as such is one of the constants of the universe.
The speed of light in a vacuum is physically constrained. It may not vary from the constant c.
#21
07-18-2017, 09:56 AM
Leo Bloom Member Join Date: Jun 2009 Location: Here Posts: 11,566
Light is always in Hammertime.
#22
07-18-2017, 10:51 AM
enalzi Guest Join Date: Apr 2008 Location: Chicago, IL Posts: 6,510
Quote:
Originally Posted by LSLGuy Halley's apehelion isn't 1/200th of c. It's 1/6000th.
Just for a point of comparison, that's roughly equivalent to a giant tortoise vs. the Concorde.
#23
07-18-2017, 11:12 AM
Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 72,722
Quote:
Quoth Darren Garrison: But whatever the speed of a neutrino is, it is fast, and it is innate.
No speed other than c is ever innate to anything. There's no reason you couldn't have a neutrino sitting at rest relative to the Earth. If it's got a lot of energy, it's because some process gave it that energy.
I suppose you could say that it's "innate" in the sense that, in the typical processes we know of by which neutrinos are produced, the neutrino ends up with a large amount of energy relative to the zero-momentum frame of the process that creates it. But it's tough to compare that to the Oh-my-god particle: What's the typical process by which a proton is produced? The vast majority of protons in the Universe date back to a time when our knowledge of the Universe is sketchy at best, and the production process of those protons was surely highly energetic. If anything, we could say that we observe slow protons only because external processes have acted to slow them down.
#24
07-18-2017, 11:41 AM
ZonexandScout Guest Join Date: Oct 2016 Location: Southeast US Posts: 689
Quote:
Originally Posted by Bill Door The speed of gravity is the same as the speed of light, so tied for first, or tied for second, take your pick. The wave propagation speed of electricity varies with the medium, but can get up to 0.99c, so a close second. If you're talking about the speed of an actual electron in a wire, that's a lot slower.
Knowing the velocity of propagation (VoP) for an electrical pulse in a cable or conductor is essential to using a time domain reflectometer (TDR). We normally test a cable to ascertain the VoP before working on it or performing measurements.
The major contributing factor for the VoP is the construction of the cable. The distance between the conductors and the insulating material used can make a huge difference. For standard electrical cables (i.e., cables that were not specially constructed to be used in scientific experiments or to achieve much faster current flow), the VoP ranges from about 72% to 88% of the speed of light. I've never worked on one that exceeded 90%, but I'm sure they exist.
#25
07-18-2017, 12:41 PM
wolfpup Guest Join Date: Jan 2014 Posts: 7,496
Quote:
Originally Posted by Pleonast The speed of light in a vacuum is physically constrained. It may not vary from the constant c.
Not to nitpick, but I would argue with that wordsmithing. This is just saying that the speed of light is subject to the laws of physics. But this is not the same as saying that it is arbitrarily constrained by material properties, like the energy level of a particle or the wave-propagation properties of a material. I was just trying to get across the idea that the speed of light in a vacuum is innate, fundamental to the nature of spacetime. You can't tweak space or add energy to make light go faster -- in effect, its speed is already infinite, since photons experience zero time and move with the speed of causality.
#26
07-18-2017, 12:45 PM
Whack-a-Mole Member Join Date: Apr 2000 Location: Chicago, IL USA Posts: 18,829
My vote is for the Oh-My-God particle:
Quote:
On the night of October 15, 1991, the Fly's Eye detected a proton with an energy of 3.2±0.9×1020 electron volts.[1,2] By comparison, the recently-canceled Superconducting Super Collider (SSC) would have accelerated protons to an energy of 20 TeV, or 2×1013 electron volts—ten million times less. The energy of the Oh My God particle seen by the Fly's Eye is equivalent to 51 joules—enough to light a 40 watt light bulb for more than a second—equivalent, in the words of Utah physicist Pierre Sokolsky, to “a brick falling on your toe.” The particle's energy is equivalent to an American baseball travelling fifty-five miles an hour. How fast was it going? Pretty fast. So taking 3×108 metres per second as the speed of light, we find that the particle was traveling 2.9999999999999999999999853×108 metres per second, thus 1.467×10−15 metres per second slower than light—one and a half femtometres per second slower than light. If God's radar gun is slightly out of calibration, this puppy's gonna be doin' hard time for speeding. After traveling one light year, the particle would be only 0.15 femtoseconds—46 nanometres—behind a photon that left at the same time. SOURCE: https://www.fourmilab.ch/documents/OhMyGodParticle/
Last edited by Whack-a-Mole; 07-18-2017 at 12:46 PM.
#27
07-18-2017, 01:07 PM
Stranger On A Train Guest Join Date: May 2003 Location: Manor Farm Posts: 16,536
Quote:
Originally Posted by Palooka Electrons in a wire are going to be way slower than random things flying around in outer space.
The drift velocity of electrons in a conductor is on the order of millimeters per hour. The phase velocity of electrical current depends upon the construction of the cable and insulation, but it is on the order of 80% of c, which is fast enough to not matter for normal purposes but does result in enough latency for submarine communications cables that fiber optic cables are necessary for high data rate applications. (Using optical signals also provides substantially more bandwidth per cable than electrical signals.)
As a class of particles, neutrinos are probably a reasonable guess as the "next fastest particle", but particles accelerated by the Penrose or Blandford–Znajek process can increase in momentum to any arbitrary fraction of c, and this is hypothesized to be a mechanism by which quasars continually produce supernova levels of radiation. On the other hand, the speed of neutrinos is limited to the energy of the beta particle it decays from and mass of the neutrino. You could, of course, accelerate a beta particle to any speed before it decays, but "second fastest thing" is going to be an arbitary selection of some object with mass being accelerated to c minus some infinitesimal amount. While the so-called "Oh My God particle" was moving very fast, statistically it is unlikely to a point of certainty that the Earth has encountered anything near the fastest particle in the universe.
Note that while we refer to c colloqually as "the speed of light", it is actually the speed of propagation of interactions in Minkowski spacetime. All gauge bosons (the photon, the gluon, and the hypothetical graviton) and any other massless particles that may exist beyond the Standard Model of particle physics have to move exactly at c by fundamental nature.
Stranger
Last edited by Stranger On A Train; 07-18-2017 at 01:12 PM.
#28
07-18-2017, 01:20 PM
Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 72,722
Hm... I did some more digging, and apparently there's been a neutrino event detected with an energy of 2e15 eV. Given a maximum neutrino mass of ~1 eV (though I think there are tighter bounds than that, even, now), that means a gamma factor of 2e15, much greater than OMG's 3e11. If you really want, you can work out to what that means as a speed, but at that point you're writing so many nines that it's much more sensible to just compare gammas.
#29
07-18-2017, 02:10 PM
Whack-a-Mole Member Join Date: Apr 2000 Location: Chicago, IL USA Posts: 18,829
Quote:
Originally Posted by Chronos Hm... I did some more digging, and apparently there's been a neutrino event detected with an energy of 2e15 eV. Given a maximum neutrino mass of ~1 eV (though I think there are tighter bounds than that, even, now), that means a gamma factor of 2e15, much greater than OMG's 3e11.
The OMG particle had an energy of 3.2±0.9×1020 eV.
Last edited by Whack-a-Mole; 07-18-2017 at 02:10 PM.
#30
07-18-2017, 02:39 PM
wolfpup Guest Join Date: Jan 2014 Posts: 7,496
Quote:
Originally Posted by Whack-a-Mole The OMG particle had an energy of 3.2±0.9×1020 eV.
That proton could not only smash your window, it could also knock your grandmother's framed family portrait off the mantelpiece and smash it to bits!
#31
07-18-2017, 02:50 PM
Lemur866 Charter Member Join Date: Jul 2000 Location: The Middle of Puget Sound Posts: 21,022
But the thing about measure the speed of various objects and particles in the universe is that we're always measuring their speed relative to some other reference frame. So some proton hits a detector going 99.999999999% of c. So that's the fastest thing? No, because we can always say that the the whole goddam Earth was traveling at 99.9999999% of c, and it was the particle that was stationary.
So any measured speed of anything in the universe could always be higher or lower if we just change the reference frame. Unless that thing is traveling at c, in which case it doesn't matter what crazy reference frame we use, we always measure it at c.
A neutrino emitted from the Sun is traveling really really fast relative to the Sun. But if we measure it relative to another neutrino emitted in the opposite direction, it's traveling even faster. It doesn't matter what we measure, there's always going to be something faster.
#32
07-18-2017, 02:58 PM
Pleonast Charter Member Join Date: Aug 1999 Location: Los 'Kamala'ngeles Posts: 6,346
Quote:
Originally Posted by wolfpup Not to nitpick, but I would argue with that wordsmithing. This is just saying that the speed of light is subject to the laws of physics. But this is not the same as saying that it is arbitrarily constrained by material properties, like the energy level of a particle or the wave-propagation properties of a material. I was just trying to get across the idea that the speed of light in a vacuum is innate, fundamental to the nature of spacetime. You can't tweak space or add energy to make light go faster -- in effect, its speed is already infinite, since photons experience zero time and move with the speed of causality.
I think the trouble is you're conflating "physical" and "material". There is no material constraint on the speed of light in a vacuum (since there is no matter involved). But, as you explain, there is a physical constraint on the speed of light in a vacuum--the physics of the universe permit no other speed.
#33
07-18-2017, 05:07 PM
Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 72,722
Quote:
Quoth Whack-a-Mole: The OMG particle had an energy of 3.2±0.9×1020 eV.
And a rest mass (assuming it was a proton) of approximately 1e9 eV. The energy divided by the rest mass gives you the gamma.
In other words, the OMG particle had more energy than the highest-energy neutrino, but only because protons are much more massive than neutrinos.
#34
07-18-2017, 06:35 PM
Chopsticks Guest Join Date: Mar 2015 Location: Ohio Posts: 45
For practical definitions of fast, I'd agree with OMG, because we can calculate how fast it was going. For more oddball situations...
AFAIK, Kamiokande demonstrated that neutrinos have mass and therefore can't travel at c, but couldn't calculate what the mass was. I don't know if any more recent experiments got us any closer to a value, so it's not really possible to say how close to c it can go. But, watch supernovae for long enough and you might find a super-OMG neutrino.
Cherenkov radiation. It may not be anywhere near 3e8 m/s, but being able to say "this particle traveled through air faster than the speed of light in air" is fast and cool.
With quantum effects, speed is probably closer to NaN than 'very fast', but quantum tunneling, and general effects like electron jumping between atomic orbitals, are traveling a non-zero distance in a very very very small time. Hawking radiation from the "photon travels faster than c to escape the black hole" point of view.
Under sort-of-illusionary speed, you have expansion of space and galaxy-sized pairs of scissors, where two objects move faster than c relative to a third observer, but each object itself is slower than c.
For slowest speed, I nominate 'how long is it until lunch'.
#35
07-18-2017, 07:18 PM
Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 72,722
Even if we can't say precisely how fast that super-neutrino was moving, we can still say that its gamma factor was at least four orders of magnitude greater than OMG's. Given that, I think that it's probably safe to say that the fastest neutrino in the Universe is probably faster than the fastest proton in the Universe, even though we've (probably) never observed either.
#36
07-18-2017, 08:38 PM
markn+ Guest Join Date: Feb 2015 Posts: 1,078
Chronos, I know you're too smart to forget Gallilean relativity, so I'm puzzled by your statements about one thing being "faster" than another. In some reference frame the proton is faster than the neutrino, and in fact in some reference frame the neutrino is standing still. So how can you talk about one thing being absolutely faster than another thing?
#37
07-18-2017, 10:15 PM
Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 72,722
Would you be happier if I said "in any reasonable choice of reference frame"? Or "Any reference frame that anyone would ever use for any purpose other than making the notion of 'motion' seem absurd"?
#38
07-18-2017, 11:39 PM
Claude Remains Guest Join Date: Oct 2004 Posts: 1,324
I think Chronos is post padding now that he's hit 70,000.
- Ducks and runs-
Last edited by Claude Remains; 07-18-2017 at 11:40 PM.
#39
07-19-2017, 07:07 AM
Pasta Charter Member Join Date: Sep 1999 Posts: 2,102
Note that beta decay and fusion aren't the only way to make neutrinos. Whatever cosmic furnace is able to churn out OMG protons will very likely churn out mesons that will decay to produce OMG neutrinos of similar energies and therefore much higher speeds. In other words, it is difficult to imagine making crazy high-speed protons without also making even crazier higher-speed neutrinos at the same time.
#40
07-19-2017, 07:24 AM
buddha_david Guest Join Date: Apr 2011 Location: Beyond The Fringe Posts: 26,653
What is the 2nd fastest thing?
#41
07-19-2017, 07:45 AM
Darren Garrison Guest Join Date: Oct 2016 Posts: 4,658
Quote:
Originally Posted by Pasta and therefore much higher speeds
Well, not exactly "much." Shaving this close to the speed of light, we are arguing about speed differences of something like millimeters per second.
#42
07-19-2017, 07:49 AM
Pasta Charter Member Join Date: Sep 1999 Posts: 2,102
Quote:
Originally Posted by Darren Garrison Well, not exactly "much." Shaving this close to the speed of light, we are arguing about speed differences of something like millimeters per second.
*shrug*. Relative terms are... relative. "s/speed/gamma/g" if you'd rather that.
#43
07-19-2017, 07:53 AM
Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 72,722
Probably true, with the caveat that we have no idea what the OMG furnace is, or how it works. But whatever it is, that's certainly a reasonable guess.
#44
07-19-2017, 07:56 AM
Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 72,722
Darren Garrison, put it this way: If you had an OMG proton and an OMG neutrino, emitted from the same source and traveling in the same direction, and if you were riding along in the proton's reference frame, the neutrino would still zip past you very quickly.
#45
07-19-2017, 01:24 PM
Stranger On A Train Guest Join Date: May 2003 Location: Manor Farm Posts: 16,536
Of course, from the reference frame of the neutrino, the proton is moving at the same speed. It is probably more sensible to talk in terms of a maximum gamma (that is the Lorentz contraction factor for anybody who is confused) from an arbitrary reference frame, but as a practical matter, any no-zero-mass particle could be accelerated to any speed short of c by successive Penrose or other gravitational momentum transfer processes regardless of the energy it had when it was produced, so there is no upper infinitesimal ceiling to particle kinetic energy or speed short of conservation of the total momentum of the universe. There is probably a practical upper limit of potential momentum transfer given the mass of the 'local' universe and time, and it is statistically certain that the cosmic particles we have observed do not even approach that by many orders of magnitude.
Stranger
#46
07-19-2017, 02:04 PM
Hocus Pocus Guest Join Date: Dec 2015 Location: South King County, WA Posts: 340
#2 Gossip
#3 Cars breaking down when the warranty is up
#47
07-19-2017, 02:29 PM
Really Not All That Bright Guest Join Date: May 2003 Location: Florida Posts: 67,099
I'm not very good at physics. So bear with me. I've read that some particles come in twins that have the same "spin" - and that each will maintain the same spin even if the particles are separated. As soon as the spin of one is affected, the spin of the other will be too. So wouldn't whatever force transmits the spin between the twin particles be the fastest thing?
#48
07-19-2017, 02:40 PM
DPRK Guest Join Date: May 2016 Posts: 819
There is no instantaneous force or transmission of information between separated particles. You are thinking of entangled states.
#49
07-19-2017, 02:55 PM
Stranger On A Train Guest Join Date: May 2003 Location: Manor Farm Posts: 16,536
You are referring to the quantum entaglement, in which particles have innately complementary states regardless of the separation distance between them. The change of states is apparently instantaneous (if you observe the change of state of a local particle, you will see the complementary change with a delay appropriate to the distance it takes light to travel from the distant particle) but it is wrong to think of this as having a speed or rate, and there is no 'force' or other exchange between them. They're just tied together in an apparently nonlocal but causal fashion. Although there are various interpretations of quantum mechanics which attempt to rationalize this behavior there is as yet no falsifiable hypothesis for how it works other than that it is a characteristic of nature at the level of single particles and very small systems where quantum mechanics dominates.
Stranger
#50
07-19-2017, 03:24 PM
Really Not All That Bright Guest Join Date: May 2003 Location: Florida Posts: 67,099
That's the thing. So there is actually an appreciable delay before the change in the state of the distant particle?
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# Schwaches Erdbeben Stärke 2.3 - Iceland: 11.2 Km E of Grímsey, am Donnerstag, 29. Sep 2022 um 06:01 Lokalzeit - writeAge(1664431267)
Quakes around its antipode during the following 3 Tage
Check different time intervals: 24 hours | 2 days
Datum & Zeit: 29 Sep 2022 06:01:07 UTC -
Ortszeit am Epizentrum: Donnerstag, 29. Sep. 2022 06:01 (GMT +0)
Magnitude: 2.3
Tiefe: 14 km
Epizentrum Breite / Länge: 66.55°N / 17.758°W (Island)
Antipode Breite / Länge: 66.55°S / 162.242°E (Antarktis)
Primäre Datenquelle: IMO (Icelandic Met Office )
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## Quakes within 30 degrees from seismic antipode up to 3 Tage after the quake
The area around the antipode is marked with the big circle on the map.
The total number of quakes that occurred within 3 Tage around the antipode of the quake was 0. This is compared to the average number of 0 quakes during preceding intervals of 3 Tage with a standard deviation (σ) of 0, calculated by analzing all 100 preceding intervals of 3 Tage for the same region.
Interpretation:
Note: At a magnitude of 2.3, it is unlikely that this quake could affect areas around its antipode. If an increase or decrease of quake is detected, it is likely caused by other factors.
There is no significant variation in the number of quakes that occurred around the antipode within 3 Tage.
Mag class Total # Average σ Observed deviation 6+ 0 0 0 n/a 5+ 0 0 0 n/a 4+ 0 0 0 n/a 3+ 0 0 0 n/a Total 0 0 0 n/a
Explanation:
BLACK = Number of quakes within normal statistical range
RED = Significantly lower than normal number of quakes
GREEN = Significantly higher than normal number of quakes
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Volcanoes & Earthquakes | 713 | 2,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.404259 |
https://subversion.xray.aps.anl.gov/trac/pyGSAS/browser/trunk/GSASIIlattice.py?rev=2038 | 1,686,247,736,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655092.36/warc/CC-MAIN-20230608172023-20230608202023-00594.warc.gz | 594,420,526 | 43,792 | # source:trunk/GSASIIlattice.py@2038
Last change on this file since 2038 was 2038, checked in by vondreele, 8 years ago
revisions to SS structure factor calcs. Use hklt proj to hkl is several places
fxn works for thiourea derivs OK except X,Y,Zcos modulations; no Uijsin/cos derivatives yet
adj scaling of 4D charge flip maps
convert betaij vals from Jana2K files to Uij
start on SS read phase from cif
added a hklt F import (might vanish)
• Property svn:eol-style set to native
• Property svn:keywords set to Date Author Revision URL Id
File size: 62.1 KB
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1# -*- coding: utf-8 -*-
2'''
3*GSASIIlattice: Unit cells*
4---------------------------
5
6Perform lattice-related computations
7
8Note that *g* is the reciprocal lattice tensor, and *G* is its inverse,
9:math:G = g^{-1}, where
10
11 .. math::
12
13 G = \\left( \\begin{matrix}
14 a^2 & a b\\cos\gamma & a c\\cos\\beta \\\\
15 a b\\cos\\gamma & b^2 & b c \cos\\alpha \\\\
16 a c\\cos\\beta & b c \\cos\\alpha & c^2
17 \\end{matrix}\\right)
18
19The "*A* tensor" terms are defined as
20:math:A = (\\begin{matrix} G_{11} & G_{22} & G_{33} & 2G_{12} & 2G_{13} & 2G_{23}\\end{matrix}) and *A* can be used in this fashion:
21:math:d^* = \sqrt {A_1 h^2 + A_2 k^2 + A_3 l^2 + A_4 hk + A_5 hl + A_6 kl}, where
22*d* is the d-spacing, and :math:d^* is the reciprocal lattice spacing,
23:math:Q = 2 \\pi d^* = 2 \\pi / d
24'''
25########### SVN repository information ###################
26# $Date: 2015-10-30 21:02:02 +0000 (Fri, 30 Oct 2015)$
27# $Author: vondreele$
28# $Revision: 2038$
29# $URL: trunk/GSASIIlattice.py$
30# $Id: GSASIIlattice.py 2038 2015-10-30 21:02:02Z vondreele$
31########### SVN repository information ###################
32import math
33import numpy as np
34import numpy.linalg as nl
35import GSASIIpath
36import GSASIImath as G2mth
37import GSASIIspc as G2spc
38GSASIIpath.SetVersionNumber("$Revision: 2038$")
39# trig functions in degrees
40sind = lambda x: np.sin(x*np.pi/180.)
41asind = lambda x: 180.*np.arcsin(x)/np.pi
42tand = lambda x: np.tan(x*np.pi/180.)
43atand = lambda x: 180.*np.arctan(x)/np.pi
44atan2d = lambda y,x: 180.*np.arctan2(y,x)/np.pi
45cosd = lambda x: np.cos(x*np.pi/180.)
46acosd = lambda x: 180.*np.arccos(x)/np.pi
47rdsq2d = lambda x,p: round(1.0/np.sqrt(x),p)
48rpd = np.pi/180.
49RSQ2PI = 1./np.sqrt(2.*np.pi)
50SQ2 = np.sqrt(2.)
51RSQPI = 1./np.sqrt(np.pi)
52R2pisq = 1./(2.*np.pi**2)
53
54def sec2HMS(sec):
55 """Convert time in sec to H:M:S string
56
57 :param sec: time in seconds
58 :return: H:M:S string (to nearest 100th second)
59
60 """
61 H = int(sec/3600)
62 M = int(sec/60-H*60)
63 S = sec-3600*H-60*M
64 return '%d:%2d:%.2f'%(H,M,S)
65
66def rotdMat(angle,axis=0):
67 """Prepare rotation matrix for angle in degrees about axis(=0,1,2)
68
69 :param angle: angle in degrees
70 :param axis: axis (0,1,2 = x,y,z) about which for the rotation
71 :return: rotation matrix - 3x3 numpy array
72
73 """
74 if axis == 2:
75 return np.array([[cosd(angle),-sind(angle),0],[sind(angle),cosd(angle),0],[0,0,1]])
76 elif axis == 1:
77 return np.array([[cosd(angle),0,-sind(angle)],[0,1,0],[sind(angle),0,cosd(angle)]])
78 else:
79 return np.array([[1,0,0],[0,cosd(angle),-sind(angle)],[0,sind(angle),cosd(angle)]])
80
81def rotdMat4(angle,axis=0):
82 """Prepare rotation matrix for angle in degrees about axis(=0,1,2) with scaling for OpenGL
83
84 :param angle: angle in degrees
85 :param axis: axis (0,1,2 = x,y,z) about which for the rotation
86 :return: rotation matrix - 4x4 numpy array (last row/column for openGL scaling)
87
88 """
89 Mat = rotdMat(angle,axis)
90 return np.concatenate((np.concatenate((Mat,[[0],[0],[0]]),axis=1),[[0,0,0,1],]),axis=0)
91
92def fillgmat(cell):
93 """Compute lattice metric tensor from unit cell constants
94
95 :param cell: tuple with a,b,c,alpha, beta, gamma (degrees)
96 :return: 3x3 numpy array
97
98 """
99 a,b,c,alp,bet,gam = cell
100 g = np.array([
101 [a*a, a*b*cosd(gam), a*c*cosd(bet)],
102 [a*b*cosd(gam), b*b, b*c*cosd(alp)],
103 [a*c*cosd(bet) ,b*c*cosd(alp), c*c]])
104 return g
105
106def cell2Gmat(cell):
107 """Compute real and reciprocal lattice metric tensor from unit cell constants
108
109 :param cell: tuple with a,b,c,alpha, beta, gamma (degrees)
110 :return: reciprocal (G) & real (g) metric tensors (list of two numpy 3x3 arrays)
111
112 """
113 g = fillgmat(cell)
114 G = nl.inv(g)
115 return G,g
116
117def A2Gmat(A,inverse=True):
118 """Fill real & reciprocal metric tensor (G) from A.
119
120 :param A: reciprocal metric tensor elements as [G11,G22,G33,2*G12,2*G13,2*G23]
121 :param bool inverse: if True return both G and g; else just G
122 :return: reciprocal (G) & real (g) metric tensors (list of two numpy 3x3 arrays)
123
124 """
125 G = np.zeros(shape=(3,3))
126 G = [
127 [A[0], A[3]/2., A[4]/2.],
128 [A[3]/2.,A[1], A[5]/2.],
129 [A[4]/2.,A[5]/2., A[2]]]
130 if inverse:
131 g = nl.inv(G)
132 return G,g
133 else:
134 return G
135
136def Gmat2A(G):
137 """Extract A from reciprocal metric tensor (G)
138
139 :param G: reciprocal maetric tensor (3x3 numpy array
140 :return: A = [G11,G22,G33,2*G12,2*G13,2*G23]
141
142 """
143 return [G[0][0],G[1][1],G[2][2],2.*G[0][1],2.*G[0][2],2.*G[1][2]]
144
145def cell2A(cell):
146 """Obtain A = [G11,G22,G33,2*G12,2*G13,2*G23] from lattice parameters
147
148 :param cell: [a,b,c,alpha,beta,gamma] (degrees)
149 :return: G reciprocal metric tensor as 3x3 numpy array
150
151 """
152 G,g = cell2Gmat(cell)
153 return Gmat2A(G)
154
155def A2cell(A):
156 """Compute unit cell constants from A
157
158 :param A: [G11,G22,G33,2*G12,2*G13,2*G23] G - reciprocal metric tensor
159 :return: a,b,c,alpha, beta, gamma (degrees) - lattice parameters
160
161 """
162 G,g = A2Gmat(A)
163 return Gmat2cell(g)
164
165def Gmat2cell(g):
166 """Compute real/reciprocal lattice parameters from real/reciprocal metric tensor (g/G)
167 The math works the same either way.
168
169 :param g (or G): real (or reciprocal) metric tensor 3x3 array
170 :return: a,b,c,alpha, beta, gamma (degrees) (or a*,b*,c*,alpha*,beta*,gamma* degrees)
171
172 """
173 oldset = np.seterr('raise')
174 a = np.sqrt(max(0,g[0][0]))
175 b = np.sqrt(max(0,g[1][1]))
176 c = np.sqrt(max(0,g[2][2]))
177 alp = acosd(g[2][1]/(b*c))
178 bet = acosd(g[2][0]/(a*c))
179 gam = acosd(g[0][1]/(a*b))
180 np.seterr(**oldset)
181 return a,b,c,alp,bet,gam
182
183def invcell2Gmat(invcell):
184 """Compute real and reciprocal lattice metric tensor from reciprocal
185 unit cell constants
186
187 :param invcell: [a*,b*,c*,alpha*, beta*, gamma*] (degrees)
188 :return: reciprocal (G) & real (g) metric tensors (list of two 3x3 arrays)
189
190 """
191 G = fillgmat(invcell)
192 g = nl.inv(G)
193 return G,g
194
195def calc_rVsq(A):
196 """Compute the square of the reciprocal lattice volume (1/V**2) from A'
197
198 """
199 G,g = A2Gmat(A)
200 rVsq = nl.det(G)
201 if rVsq < 0:
202 return 1
203 return rVsq
204
205def calc_rV(A):
206 """Compute the reciprocal lattice volume (V*) from A
207 """
208 return np.sqrt(calc_rVsq(A))
209
210def calc_V(A):
211 """Compute the real lattice volume (V) from A
212 """
213 return 1./calc_rV(A)
214
215def A2invcell(A):
216 """Compute reciprocal unit cell constants from A
217 returns tuple with a*,b*,c*,alpha*, beta*, gamma* (degrees)
218 """
219 G,g = A2Gmat(A)
220 return Gmat2cell(G)
221
222def Gmat2AB(G):
223 """Computes orthogonalization matrix from reciprocal metric tensor G
224
225 :returns: tuple of two 3x3 numpy arrays (A,B)
226
227 * A for crystal to Cartesian transformations A*x = np.inner(A,x) = X
228 * B (= inverse of A) for Cartesian to crystal transformation B*X = np.inner(B,X) = x
229
230 """
231 cellstar = Gmat2cell(G)
232 g = nl.inv(G)
233 cell = Gmat2cell(g)
234 A = np.zeros(shape=(3,3))
235 # from Giacovazzo (Fundamentals 2nd Ed.) p.75
236 A[0][0] = cell[0] # a
237 A[0][1] = cell[1]*cosd(cell[5]) # b cos(gamma)
238 A[0][2] = cell[2]*cosd(cell[4]) # c cos(beta)
239 A[1][1] = cell[1]*sind(cell[5]) # b sin(gamma)
240 A[1][2] = -cell[2]*cosd(cellstar[3])*sind(cell[4]) # - c cos(alpha*) sin(beta)
241 A[2][2] = 1/cellstar[2] # 1/c*
242 B = nl.inv(A)
243 return A,B
244
245
246def cell2AB(cell):
247 """Computes orthogonalization matrix from unit cell constants
248
249 :param tuple cell: a,b,c, alpha, beta, gamma (degrees)
250 :returns: tuple of two 3x3 numpy arrays (A,B)
251 A for crystal to Cartesian transformations A*x = np.inner(A,x) = X
252 B (= inverse of A) for Cartesian to crystal transformation B*X = np.inner(B,X) = x
253 """
254 G,g = cell2Gmat(cell)
255 cellstar = Gmat2cell(G)
256 A = np.zeros(shape=(3,3))
257 # from Giacovazzo (Fundamentals 2nd Ed.) p.75
258 A[0][0] = cell[0] # a
259 A[0][1] = cell[1]*cosd(cell[5]) # b cos(gamma)
260 A[0][2] = cell[2]*cosd(cell[4]) # c cos(beta)
261 A[1][1] = cell[1]*sind(cell[5]) # b sin(gamma)
262 A[1][2] = -cell[2]*cosd(cellstar[3])*sind(cell[4]) # - c cos(alpha*) sin(beta)
263 A[2][2] = 1/cellstar[2] # 1/c*
264 B = nl.inv(A)
265 return A,B
266
267def U6toUij(U6):
268 """Fill matrix (Uij) from U6 = [U11,U22,U33,U12,U13,U23]
269 NB: there is a non numpy version in GSASIIspc: U2Uij
270
271 :param list U6: 6 terms of u11,u22,...
272 :returns:
273 Uij - numpy [3][3] array of uij
274 """
275 U = np.array([
276 [U6[0], U6[3], U6[4]],
277 [U6[3], U6[1], U6[5]],
278 [U6[4], U6[5], U6[2]]])
279 return U
280
281def UijtoU6(U):
282 """Fill vector [U11,U22,U33,U12,U13,U23] from Uij
283 NB: there is a non numpy version in GSASIIspc: Uij2U
284 """
285 U6 = np.array([U[0][0],U[1][1],U[2][2],U[0][1],U[0][2],U[1][2]])
286 return U6
287
288def betaij2Uij(betaij,G):
289 """
290 Convert beta-ij to Uij tensors
291
292 :param beta-ij - numpy array [beta-ij]
293 :param G: reciprocal metric tensor
294 :returns: Uij: numpy array [Uij]
295 """
296 ast = np.sqrt(np.diag(G)) #a*, b*, c*
297 Mast = np.multiply.outer(ast,ast)
298 return R2pisq*UijtoU6(U6toUij(betaij)/Mast)
299
300def Uij2betaij(Uij,G):
301 """
302 Convert Uij to beta-ij tensors -- stub for eventual completion
303
304 :param Uij: numpy array [Uij]
305 :param G: reciprocal metric tensor
306 :returns: beta-ij - numpy array [beta-ij]
307 """
308 pass
309
310def cell2GS(cell):
311 ''' returns Uij to betaij conversion matrix'''
312 G,g = cell2Gmat(cell)
313 GS = G
314 GS[0][1] = GS[1][0] = math.sqrt(GS[0][0]*GS[1][1])
315 GS[0][2] = GS[2][0] = math.sqrt(GS[0][0]*GS[2][2])
316 GS[1][2] = GS[2][1] = math.sqrt(GS[1][1]*GS[2][2])
317 return GS
318
319def Uij2Ueqv(Uij,GS,Amat):
320 ''' returns 1/3 trace of diagonalized U matrix'''
321 U = np.multiply(U6toUij(Uij),GS)
322 U = np.inner(Amat,np.inner(U,Amat).T)
323 E,R = nl.eigh(U)
324 return np.sum(E)/3.
325
326def CosAngle(U,V,G):
327 """ calculate cos of angle between U & V in generalized coordinates
328 defined by metric tensor G
329
330 :param U: 3-vectors assume numpy arrays, can be multiple reflections as (N,3) array
331 :param V: 3-vectors assume numpy arrays, only as (3) vector
332 :param G: metric tensor for U & V defined space assume numpy array
333 :returns:
334 cos(phi)
335 """
336 u = (U.T/np.sqrt(np.sum(np.inner(U,G)*U,axis=1))).T
337 v = V/np.sqrt(np.inner(V,np.inner(G,V)))
338 cosP = np.inner(u,np.inner(G,v))
339 return cosP
340
341def CosSinAngle(U,V,G):
342 """ calculate sin & cos of angle between U & V in generalized coordinates
343 defined by metric tensor G
344
345 :param U: 3-vectors assume numpy arrays
346 :param V: 3-vectors assume numpy arrays
347 :param G: metric tensor for U & V defined space assume numpy array
348 :returns:
349 cos(phi) & sin(phi)
350 """
351 u = U/np.sqrt(np.inner(U,np.inner(G,U)))
352 v = V/np.sqrt(np.inner(V,np.inner(G,V)))
353 cosP = np.inner(u,np.inner(G,v))
354 sinP = np.sqrt(max(0.0,1.0-cosP**2))
355 return cosP,sinP
356
357def criticalEllipse(prob):
358 """
359 Calculate critical values for probability ellipsoids from probability
360 """
361 if not ( 0.01 <= prob < 1.0):
362 return 1.54
363 coeff = np.array([6.44988E-09,4.16479E-07,1.11172E-05,1.58767E-04,0.00130554,
364 0.00604091,0.0114921,-0.040301,-0.6337203,1.311582])
365 llpr = math.log(-math.log(prob))
366 return np.polyval(coeff,llpr)
367
368def CellBlock(nCells):
369 """
370 Generate block of unit cells n*n*n on a side; [0,0,0] centered, n = 2*nCells+1
371 currently only works for nCells = 0 or 1 (not >1)
372 """
373 if nCells:
374 N = 2*nCells+1
375 N2 = N*N
376 N3 = N*N*N
377 cellArray = []
378 A = np.array(range(N3))
379 cellGen = np.array([A/N2-1,A/N%N-1,A%N-1]).T
380 for cell in cellGen:
381 cellArray.append(cell)
382 return cellArray
383 else:
384 return [0,0,0]
385
386def CellAbsorption(ElList,Volume):
387 '''Compute unit cell absorption
388
389 :param dict ElList: dictionary of element contents including mu and
390 number of atoms be cell
391 :param float Volume: unit cell volume
392 :returns: mu-total/Volume
393 '''
394 muT = 0
395 for El in ElList:
396 muT += ElList[El]['mu']*ElList[El]['FormulaNo']
397 return muT/Volume
398
399#Permutations and Combinations
400# Four routines: combinations,uniqueCombinations, selections & permutations
401#These taken from Python Cookbook, 2nd Edition. 19.15 p724-726
402#
403def _combinators(_handle, items, n):
404 """ factored-out common structure of all following combinators """
405 if n==0:
406 yield [ ]
407 return
408 for i, item in enumerate(items):
409 this_one = [ item ]
410 for cc in _combinators(_handle, _handle(items, i), n-1):
411 yield this_one + cc
412def combinations(items, n):
413 """ take n distinct items, order matters """
414 def skipIthItem(items, i):
415 return items[:i] + items[i+1:]
416 return _combinators(skipIthItem, items, n)
417def uniqueCombinations(items, n):
418 """ take n distinct items, order is irrelevant """
419 def afterIthItem(items, i):
420 return items[i+1:]
421 return _combinators(afterIthItem, items, n)
422def selections(items, n):
423 """ take n (not necessarily distinct) items, order matters """
424 def keepAllItems(items, i):
425 return items
426 return _combinators(keepAllItems, items, n)
427def permutations(items):
428 """ take all items, order matters """
429 return combinations(items, len(items))
430
431#reflection generation routines
432#for these: H = [h,k,l]; A is as used in calc_rDsq; G - inv metric tensor, g - metric tensor;
433# cell - a,b,c,alp,bet,gam in A & deg
434
435def Pos2dsp(Inst,pos):
436 ''' convert powder pattern position (2-theta or TOF, musec) to d-spacing
437 '''
438 if 'C' in Inst['Type'][0] or 'PKS' in Inst['Type'][0]:
439 wave = G2mth.getWave(Inst)
440 return wave/(2.0*sind((pos-Inst.get('Zero',[0,0])[1])/2.0))
441 else: #'T'OF - ignore difB
442 return TOF2dsp(Inst,pos)
443
444def TOF2dsp(Inst,Pos):
445 ''' convert powder pattern TOF, musec to d-spacing by successive approximation
446 Pos can be numpy array
447 '''
448 def func(d,pos,Inst):
449 return (pos-Inst['difA'][1]*d**2-Inst['Zero'][1]-Inst['difB'][1]/d)/Inst['difC'][1]
450 dsp0 = np.ones_like(Pos)
451 while True: #successive approximations
452 dsp = func(dsp0,Pos,Inst)
453 if np.allclose(dsp,dsp0,atol=0.000001):
454 return dsp
455 dsp0 = dsp
456
457def Dsp2pos(Inst,dsp):
458 ''' convert d-spacing to powder pattern position (2-theta or TOF, musec)
459 '''
460 if 'C' in Inst['Type'][0] or 'PKS' in Inst['Type'][0]:
461 wave = G2mth.getWave(Inst)
462 pos = 2.0*asind(wave/(2.*dsp))+Inst.get('Zero',[0,0])[1]
463 else: #'T'OF
464 pos = Inst['difC'][1]*dsp+Inst['Zero'][1]+Inst['difA'][1]*dsp**2+Inst.get('difB',[0,0,False])[1]/dsp
465 return pos
466
467def getPeakPos(dataType,parmdict,dsp):
468 ''' convert d-spacing to powder pattern position (2-theta or TOF, musec)
469 '''
470 if 'C' in dataType:
471 pos = 2.0*asind(parmdict['Lam']/(2.*dsp))+parmdict['Zero']
472 else: #'T'OF
473 pos = parmdict['difC']*dsp+parmdict['difA']*dsp**2+parmdict['difB']/dsp+parmdict['Zero']
474 return pos
475
476def calc_rDsq(H,A):
477 'needs doc string'
478 rdsq = H[0]*H[0]*A[0]+H[1]*H[1]*A[1]+H[2]*H[2]*A[2]+H[0]*H[1]*A[3]+H[0]*H[2]*A[4]+H[1]*H[2]*A[5]
479 return rdsq
480
481def calc_rDsq2(H,G):
482 'needs doc string'
483 return np.inner(H,np.inner(G,H))
484
485def calc_rDsqSS(H,A,vec):
486 'needs doc string'
487 rdsq = calc_rDsq(H[:3]+(H[3]*vec).T,A)
488 return rdsq
489
490def calc_rDsqZ(H,A,Z,tth,lam):
491 'needs doc string'
492 rdsq = calc_rDsq(H,A)+Z*sind(tth)*2.0*rpd/lam**2
493 return rdsq
494
495def calc_rDsqZSS(H,A,vec,Z,tth,lam):
496 'needs doc string'
497 rdsq = calc_rDsq(H[:3]+(H[3][:,np.newaxis]*vec).T,A)+Z*sind(tth)*2.0*rpd/lam**2
498 return rdsq
499
500def calc_rDsqT(H,A,Z,tof,difC):
501 'needs doc string'
502 rdsq = calc_rDsq(H,A)+Z/difC
503 return rdsq
504
505def calc_rDsqTSS(H,A,vec,Z,tof,difC):
506 'needs doc string'
507 rdsq = calc_rDsq(H[:3]+(H[3][:,np.newaxis]*vec).T,A)+Z/difC
508 return rdsq
509
510def MaxIndex(dmin,A):
511 'needs doc string'
512 Hmax = [0,0,0]
513 try:
514 cell = A2cell(A)
515 except:
516 cell = [1,1,1,90,90,90]
517 for i in range(3):
518 Hmax[i] = int(round(cell[i]/dmin))
519 return Hmax
520
521def sortHKLd(HKLd,ifreverse,ifdup,ifSS=False):
522 '''needs doc string
523
524 :param HKLd: a list of [h,k,l,d,...];
525 :param ifreverse: True for largest d first
526 :param ifdup: True if duplicate d-spacings allowed
527 '''
528 T = []
529 N = 3
530 if ifSS:
531 N = 4
532 for i,H in enumerate(HKLd):
533 if ifdup:
534 T.append((H[N],i))
535 else:
536 T.append(H[N])
537 D = dict(zip(T,HKLd))
538 T.sort()
539 if ifreverse:
540 T.reverse()
541 X = []
542 okey = ''
543 for key in T:
544 if key != okey: X.append(D[key]) #remove duplicate d-spacings
545 okey = key
546 return X
547
548def SwapIndx(Axis,H):
549 'needs doc string'
550 if Axis in [1,-1]:
551 return H
552 elif Axis in [2,-3]:
553 return [H[1],H[2],H[0]]
554 else:
555 return [H[2],H[0],H[1]]
556
557def Rh2Hx(Rh):
558 'needs doc string'
559 Hx = [0,0,0]
560 Hx[0] = Rh[0]-Rh[1]
561 Hx[1] = Rh[1]-Rh[2]
562 Hx[2] = np.sum(Rh)
563 return Hx
564
565def Hx2Rh(Hx):
566 'needs doc string'
567 Rh = [0,0,0]
568 itk = -Hx[0]+Hx[1]+Hx[2]
569 if itk%3 != 0:
570 return 0 #error - not rhombohedral reflection
571 else:
572 Rh[1] = itk/3
573 Rh[0] = Rh[1]+Hx[0]
574 Rh[2] = Rh[1]-Hx[1]
575 if Rh[0] < 0:
576 for i in range(3):
577 Rh[i] = -Rh[i]
578 return Rh
579
580def CentCheck(Cent,H):
581 'needs doc string'
582 h,k,l = H
583 if Cent == 'A' and (k+l)%2:
584 return False
585 elif Cent == 'B' and (h+l)%2:
586 return False
587 elif Cent == 'C' and (h+k)%2:
588 return False
589 elif Cent == 'I' and (h+k+l)%2:
590 return False
591 elif Cent == 'F' and ((h+k)%2 or (h+l)%2 or (k+l)%2):
592 return False
593 elif Cent == 'R' and (-h+k+l)%3:
594 return False
595 else:
596 return True
597
598def GetBraviasNum(center,system):
599 """Determine the Bravais lattice number, as used in GenHBravais
600
601 :param center: one of: 'P', 'C', 'I', 'F', 'R' (see SGLatt from GSASIIspc.SpcGroup)
602 :param system: one of 'cubic', 'hexagonal', 'tetragonal', 'orthorhombic', 'trigonal' (for R)
603 'monoclinic', 'triclinic' (see SGSys from GSASIIspc.SpcGroup)
604 :return: a number between 0 and 13
605 or throws a ValueError exception if the combination of center, system is not found (i.e. non-standard)
606
607 """
608 if center.upper() == 'F' and system.lower() == 'cubic':
609 return 0
610 elif center.upper() == 'I' and system.lower() == 'cubic':
611 return 1
612 elif center.upper() == 'P' and system.lower() == 'cubic':
613 return 2
614 elif center.upper() == 'R' and system.lower() == 'trigonal':
615 return 3
616 elif center.upper() == 'P' and system.lower() == 'hexagonal':
617 return 4
618 elif center.upper() == 'I' and system.lower() == 'tetragonal':
619 return 5
620 elif center.upper() == 'P' and system.lower() == 'tetragonal':
621 return 6
622 elif center.upper() == 'F' and system.lower() == 'orthorhombic':
623 return 7
624 elif center.upper() == 'I' and system.lower() == 'orthorhombic':
625 return 8
626 elif center.upper() == 'C' and system.lower() == 'orthorhombic':
627 return 9
628 elif center.upper() == 'P' and system.lower() == 'orthorhombic':
629 return 10
630 elif center.upper() == 'C' and system.lower() == 'monoclinic':
631 return 11
632 elif center.upper() == 'P' and system.lower() == 'monoclinic':
633 return 12
634 elif center.upper() == 'P' and system.lower() == 'triclinic':
635 return 13
636 raise ValueError,'non-standard Bravais lattice center=%s, cell=%s' % (center,system)
637
638def GenHBravais(dmin,Bravais,A):
639 """Generate the positionally unique powder diffraction reflections
640
641 :param dmin: minimum d-spacing in A
642 :param Bravais: lattice type (see GetBraviasNum). Bravais is one of::
643 0 F cubic
644 1 I cubic
645 2 P cubic
646 3 R hexagonal (trigonal not rhombohedral)
647 4 P hexagonal
648 5 I tetragonal
649 6 P tetragonal
650 7 F orthorhombic
651 8 I orthorhombic
652 9 C orthorhombic
653 10 P orthorhombic
654 11 C monoclinic
655 12 P monoclinic
656 13 P triclinic
657
658 :param A: reciprocal metric tensor elements as [G11,G22,G33,2*G12,2*G13,2*G23]
659 :return: HKL unique d list of [h,k,l,d,-1] sorted with largest d first
660
661 """
662 import math
663 if Bravais in [9,11]:
664 Cent = 'C'
665 elif Bravais in [1,5,8]:
666 Cent = 'I'
667 elif Bravais in [0,7]:
668 Cent = 'F'
669 elif Bravais in [3]:
670 Cent = 'R'
671 else:
672 Cent = 'P'
673 Hmax = MaxIndex(dmin,A)
674 dminsq = 1./(dmin**2)
675 HKL = []
676 if Bravais == 13: #triclinic
677 for l in range(-Hmax[2],Hmax[2]+1):
678 for k in range(-Hmax[1],Hmax[1]+1):
679 hmin = 0
680 if (k < 0): hmin = 1
681 if (k ==0 and l < 0): hmin = 1
682 for h in range(hmin,Hmax[0]+1):
683 H=[h,k,l]
684 rdsq = calc_rDsq(H,A)
685 if 0 < rdsq <= dminsq:
686 HKL.append([h,k,l,rdsq2d(rdsq,6),-1])
687 elif Bravais in [11,12]: #monoclinic - b unique
688 Hmax = SwapIndx(2,Hmax)
689 for h in range(Hmax[0]+1):
690 for k in range(-Hmax[1],Hmax[1]+1):
691 lmin = 0
692 if k < 0:lmin = 1
693 for l in range(lmin,Hmax[2]+1):
694 [h,k,l] = SwapIndx(-2,[h,k,l])
695 H = []
696 if CentCheck(Cent,[h,k,l]): H=[h,k,l]
697 if H:
698 rdsq = calc_rDsq(H,A)
699 if 0 < rdsq <= dminsq:
700 HKL.append([h,k,l,rdsq2d(rdsq,6),-1])
701 [h,k,l] = SwapIndx(2,[h,k,l])
702 elif Bravais in [7,8,9,10]: #orthorhombic
703 for h in range(Hmax[0]+1):
704 for k in range(Hmax[1]+1):
705 for l in range(Hmax[2]+1):
706 H = []
707 if CentCheck(Cent,[h,k,l]): H=[h,k,l]
708 if H:
709 rdsq = calc_rDsq(H,A)
710 if 0 < rdsq <= dminsq:
711 HKL.append([h,k,l,rdsq2d(rdsq,6),-1])
712 elif Bravais in [5,6]: #tetragonal
713 for l in range(Hmax[2]+1):
714 for k in range(Hmax[1]+1):
715 for h in range(k,Hmax[0]+1):
716 H = []
717 if CentCheck(Cent,[h,k,l]): H=[h,k,l]
718 if H:
719 rdsq = calc_rDsq(H,A)
720 if 0 < rdsq <= dminsq:
721 HKL.append([h,k,l,rdsq2d(rdsq,6),-1])
722 elif Bravais in [3,4]:
723 lmin = 0
724 if Bravais == 3: lmin = -Hmax[2] #hexagonal/trigonal
725 for l in range(lmin,Hmax[2]+1):
726 for k in range(Hmax[1]+1):
727 hmin = k
728 if l < 0: hmin += 1
729 for h in range(hmin,Hmax[0]+1):
730 H = []
731 if CentCheck(Cent,[h,k,l]): H=[h,k,l]
732 if H:
733 rdsq = calc_rDsq(H,A)
734 if 0 < rdsq <= dminsq:
735 HKL.append([h,k,l,rdsq2d(rdsq,6),-1])
736
737 else: #cubic
738 for l in range(Hmax[2]+1):
739 for k in range(l,Hmax[1]+1):
740 for h in range(k,Hmax[0]+1):
741 H = []
742 if CentCheck(Cent,[h,k,l]): H=[h,k,l]
743 if H:
744 rdsq = calc_rDsq(H,A)
745 if 0 < rdsq <= dminsq:
746 HKL.append([h,k,l,rdsq2d(rdsq,6),-1])
747 return sortHKLd(HKL,True,False)
748
749def getHKLmax(dmin,SGData,A):
750 'finds maximum allowed hkl for given A within dmin'
751 SGLaue = SGData['SGLaue']
752 if SGLaue in ['3R','3mR']: #Rhombohedral axes
753 Hmax = [0,0,0]
754 cell = A2cell(A)
755 aHx = cell[0]*math.sqrt(2.0*(1.0-cosd(cell[3])))
756 cHx = cell[0]*math.sqrt(3.0*(1.0+2.0*cosd(cell[3])))
757 Hmax[0] = Hmax[1] = int(round(aHx/dmin))
758 Hmax[2] = int(round(cHx/dmin))
759 #print Hmax,aHx,cHx
760 else: # all others
761 Hmax = MaxIndex(dmin,A)
762 return Hmax
763
764def GenHLaue(dmin,SGData,A):
765 """Generate the crystallographically unique powder diffraction reflections
766 for a lattice and Bravais type
767
768 :param dmin: minimum d-spacing
769 :param SGData: space group dictionary with at least
770
771 * 'SGLaue': Laue group symbol: one of '-1','2/m','mmm','4/m','6/m','4/mmm','6/mmm', '3m1', '31m', '3', '3R', '3mR', 'm3', 'm3m'
772 * 'SGLatt': lattice centering: one of 'P','A','B','C','I','F'
773 * 'SGUniq': code for unique monoclinic axis one of 'a','b','c' (only if 'SGLaue' is '2/m') otherwise an empty string
774
775 :param A: reciprocal metric tensor elements as [G11,G22,G33,2*G12,2*G13,2*G23]
776 :return: HKL = list of [h,k,l,d] sorted with largest d first and is unique
777 part of reciprocal space ignoring anomalous dispersion
778
779 """
780 import math
781 SGLaue = SGData['SGLaue']
782 SGLatt = SGData['SGLatt']
783 SGUniq = SGData['SGUniq']
784 #finds maximum allowed hkl for given A within dmin
785 Hmax = getHKLmax(dmin,SGData,A)
786
787 dminsq = 1./(dmin**2)
788 HKL = []
789 if SGLaue == '-1': #triclinic
790 for l in range(-Hmax[2],Hmax[2]+1):
791 for k in range(-Hmax[1],Hmax[1]+1):
792 hmin = 0
793 if (k < 0) or (k ==0 and l < 0): hmin = 1
794 for h in range(hmin,Hmax[0]+1):
795 H = []
796 if CentCheck(SGLatt,[h,k,l]): H=[h,k,l]
797 if H:
798 rdsq = calc_rDsq(H,A)
799 if 0 < rdsq <= dminsq:
800 HKL.append([h,k,l,1/math.sqrt(rdsq)])
801 elif SGLaue == '2/m': #monoclinic
802 axisnum = 1 + ['a','b','c'].index(SGUniq)
803 Hmax = SwapIndx(axisnum,Hmax)
804 for h in range(Hmax[0]+1):
805 for k in range(-Hmax[1],Hmax[1]+1):
806 lmin = 0
807 if k < 0:lmin = 1
808 for l in range(lmin,Hmax[2]+1):
809 [h,k,l] = SwapIndx(-axisnum,[h,k,l])
810 H = []
811 if CentCheck(SGLatt,[h,k,l]): H=[h,k,l]
812 if H:
813 rdsq = calc_rDsq(H,A)
814 if 0 < rdsq <= dminsq:
815 HKL.append([h,k,l,1/math.sqrt(rdsq)])
816 [h,k,l] = SwapIndx(axisnum,[h,k,l])
817 elif SGLaue in ['mmm','4/m','6/m']: #orthorhombic
818 for l in range(Hmax[2]+1):
819 for h in range(Hmax[0]+1):
820 kmin = 1
821 if SGLaue == 'mmm' or h ==0: kmin = 0
822 for k in range(kmin,Hmax[1]+1):
823 H = []
824 if CentCheck(SGLatt,[h,k,l]): H=[h,k,l]
825 if H:
826 rdsq = calc_rDsq(H,A)
827 if 0 < rdsq <= dminsq:
828 HKL.append([h,k,l,1/math.sqrt(rdsq)])
829 elif SGLaue in ['4/mmm','6/mmm']: #tetragonal & hexagonal
830 for l in range(Hmax[2]+1):
831 for h in range(Hmax[0]+1):
832 for k in range(h+1):
833 H = []
834 if CentCheck(SGLatt,[h,k,l]): H=[h,k,l]
835 if H:
836 rdsq = calc_rDsq(H,A)
837 if 0 < rdsq <= dminsq:
838 HKL.append([h,k,l,1/math.sqrt(rdsq)])
839 elif SGLaue in ['3m1','31m','3','3R','3mR']: #trigonals
840 for l in range(-Hmax[2],Hmax[2]+1):
841 hmin = 0
842 if l < 0: hmin = 1
843 for h in range(hmin,Hmax[0]+1):
844 if SGLaue in ['3R','3']:
845 kmax = h
846 kmin = -int((h-1.)/2.)
847 else:
848 kmin = 0
849 kmax = h
850 if SGLaue in ['3m1','3mR'] and l < 0: kmax = h-1
851 if SGLaue == '31m' and l < 0: kmin = 1
852 for k in range(kmin,kmax+1):
853 H = []
854 if CentCheck(SGLatt,[h,k,l]): H=[h,k,l]
855 if SGLaue in ['3R','3mR']:
856 H = Hx2Rh(H)
857 if H:
858 rdsq = calc_rDsq(H,A)
859 if 0 < rdsq <= dminsq:
860 HKL.append([H[0],H[1],H[2],1/math.sqrt(rdsq)])
861 else: #cubic
862 for h in range(Hmax[0]+1):
863 for k in range(h+1):
864 lmin = 0
865 lmax = k
866 if SGLaue =='m3':
867 lmax = h-1
868 if h == k: lmax += 1
869 for l in range(lmin,lmax+1):
870 H = []
871 if CentCheck(SGLatt,[h,k,l]): H=[h,k,l]
872 if H:
873 rdsq = calc_rDsq(H,A)
874 if 0 < rdsq <= dminsq:
875 HKL.append([h,k,l,1/math.sqrt(rdsq)])
876 return sortHKLd(HKL,True,True)
877
878def GenPfHKLs(nMax,SGData,A):
879 """Generate the unique pole figure reflections for a lattice and Bravais type.
880 Min d-spacing=1.0A & no more than nMax returned
881
882 :param nMax: maximum number of hkls returned
883 :param SGData: space group dictionary with at least
884
885 * 'SGLaue': Laue group symbol: one of '-1','2/m','mmm','4/m','6/m','4/mmm','6/mmm', '3m1', '31m', '3', '3R', '3mR', 'm3', 'm3m'
886 * 'SGLatt': lattice centering: one of 'P','A','B','C','I','F'
887 * 'SGUniq': code for unique monoclinic axis one of 'a','b','c' (only if 'SGLaue' is '2/m') otherwise an empty string
888
889 :param A: reciprocal metric tensor elements as [G11,G22,G33,2*G12,2*G13,2*G23]
890 :return: HKL = list of 'h k l' strings sorted with largest d first; no duplicate zones
891
892 """
893 HKL = np.array(GenHLaue(1.0,SGData,A)).T[:3].T #strip d-spacings
894 N = min(nMax,len(HKL))
895 return ['%d %d %d'%(h[0],h[1],h[2]) for h in HKL[:N]]
896
897
898def GenSSHLaue(dmin,SGData,SSGData,Vec,maxH,A):
899 'needs a doc string'
900 HKLs = []
901 vec = np.array(Vec)
902 vstar = np.sqrt(calc_rDsq(vec,A)) #find extra needed for -n SS reflections
903 dvec = 1./(maxH*vstar+1./dmin)
904 HKL = GenHLaue(dvec,SGData,A)
905 SSdH = [vec*h for h in range(-maxH,maxH+1)]
906 SSdH = dict(zip(range(-maxH,maxH+1),SSdH))
907 for h,k,l,d in HKL:
908 ext = G2spc.GenHKLf([h,k,l],SGData)[0] #h,k,l must be integral values here
909 if not ext and d >= dmin:
910 HKLs.append([h,k,l,0,d])
911 for dH in SSdH:
912 if dH:
913 DH = SSdH[dH]
914 H = [h+DH[0],k+DH[1],l+DH[2]]
915 d = 1/np.sqrt(calc_rDsq(H,A))
916 if d >= dmin:
917 HKLM = np.array([h,k,l,dH])
918 if G2spc.checkSSLaue([h,k,l,dH],SGData,SSGData) and G2spc.checkSSextc(HKLM,SSGData):
919 HKLs.append([h,k,l,dH,d])
920 return HKLs
921
922#Spherical harmonics routines
923def OdfChk(SGLaue,L,M):
924 'needs doc string'
925 if not L%2 and abs(M) <= L:
926 if SGLaue == '0': #cylindrical symmetry
927 if M == 0: return True
928 elif SGLaue == '-1':
929 return True
930 elif SGLaue == '2/m':
931 if not abs(M)%2: return True
932 elif SGLaue == 'mmm':
933 if not abs(M)%2 and M >= 0: return True
934 elif SGLaue == '4/m':
935 if not abs(M)%4: return True
936 elif SGLaue == '4/mmm':
937 if not abs(M)%4 and M >= 0: return True
938 elif SGLaue in ['3R','3']:
939 if not abs(M)%3: return True
940 elif SGLaue in ['3mR','3m1','31m']:
941 if not abs(M)%3 and M >= 0: return True
942 elif SGLaue == '6/m':
943 if not abs(M)%6: return True
944 elif SGLaue == '6/mmm':
945 if not abs(M)%6 and M >= 0: return True
946 elif SGLaue == 'm3':
947 if M > 0:
948 if L%12 == 2:
949 if M <= L/12: return True
950 else:
951 if M <= L/12+1: return True
952 elif SGLaue == 'm3m':
953 if M > 0:
954 if L%12 == 2:
955 if M <= L/12: return True
956 else:
957 if M <= L/12+1: return True
958 return False
959
960def GenSHCoeff(SGLaue,SamSym,L,IfLMN=True):
961 'needs doc string'
962 coeffNames = []
963 for iord in [2*i+2 for i in range(L/2)]:
964 for m in [i-iord for i in range(2*iord+1)]:
965 if OdfChk(SamSym,iord,m):
966 for n in [i-iord for i in range(2*iord+1)]:
967 if OdfChk(SGLaue,iord,n):
968 if IfLMN:
969 coeffNames.append('C(%d,%d,%d)'%(iord,m,n))
970 else:
971 coeffNames.append('C(%d,%d)'%(iord,n))
972 return coeffNames
973
974def CrsAng(H,cell,SGData):
975 'needs doc string'
976 a,b,c,al,be,ga = cell
977 SQ3 = 1.732050807569
978 H1 = np.array([1,0,0])
979 H2 = np.array([0,1,0])
980 H3 = np.array([0,0,1])
981 H4 = np.array([1,1,1])
982 G,g = cell2Gmat(cell)
983 Laue = SGData['SGLaue']
984 Naxis = SGData['SGUniq']
985 if len(H.shape) == 1:
986 DH = np.inner(H,np.inner(G,H))
987 else:
988 DH = np.array([np.inner(h,np.inner(G,h)) for h in H])
989 if Laue == '2/m':
990 if Naxis == 'a':
991 DR = np.inner(H1,np.inner(G,H1))
992 DHR = np.inner(H,np.inner(G,H1))
993 elif Naxis == 'b':
994 DR = np.inner(H2,np.inner(G,H2))
995 DHR = np.inner(H,np.inner(G,H2))
996 else:
997 DR = np.inner(H3,np.inner(G,H3))
998 DHR = np.inner(H,np.inner(G,H3))
999 elif Laue in ['R3','R3m']:
1000 DR = np.inner(H4,np.inner(G,H4))
1001 DHR = np.inner(H,np.inner(G,H4))
1002 else:
1003 DR = np.inner(H3,np.inner(G,H3))
1004 DHR = np.inner(H,np.inner(G,H3))
1005 DHR /= np.sqrt(DR*DH)
1006 phi = np.where(DHR <= 1.0,acosd(DHR),0.0)
1007 if Laue == '-1':
1008 BA = H.T[1]*a/(b-H.T[0]*cosd(ga))
1009 BB = H.T[0]*sind(ga)**2
1010 elif Laue == '2/m':
1011 if Naxis == 'a':
1012 BA = H.T[2]*b/(c-H.T[1]*cosd(al))
1013 BB = H.T[1]*sind(al)**2
1014 elif Naxis == 'b':
1015 BA = H.T[0]*c/(a-H.T[2]*cosd(be))
1016 BB = H.T[2]*sind(be)**2
1017 else:
1018 BA = H.T[1]*a/(b-H.T[0]*cosd(ga))
1019 BB = H.T[0]*sind(ga)**2
1020 elif Laue in ['mmm','4/m','4/mmm']:
1021 BA = H.T[1]*a
1022 BB = H.T[0]*b
1023 elif Laue in ['3R','3mR']:
1024 BA = H.T[0]+H.T[1]-2.0*H.T[2]
1025 BB = SQ3*(H.T[0]-H.T[1])
1026 elif Laue in ['m3','m3m']:
1027 BA = H.T[1]
1028 BB = H.T[0]
1029 else:
1030 BA = H.T[0]+2.0*H.T[1]
1031 BB = SQ3*H.T[0]
1032 beta = atan2d(BA,BB)
1033 return phi,beta
1034
1035def SamAng(Tth,Gangls,Sangl,IFCoup):
1036 """Compute sample orientation angles vs laboratory coord. system
1037
1038 :param Tth: Signed theta
1039 :param Gangls: Sample goniometer angles phi,chi,omega,azmuth
1040 :param Sangl: Sample angle zeros om-0, chi-0, phi-0
1041 :param IFCoup: True if omega & 2-theta coupled in CW scan
1042 :returns:
1043 psi,gam: Sample odf angles
1044 dPSdA,dGMdA: Angle zero derivatives
1045 """
1046
1047 if IFCoup:
1048 GSomeg = sind(Gangls[2]+Tth)
1049 GComeg = cosd(Gangls[2]+Tth)
1050 else:
1051 GSomeg = sind(Gangls[2])
1052 GComeg = cosd(Gangls[2])
1053 GSTth = sind(Tth)
1054 GCTth = cosd(Tth)
1055 GSazm = sind(Gangls[3])
1056 GCazm = cosd(Gangls[3])
1057 GSchi = sind(Gangls[1])
1058 GCchi = cosd(Gangls[1])
1059 GSphi = sind(Gangls[0]+Sangl[2])
1060 GCphi = cosd(Gangls[0]+Sangl[2])
1061 SSomeg = sind(Sangl[0])
1062 SComeg = cosd(Sangl[0])
1063 SSchi = sind(Sangl[1])
1064 SCchi = cosd(Sangl[1])
1065 AT = -GSTth*GComeg+GCTth*GCazm*GSomeg
1066 BT = GSTth*GSomeg+GCTth*GCazm*GComeg
1067 CT = -GCTth*GSazm*GSchi
1068 DT = -GCTth*GSazm*GCchi
1069
1070 BC1 = -AT*GSphi+(CT+BT*GCchi)*GCphi
1071 BC2 = DT-BT*GSchi
1072 BC3 = AT*GCphi+(CT+BT*GCchi)*GSphi
1073
1074 BC = BC1*SComeg*SCchi+BC2*SComeg*SSchi-BC3*SSomeg
1075 psi = acosd(BC)
1076
1077 BD = 1.0-BC**2
1078 C = np.where(BD>1.e-6,rpd/np.sqrt(BD),0.)
1079 dPSdA = [-C*(-BC1*SSomeg*SCchi-BC2*SSomeg*SSchi-BC3*SComeg),
1080 -C*(-BC1*SComeg*SSchi+BC2*SComeg*SCchi),
1081 -C*(-BC1*SSomeg-BC3*SComeg*SCchi)]
1082
1083 BA = -BC1*SSchi+BC2*SCchi
1084 BB = BC1*SSomeg*SCchi+BC2*SSomeg*SSchi+BC3*SComeg
1085 gam = atan2d(BB,BA)
1086
1087 BD = (BA**2+BB**2)/rpd
1088
1092
1093 dBBdO = BC1*SComeg*SCchi+BC2*SComeg*SSchi-BC3*SSomeg
1094 dBBdC = -BC1*SSomeg*SSchi+BC2*SSomeg*SCchi
1095 dBBdF = BC1*SComeg-BC3*SSomeg*SCchi
1096
1099
1100 return psi,gam,dPSdA,dGMdA
1101
1102BOH = {
1103'L=2':[[],[],[]],
1104'L=4':[[0.30469720,0.36418281],[],[]],
1105'L=6':[[-0.14104740,0.52775103],[],[]],
1106'L=8':[[0.28646862,0.21545346,0.32826995],[],[]],
1107'L=10':[[-0.16413497,0.33078546,0.39371345],[],[]],
1108'L=12':[[0.26141975,0.27266871,0.03277460,0.32589402],
1109 [0.09298802,-0.23773812,0.49446631,0.0],[]],
1110'L=14':[[-0.17557309,0.25821932,0.27709173,0.33645360],[],[]],
1111'L=16':[[0.24370673,0.29873515,0.06447688,0.00377,0.32574495],
1112 [0.12039646,-0.25330128,0.23950998,0.40962508,0.0],[]],
1113'L=18':[[-0.16914245,0.17017340,0.34598142,0.07433932,0.32696037],
1114 [-0.06901768,0.16006562,-0.24743528,0.47110273,0.0],[]],
1115'L=20':[[0.23067026,0.31151832,0.09287682,0.01089683,0.00037564,0.32573563],
1116 [0.13615420,-0.25048007,0.12882081,0.28642879,0.34620433,0.0],[]],
1117'L=22':[[-0.16109560,0.10244188,0.36285175,0.13377513,0.01314399,0.32585583],
1118 [-0.09620055,0.20244115,-0.22389483,0.17928946,0.42017231,0.0],[]],
1119'L=24':[[0.22050742,0.31770654,0.11661736,0.02049853,0.00150861,0.00003426,0.32573505],
1120 [0.13651722,-0.21386648,0.00522051,0.33939435,0.10837396,0.32914497,0.0],
1121 [0.05378596,-0.11945819,0.16272298,-0.26449730,0.44923956,0.0,0.0]],
1122'L=26':[[-0.15435003,0.05261630,0.35524646,0.18578869,0.03259103,0.00186197,0.32574594],
1123 [-0.11306511,0.22072681,-0.18706142,0.05439948,0.28122966,0.35634355,0.0],[]],
1124'L=28':[[0.21225019,0.32031716,0.13604702,0.03132468,0.00362703,0.00018294,0.00000294,0.32573501],
1125 [0.13219496,-0.17206256,-0.08742608,0.32671661,0.17973107,0.02567515,0.32619598,0.0],
1126 [0.07989184,-0.16735346,0.18839770,-0.20705337,0.12926808,0.42715602,0.0,0.0]],
1127'L=30':[[-0.14878368,0.01524973,0.33628434,0.22632587,0.05790047,0.00609812,0.00022898,0.32573594],
1128 [-0.11721726,0.20915005,-0.11723436,-0.07815329,0.31318947,0.13655742,0.33241385,0.0],
1129 [-0.04297703,0.09317876,-0.11831248,0.17355132,-0.28164031,0.42719361,0.0,0.0]],
1130'L=32':[[0.20533892,0.32087437,0.15187897,0.04249238,0.00670516,0.00054977,0.00002018,0.00000024,0.32573501],
1131 [0.12775091,-0.13523423,-0.14935701,0.28227378,0.23670434,0.05661270,0.00469819,0.32578978,0.0],
1132 [0.09703829,-0.19373733,0.18610682,-0.14407046,0.00220535,0.26897090,0.36633402,0.0,0.0]],
1133'L=34':[[-0.14409234,-0.01343681,0.31248977,0.25557722,0.08571889,0.01351208,0.00095792,0.00002550,0.32573508],
1134 [-0.11527834,0.18472133,-0.04403280,-0.16908618,0.27227021,0.21086614,0.04041752,0.32688152,0.0],
1135 [-0.06773139,0.14120811,-0.15835721,0.18357456,-0.19364673,0.08377174,0.43116318,0.0,0.0]]
1136}
1137
1138Lnorm = lambda L: 4.*np.pi/(2.0*L+1.)
1139
1140def GetKcl(L,N,SGLaue,phi,beta):
1141 'needs doc string'
1142 import pytexture as ptx
1143 if SGLaue in ['m3','m3m']:
1144 if 'array' in str(type(phi)) and np.any(phi.shape):
1145 Kcl = np.zeros_like(phi)
1146 else:
1147 Kcl = 0.
1148 for j in range(0,L+1,4):
1149 im = j/4
1150 if 'array' in str(type(phi)) and np.any(phi.shape):
1151 pcrs = ptx.pyplmpsi(L,j,len(phi),phi)[0]
1152 else:
1153 pcrs = ptx.pyplmpsi(L,j,1,phi)[0]
1154 Kcl += BOH['L=%d'%(L)][N-1][im]*pcrs*cosd(j*beta)
1155 else:
1156 if 'array' in str(type(phi)) and np.any(phi.shape):
1157 pcrs = ptx.pyplmpsi(L,N,len(phi),phi)[0]
1158 else:
1159 pcrs = ptx.pyplmpsi(L,N,1,phi)[0]
1160 pcrs *= RSQ2PI
1161 if N:
1162 pcrs *= SQ2
1163 if SGLaue in ['mmm','4/mmm','6/mmm','R3mR','3m1','31m']:
1164 if SGLaue in ['3mR','3m1','31m']:
1165 if N%6 == 3:
1166 Kcl = pcrs*sind(N*beta)
1167 else:
1168 Kcl = pcrs*cosd(N*beta)
1169 else:
1170 Kcl = pcrs*cosd(N*beta)
1171 else:
1172 Kcl = pcrs*(cosd(N*beta)+sind(N*beta))
1173 return Kcl
1174
1175def GetKsl(L,M,SamSym,psi,gam):
1176 'needs doc string'
1177 import pytexture as ptx
1178 if 'array' in str(type(psi)) and np.any(psi.shape):
1179 psrs,dpdps = ptx.pyplmpsi(L,M,len(psi),psi)
1180 else:
1181 psrs,dpdps = ptx.pyplmpsi(L,M,1,psi)
1182 psrs *= RSQ2PI
1183 dpdps *= RSQ2PI
1184 if M:
1185 psrs *= SQ2
1186 dpdps *= SQ2
1187 if SamSym in ['mmm',]:
1188 dum = cosd(M*gam)
1189 Ksl = psrs*dum
1190 dKsdp = dpdps*dum
1191 dKsdg = -psrs*M*sind(M*gam)
1192 else:
1193 dum = cosd(M*gam)+sind(M*gam)
1194 Ksl = psrs*dum
1195 dKsdp = dpdps*dum
1196 dKsdg = psrs*M*(-sind(M*gam)+cosd(M*gam))
1197 return Ksl,dKsdp,dKsdg
1198
1199def GetKclKsl(L,N,SGLaue,psi,phi,beta):
1200 """
1201 This is used for spherical harmonics description of preferred orientation;
1202 cylindrical symmetry only (M=0) and no sample angle derivatives returned
1203 """
1204 import pytexture as ptx
1205 Ksl,x = ptx.pyplmpsi(L,0,1,psi)
1206 Ksl *= RSQ2PI
1207 if SGLaue in ['m3','m3m']:
1208 Kcl = 0.0
1209 for j in range(0,L+1,4):
1210 im = j/4
1211 pcrs,dum = ptx.pyplmpsi(L,j,1,phi)
1212 Kcl += BOH['L=%d'%(L)][N-1][im]*pcrs*cosd(j*beta)
1213 else:
1214 pcrs,dum = ptx.pyplmpsi(L,N,1,phi)
1215 pcrs *= RSQ2PI
1216 if N:
1217 pcrs *= SQ2
1218 if SGLaue in ['mmm','4/mmm','6/mmm','R3mR','3m1','31m']:
1219 if SGLaue in ['3mR','3m1','31m']:
1220 if N%6 == 3:
1221 Kcl = pcrs*sind(N*beta)
1222 else:
1223 Kcl = pcrs*cosd(N*beta)
1224 else:
1225 Kcl = pcrs*cosd(N*beta)
1226 else:
1227 Kcl = pcrs*(cosd(N*beta)+sind(N*beta))
1228 return Kcl*Ksl,Lnorm(L)
1229
1230def Glnh(Start,SHCoef,psi,gam,SamSym):
1231 'needs doc string'
1232 import pytexture as ptx
1233
1234 if Start:
1235 ptx.pyqlmninit()
1236 Start = False
1237 Fln = np.zeros(len(SHCoef))
1238 for i,term in enumerate(SHCoef):
1239 l,m,n = eval(term.strip('C'))
1240 pcrs,dum = ptx.pyplmpsi(l,m,1,psi)
1241 pcrs *= RSQPI
1242 if m == 0:
1243 pcrs /= SQ2
1244 if SamSym in ['mmm',]:
1245 Ksl = pcrs*cosd(m*gam)
1246 else:
1247 Ksl = pcrs*(cosd(m*gam)+sind(m*gam))
1248 Fln[i] = SHCoef[term]*Ksl*Lnorm(l)
1249 ODFln = dict(zip(SHCoef.keys(),list(zip(SHCoef.values(),Fln))))
1250 return ODFln
1251
1252def Flnh(Start,SHCoef,phi,beta,SGData):
1253 'needs doc string'
1254 import pytexture as ptx
1255
1256 if Start:
1257 ptx.pyqlmninit()
1258 Start = False
1259 Fln = np.zeros(len(SHCoef))
1260 for i,term in enumerate(SHCoef):
1261 l,m,n = eval(term.strip('C'))
1262 if SGData['SGLaue'] in ['m3','m3m']:
1263 Kcl = 0.0
1264 for j in range(0,l+1,4):
1265 im = j/4
1266 pcrs,dum = ptx.pyplmpsi(l,j,1,phi)
1267 Kcl += BOH['L='+str(l)][n-1][im]*pcrs*cosd(j*beta)
1268 else: #all but cubic
1269 pcrs,dum = ptx.pyplmpsi(l,n,1,phi)
1270 pcrs *= RSQPI
1271 if n == 0:
1272 pcrs /= SQ2
1273 if SGData['SGLaue'] in ['mmm','4/mmm','6/mmm','R3mR','3m1','31m']:
1274 if SGData['SGLaue'] in ['3mR','3m1','31m']:
1275 if n%6 == 3:
1276 Kcl = pcrs*sind(n*beta)
1277 else:
1278 Kcl = pcrs*cosd(n*beta)
1279 else:
1280 Kcl = pcrs*cosd(n*beta)
1281 else:
1282 Kcl = pcrs*(cosd(n*beta)+sind(n*beta))
1283 Fln[i] = SHCoef[term]*Kcl*Lnorm(l)
1284 ODFln = dict(zip(SHCoef.keys(),list(zip(SHCoef.values(),Fln))))
1285 return ODFln
1286
1287def polfcal(ODFln,SamSym,psi,gam):
1288 '''Perform a pole figure computation.
1289 Note that the the number of gam values must either be 1 or must
1290 match psi. Updated for numpy 1.8.0
1291 '''
1292 import pytexture as ptx
1293 PolVal = np.ones_like(psi)
1294 for term in ODFln:
1295 if abs(ODFln[term][1]) > 1.e-3:
1296 l,m,n = eval(term.strip('C'))
1297 psrs,dum = ptx.pyplmpsi(l,m,len(psi),psi)
1298 if SamSym in ['-1','2/m']:
1299 if m:
1300 Ksl = RSQPI*psrs*(cosd(m*gam)+sind(m*gam))
1301 else:
1302 Ksl = RSQPI*psrs/SQ2
1303 else:
1304 if m:
1305 Ksl = RSQPI*psrs*cosd(m*gam)
1306 else:
1307 Ksl = RSQPI*psrs/SQ2
1308 PolVal += ODFln[term][1]*Ksl
1309 return PolVal
1310
1311def invpolfcal(ODFln,SGData,phi,beta):
1312 'needs doc string'
1313 import pytexture as ptx
1314
1315 invPolVal = np.ones_like(beta)
1316 for term in ODFln:
1317 if abs(ODFln[term][1]) > 1.e-3:
1318 l,m,n = eval(term.strip('C'))
1319 if SGData['SGLaue'] in ['m3','m3m']:
1320 Kcl = 0.0
1321 for j in range(0,l+1,4):
1322 im = j/4
1323 pcrs,dum = ptx.pyplmpsi(l,j,len(beta),phi)
1324 Kcl += BOH['L=%d'%(l)][n-1][im]*pcrs*cosd(j*beta)
1325 else: #all but cubic
1326 pcrs,dum = ptx.pyplmpsi(l,n,len(beta),phi)
1327 pcrs *= RSQPI
1328 if n == 0:
1329 pcrs /= SQ2
1330 if SGData['SGLaue'] in ['mmm','4/mmm','6/mmm','R3mR','3m1','31m']:
1331 if SGData['SGLaue'] in ['3mR','3m1','31m']:
1332 if n%6 == 3:
1333 Kcl = pcrs*sind(n*beta)
1334 else:
1335 Kcl = pcrs*cosd(n*beta)
1336 else:
1337 Kcl = pcrs*cosd(n*beta)
1338 else:
1339 Kcl = pcrs*(cosd(n*beta)+sind(n*beta))
1340 invPolVal += ODFln[term][1]*Kcl
1341 return invPolVal
1342
1343
1344def textureIndex(SHCoef):
1345 'needs doc string'
1346 Tindx = 1.0
1347 for term in SHCoef:
1348 l = eval(term.strip('C'))[0]
1349 Tindx += SHCoef[term]**2/(2.0*l+1.)
1350 return Tindx
1351
1352# self-test materials follow.
1353selftestlist = []
1354'''Defines a list of self-tests'''
1355selftestquiet = True
1356def _ReportTest():
1357 'Report name and doc string of current routine when selftestquiet is False'
1358 if not selftestquiet:
1359 import inspect
1360 caller = inspect.stack()[1][3]
1361 doc = eval(caller).__doc__
1362 if doc is not None:
1363 print('testing '+__file__+' with '+caller+' ('+doc+')')
1364 else:
1365 print('testing '+__file__()+" with "+caller)
1366NeedTestData = True
1367def TestData():
1368 array = np.array
1369 global NeedTestData
1370 NeedTestData = False
1371 global CellTestData
1372 # output from uctbx computed on platform darwin on 2010-05-28
1373 CellTestData = [
1374# cell, g, G, cell*, V, V*
1375 [(4, 4, 4, 90, 90, 90),
1376 array([[ 1.60000000e+01, 9.79717439e-16, 9.79717439e-16],
1377 [ 9.79717439e-16, 1.60000000e+01, 9.79717439e-16],
1378 [ 9.79717439e-16, 9.79717439e-16, 1.60000000e+01]]), array([[ 6.25000000e-02, 3.82702125e-18, 3.82702125e-18],
1379 [ 3.82702125e-18, 6.25000000e-02, 3.82702125e-18],
1380 [ 3.82702125e-18, 3.82702125e-18, 6.25000000e-02]]), (0.25, 0.25, 0.25, 90.0, 90.0, 90.0), 64.0, 0.015625],
1381# cell, g, G, cell*, V, V*
1382 [(4.0999999999999996, 5.2000000000000002, 6.2999999999999998, 100, 80, 130),
1383 array([[ 16.81 , -13.70423184, 4.48533243],
1384 [-13.70423184, 27.04 , -5.6887143 ],
1385 [ 4.48533243, -5.6887143 , 39.69 ]]), array([[ 0.10206349, 0.05083339, -0.00424823],
1386 [ 0.05083339, 0.06344997, 0.00334956],
1387 [-0.00424823, 0.00334956, 0.02615544]]), (0.31947376387537696, 0.25189277536327803, 0.16172643497798223, 85.283666420376008, 94.716333579624006, 50.825714168082683), 100.98576357983838, 0.0099023858863968445],
1388# cell, g, G, cell*, V, V*
1389 [(3.5, 3.5, 6, 90, 90, 120),
1390 array([[ 1.22500000e+01, -6.12500000e+00, 1.28587914e-15],
1391 [ -6.12500000e+00, 1.22500000e+01, 1.28587914e-15],
1392 [ 1.28587914e-15, 1.28587914e-15, 3.60000000e+01]]), array([[ 1.08843537e-01, 5.44217687e-02, 3.36690552e-18],
1393 [ 5.44217687e-02, 1.08843537e-01, 3.36690552e-18],
1394 [ 3.36690552e-18, 3.36690552e-18, 2.77777778e-02]]), (0.32991443953692895, 0.32991443953692895, 0.16666666666666669, 90.0, 90.0, 60.000000000000021), 63.652867178156257, 0.015710211406520427],
1395 ]
1396 global CoordTestData
1397 CoordTestData = [
1398# cell, ((frac, ortho),...)
1399 ((4,4,4,90,90,90,), [
1400 ((0.10000000000000001, 0.0, 0.0),(0.40000000000000002, 0.0, 0.0)),
1401 ((0.0, 0.10000000000000001, 0.0),(2.4492935982947065e-17, 0.40000000000000002, 0.0)),
1402 ((0.0, 0.0, 0.10000000000000001),(2.4492935982947065e-17, -2.4492935982947065e-17, 0.40000000000000002)),
1403 ((0.10000000000000001, 0.20000000000000001, 0.29999999999999999),(0.40000000000000013, 0.79999999999999993, 1.2)),
1404 ((0.20000000000000001, 0.29999999999999999, 0.10000000000000001),(0.80000000000000016, 1.2, 0.40000000000000002)),
1405 ((0.29999999999999999, 0.20000000000000001, 0.10000000000000001),(1.2, 0.80000000000000004, 0.40000000000000002)),
1406 ((0.5, 0.5, 0.5),(2.0, 1.9999999999999998, 2.0)),
1407]),
1408# cell, ((frac, ortho),...)
1409 ((4.1,5.2,6.3,100,80,130,), [
1410 ((0.10000000000000001, 0.0, 0.0),(0.40999999999999998, 0.0, 0.0)),
1411 ((0.0, 0.10000000000000001, 0.0),(-0.33424955703700043, 0.39834311042186865, 0.0)),
1412 ((0.0, 0.0, 0.10000000000000001),(0.10939835193016617, -0.051013289294572106, 0.6183281045774256)),
1413 ((0.10000000000000001, 0.20000000000000001, 0.29999999999999999),(0.069695941716497567, 0.64364635296002093, 1.8549843137322766)),
1414 ((0.20000000000000001, 0.29999999999999999, 0.10000000000000001),(-0.073350319180835066, 1.1440160419710339, 0.6183281045774256)),
1415 ((0.29999999999999999, 0.20000000000000001, 0.10000000000000001),(0.67089923785616512, 0.74567293154916525, 0.6183281045774256)),
1416 ((0.5, 0.5, 0.5),(0.92574397446582857, 1.7366491056364828, 3.0916405228871278)),
1417]),
1418# cell, ((frac, ortho),...)
1419 ((3.5,3.5,6,90,90,120,), [
1420 ((0.10000000000000001, 0.0, 0.0),(0.35000000000000003, 0.0, 0.0)),
1421 ((0.0, 0.10000000000000001, 0.0),(-0.17499999999999993, 0.3031088913245536, 0.0)),
1422 ((0.0, 0.0, 0.10000000000000001),(3.6739403974420595e-17, -3.6739403974420595e-17, 0.60000000000000009)),
1423 ((0.10000000000000001, 0.20000000000000001, 0.29999999999999999),(2.7675166561703527e-16, 0.60621778264910708, 1.7999999999999998)),
1424 ((0.20000000000000001, 0.29999999999999999, 0.10000000000000001),(0.17500000000000041, 0.90932667397366063, 0.60000000000000009)),
1425 ((0.29999999999999999, 0.20000000000000001, 0.10000000000000001),(0.70000000000000018, 0.6062177826491072, 0.60000000000000009)),
1426 ((0.5, 0.5, 0.5),(0.87500000000000067, 1.5155444566227676, 3.0)),
1427]),
1428]
1429 global LaueTestData #generated by GSAS
1430 LaueTestData = {
1431 'R 3 m':[(4.,4.,6.,90.,90.,120.),((1,0,1,6),(1,0,-2,6),(0,0,3,2),(1,1,0,6),(2,0,-1,6),(2,0,2,6),
1432 (1,1,3,12),(1,0,4,6),(2,1,1,12),(2,1,-2,12),(3,0,0,6),(1,0,-5,6),(2,0,-4,6),(3,0,-3,6),(3,0,3,6),
1433 (0,0,6,2),(2,2,0,6),(2,1,4,12),(2,0,5,6),(3,1,-1,12),(3,1,2,12),(1,1,6,12),(2,2,3,12),(2,1,-5,12))],
1434 'R 3':[(4.,4.,6.,90.,90.,120.),((1,0,1,6),(1,0,-2,6),(0,0,3,2),(1,1,0,6),(2,0,-1,6),(2,0,2,6),(1,1,3,6),
1435 (1,1,-3,6),(1,0,4,6),(3,-1,1,6),(2,1,1,6),(3,-1,-2,6),(2,1,-2,6),(3,0,0,6),(1,0,-5,6),(2,0,-4,6),
1436 (2,2,0,6),(3,0,3,6),(3,0,-3,6),(0,0,6,2),(3,-1,4,6),(2,0,5,6),(2,1,4,6),(4,-1,-1,6),(3,1,-1,6),
1437 (3,1,2,6),(4,-1,2,6),(2,2,-3,6),(1,1,-6,6),(1,1,6,6),(2,2,3,6),(2,1,-5,6),(3,-1,-5,6))],
1438 'P 3':[(4.,4.,6.,90.,90.,120.),((0,0,1,2),(1,0,0,6),(1,0,1,6),(0,0,2,2),(1,0,-1,6),(1,0,2,6),(1,0,-2,6),
1439 (1,1,0,6),(0,0,3,2),(1,1,1,6),(1,1,-1,6),(1,0,3,6),(1,0,-3,6),(2,0,0,6),(2,0,-1,6),(1,1,-2,6),
1440 (1,1,2,6),(2,0,1,6),(2,0,-2,6),(2,0,2,6),(0,0,4,2),(1,1,-3,6),(1,1,3,6),(1,0,-4,6),(1,0,4,6),
1441 (2,0,-3,6),(2,1,0,6),(2,0,3,6),(3,-1,0,6),(2,1,1,6),(3,-1,-1,6),(2,1,-1,6),(3,-1,1,6),(1,1,4,6),
1442 (3,-1,2,6),(3,-1,-2,6),(1,1,-4,6),(0,0,5,2),(2,1,2,6),(2,1,-2,6),(3,0,0,6),(3,0,1,6),(2,0,4,6),
1443 (2,0,-4,6),(3,0,-1,6),(1,0,-5,6),(1,0,5,6),(3,-1,-3,6),(2,1,-3,6),(2,1,3,6),(3,-1,3,6),(3,0,-2,6),
1444 (3,0,2,6),(1,1,5,6),(1,1,-5,6),(2,2,0,6),(3,0,3,6),(3,0,-3,6),(0,0,6,2),(2,0,-5,6),(2,1,-4,6),
1445 (2,2,-1,6),(3,-1,-4,6),(2,2,1,6),(3,-1,4,6),(2,1,4,6),(2,0,5,6),(1,0,-6,6),(1,0,6,6),(4,-1,0,6),
1446 (3,1,0,6),(3,1,-1,6),(3,1,1,6),(4,-1,-1,6),(2,2,2,6),(4,-1,1,6),(2,2,-2,6),(3,1,2,6),(3,1,-2,6),
1447 (3,0,4,6),(3,0,-4,6),(4,-1,-2,6),(4,-1,2,6),(2,2,-3,6),(1,1,6,6),(1,1,-6,6),(2,2,3,6),(3,-1,5,6),
1448 (2,1,5,6),(2,1,-5,6),(3,-1,-5,6))],
1449 'P 3 m 1':[(4.,4.,6.,90.,90.,120.),((0,0,1,2),(1,0,0,6),(1,0,-1,6),(1,0,1,6),(0,0,2,2),(1,0,-2,6),
1450 (1,0,2,6),(1,1,0,6),(0,0,3,2),(1,1,1,12),(1,0,-3,6),(1,0,3,6),(2,0,0,6),(1,1,2,12),(2,0,1,6),
1451 (2,0,-1,6),(0,0,4,2),(2,0,-2,6),(2,0,2,6),(1,1,3,12),(1,0,-4,6),(1,0,4,6),(2,0,3,6),(2,1,0,12),
1452 (2,0,-3,6),(2,1,1,12),(2,1,-1,12),(1,1,4,12),(2,1,2,12),(0,0,5,2),(2,1,-2,12),(3,0,0,6),(1,0,-5,6),
1453 (3,0,1,6),(3,0,-1,6),(1,0,5,6),(2,0,4,6),(2,0,-4,6),(2,1,3,12),(2,1,-3,12),(3,0,-2,6),(3,0,2,6),
1454 (1,1,5,12),(3,0,-3,6),(0,0,6,2),(2,2,0,6),(3,0,3,6),(2,1,4,12),(2,2,1,12),(2,0,5,6),(2,1,-4,12),
1455 (2,0,-5,6),(1,0,-6,6),(1,0,6,6),(3,1,0,12),(3,1,-1,12),(3,1,1,12),(2,2,2,12),(3,1,2,12),
1456 (3,0,4,6),(3,1,-2,12),(3,0,-4,6),(1,1,6,12),(2,2,3,12))],
1457 'P 3 1 m':[(4.,4.,6.,90.,90.,120.),((0,0,1,2),(1,0,0,6),(0,0,2,2),(1,0,1,12),(1,0,2,12),(1,1,0,6),
1458 (0,0,3,2),(1,1,-1,6),(1,1,1,6),(1,0,3,12),(2,0,0,6),(2,0,1,12),(1,1,2,6),(1,1,-2,6),(2,0,2,12),
1459 (0,0,4,2),(1,1,-3,6),(1,1,3,6),(1,0,4,12),(2,1,0,12),(2,0,3,12),(2,1,1,12),(2,1,-1,12),(1,1,-4,6),
1460 (1,1,4,6),(0,0,5,2),(2,1,-2,12),(2,1,2,12),(3,0,0,6),(1,0,5,12),(2,0,4,12),(3,0,1,12),(2,1,-3,12),
1461 (2,1,3,12),(3,0,2,12),(1,1,5,6),(1,1,-5,6),(3,0,3,12),(0,0,6,2),(2,2,0,6),(2,1,-4,12),(2,0,5,12),
1462 (2,2,-1,6),(2,2,1,6),(2,1,4,12),(3,1,0,12),(1,0,6,12),(2,2,2,6),(3,1,-1,12),(2,2,-2,6),(3,1,1,12),
1463 (3,1,-2,12),(3,0,4,12),(3,1,2,12),(1,1,-6,6),(2,2,3,6),(2,2,-3,6),(1,1,6,6))],
1464 }
1465
1466 global FLnhTestData
1467 FLnhTestData = [{
1468 'C(4,0,0)': (0.965, 0.42760447),
1469 'C(2,0,0)': (1.0122, -0.80233610),
1470 'C(2,0,2)': (0.0061, 8.37491546E-03),
1471 'C(6,0,4)': (-0.0898, 4.37985696E-02),
1472 'C(6,0,6)': (-0.1369, -9.04081762E-02),
1473 'C(6,0,0)': (0.5935, -0.18234928),
1474 'C(4,0,4)': (0.1872, 0.16358127),
1475 'C(6,0,2)': (0.6193, 0.27573633),
1476 'C(4,0,2)': (-0.1897, 0.12530720)},[1,0,0]]
1477def test0():
1478 if NeedTestData: TestData()
1479 msg = 'test cell2Gmat, fillgmat, Gmat2cell'
1480 for (cell, tg, tG, trcell, tV, trV) in CellTestData:
1481 G, g = cell2Gmat(cell)
1482 assert np.allclose(G,tG),msg
1483 assert np.allclose(g,tg),msg
1484 tcell = Gmat2cell(g)
1485 assert np.allclose(cell,tcell),msg
1486 tcell = Gmat2cell(G)
1487 assert np.allclose(tcell,trcell),msg
1488selftestlist.append(test0)
1489
1490def test1():
1491 'test cell2A and A2Gmat'
1492 _ReportTest()
1493 if NeedTestData: TestData()
1494 msg = 'test cell2A and A2Gmat'
1495 for (cell, tg, tG, trcell, tV, trV) in CellTestData:
1496 G, g = A2Gmat(cell2A(cell))
1497 assert np.allclose(G,tG),msg
1498 assert np.allclose(g,tg),msg
1499selftestlist.append(test1)
1500
1501def test2():
1502 'test Gmat2A, A2cell, A2Gmat, Gmat2cell'
1503 _ReportTest()
1504 if NeedTestData: TestData()
1505 msg = 'test Gmat2A, A2cell, A2Gmat, Gmat2cell'
1506 for (cell, tg, tG, trcell, tV, trV) in CellTestData:
1507 G, g = cell2Gmat(cell)
1508 tcell = A2cell(Gmat2A(G))
1509 assert np.allclose(cell,tcell),msg
1510selftestlist.append(test2)
1511
1512def test3():
1513 'test invcell2Gmat'
1514 _ReportTest()
1515 if NeedTestData: TestData()
1516 msg = 'test invcell2Gmat'
1517 for (cell, tg, tG, trcell, tV, trV) in CellTestData:
1518 G, g = invcell2Gmat(trcell)
1519 assert np.allclose(G,tG),msg
1520 assert np.allclose(g,tg),msg
1521selftestlist.append(test3)
1522
1523def test4():
1524 'test calc_rVsq, calc_rV, calc_V'
1525 _ReportTest()
1526 if NeedTestData: TestData()
1527 msg = 'test calc_rVsq, calc_rV, calc_V'
1528 for (cell, tg, tG, trcell, tV, trV) in CellTestData:
1529 assert np.allclose(calc_rV(cell2A(cell)),trV), msg
1530 assert np.allclose(calc_V(cell2A(cell)),tV), msg
1531selftestlist.append(test4)
1532
1533def test5():
1534 'test A2invcell'
1535 _ReportTest()
1536 if NeedTestData: TestData()
1537 msg = 'test A2invcell'
1538 for (cell, tg, tG, trcell, tV, trV) in CellTestData:
1539 rcell = A2invcell(cell2A(cell))
1540 assert np.allclose(rcell,trcell),msg
1541selftestlist.append(test5)
1542
1543def test6():
1544 'test cell2AB'
1545 _ReportTest()
1546 if NeedTestData: TestData()
1547 msg = 'test cell2AB'
1548 for (cell,coordlist) in CoordTestData:
1549 A,B = cell2AB(cell)
1550 for (frac,ortho) in coordlist:
1551 to = np.inner(A,frac)
1552 tf = np.inner(B,to)
1553 assert np.allclose(ortho,to), msg
1554 assert np.allclose(frac,tf), msg
1555 to = np.sum(A*frac,axis=1)
1556 tf = np.sum(B*to,axis=1)
1557 assert np.allclose(ortho,to), msg
1558 assert np.allclose(frac,tf), msg
1559selftestlist.append(test6)
1560
1561def test7():
1562 'test GetBraviasNum(...) and GenHBravais(...)'
1563 _ReportTest()
1564 import os.path
1565 import sys
1566 import GSASIIspc as spc
1567 testdir = os.path.join(os.path.split(os.path.abspath( __file__ ))[0],'testinp')
1568 if os.path.exists(testdir):
1569 if testdir not in sys.path: sys.path.insert(0,testdir)
1570 import sgtbxlattinp
1571 derror = 1e-4
1572 def indexmatch(hklin, hkllist, system):
1573 for hklref in hkllist:
1574 hklref = list(hklref)
1575 # these permutations are far from complete, but are sufficient to
1576 # allow the test to complete
1577 if system == 'cubic':
1578 permlist = [(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1),]
1579 elif system == 'monoclinic':
1580 permlist = [(1,2,3),(-1,2,-3)]
1581 else:
1582 permlist = [(1,2,3)]
1583
1584 for perm in permlist:
1585 hkl = [abs(i) * hklin[abs(i)-1] / i for i in perm]
1586 if hkl == hklref: return True
1587 if [-i for i in hkl] == hklref: return True
1588 else:
1589 return False
1590
1591 for key in sgtbxlattinp.sgtbx7:
1592 spdict = spc.SpcGroup(key)
1593 cell = sgtbxlattinp.sgtbx7[key][0]
1594 system = spdict[1]['SGSys']
1595 center = spdict[1]['SGLatt']
1596
1597 bravcode = GetBraviasNum(center, system)
1598
1599 g2list = GenHBravais(sgtbxlattinp.dmin, bravcode, cell2A(cell))
1600
1601 assert len(sgtbxlattinp.sgtbx7[key][1]) == len(g2list), 'Reflection lists differ for %s' % key
1602 for h,k,l,d,num in g2list:
1603 for hkllist,dref in sgtbxlattinp.sgtbx7[key][1]:
1604 if abs(d-dref) < derror:
1605 if indexmatch((h,k,l,), hkllist, system):
1606 break
1607 else:
1608 assert 0,'No match for %s at %s (%s)' % ((h,k,l),d,key)
1609selftestlist.append(test7)
1610
1611def test8():
1612 'test GenHLaue'
1613 _ReportTest()
1614 import GSASIIspc as spc
1615 import sgtbxlattinp
1616 derror = 1e-4
1617 dmin = sgtbxlattinp.dmin
1618
1619 def indexmatch(hklin, hklref, system, axis):
1620 # these permutations are far from complete, but are sufficient to
1621 # allow the test to complete
1622 if system == 'cubic':
1623 permlist = [(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1),]
1624 elif system == 'monoclinic' and axis=='b':
1625 permlist = [(1,2,3),(-1,2,-3)]
1626 elif system == 'monoclinic' and axis=='a':
1627 permlist = [(1,2,3),(1,-2,-3)]
1628 elif system == 'monoclinic' and axis=='c':
1629 permlist = [(1,2,3),(-1,-2,3)]
1630 elif system == 'trigonal':
1631 permlist = [(1,2,3),(2,1,3),(-1,-2,3),(-2,-1,3)]
1632 elif system == 'rhombohedral':
1633 permlist = [(1,2,3),(2,3,1),(3,1,2)]
1634 else:
1635 permlist = [(1,2,3)]
1636
1637 hklref = list(hklref)
1638 for perm in permlist:
1639 hkl = [abs(i) * hklin[abs(i)-1] / i for i in perm]
1640 if hkl == hklref: return True
1641 if [-i for i in hkl] == hklref: return True
1642 return False
1643
1644 for key in sgtbxlattinp.sgtbx8:
1645 spdict = spc.SpcGroup(key)[1]
1646 cell = sgtbxlattinp.sgtbx8[key][0]
1647 center = spdict['SGLatt']
1648 Laue = spdict['SGLaue']
1649 Axis = spdict['SGUniq']
1650 system = spdict['SGSys']
1651
1652 g2list = GenHLaue(dmin,spdict,cell2A(cell))
1653 #if len(g2list) != len(sgtbxlattinp.sgtbx8[key][1]):
1654 # print 'failed',key,':' ,len(g2list),'vs',len(sgtbxlattinp.sgtbx8[key][1])
1655 # print 'GSAS-II:'
1656 # for h,k,l,d in g2list: print ' ',(h,k,l),d
1657 # print 'SGTBX:'
1658 # for hkllist,dref in sgtbxlattinp.sgtbx8[key][1]: print ' ',hkllist,dref
1659 assert len(g2list) == len(sgtbxlattinp.sgtbx8[key][1]), (
1660 'Reflection lists differ for %s' % key
1661 )
1662 #match = True
1663 for h,k,l,d in g2list:
1664 for hkllist,dref in sgtbxlattinp.sgtbx8[key][1]:
1665 if abs(d-dref) < derror:
1666 if indexmatch((h,k,l,), hkllist, system, Axis): break
1667 else:
1668 assert 0,'No match for %s at %s (%s)' % ((h,k,l),d,key)
1669 #match = False
1670 #if not match:
1671 #for hkllist,dref in sgtbxlattinp.sgtbx8[key][1]: print ' ',hkllist,dref
1672 #print center, Laue, Axis, system
1673selftestlist.append(test8)
1674
1675def test9():
1676 'test GenHLaue'
1677 _ReportTest()
1678 import GSASIIspc as G2spc
1679 if NeedTestData: TestData()
1680 for spc in LaueTestData:
1681 data = LaueTestData[spc]
1682 cell = data[0]
1683 hklm = np.array(data[1])
1684 H = hklm[-1][:3]
1685 hklO = hklm.T[:3].T
1686 A = cell2A(cell)
1687 dmin = 1./np.sqrt(calc_rDsq(H,A))
1688 SGData = G2spc.SpcGroup(spc)[1]
1689 hkls = np.array(GenHLaue(dmin,SGData,A))
1690 hklN = hkls.T[:3].T
1691 #print spc,hklO.shape,hklN.shape
1692 err = True
1693 for H in hklO:
1694 if H not in hklN:
1695 print H,' missing from hkl from GSASII'
1696 err = False
1697 assert(err)
1698selftestlist.append(test9)
1699
1700
1701
1702
1703if __name__ == '__main__':
1704 # run self-tests
1705 selftestquiet = False
1706 for test in selftestlist:
1707 test()
1708 print "OK"
Note: See TracBrowser for help on using the repository browser. | 28,569 | 69,021 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-23 | latest | en | 0.4014 |
https://physics.stackexchange.com/questions/687940/rolling-motion-sliding-contact-point-friction | 1,721,358,852,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00257.warc.gz | 399,281,327 | 39,389 | Rolling motion, sliding, contact point, friction
I have a question about rolling motion. So suppose we throw a bowling ball with no initial angular velocity, only linear velocity $$v_0$$. At first it will be sliding on the ground. Let the kinetic friction between the bowling ball and the ground be $$\mu_k > 0$$. The friction force opposes the direction of linear motion which will act as a torque about the bowling ball's axis and increase its angular velocity. Now my confusion stems from how to handle the following cases:
Assume the ball is moving in the $$xy$$-plane only and to the right (positive $$x$$-direction). CM = center of mass
$$(1)$$ As long as the bowling ball is not in pure rolling motion, the contact point $$P$$ moves relative to the surface, right? Its velocity is the sum of the velocity of the center of mass and the velocity due to its rotation in the opposite direction. $$v_P = v_\mathrm{CM} - R\omega$$ Initially, $$v_P = v_\mathrm{CM}$$. To increase $$P$$'s angular velocity, there needs to be a torque, which should be the kinetic friction force. In effect, the contact point still moves more to the right (in the direction of the CM velocity, i.e. $$v_P > 0$$) and the net friction will be to the left, still, causing an increase in angular velocity. Does the friction only apply to the contact point, thus only increasing its linear velocity opposing the motion of the CM, or does it also act to decrease the bowling ball's linear velocity until the the rolling condition $$v_\mathrm{CM} = R \omega$$ is satisfied? Will there be energy lost or just converted form linear kinetic energy to rotational kinetic energy until they "match" (pure rolling)?
$$(2)$$ What exactly happens at the transition from sliding to rolling? Why is friction not doing any work at that point, given that we still have static friction? Mathematically, $$v_P = v_\mathrm{Cm} - R\omega = 0$$ for pure rolling, so that there's "no motion" of the contact point, which doesn't make sense intuitively to me, yet. Couldn't we still say that the contact point "rubs" the surface?
• You've got several questions in one post. On the question of why there is 'no motion' of the contact point during pure rolling, the key thing is that there is no relative (tangential) motion between the floor and the surface of the ball, because they are both moving at the same speed. Commented Jan 10, 2022 at 20:03
1 Answer
(1) In a horizontal surface, until reaching the situation of no slip, if we disregard air drag, the kinetic friction is the net force on the ball. So the COM decelerates following $$F_f = ma$$. This force also generates a torque: $$\tau = RF_f = I\alpha$$. While $$F_f$$ in general changes with time, at any instant $$t$$ during the slip deceleration period: $$m \frac{dv}{dt} = -\frac{I}{R}\frac{d \omega}{dt} \implies \frac{dv}{d\omega} = -\frac{I}{mR}$$ The minus sign is there because $$v$$ decreases as $$\omega$$ increases. In the case of a massive ball, with $$I = \frac{2mR^2}{5}$$: $$\frac{dv}{d\omega} = -\frac{2R}{5} \implies v = -\frac{2R}{5}\omega + v_0$$
When the ball is close to the situation of no slip, $$\omega \to \frac{v}{R}$$
$$v = -\frac{2v}{5} + v_0 \implies v = \frac{5}{7}v_0$$
I suppose that initially the ball has a velocity $$v_0$$ without rolling, so $$\omega = 0 \implies v = v_0$$.
(2) Note that rolling without slip is a pure inertial situation. Theoretically, once this situation is reached, friction is no longer necessary. If the ground is suddenly without friction, the ball keeps rotating and moving the same way. | 936 | 3,574 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 26, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-30 | latest | en | 0.91926 |
http://www.jiskha.com/display.cgi?id=1251479966 | 1,498,480,568,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320736.82/warc/CC-MAIN-20170626115614-20170626135614-00536.warc.gz | 572,423,256 | 3,650 | # math
posted by .
write a word problem for 1/2 - 2/3 = and write a word problem for 5/12 - 1/3 =
• math -
Have a person get a portion of a pie and give part of it to someone else. What proportion of the whole pie did he still have? | 70 | 236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-26 | latest | en | 0.949827 |
https://fenicsproject.discourse.group/t/how-to-get-vertex-values-in-dolfinx/4520 | 1,696,135,177,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510781.66/warc/CC-MAIN-20231001041719-20231001071719-00197.warc.gz | 272,782,835 | 6,971 | # How to get vertex values in dolfinx?
Hi,
I am transferring fenics to dolfinx but having a trouble now. I was wondering how can I get the array of vertex values coinciding with the nodal order like compute_vertex_values() in fenics? The reason why I need this function is to verify my solutions.
Looking at dolfinx/function.py, compute_point_values seems to be the dolfinx equivalent of compute_vertex_values.
2 Likes
Thanks for your prompt reply! But it seems that compute_point_values still generates the same values as the original ones but in a tuple form. Maybe I need to write a script to fulfill my needs.
Compute point values does the same thing as compute vertex values for a first order nesh. For a second order mesh it computes the values at every node on the mesh geometry.
If your mesh is first order, it should return you the values at each vertex.
2 Likes
Thank you! I figure it out this works and the original result coincides with the nodal order as well.
But I found something very strange, if I read the mesh from the xdmf file, the nodal order of the mesh in dolfinx is actually changed after I comparing with the input meshing file. So I was wondering how could this happen?
Please produce a minimal working code example.
Hi dokken, here is an example as what I mentioned before. In this script, I tried to generate and save a simple mesh with hexahedral elements and then use this mesh file again to check if there is any change.
import numpy as np
from mpi4py import MPI
from dolfinx import Function, FunctionSpace, Constant, BoxMesh, VectorFunctionSpace
from dolfinx.io import XDMFFile
# Create the mesh file
Mesh =BoxMesh(
MPI.COMM_WORLD, [np.array([-0.5, -5, 0]),
np.array([0.5, 5, 0.5])], [2, 2, 2],
CellType.hexahedron, dolfinx.cpp.mesh.GhostMode.none)
# Save solution in XDMF format
with XDMFFile(MPI.COMM_WORLD, "mesh_hex.xdmf", "w") as file:
file.write_mesh(Mesh)
def expr(x):
return x[0] + x[1] + x[2]
# use "Mesh" to generate function u
W = FunctionSpace(Mesh, ("Lagrange", 1))
w = Function(W)
w.interpolate(expr)
vertex_w = w.compute_point_values()
print(vertex_w)
# use "mesh" to generate function u
with XDMFFile(MPI.COMM_WORLD,"mesh_hex.xdmf", "r") as infile:
U = FunctionSpace(mesh, ("Lagrange", 1))
u = Function(U)
u.interpolate(expr)
vertex_u = u.compute_point_values()
# compare the difference
print(vertex_u - vertex_w)
print(np.linalg.norm((vertex_u - vertex_w)))
with XDMFFile(MPI.COMM_WORLD, "test_hex.xdmf", "w") as out_file:
out_file.write_mesh(mesh)
out_file.write_function(u)
And the difference of the two functions is not zero, which looks weird to me. Thanks in advance for your any comments.
If you write w to file, you will observe that the function is the same as the test_hex.xdmf output.
The reason for the vertex values not matching, is that the mesh geometry is ordered differently in the read-in mesh.
This can be shown by:
for coord, Coord in zip(mesh.geometry.x, Mesh.geometry.x):
print(coord - Coord)
Thanks. I do observe that the function w and u are exactly the same but in different orders. So I was wondering if there is anyway to generate an array of function u with the same order to w i.e. the read-in mesh, as I expected compute_point_values will do the job but it did not.
If you are not using a built mesh (i.e. loading the original mesh with XDMF) I thing it should Give the same order. You should verify this.
Thanks for your quick reply. Yes I agree with that, but I guess I do need to use a mesh with the help of a XDMF input file. So does that mean I should mesh the geometry directly in dolfinx if I want to keep the same order?
Is there a particular reason for wanting to use multiple meshes of the same mesh in the same file with IO? What is the application?
I do not intend to use multiple meshes. I would like to verify the numerical solution in dolfinx, so I need to use the same mesh file. This works well when I was using fenics, but after transitioning to dolfinx I haven’t figured out how to accomplish that.
So do you want to compare dolfin and dolfinx solutions? Why not work with a manufactured problem to verify the error norms with the exact solutions? Using compute point values is at least not the way to go. If anything, you should tabulate the dof coordinates, and create a map from the dof coordinates in dolfinx to those in dolfin
Thanks dokken. Actually I want to compare with the solution computed in ABAQUS, since there is not a closed form for my problem. But I believe I can do it with “dofmap” as you suggested.
Has the compute_point_values() function been removed? I get the Error 'Function' object has no attribute 'compute_point_values'…
From the change log of the dolfinx-tutorial,
dolfinx.fem.Function.compute_point_values has been deprecated. Interpolation into a CG-1 is now the way of getting vertex values.
1 Like | 1,199 | 4,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-40 | latest | en | 0.830662 |
https://studydaddy.com/question/how-do-you-convert-16-weeks-to-seconds | 1,563,861,184,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528869.90/warc/CC-MAIN-20190723043719-20190723065719-00262.warc.gz | 576,488,348 | 8,003 | Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
# How do you convert 16 weeks to seconds?
You use a series of conversion factors.
Every time you need to convert from one unit to another, you need to focus on finding the appropriate series of conversion factors that can be used to convert between the two units.
In your case, you will need to write conversion factors to take you from weeks to days, from days to hours, from hours to minutes, and finally, from minuted to seconds.
This is how your conversion will look like
##16color(red)(cancel(color(black)("weeks"))) * (7color(red)(cancel(color(black)("days"))))/(1color(red)(cancel(color(black)("week")))) * (24color(red)(cancel(color(black)("hours"))))/(1color(red)(cancel(color(black)("day")))) * (60color(red)(cancel(color(black)("minutes"))))/(1color(red)(cancel(color(black)("hour")))) * "60 seconds"/(1color(red)(cancel(color(black)("minute")))) = "9,676,800 seconds"##
If you want, you can round this off to two
##"16 weeks" = "9,700,000 seconds"## | 271 | 1,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.877369 |
https://statistical-programming.com/convert-data-frame-column-to-numeric-in-r | 1,561,578,863,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000545.97/warc/CC-MAIN-20190626194744-20190626220744-00345.warc.gz | 603,816,943 | 22,313 | # Convert Data Frame Column to Numeric in R (2 Examples) | Change Factor, Character & Integer
In this R tutorial, I’ll explain how to convert a data frame column to numeric in R. No matter if you need to change the class of factors, characters, or integers, this tutorial will show you how to do it.
The article is structured as follows:
Let’s dive right in!
## Create Example Data
First we need to create some data in R that we can use in the examples later on:
```data <- data.frame(x1 = c(1, 5, 8, 2), # Create example data frame x2 = c(3, 2, 5, 2), x3 = c(2, 7, 1, 2)) data\$x1 <- as.factor(data\$x1) # First column is a factor data\$x2 <- as.character(data\$x2) # Second column is a character data\$x3 <- as.integer(data\$x3) # Third column is an integer data # Print data to RStudio console```
You can see the structure of our example data frame in Table 1. The data contains three columns: a factor variable, a character variable, and an integer variable.
Table 1: Example Data Frame with Factor, Character & Integer Variables.
We can check the class of each column of our data table with the sapply function:
```sapply(data, class) # Get classes of all columns # x1 x2 x3 # "factor" "character" "integer"```
The data is set up, so let’s move on to the examples…
## Example 1: Convert One Variable of Data Frame to Numeric
In the first example I’m going to convert only one variable to numeric. For this task, we can use the following R code:
`data\$x1 <- as.numeric(as.character(data\$x1)) # Convert one variable to numeric`
Note: The previous code converts our factor variable to character first and then it converts the character to numeric. This is important in order to retain the values (i.e. the numbers) of the factor variable. You can learn more about that in this tutorial.
However, let’s check the classes of our columns again to see how our data has changed:
```sapply(data, class) # Get classes of all columns # x1 x2 x3 # "numeric" "character" "integer"```
As we wanted: The factor column was converted to numeric.
## Example 2: Change Multiple Columns to Numeric
In Example 1 we used the as.numeric and the as.character functions to modify one variable of our example data. However, when we want to change several variables to numeric simultaneously, the approach of Example 1 might be too slow (i.e. too much programming). In this example, I’m therefore going to show you how to change as many columns as you want at the same time.
First, we need to specify which columns we want to modify. In this example, we are converting columns 2 and 3 (i.e. the character string and the integer):
`i <- c(2, 3) # Specify columns you want to change`
We can now use the apply function to change columns 2 and 3 to numeric:
```data[ , i] <- apply(data[ , i], 2, # Specify own function within apply function(x) as.numeric(as.character(x)))```
Let’s check the classes of the variables of our data frame:
```sapply(data, class) # Get classes of all columns # x1 x2 x3 # "numeric" "numeric" "numeric"```
The whole data frame was converted to numeric!
## Further Resources
Converting variable classes in R is a complex topic. I have therefore listed some additional resources about the Modification of R data classes in the following.
If you want to learn more about the basic data types in R, I can recommend the following video of the Data Camp YouTube channel:
Also, you could have a look at the following R tutorials of this homepage:
I hope you liked this tutorial! Let me know in the comments if you have any further questions and of cause I am also happy about general feedback. | 879 | 3,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.758213 |
http://mathhelpforum.com/pre-calculus/123133-show-ln-4-ln-0-25-a.html | 1,529,880,451,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867095.70/warc/CC-MAIN-20180624215228-20180624235228-00406.warc.gz | 203,880,827 | 10,639 | # Thread: Show that -ln(4) = ln(0.25)???
1. ## Show that -ln(4) = ln(0.25)???
How can I show that -ln(4) = ln(0.25).
A hint would be fine...
2. Originally Posted by Henryt999
How can I show that -ln(4) = ln(0.25).
A hint would be fine...
Hint
Spoiler:
Lose the logarithms
3. Originally Posted by Henryt999
How can I show that -ln(4) = ln(0.25).
A hint would be fine...
$\displaystyle -\ln(4) = 0-\ln(4) = \ln(1)-\ln(4)$
I'll leave the rest for you.
4. It's a basic property of logarithms: $\displaystyle a\cdot \ln(b) = \ln(b^a)$
So if $\displaystyle a = -1, b= 4$? what is $\displaystyle b^{a}$
5. ## ops
That was very very easy ops ops ops
6. Hello, Henryt999!
How can I show that: .$\displaystyle -\ln(4) \:=\: \ln(0.25)$
The left side is: .$\displaystyle -1\cdot\ln(4) \;=\;\ln\left(4^{-1}\right) \;=\;\ln\left(\frac{1}{4}\right) \;=\;\ln(0.25)$ | 333 | 863 | {"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} | 3.875 | 4 | CC-MAIN-2018-26 | latest | en | 0.7049 |
https://fuxuemingzhu.cn/leetcode/35.html | 1,726,806,897,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652130.6/warc/CC-MAIN-20240920022257-20240920052257-00696.warc.gz | 232,321,679 | 9,845 | 35. Search Insert Position 搜索插入位置
@TOC
# 题目描述
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Example 1:
``````Input: [1,3,5,6], 5
Output: 2
``````
Example 2:
``````Input: [1,3,5,6], 2
Output: 1
``````
Example 3:
``````Input: [1,3,5,6], 7
Output: 4
``````
Example 4:
``````Input: [1,3,5,6], 0
Output: 0
``````
# 解题方法
# 二分查找
``````class Solution(object):
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
return bisect.bisect_left(nums, target)
``````
``````class Solution(object):
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
N = len(nums)
left, right = 0, N #[left, right)
while left < right:
mid = left + (right - left) / 2
if nums[mid] == target:
return mid
elif nums[mid] > target:
right = mid
else:
left = mid + 1
return left
``````
Java代码如下：
``````public class Solution {
public int searchInsert(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
int mid = 0;
while(left <= right){
mid = (left + right) / 2;
if(nums[mid] == target){
return mid;
}else if(nums[mid] > target){
right--;
}else{
left++;
}
}
return left;
}
}
``````
# 日期
2017 年 4 月 25 日 2018 年 11 月 21 日 —— 又是一个美好的开始 | 477 | 1,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.468105 |
https://www.boomfoto.pl/39422/12/2020.html | 1,642,390,496,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300289.37/warc/CC-MAIN-20220117031001-20220117061001-00667.warc.gz | 740,188,115 | 7,340 | # Physics Sand On Conveyor Belt Pictures
## DESIGN AND APPLICATION OF FEEDERS FOR THE …
The success of belt conveyors depends on a number of factors, not the least of which is the initial feeding of the bulk solid onto the belt and to the efficient transfer of solids from one belt to another at conveyor transfer stations. With the future trend towards higher belt speeds and narrower belts in order to achieve higher economic ...
## Friction on a conveyor belt
A block of mass 1 kg is stationary with respect to a conveyor belt that is accelerating with \$1, tfrac{m}{s^2}\$ upwards at an angle of 30° as shown in figure. Determine force of friction on block and contact force between the block & belt. (I don't have enough reputation to post a diagram)
## Conveyor Stock Photos And Images
Download Conveyor stock photos. Affordable and search from millions of royalty free images, photos and vectors. Photos. Vectors. FOOTAGE. AUDIO. SEE PRICING & PLANS. Support. en ... #100477345 - Conveyor belt with canned fish, seafood production, fish industry.. Vector. Similar Images
## Solved: 3 Points A Conveyor Belt Is Pulling Sand At A Spee ...
3 points A conveyor belt is pulling sand at a speed of 6.0 m/s. The sand is dumping onto the belt at a rate of 100 kg/ 6.0 m/s In the previous problem you found the rate of work done, power, by the non-conservative force pulling the conveyor belt At whatrate in J, is the thermal energy of the Sand-belt system increasing?
## [Solved] Sand falls vertically on a conveyor belt at a ...
PHYSICS PHY001. Sand falls vertically on a conveyor belt at a rate of m kg s - 1 . mv 2 . Answered Subscribe to unlock Question 7. Sand falls vertically on a conveyor belt at a …
## Company Eyes Texas
Jan 08, 2020· The Atlas Sand Company wants to erect a 16.5-mile (10.4 kilometer) covered conveyor belt system to carry the sand from an offloading facility in Loving County, Texas, to a …
## 14 Best Conveyor belt images | Conveyor, Conveyor belt, Belt
Dec 6, 2017 - Explore fangzihao1215's board "Conveyor belt" on Pinterest. See more ideas about Conveyor, Conveyor belt, Belt.
## Conveyor Belt Stock Pictures, Royalty
15,919 Conveyor Belt stock pictures and images Browse 15,919 conveyor belt stock photos and images available, or search for production line or factory to find more great stock photos and pictures. Explore {{searchView.params.phrase}} by color family
## 100+ Free Conveyor & Conveyor Belt Images
Images Photos Vector graphics Illustrations Videos. ... Category . Size . Larger than × px Color . Transparent Black and white. Related Images: conveyor belt industry mining carbon factory. 148 Free images of Conveyor. 51 67 4. Factory Industrial Hall. 15 7 4. Transport ... Open Pit Mining Sand. 19 17 0. Counter Sales Man. 6 12 0. Airport ...
## Sand is being dropped on a conveyor belt at the rate of M ...
Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v m/s will be: asked Aug 3, 2017 in Physics …
## 5.8 Inherently inelastic processes
mass of the conveyor belt plus the sand on the belt, then F = dp dt = d(mv) dt = m dv dt + dm dt v = 0 +σv, (5.75) where we have used the fact that v is constant. (b) The kinetic energy gained per unit time is d dt mv2 2 = dm dt v2 2 = σv2 2. (5.76) From "Introduction to Classical Mechanics" by David Morin.
## homework and exercises
The conservation of momentum approach is the correct one as the "collision" between the conveyor belt and the sand is "inelestic" for the following reason. Just before hitting the belt the horizontal speed of the sand …
## crusher belt conveyor pictures
image of conveyor belts for transporting concrete TY is one of the biggest manufacturers in Aggregate Processing Machinery for the physics sand on conveyor belt pictures sand gravel concrete crushing by stone Read more aggregate transporting conveyor belts Sand is dropped on a conveyor belt at rate of M kg/s Force Sand is dropped on a
## Conveyor belt question | Physics Forums
Apr 17, 2011· sand is pouring on a conveyor belt in a rate of rate of dm/dt kgs-1 if the velocity of the conveyor belt is v ms-1, what is the power needed to run the conveyor belt. My friend arrived at his answer in this way F = dp/dt = d(mv)/dt = v (dm/dt) N Power = Fv = (dm/dt)v2 Watts...
## 02. Sand Spilling Problem | high School physics problem ...
Aug 28, 2017· physics by deepak shukla 601 views 6:42 Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min, and its coarseness is such that - Duration: 5:11.
## NEET Physics Laws of Motion Questions Solved
A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is μ = 0. 5. The distance that the box will move relative to the belt before coming to rest on it taking g = 10 ms-2, is . 1. 1. 2 m 2. 0. 6 m
## Conveyor Belt Problem Changing Mass.mp4
Sep 17, 2012· Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min, and its coarseness is such that - Duration: 5:11. WNY Tutor 31,169 views. 5:11. For the Love of Physics - Walter Lewin - …
## Dropping sand onto a conveyor belt | Physics Forums
Jan 31, 2012· Homework Statement Sand is placed onto a horizontal conveyor belt. rate of sand placed = 0.5 kg s-1 velocity of conveyor belt = 2 m s-1 a) what is the power supplied by the system to keep the velocity constant? b) what is the rate of change of KE by the sand …
## free download video cement plant conveyors animation
Image Of Conveyor Belts For Transporting Concrete Read more physics sand on conveyor belt pictures - Concrete . Belt Conveyor - cnextrudermachine.com Hongrun belt conveyor is a kind of machinery that can transfer the material continuously by friction drive.
## Power problem
Nov 27, 2007· Sand is dropped straight down on a moving conveyor belt at the rate of 3.0kg/s. If friction in the bearings can be ignored the power that must be expended to keep the belt …
## Conveyor Belt common problem trouble shooting guide 1 ...
6. Belt runs to one side for a considerable distance, or the entire conveyor. A. The belt is running off centre as it comes around the tail pulley and/or through the load point. Re-track belt, install training idlers on the return prior to the tail pulley. B. Build up of material on idler rollers. Clean and maintain. Install scrapers, brushes ...
## Ice Block Sliding on a Belt
Jan 27, 2014· For the Love of Physics - Walter Lewin - May 16, 2011 - Duration ... Belt Tracking & Conveyor Maintenance - Duration: 5:01. Kinder Australia Pty Ltd Recommended for you. 5:01. physics …
## Rail/Truck/Barge Loading and Unloading Conveyors
Couple with overland and stockpiling conveyors to reduce heavy equipment fuel charges and maximize stockpiling height. Available in a range of sizes to fit your needs. Standard belt widths available from 24" up to 60"+ depending on conveyor type; Quick to set up and maneuver around site
## Sand from a stationary hopper falls onto a moving conveyor ...
Sand from a stationary hopper falls onto a moving conveyor belt at the rate of 5.00 kg/s as shown in Figure P9.52. The conveyor belt is supported by frictionless rollers and moves at a constant speed of v = 0.750 m/s under the action of a constant horizontal external force F → ext supplied by the motor that drives the belt. Find (a) the sand's rate of change of momentum in the horizontal ...
## Conveyor belt problem | Physics Forums
May 04, 2017· Sand falls on a conveyor belt at 60 kg per second and the belt moves at 2 meters per second. a) what is the force moving the sand? b) what is the magnitude of the force? c) force exerted on belt by the engine. d) the engine has 40% efficiency. how much is the input power?
## Sand from a stationary hopper falls onto a moving conveyor ...
Problem 94. Sand from a stationary hopper falls onto a moving conveyor belt at the rate of 5.00 kg/s as shown in Figure P9.94. The conveyor belt is supported by frictionless rollers and moves at a constant speed of v 5 0.750 m/s under the action of a constant horizontal external force F S ext supplied by the motor that drives the belt.
## Solutions
The force allows the sand to be carried at a constant speed, even if the changing mass on top of the conveyor belt is increasing. Because of the conservation of the momentum, the rate of change in the momentum of the conveyor belt induced by the falling sand is also equal to the rate of change of the falling sand.
## Physics Problem please help sand drops from a stationary ...
Jun 12, 2013· Physics Problem please help sand drops from a stationary hopper onto a belt calculate the the power dissipated? sand drops from stationary hopper at the rate of 5kg/s on to a conveyor belt moving with a constant speed of 2m/s. | 2,143 | 8,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | latest | en | 0.86654 |
https://www.esaral.com/q/multiply-13558 | 1,725,812,107,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00442.warc.gz | 726,950,442 | 11,525 | # Multiply:
Question:
Multiply:
(x6 − y6) by (x2 y2)
Solution:
To multiply, we will use distributive law as follows:
$\left(x^{6}-y^{6}\right)\left(x^{2}+y^{2}\right)$
$=x^{6}\left(x^{2}+y^{2}\right)-y^{6}\left(x^{2}+y^{2}\right)$
$=\left(x^{8}+x^{6} y^{2}\right)-\left(y^{6} x^{2}+y^{8}\right)$
$=x^{8}+x^{6} y^{2}-y^{6} x^{2}-y^{8}$
Thus, the answer is $x^{8}+x^{6} y^{2}-y^{6} x^{2}-y^{8}$. | 194 | 402 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.464471 |
http://mathhelpforum.com/discrete-math/205688-cauchy-sequence-print.html | 1,529,324,084,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859766.6/warc/CC-MAIN-20180618105733-20180618125733-00430.warc.gz | 194,991,822 | 2,934 | # Cauchy Sequence
• Oct 19th 2012, 04:10 PM
lovesmath
Cauchy Sequence
I need to prove that every Cauchy sequence is bounded. Here is my work thus far:
{a(sub n)}n=1 to infinity is a Cauchy sequence. A sequence is said to be Cauchy if for all E>0 there exists N in the natural numbers such that if m,n>=N, then |a(sub n)-a (sub m)|<E.
Proof: Let E=1. Then there exists N in the natural numbers such that m,n>=N implies that |a(sub n)-a(sub m)|<1. By the reverse triangle inequality, ||a(sub n)|-|a(sub m)||<=|a(sub n)-a(sub m)|<1 for m,n>=N, so |a(sub n)|<=|a(sub N+1)|+1 for n>=N. (Fix m=N+1). Let M=max{|a(sub 1)|, |a(sub 2)|, ..., |a(sub N+1)|+1} then |a(sub n)|<=M.
Can you tell me if I am missing any information?
• Oct 19th 2012, 04:30 PM
Plato
Re: Cauchy Sequence
Quote:
Originally Posted by lovesmath
I need to prove that every Cauchy sequence is bounded.
Because this may well be a most important theorem in sequences, I think its proof needs to done correctly.
$\displaystyle 1>0$ so $\displaystyle \exists N$ such that if $\displaystyle n\ge N$ then $\displaystyle |a_n-a_N|<1$
Of course that means $\displaystyle |a_n|\le 1+|a_N|$ if $\displaystyle n\ge N$
Let $\displaystyle B = 1 + \sum\limits_{k = 1}^N {|a_k |}$.
Thus $\displaystyle |a_n|\le B$ for $\displaystyle \forall n~.$ | 439 | 1,299 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-26 | latest | en | 0.771354 |
http://www.expertsmind.com/questions/investment-value-and-rate-30140155.aspx | 1,607,182,233,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747887.95/warc/CC-MAIN-20201205135106-20201205165106-00251.warc.gz | 115,192,567 | 13,937 | ## Investment value and rate, Taxation
Assignment Help:
The second task of the program is to calculate/display a possible investment. First, the program should ask
the amount of money that he/she requires (on an average) per week as living expenditure. You are to assume this expenditure value does not change for the individual over time. (the amount needs to be a positive number and different from zero; if not, the program will show an error message and ask the salary again)
On the basis of this information the program should advise the user whether or not the amount he/she is earning each week is sufficient to cover his/her weekly expenses. If the user earns less than the amount that is required to live, you should show a warning message and ask if the user would like to enter a new amount or terminate the program.
If an amount is sufficient to cover his/her weekly expenses, the program should ask if the user would like to invest some money. If yes, the program should ask
Investment value - the amount of money that he/she would like to invest per week (it is necessary to verify if this amount is feasible. If not, show an error message and ask again)
Interest rate - the interest rate of the investment account per annum (the interest rate needs to be between 1% and 100%; otherwise the program will show an error message and ask the user for the interest rate again)
Investment length - number of weeks that the user will invest the money (it needs to be a positive number and different from zero; otherwise the program will show an error message and ask the number of weeks again)
The program is then expected to display:
a table with details of the investment at four week intervals for the length of the investment. For the purposes of this assignment you will assume that interest is only applied at the end of a complete four week period, see the example below.
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View Full Version : Normal Values
Big Ern
10-26-2007, 08:44 PM
With the risk of sounding stupid, do we get a table of normal values during the exam? Thanks in advance.
yanz
10-26-2007, 08:46 PM
I really hope so. I didn't take 8V, so I'm not sure, but maybe someone else did and will know (I didn't see a table in any of the old exams, but it could've been a separate handout). I'll try to dig through the old 8V threads...
ETA: this thread (http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?t=88773&highlight=normal+table)suggests that in the past, a normal table has been provided
howdy25th
10-28-2007, 09:33 AM
I was just about to post a qn on this. Somebody who knows the answer, please respond. I am assuming so too, but please confirm...
rekrap
10-28-2007, 10:26 AM
If there's not an entire table made available, select values should be given in a smaller table within any question requiring values be determined.
http://www.actuarialoutpost.com/actuarial_discussion_forum/showpost.php?p=2181737&postcount=4
TiderInsider
11-01-2007, 06:43 PM
:bump:
rekrap
11-02-2007, 06:41 AM
:lol: This thread was the first thing I thought of when that question came up... :judge:
statghost
11-02-2007, 09:23 AM
You can interpolate the N(d1) values from the table in part (e). The second column is d1, the third column is delta which is N(d1) - 1 for European put option.
yanz
11-02-2007, 09:49 AM
You can interpolate the N(d1) values from the table in part (e). The second column is d1, the third column is delta which is N(d1) - 1 for European put option.
I'm not sure they could rely on us to use info from part e to answer part c...
yanz
11-02-2007, 09:49 AM
:lol: This thread was the first thing I thought of when that question came up... :judge:
did you take the exam?!
Car'a'carn
11-02-2007, 10:28 AM
You can interpolate the N(d1) values from the table in part (e). The second column is d1, the third column is delta which is N(d1) - 1 for European put option.
But we still had to find N(d2). :shrug:
beeitt
11-02-2007, 02:11 PM
In the big 14 marks question,
The part where you need to fill in blank value is very similar to FE assignment
(opps is this considered as discussion or module assignment?if it does, delete it, but i don't think this is)
Anyway, you also need to find N(-d2) to find the bond part of hedge. Honestly, fill in the blank alone could take a lot of time, calculating d2 and N(-d2) and others. I don't know. maybe someone know a short cut. I know i could find the stock and delta part but the bond part, i blew it
Car'a'carn
11-02-2007, 02:19 PM
I noticed that the cash flow was the sum of the stock and bond columns, since there were enough values in the table filling up the blanks was easy.
yanz
11-02-2007, 02:44 PM
I noticed that the cash flow was the sum of the stock and bond columns, since there were enough values in the table filling up the blanks was easy.
I thought that was true as well (since it worked for the first row), but it wasn't true for one of the other rows (row 3 or 4), so I bailed on that strategy.
Car'a'carn
11-02-2007, 02:51 PM
I thought that was true as well (since it worked for the first row), but it wasn't true for one of the other rows (row 3 or 4), so I bailed on that strategy.
So I did not get all of it. :crying:
qqpp
11-02-2007, 03:02 PM
I thought that was true as well (since it worked for the first row), but it wasn't true for one of the other rows (row 3 or 4), so I bailed on that strategy.
That's what I thought. The first row was clear, but I wasn't sure if there is any interests that we need to account for for the later periods (thinking about that one question on a right-hand-side page from Goldfarb's manual with cumulative cost, interests, etc... can't remember the exact question and whether it's different enough than the one on the exam).
carzymathematician
11-02-2007, 04:28 PM
Recall all you're doing is buying a put = Ke^-rt*N(-d2) - S0N(-d1). This is equivalent to shorting N(-d1) of the stock and investing in the risk-free asset.
Allacalander
11-02-2007, 04:39 PM
The stock portion was given by the value of the index divided by delta. I think that's right. There was a simple relationship of some kind like that.
Allacalander
11-02-2007, 04:42 PM
There was another problem I needed the normal table for. The one where we had to determine the probability that the VA account was in the money in 2 years. I was pretty sure that was a basic statistics question, but without the normal table, I was sunk.
yanz
11-02-2007, 04:43 PM
There was another problem I needed the normal table for. The one where we had to determine the probability that the VA account was in the money in 2 years. I was pretty sure that was a basic statistics question, but without the normal table, I was sunk.
I wrote that the answer was N(d1) but I didn't have a normal table, so if I were to approximate the result with a binomial tree, I'd get [whatever].
Who knows what they were looking for there...
sundwarf
11-02-2007, 05:38 PM
looks like I start forgetting stuff already, but isn't N(d1) delta? the probability to be in-the-money is something related to N(d2)???
McMoM
11-02-2007, 06:21 PM
I used N(d2), thinking of it like a cash-or-nothing call, but looking back at the JAM practice exams the asset-or-nothing probably would have been a better comparison - it uses N(-d1).
clip77
11-02-2007, 09:19 PM
I used N(d2), thinking of it like a cash-or-nothing call, but looking back at the JAM practice exams the asset-or-nothing probably would have been a better comparison - it uses N(-d1).
N(d2) is the risk neutral probabilty. I don't know if the question is asking risk neutral probability - there should be a formula for assets following Geometric Brownian Motion...it was something like Ln(So) + (Mu - 0.5*Var)T?
I put both in the exam, and made a note about one being risk neutral and the other is the probability that follows the model. i don't know if they will consider my answer self-conflicting.
carzymathematician
11-07-2007, 04:29 PM
Was there a consensus as to whether this question could be considered defective or not?
yanz
11-07-2007, 04:33 PM
I was told to be safe and write in a complaint. Even if it doesn't get thrown out, they'll take the complaint into consideration when grading [supposedly].
carzymathematician
11-08-2007, 09:31 AM
I was told to be safe and write in a complaint. Even if it doesn't get thrown out, they'll take the complaint into consideration when grading [supposedly].
So have you written to them?
yanz
11-08-2007, 09:53 AM
So have you written to them?
no. I, personally, have not.
carzymathematician
11-08-2007, 10:19 AM
no. I, personally, have not.
Ok, I will go ahead and send an email. Do you remember what question this was? Or will it be sufficient to write "RE: that 14-pt question"? :)
rekrap
11-08-2007, 10:30 AM
Ok, I will go ahead and send an email. Do you remember what question this was? Or will it be sufficient to write "RE: that 14-pt question"? :)
The test should probably be out today or tomorrow, if you would like to wait.
yanz
11-08-2007, 10:31 AM
I believe it was #6. The parts that required a normal table were a, c, [and possibly e], and parts d and f relied on comparing numerical answers received in the above parts.
carzymathematician
11-08-2007, 02:22 PM
Just sent a note. I guess we'll just have to wait and see what happens!
carzymathematician
11-08-2007, 03:31 PM
This was the response that I got
Thank you for taking the time to share your thoughts and concerns with regard to Exam FET. Your comments have been forwarded to the Exam committee for review. Your patience in awaiting a response is appreciated. We will forward the Exam committee’s response to your comments after grades are released. | 2,183 | 7,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2013-20 | latest | en | 0.895315 |
https://cs.stackexchange.com/questions/75160/higher-order-empirical-entropy-is-not-the-entropy-of-the-empirical-distribution/75165#75165 | 1,638,725,216,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00119.warc.gz | 259,837,308 | 34,275 | Higher order empirical entropy is not the entropy of the empirical distribution?
Basically, the problem is that I always thought that the (unnormalized) $k$th order empirical entropy $n\cdot H_k(x)$ (see "Backround" at the end of this post for more information) for a given string $x$ of length $n$ equals the entropy of the empirical distribution on the set of all words of length $n$ (the conditional probabilities are taken from the observed occurrences in $x$). Until today, I was totally sure that this is true, but it seems to be wrong for very easy examples (e.g. for $x=1011$).
To be more accurate, consider $\Sigma=\{0,1\}$, $x=1011$ and $k=1$ such that $$n\cdot H_k(x)=4\cdot H_1(x)=2.$$ On the other hand, the corresponding empirical distribution is a First-order Markov source described by
• $P(\text{The first letter is }0)=0$
• $P(\text{The first letter is }1)=1$
• $P(\text{Letter }0 \text{ immediately follows after letter }0)=0$
• $P(\text{Letter }1 \text{ immediately follows after letter }0)=1$
• $P(\text{Letter }0 \text{ immediately follows after letter }1)=0.5$
• $P(\text{Letter }1 \text{ immediately follows after letter }1)=0.5$
Now this probabilities yield a random variable $X$ on words of length $n=4$, where the Shannon entropy is $$H(X)=-\sum_{w\in\Sigma^4}P(w)\log P(w)=\frac{9}{4}.$$ (You only need to consider words of length $4$ which start with $1$ and do not contain $00$.)
But shouldn't be both measures intuitively give the same value? I did the calculations more than once, so I am pretty sure the mistake is in my understanding, not in the calculation. Or did I always thought something wrong? But it seems so clear that the empirical entropy is the entropy of the empirical distribution (also for higher order Markov sources).
Backround
For a given string $x\in\Sigma^n$ over an alphabet of size $k$, the ($0$-th order) empirical entropy is defined as $$H_0(x)=\frac{1}{n}\sum_{i=1}^{k}n_i\cdot\log \frac{n}{n_i},$$ where $n_i$ is the number of occurrences of the $i$th alphabet symbol. The $k$-order empirical entropy ($k\ge 1$) is then defined as $$H_k(x)=\frac{1}{n}\sum_{|w|=k}|S_w|H_0(S_w),$$ where $S_w$ is the string obtained by concatenating the characters immediately following occurrences of $w$ in $x$.
1 Answer
No, as your counterexample shows, it is not true. I don't understand why you previously thought it was true, either, so it's hard to say what might have gone wrong with your previous intuition.
We can say that
$$H(X) = H(X_1) + H(X_2 | X_1) + H(X_3 | X_1,X_2) + H(X_4 | X_1,X_2,X_3),$$
where $X_i$ denotes the $i$th character in $X$. that much is true.
In the example you gave, $H(X_1) = 0$, $H(X_2|X_1) = 1$, $H(X_3|X_1,X_2) = 0.5$, $H(X_4 | X_1,X_2,X_3) = 0.75$ (see footnote * for explanation). Thus $H(X) = 1 + 0.5 + 0.75 = 2.25$.
Do notice that $\Pr[X_i=1]$ and $H(X_{i+1}|X_1,\dots,X_i)$ both depend on $i$. This might be important.
Also, notice that $H(X_{i+1}|X_1,\dots,X_i)$ is unrelated to $H_1(x)$.
Perhaps you were expecting $\Pr[X_i=1]$ and $H(X_{i+1}|X_1,\dots,X_i)$ to be the same for all $i$? If that were true, there might be a relationship between $H_1(x)$ and $\Pr[X_i=1]$ and $H(X_{i+1}|X_1,\dots,X_i)$ (or something similar to them)... but it's not.
Footnote *: How did I calculate those? I used the formula
$$H(X_{i+1}|X_1,\dots,X_i) = \sum_v \Pr[(X_1,\dots,X_i) = v] \times H(X_{i+1}|(X_1,\dots,X_i) = v).$$
For $i=1,2$, we only need to consider two values of $v$; for $i=3$, we only need to consider three values of $v$.
• Thanks for your answer. I think my confusion is based on the fact that for the $0$th order empirical entropy, this is true, i.e. the empirical entropy is the Shannon entropy of the empirical distribution. So I thought it is true also for higher order Markov sources. May 10 '17 at 8:59
• Also, Paul Vitányi states in his paper "On empirical entropy" on side 5 that the Shannon entropy of the random variable $X$ "associated with the $k$th order empirical entropy computed from $x$" equals $n\cdot H_k(x)$. Do you have an idea which distribution on the set of words of length $n$ he could mean, i.e. which natural distribution on $\Sigma^n$ yields the Shannon entropy $n\cdot H_k(x)$ for $x\in\Sigma^n$ ? I always thought he refers to the empirical distribution as described in my question. May 10 '17 at 8:59 | 1,345 | 4,341 | {"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} | 3.65625 | 4 | CC-MAIN-2021-49 | latest | en | 0.86835 |
http://mizar.uwb.edu.pl/version/current/html/proofs/matrlin/53_1_1 | 1,571,326,147,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-43/segments/1570986675409.61/warc/CC-MAIN-20191017145741-20191017173241-00200.warc.gz | 123,958,499 | 2,889 | deffunc H1( Nat) -> FinSequence of K = (f . (b1 /. \$1)) |-- b2;
consider M being FinSequence such that
A1: len M = len b1 and
A2: for k being Nat st k in dom M holds
M . k = H1(k) from A3: dom M = Seg (len b1) by ;
ex n being Nat st
for x being object st x in rng M holds
ex s being FinSequence st
( s = x & len s = n )
proof
take n = len ((f . (b1 /. 1)) |-- b2); :: thesis: for x being object st x in rng M holds
ex s being FinSequence st
( s = x & len s = n )
let x be object ; :: thesis: ( x in rng M implies ex s being FinSequence st
( s = x & len s = n ) )
assume x in rng M ; :: thesis: ex s being FinSequence st
( s = x & len s = n )
then consider y being object such that
A4: y in dom M and
A5: x = M . y by FUNCT_1:def 3;
reconsider y = y as Nat by ;
M . y = (f . (b1 /. y)) |-- b2 by A2, A4;
then reconsider s = M . y as FinSequence ;
take s ; :: thesis: ( s = x & len s = n )
thus s = x by A5; :: thesis: len s = n
thus len s = len ((f . (b1 /. y)) |-- b2) by A2, A4
.= len b2 by Def7
.= n by Def7 ; :: thesis: verum
end;
then reconsider M = M as tabular FinSequence by MATRIX_0:def 1;
now :: thesis: for x being object st x in rng M holds
x in the carrier of K *
let x be object ; :: thesis: ( x in rng M implies x in the carrier of K * )
assume x in rng M ; :: thesis: x in the carrier of K *
then consider y being object such that
A6: y in dom M and
A7: x = M . y by FUNCT_1:def 3;
reconsider y = y as Nat by ;
M . y = (f . (b1 /. y)) |-- b2 by A2, A6;
then reconsider s = M . y as FinSequence of K ;
s = x by A7;
hence x in the carrier of K * by FINSEQ_1:def 11; :: thesis: verum
end;
then rng M c= the carrier of K * ;
then reconsider M = M as Matrix of K by FINSEQ_1:def 4;
take M1 = M; :: thesis: ( len M1 = len b1 & ( for k being Nat st k in dom b1 holds
M1 /. k = (f . (b1 /. k)) |-- b2 ) )
for k being Nat st k in dom b1 holds
M1 /. k = (f . (b1 /. k)) |-- b2
proof
let k be Nat; :: thesis: ( k in dom b1 implies M1 /. k = (f . (b1 /. k)) |-- b2 )
assume A8: k in dom b1 ; :: thesis: M1 /. k = (f . (b1 /. k)) |-- b2
then A9: k in Seg (len b1) by FINSEQ_1:def 3;
dom M1 = dom b1 by ;
hence M1 /. k = M1 . k by
.= (f . (b1 /. k)) |-- b2 by A2, A3, A9 ;
:: thesis: verum
end;
hence ( len M1 = len b1 & ( for k being Nat st k in dom b1 holds
M1 /. k = (f . (b1 /. k)) |-- b2 ) ) by A1; :: thesis: verum | 870 | 2,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.907326 |
https://physicscalculatorpro.com/ohms-law-calculator/ | 1,721,537,173,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00528.warc.gz | 396,682,554 | 8,554 | Created By : Vaibhavi Kumari
Reviewed By : Rajashekhar Valipishetty
Last Updated : May 10, 2023
The Ohm’s Law Calculator is a useful tool for determining the voltage, current, and power relationship. Simply type in the two parameters, and the unknown parameter will appear in a flash.
Choose a Calculation
Voltage (V):
Current (I):
### What is Meant by Ohm’s Law?
According to Ohm's law, the potential difference between two points in a circuit is directly proportional to the current running through the circuit and the resistance of the circuit.
### Ohms Law Formula
The formula that defines Ohm's law is; V = IR
• Where, V = difference in voltage or potential.
• I = current
• R = conductor's resistance
The electrical circuit's current, voltage, and resistance are calculated using Ohm's law. The voltage drop across electrical components is controlled using this law.
### Applications of Ohm's Law
• Ohm's law is most commonly used to determine an electric circuit's voltage, resistance, or current.
• To keep the desired voltage drop across the electronic components, Ohm's law is applied.
• Ohm's law is also utilised to divert current in DC ammeters and other DC shunts.
### Limitations of Ohm's Law
The following are some of Ohm's law's limitations
• Ohm's law does not apply to unilateral electrical elements like diodes and transistors since they only allow current to flow in one direction.
• Voltage and current will not be constant concerning time with non-linear electrical elements with factors like capacitance, resistance, and so on, making Ohm's law difficult to apply.
For more concepts check out physicscalculatorpro.com to get quick answers by using this free tool.
### Voltage Formula
Finding Voltage is one of Ohm’s Law's most basic concepts. To get the Voltage, we can rewrite the calculation above as V= IR. If you know the current and resistance, you can easily compute the voltage by replacing them in the formula above. Volts are the units of voltage.
### How to Calculate Power?
Power is another value that can be determined using Ohm's Law as a foundation. The formula P = V x I expresses power as the product of voltage and current. Watts are the units of power.
### Ohm's Law for Anisotropic Material
Ohm’s Law can also be expressed in terms of the location of electrical properties within a conductor. If the particular resistance of conductive materials is independent of the applied field value and direction. The formula for calculating Ohm's law is
ρ = E / J
• Where, ρ = specific resistance of the conductive material
• E = Electric Field of the Vector
• J = Current Density Vector
When working with anisotropic materials like wood or graphite, the above formula can be quite useful.
### How to Use the Ohms Law Calculator?
The following is the procedure how to use the Ohms law calculator
• Fill in the appropriate input fields with voltage, resistance, or current.
• To acquire the result, click the "Calculate" button.
• Finally, in the new window, the resistance, voltage, and current will be presented.
### Examples on Ohm’s Law Calculations
Question 1: A current of 4.2 A passes through an electric iron with a resistance of 60. Calculate the difference between two places in terms of voltage?
Solution:
Given: I = 4.2A
R = 60
V = IR = 4.2 x 60 = 252v
Question 2: A fully resistive electrical equipment is connected to an EMF source of 7.0 V. (a light bulb). It is powered by a 3.0 A electric current. Assume that the conducting wires have no resistance. Determine how much resistance the electrical appliance provides?
Solution:
Given: V = 7.0v
I = 3.0A
V = IR
R = V/I = 7.0/3.0 = 2.33Ω
### FAQs on Ohm’s Law Calculator
1. How many amperes does one ohm contain?
One. We'll assume you're working with amperes and volts/ohms. The ampere is the SI base unit for electric current.
2. What is the application of Ohm's Law?
Ohm's law is used to validate the static values of circuit components such as current levels, voltage supplies, and voltage drops.
3. How many ohms in 100 watts?
8 ohms. It's worth noting that the power output (100 watts) is at a set load (8 ohms). This means that the maximum output power of an 8-ohm speaker will be 100 watts.
4. Is there a difference between ohms and K ohms?
One kilo ohm is equal to 1,000 ohms, which is the resistance between two points of a conductor carrying one amp at one volt. | 1,054 | 4,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-30 | latest | en | 0.907583 |
https://la.mathworks.com/matlabcentral/cody/problems/324-7-segment-led-display/solutions/1650075 | 1,603,982,520,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904287.88/warc/CC-MAIN-20201029124628-20201029154628-00051.warc.gz | 402,385,339 | 17,364 | Cody
# Problem 324. 7 segment LED display
Solution 1650075
Submitted on 16 Oct 2018 by Majid Farzaneh
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
x = 1; y_correct = 2; assert(isequal(seven_seg_led(x),y_correct))
ans = 2
2 Pass
x = 2; y_correct = 5; assert(isequal(seven_seg_led(x),y_correct))
ans = 5
3 Pass
x = 3; y_correct = 5; assert(isequal(seven_seg_led(x),y_correct))
ans = 5
4 Pass
x = 4; y_correct = 4; assert(isequal(seven_seg_led(x),y_correct))
ans = 4
5 Pass
x = 5; y_correct = 5; assert(isequal(seven_seg_led(x),y_correct))
ans = 5
6 Pass
x = 6; y_correct = 6; assert(isequal(seven_seg_led(x),y_correct))
ans = 6
7 Pass
x = 7; y_correct = 3; assert(isequal(seven_seg_led(x),y_correct))
ans = 3
8 Pass
x = 8; y_correct = 7; assert(isequal(seven_seg_led(x),y_correct))
ans = 7
9 Pass
x = 9; y_correct = 6; assert(isequal(seven_seg_led(x),y_correct))
ans = 6
10 Pass
x = 0; y_correct = 6; assert(isequal(seven_seg_led(x),y_correct))
ans = 6
11 Pass
x = 10; y_correct = 8; assert(isequal(seven_seg_led(x),y_correct))
ans = 8
12 Pass
x = 42; y_correct = 9; assert(isequal(seven_seg_led(x),y_correct))
ans = 9
13 Pass
x = 101010101; y_correct = 34; assert(isequal(seven_seg_led(x),y_correct))
ans = 34
14 Pass
x = 1234567890; y_correct = 49; assert(isequal(seven_seg_led(x),y_correct))
ans = 49
15 Pass
% x = 12345678987654321; % y_correct = 86; % assert(isequal(seven_seg_led(x),y_correct))
16 Pass
x = 222444666888000; y_correct = 84; assert(isequal(seven_seg_led(x),y_correct))
ans = 84
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https://ischool.sg/Clones/index2/29215?url=https%3A%2F%2Fischool.sg%2Fquestions%2Fhashtag%3Ftype%3Dall%26tag%3DRate%2Bmore%2Bthan%2B1%2Bobject%26level%3DPrimary%2B6?login=1 | 1,652,809,228,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662519037.11/warc/CC-MAIN-20220517162558-20220517192558-00200.warc.gz | 373,032,742 | 38,774 | Question
Tap M can fill a tank completely in 6 hours while Tap N takes 2 hours more to fill the same tank completely. There is a hole at the bottom of the tank which can empty the full tank in 7 hours. How long does it take for the tank to be completely filled if both taps are turned on? Give your answer in whole number or decimal.
3 m
Click button first when a symbol is required. X
Tap M can fill a tank completely in 6 hours while Tap N takes 2 hours more to fill the same tank completely. There is a hole at the bottom of the tank which can empty the full tank in 7 hours. How long does it take for the tank to be completely filled if both taps are turned on? Give your answer in whole number or decimal. | 167 | 712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-21 | latest | en | 0.955818 |
http://www.massmind.org/techref/microchip/math/div/24byconst6-h.htm | 1,708,595,991,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00439.warc.gz | 62,828,010 | 6,276 | # PICMicrocontollerMathMethod
## Divide 24 bits by the constant value 6
From: Hughe
Divide a 24 bit number by 6 and round up if the remainder is 3 or more - can be used for any 8 bit constant divisor with 2 values changed.
```
CBLOCK
iA:3
iB
iC
ENDC
DivBySix24:
movlw .24 ; loop for 24 shifts
movwf iC ; iC is loop counter
clrf iB ; clear remainder reg
bcf STATUS,C ; clear result bit
DivBySixLoop:
rlf iA+0 ; roll result bit onto iA
rlf iA+1
rlf iA+2
rlf iB ; and top bit of iA onto iB
movlw D'6' ; check if remainder (iB) >= 6
subwf iB,w ; w = iB-6
btfss STATUS,C ; carry set if iB >= 6
goto DivBySixNogo
movwf iB ; iB-6 (w) into iB
; the carry gets rolled onto iA as a result
DivBySixNogo:
decfsz iC,f ; loop until all bits checked
goto DivBySixLoop
rlf iA+0 ; roll last result bit onto A
rlf iA+1
rlf iA+2
; the folowing could be left off if no rounding is required
movlw D'3' ; if remainder is 3 or more round up
subwf iB,w ; w = iB-3
btfss STATUS,C ; carry set if iB >= 3
return
incf iA+0,f ; increment Result
btfsc STATUS,Z
incf iA+1,f
btfsc STATUS,Z
incf iA+2,f
return
```
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http://www.presentica.com/ppt-presentation/autonomous-measures-theory-testing | 1,547,648,921,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657510.42/warc/CC-MAIN-20190116134421-20190116160421-00343.warc.gz | 373,849,837 | 12,031 | 1 / 23
# Autonomous Measures Theory Testing.
107 views
Description
Free Measures Theory Testing Unit 8 Ch 10: 1-5, 7, 9, 11, 13, 15, 19 (pp. 282-286) Looking at 2 sets of information 2 general examination systems information sets originate from 2 separate gatherings free examples between gatherings plan 2 sets of information from 1 bunch needy or related specimens
Transcripts
Slide 1
Autonomous Measures Hypothesis Testing Unit 8 Ch 10: 1-5, 7, 9, 11, 13, 15, 19 (pp. 282-286)
Slide 2
Comparing 2 sets of information 2 general exploration procedures information sets originate from 2 separate gatherings autonomous examples between gatherings outline 2 sets of information from 1 bunch needy or related specimens coordinated subjects (2 related gatherings) inside of subjects configuration ~
Slide 3
Independent Measures Hypothesis Test Select 2 free specimens would they say they are from same populace? Examination select 2 tests 1 gets treatment are the specimens the same? ~
Slide 4
Experimental Outcomes Do not hope to be precisely equivalent testing blunder How huge a distinction to reject H 0 ? ~
Slide 5
Hypotheses: Independent Measures Nondirectional H 0 : m 1 - m 2 = 0 H 1 : m 1 - m 2 ï¹ 0 or H 0 : m 1 = m 2 H 1 : m 1 ï¹ m 2 Directional (relies on upon forecast ) H 0 : m 1 - m 2 < 0 H 1 : m 1 - m 2 > 0 or H 0 : m 1 < m 2 H 1 : m 1 > m 2 no quality determined for either Group 1 scores = Group 2 scores ~
Slide 6
Sample measurement: t test: Independent Samples Same essential structure as single example Independent specimens [df = n 1 + n 2 - 2]
Slide 7
The Test Statistic Since m 1 - m 2 = 0 [df = n 1 + n 2 - 2]
Slide 8
Estimated Standard Error *Standard lapse of contrast between 2 test means must figure s 2 p first ~
Slide 9
Pooled Variance ( s 2 p ) Average of 2 test differences weighted normal if n 1 ï¹ n 2 if n 1 = n 2
Slide 10
The Test Statistic: Assumptions 1. Tests are free 2. Tests originate from typical populaces 3. Accept break even with change s 2 1 = s 2 does not oblige s 2 1 = s 2 homogeneity of fluctuation t test is strong infringement of suppositions Little impact on P(rejecting H 0 ) ~
Slide 11
Example: Independent Samples Is exam execution influenced by what amount of rest you get the night prior to a test? Subordinate variable? free variable? Grp 1: 4 hrs rest ( n = 6) Grp 2: 8 hrs rest ( n = 6) ~
Slide 12
Example: n 1 = n 2 1. State Hypotheses H 0 : m 1 - m 2 = 0 or H 0 : m 1 = m 2 H 1 : m 1 - m 2 ¹ 0 or H 1 : m 1 ¹ m 2. Set model for dismissing H 0 : nondirectional a = .05 df = (n 1 + n 2 - 2) = (6 + 6 - 2) = 10 t CV .05 =
Slide 13
Example: n 1 = n 2 3. select specimen, process measurements do investigation mean exam scores for every gathering Group 1: M 1 = 15 ; s 1 = 4 Group 2 : M 2 = 19; s 2 = 3 register s 2 p s M 1 - M 2 t obs ~
Slide 14
Example: n 1 = n 2 figure s 2 p
Slide 15
Example: n 1 = n 2 figure
Slide 16
Example: n 1 = n 2 figure test measurement
Slide 17
Example: n 1 = n 2 4. Choice? Is t obs in basic locale? No, neglect to reject H 0 If directional test or change level of essentialness change discriminating estimation of t ( t cv ) simply like different tests ~
Slide 18
Pooled Variance: n 1 ¹ n 2 Unequal specimen sizes weight every fluctuation greater n - > more weight
Slide 19
Example: n 1 ¹ n 2 Supplementary Material What impact does the measure of rest the night prior to an exam have on exam execution? Subordinate variable free variable Grp 1: 4 hrs rest ( n = 6) Grp 1: 8 hrs rest ( n = 7) ~
Slide 20
Example: n 1 ¹ n 2 1. State Hypotheses H 0 : m 1 = m 2 or m 1 - m 2 = 0 H 1 : m 1 ¹ m 2 or m 1 - m 2 ¹ 0 2. Set measure for dismissing H 0 : nondirectional a = .05 df = (n 1 + n 2 - 2) = (6 + 7 - 2) = 11 t CV = + 2.201 ~
Slide 21
Example: n 1 ¹ n 2 3. select specimen, process insights do test mean exam scores for every gathering Group 1: M 1 = 14 ; s 1 = 3 Group 2 : M 2 = 19; s 2 = 2 figure s 2 pooled s M 1 - M 2 t obs ~
Slide 22
Example: n 1 ¹ n 2 register s 2 pooled figure test measurement [df = n 1 + n 2 - 2]
Slide 23
Example: n 1 ¹ n 2 4. Decipher Is t obs past t CV ? In the event that yes, Reject H 0
Recommended
View more... | 1,395 | 4,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.789181 |
http://en.wikipedia.org/wiki/Diagonalization_argument | 1,417,037,741,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931007510.17/warc/CC-MAIN-20141125155647-00038-ip-10-235-23-156.ec2.internal.warc.gz | 96,125,293 | 20,468 | # Cantor's diagonal argument
(Redirected from Diagonalization argument)
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An illustration of Cantor's diagonal argument (in base 2) for the existence of uncountable sets. The sequence at the bottom cannot occur anywhere in the enumeration of sequences above.
An infinite set may have the same cardinality as a proper subset of itself, as the depicted bijection f(x)=2x from the natural to the even numbers demonstrates. Nevertheless, infinite sets of different cardinalities exist, as Cantor's diagonal argument shows.
In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument or the diagonal method, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural numbers.[1][2][3] Such sets are now known as uncountable sets, and the size of infinite sets is now treated by the theory of cardinal numbers which Cantor began.
The diagonal argument was not Cantor's first proof of the uncountability of the real numbers; it was actually published much later than his first proof, which appeared in 1874.[4][5] However, it demonstrates a powerful and general technique that has since been used in a wide range of proofs, also known as diagonal arguments by analogy with the argument used in this proof. The most famous examples are perhaps Russell's paradox, the first of Gödel's incompleteness theorems, and Turing's answer to the Entscheidungsproblem.
## An uncountable set
In his 1891 article, Cantor considered the set T of all infinite sequences of binary digits (i.e. consisting only of zeroes and ones). He begins with a constructive proof of the following theorem:
If s1, s2, … , sn, … is any enumeration of elements from T, then there is always an element s of T which corresponds to no sn in the enumeration.
To prove this, given an enumeration of arbitrary members from T, like e.g.
s1 = (0, 0, 0, 0, 0, 0, 0, ...) s2 = (1, 1, 1, 1, 1, 1, 1, ...) s3 = (0, 1, 0, 1, 0, 1, 0, ...) s4 = (1, 0, 1, 0, 1, 0, 1, ...) s5 = (1, 1, 0, 1, 0, 1, 1, ...) s6 = (0, 0, 1, 1, 0, 1, 1, ...) s7 = (1, 0, 0, 0, 1, 0, 0, ...) ...
he constructs the sequence s by choosing its nth digit as complementary to the nth digit of sn, for every n. In the example, this yields:
s1 = (0, 0, 0, 0, 0, 0, 0, ...) s2 = (1, 1, 1, 1, 1, 1, 1, ...) s3 = (0, 1, 0, 1, 0, 1, 0, ...) s4 = (1, 0, 1, 0, 1, 0, 1, ...) s5 = (1, 1, 0, 1, 0, 1, 1, ...) s6 = (0, 0, 1, 1, 0, 1, 1, ...) s7 = (1, 0, 0, 0, 1, 0, 0, ...) ... s = (1, 0, 1, 1, 1, 0, 1, ...)
By construction, s differs from each sn, since their nth digits differ (highlighted in the example). Hence, s cannot occur in the enumeration.
Based on this theorem, Cantor then uses an indirect argument to show that:
The set T is uncountable.
He assumes for contradiction that T was countable. Then (all) its elements could be written as an enumeration s1, s2, … , sn, … . Applying the previous theorem to this enumeration would produce a sequence s not belonging to the enumeration. However, s was an element of T and should therefore be in the enumeration. This contradicts the original assumption, so T must be uncountable.
### Interpretation
The interpretation of Cantor's result will depend upon one's view of mathematics. To constructivists, the argument shows no more than that there is no bijection between the natural numbers and T. It does not rule out the possibility that the latter are subcountable. In the context of classical mathematics, this is impossible, and the diagonal argument establishes that, although both sets are infinite, there are actually more infinite sequences of ones and zeros than there are natural numbers.
### Real numbers
The function tan: (−π/2,π/2) → R
The function h: (0,1) → (−π/2,π/2)
The uncountability of the real numbers was already established by Cantor's first uncountability proof, but it also follows from the above result. To see this, we will build a one-to-one correspondence between the set T of infinite binary strings and a subset of R (the set of real numbers). Since T is uncountable, this subset of R must be uncountable. Hence R is uncountable.
To build this one-to-one correspondence (or bijection), observe that the string t = 0111… appears after the binary point in the binary expansion 0.0111…. This suggests defining the function f(t) = 0.t, where t is a string in T. Unfortunately, f(1000…) = 0.1000… = 1/2, and f(0111…) = 0.0111… = 1/4 + 1/8 + 1/16 + … = 1/2. So this function is not a bijection since two strings correspond to one number—a number having two binary expansions.
However, modifying this function produces a bijection from T to the interval (0, 1)—that is, the real numbers > 0 and < 1. The idea is to remove the "problem" elements from T and (0, 1), and handle them separately. From (0, 1), remove the numbers having two binary expansions. Put these numbers in a sequence: a = (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, …). From T, remove the strings appearing after the binary point in the binary expansions of 0, 1, and the numbers in sequence a. Put these eventually-constant strings in a sequence: b = (000…, 111…, 1000…, 0111…, 01000…, 11000…, 00111…, 10111…, ...). A bijection g(t) from T to (0, 1) is defined by: If t is the nth string in sequence b, let g(t) be the nth number in sequence a; otherwise, let g(t) = 0.t.
To build a bijection from T to R: start with the tangent function tan(x), which provides a bijection from (−π/2, π/2) to R; see right picture. Next observe that the linear function h(x) = πx - π/2 provides a bijection from (0, 1) to (−π/2, π/2); see left picture. The composite function tan(h(x)) = tan(πx - π/2) provides a bijection from (0, 1) to R. Compose this function with g(t) to obtain tan(h(g(t))) = tan(πg(t) - π/2), which is a bijection from T to R. This means that T and R have the same cardinality—this cardinality is called the "cardinality of the continuum."
## General sets
Illustration of the generalized diagonal argument: The set T = {n∈ℕ: nf(n)} at the bottom cannot occur anywhere in the range of f:P(ℕ). The example mapping f happens to correspond to the example enumeration s in the above picture.
A generalized form of the diagonal argument was used by Cantor to prove Cantor's theorem: for every set S the power set of S, i.e., the set of all subsets of S (here written as P(S)), has a larger cardinality than S itself. This proof proceeds as follows:
Let f be any function from S to P(S). It suffices to prove f cannot be surjective. That means that some member T of P(S), i.e., some subset of S, is not in the image of f. As a candidate consider the set:
T = { sS: sf(s) }.
For every s in S, either s is in T or not. If s is in T, then by definition of T, s is not in f(s), so T is not equal to f(s). On the other hand, if s is not in T, then by definition of T, s is in f(s), so again T is not equal to f(s); cf. picture. For a more complete account of this proof, see Cantor's theorem.
### Consequences
This result implies that the notion of the set of all sets is an inconsistent notion. If S were the set of all sets then P(S) would at the same time be bigger than S and a subset of S.
Russell's Paradox has shown us that naive set theory, based on an unrestricted comprehension scheme, is contradictory. Note that there is a similarity between the construction of T and the set in Russell's paradox. Therefore, depending on how we modify the axiom scheme of comprehension in order to avoid Russell's paradox, arguments such as the non-existence of a set of all sets may or may not remain valid.
The diagonal argument shows that the set of real numbers is "bigger" than the set of natural numbers (and therefore, the integers and rationals as well). Therefore, we can ask if there is a set whose cardinality is "between" that of the integers and that of the reals. This question leads to the famous continuum hypothesis. Similarly, the question of whether there exists a set whose cardinality is between |S| and |P(S)| for some infinite S leads to the generalized continuum hypothesis.
Analogues of the diagonal argument are widely used in mathematics to prove the existence or nonexistence of certain objects. For example, the conventional proof of the unsolvability of the halting problem is essentially a diagonal argument. Also, diagonalization was originally used to show the existence of arbitrarily hard complexity classes and played a key role in early attempts to prove P does not equal NP. In 2008, diagonalization was used to "slam the door" on Laplace's demon.[6]
### Version for Quine's New Foundations
The above proof fails for W. V. Quine's "New Foundations" set theory (NF). In NF, the naive axiom scheme of comprehension is modified to avoid the paradoxes by introducing a kind of "local" type theory. In this axiom scheme,
{ sS: sf(s) }
is not a set — i.e., does not satisfy the axiom scheme. On the other hand, we might try to create a modified diagonal argument by noticing that
{ sS: sf({s}) }
is a set in NF. In which case, if P1(S) is the set of one-element subsets of S and f is a proposed bijection from P1(S) to P(S), one is able to use proof by contradiction to prove that |P1(S)| < |P(S)|.
The proof follows by the fact that if f were indeed a map onto P(S), then we could find r in S, such that f({r}) coincides with the modified diagonal set, above. We would conclude that if r is not in f({r}), then r is in f({r}) and vice-versa.
It is not possible to put P1(S) in a one-to-one relation with S, as the two have different types, and so any function so defined would violate the typing rules for the comprehension scheme.
## References
1. ^ Georg Cantor (1892). "Ueber eine elementare Frage der Mannigfaltigkeitslehre". Jahresbericht der Deutschen Mathematiker-Vereinigung 1890–1891 1: 75–78 (84–87 in pdf file). (in german)
2. ^ Keith Simmons (30 July 1993). Universality and the Liar: An Essay on Truth and the Diagonal Argument. Cambridge University Press. pp. 20–. ISBN 978-0-521-43069-2.
3. ^ Rudin, Walter (1976). Principles of Mathematical Analysis (3rd ed.). New York: McGraw-Hill. p. 30. ISBN 0070856133.
4. ^ Gray, Robert (1994), Georg Cantor and Transcendental Numbers, American Mathematical Monthly 101: 819–832, doi:10.2307/2975129
5. ^ Bloch, Ethan D. (2011). The Real Numbers and Real Analysis. New York: Springer. p. 429. ISBN 978-0-387-72176-7.
6. ^ P. Binder (2008). "Theories of almost everything". Nature 455: 884–885. doi:10.1038/455884a. | 2,949 | 10,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2014-49 | longest | en | 0.94741 |
http://biblioteca.posgraduacaoredentor.com.br/?q=Rational+numbers | 1,534,568,054,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213286.48/warc/CC-MAIN-20180818040558-20180818060558-00493.warc.gz | 57,078,575 | 17,107 | Página 1 dos resultados de 1512 itens digitais encontrados em 0.009 segundos
Uma abordagem diferenciada dos números racionais na forma fracionária
Prochnow, Karine Zangalli Schiling
Tipo: Trabalho de Conclusão de Curso Formato: application/pdf
POR
Relevância na Pesquisa
46.26%
As diferentes “personalidades” do número racional trabalhadas através da resolução de problemas
Onuchic, Lourdes de la Rosa; Gomes Allevato, Norma Suely
Tipo: Artigo de Revista Científica Formato: 79-102
POR
Relevância na Pesquisa
46.22%
We address the different "personalities" of the rational number and the concept of proportionality, analyzing the possibilities for using the Mathematics Teaching and Learning through Problem-solving Method. This method is based on the principle that knowledge can be constructed through the use of problems that generate new concepts and new contents. The different meanings of rational number - rational point, quotient, fraction, ratio, and operator - are constructs that depend on mathematical theories in which they are imbedded and the situations that evoke them in problem-solving. Some data will be presented from continuing education courses for teachers, aiming to contribute to understanding regarding the different "personalities" of the rational number. In general, these "personalities" are not easily identified by teachers and students, which is the reason for the many difficulties encountered during problem-solving involving rational numbers. One of these "personalities", the ratio, provides the basis for the concept of proportionality, which is relevant because it is a unifying idea in mathematics.
Prática de Ensino Supervisionada no 2.º Ciclo do Ensino Básico – Matemática e Ciências da Natureza :o ensino e a aprendizagem dos números racionais e a utilização de materiais manipuláveis
Casaca, Ana
Fonte: Instituto Politécnico de Santarém Publicador: Instituto Politécnico de Santarém
Relevância na Pesquisa
46.26%
The Cognitive Predictors of Computational Skill with Whole versus Rational Numbers: An Exploratory Study
Seethaler, Pamela M.; Fuchs, Lynn S.; Star, Jon R.; Bryant, Joan
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
46.26%
The purpose of the present study was to explore the 3rd-grade cognitive predictors of 5th-grade computational skill with rational numbers and how those are similar to and different from the cognitive predictors of whole-number computational skill. Students (n = 688) were assessed on incoming whole-number calculation skill, language, nonverbal reasoning, concept formation, processing speed, and working memory in the fall of 3rd grade. Students were followed longitudinally and assessed on calculation skill with whole numbers and with rational numbers in the spring of 5th grade. The unique predictors of skill with whole-number computation were incoming whole-number calculation skill, nonverbal reasoning, concept formation, and working memory (numerical executive control). In addition to these cognitive abilities, language emerged as a unique predictor of rational-number computational skill.
A Compreensão da idéia do número racional e suas operações na EJA: uma forma de inclusão em sala de aula
Silva, Tácio Vitaliano da
Fonte: Universidade Federal do Rio Grande do Norte; BR; UFRN; Programa de Pós-Graduação em Ensino de Ciências Naturais e Matemática; Ensino de Ciências Naturais e Matemática Publicador: Universidade Federal do Rio Grande do Norte; BR; UFRN; Programa de Pós-Graduação em Ensino de Ciências Naturais e Matemática; Ensino de Ciências Naturais e Matemática
Tipo: Dissertação Formato: application/pdf
POR
Relevância na Pesquisa
46.29%
The awareness of the difficulty which pupils, in general have in understanding the concept and operations with Rational numbers, it made to develop this study which searches to collaborate for such understanding. Our intuition was to do with that the pupils of the Education of Young and Adults, with difficulty in understanding the Rational numbers, feel included in the learning-teaching process of mathematics. It deals with a classroom research in a qualitative approach with analysis of the activities resolved for a group of pupils in classroom of a municipal school of Natal. For us elaborate such activities we accomplished the survey difficulties and obstacles that the pupils experience, when inserted in the learning-teaching process of the Rational numbers. The results indicate that the sequence of activities applied in classroom collaborated so that the pupils to overcome some impediments in the learning of this numbers; A consciência da dificuldade que alunos, em geral, têm em compreender o conceito e operações com Números Racionais, nos fez desenvolver este estudo que busca colaborar para tal compreensão. Nosso intuito foi fazer com que os alunos da Educação de Jovens e Adultos, com dificuldade em compreender os Números Racionais...
Visualisation, navigation and mathematical perception: a visual notation for rational numbers mod1
Tolmie, Julie
Tipo: Thesis (PhD); Doctor of Philosophy (PhD)
EN_AU
Relevância na Pesquisa
66.41%
There are three main results in this dissertation. The first result is the construction of an abstract visual space for rational numbers mod1, based on the visual primitives, colour, and rational radial direction. Mathematics is performed in this visual notation by defining increasingly refined visual objects from these primitives. In particular, the existence of the Farey tree enumeration of rational numbers mod1 is identified in the texture of a two-dimensional animation. ¶ The second result is a new enumeration of the rational numbers mod1, obtained, and expressed, in abstract visual space, as the visual object coset waves of coset fans on the torus. Its geometry is shown to encode a countably infinite tree structure, whose branches are cosets, nZ+m, where n, m (and k) are integers. These cosets are in geometrical 1-1 correspondence with sequences kn+m, (of denominators) of rational numbers, and with visual subobjects of the torus called coset fans. ¶ The third result is an enumeration in time of the visual hierarchy of the discrete buds of the Mandelbrot boundary by coset waves of coset fans. It is constructed by embedding the circular Farey tree geometrically into the empty internal region of the Mandelbrot set. In particular...
Cálculo mental com números racionais não negativos: um estudo sobre as estratégias utilizadas por alunos do 4.º ano de escolaridade
Santos, Ana Catarina Granado Rebelo dos
Fonte: Instituto Politécnico de Lisboa Publicador: Instituto Politécnico de Lisboa
Relevância na Pesquisa
46.26%
Acculturation institutionnelle du chercheur, de l’enseignant et des élèves de 1re secondaire présentant des difficultés d’apprentissage dans la conception et la gestion de situations-problèmes impliquant des nombres rationnels
Lessard, Geneviève
Fonte: Université de Montréal Publicador: Université de Montréal
Tipo: Thèse ou Mémoire numérique / Electronic Thesis or Dissertation
FR
Relevância na Pesquisa
46.43%
Notre recherche s’intéresse à la transformation des rapports aux nombres rationnels d’élèves de 1re secondaire présentant des difficultés d’apprentissage. Comme le montrent plusieurs recherches, le défi majeur auquel sont confrontés les enseignants, ainsi que les chercheurs, est de ne pas s’enliser dans le cercle vicieux d’une réduction des enjeux de l’apprentissage des nombres rationnels et des possibilités d’apprentissage de l’élève en difficultés d’apprentissage, cet élève n’ayant pas ainsi la chance de mettre à l’épreuve ses connaissances, d’oser s’engager dans une démarche de construction de connaissances et d’apprécier les effets de son engagement cognitif. Afin de relever ce défi, nous avons misé sur l’intégration harmonieuse de situations problèmes. Il nous a semblé que, dans une démarche d’acculturation, l’approche écologique soit tout indiquée pour penser une «dé-transposition/re-transposition didactique» (Antibi et Brousseau, 2000) et reconstruire une mémoire porteuse d’espoirs (Brousseau et Centeno, 1998). Notre recherche vise à: 1) caractériser la progression des démarches d’acculturation institutionnelle de l’enseignant, du chercheur et des élèves et leurs effets sur les processus d’élaboration et de gestion des situations d’enseignement; 2) préciser l’évolution des connaissances...
Um estudo sobre construções dos Números Reais; A study on construction of the Real Numbers
Queiroz, Fabiana Moura de
Fonte: Universidade Federal de Goiás; Brasil; UFG; Programa de Pós-graduação em Matemática (IME); Instituto de Matemática e Estatística - IME (RG) Publicador: Universidade Federal de Goiás; Brasil; UFG; Programa de Pós-graduação em Matemática (IME); Instituto de Matemática e Estatística - IME (RG)
Tipo: Dissertação Formato: application/pdf
POR
Relevância na Pesquisa
56.3%
The main objective of this paper is to present the subtle passage of rational numbers to the real numbers, using a construction via Dedekind cuts and other by Cauchy sequences .We present a construction of rational numbers by equivalence classes, so that the reader has a foundation that serves as a support for a good understanding of proposed constructions of real numbers . We use the axiomatic method for buildings that are made on real numbers, in order to show the existence of an orderly and complete field and characterize it. It is also discussed, and a more synthesized form, the real numbers and its application to elementary and high school students.; O objetivo central deste trabalho é apresentar a sutil passagem dos números racionais aos números reais, utilizando uma construção via Cortes de Dedekind e outra por sequências de Cauchy. Apresenta-se uma construção dos números racionais por classes de equivalência, para que o leitor tenha um alicerce que sirva de apoio para um bom entendimento das construções propostas dos números reais. Utiliza-se o método axiomático para as construções que são feitas sobre números reais, com o intuito de mostrar a existência de um corpo ordenado e completo e caracterizá-lo. Discute-se ainda...
A divisão e os números racionais : uma pesquisa de intervenção psicopedagógica sobre o desenvolvimento de competências conceituais de alunos e professores; Division and rational numbers : a survey of psychopedagogical intervention on the development of conceptual competencies among students and teachers
Neves, Regina da Silva Pina
Tipo: Tese
POR
Relevância na Pesquisa
46.26%
Conhecimento do professor do 1º ciclo sobre números racionais
Perfeito, Margarida de Jesus Lucas
Fonte: Instituto Politécnico de Lisboa Publicador: Instituto Politécnico de Lisboa
Relevância na Pesquisa
46.43%
Dissertação apresentada à Escola Superior de Educação de Lisboa para obtenção do grau de mestre em Educação Matemática na Educação Pré-escolar e nos 1º e 2º Ciclos do Ensino Básico; Este estudo visa analisar o conhecimento matemático dos professores do 1º ciclo sobre números racionais, procurando responder às seguintes questões: (1) Que conhecimento revelam os professores do 1º ciclo sobre os números racionais e as suas várias representações?; (2) Como avaliam os professores o conhecimento que têm sobre os números racionais?; e (3) Que dificuldades manifestam os professores relativamente ao trabalho dos números racionais com os alunos? O estudo seguiu uma abordagem metodológica mista, reunindo componentes da investigação quantitativa e qualitativa. A recolha de dados decorreu entre 3 e 10 de janeiro de 2014 e foi feita a partir da aplicação de um questionário impresso a 18 professores do 1º ciclo de três escolas públicas, de um agrupamento situado numa zona limítrofe de Lisboa. O questionário pretendeu analisar o conhecimento matemático dos professores sobre números racionais e simultaneamente recolher informações sobre o modo como analisam o seu conhecimento e as dificuldades que sentem no seu ensino. Os dados indicam que a maioria dos professores não possui conhecimentos sólidos sobre os números racionais...
Conhecimento e formação de futuros professores dos primeiros anos – o sentido de número racional
Pinto, Hélia; Ribeiro, C. Miguel
Fonte: CIED – Centro Interdisciplinar de Estudos Educacionais/Escola Superior de Educação de Lisboa Publicador: CIED – Centro Interdisciplinar de Estudos Educacionais/Escola Superior de Educação de Lisboa
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
46.22%
Nos últimos anos o conhecimento do professor tem vindo a ser reconhecido como um dos aspetos nucleares no, e para o, desenvolvimento do conhecimento matemático dos alunos. Atendendo a essa centralidade, a formação deverá focar-se onde é, efetivamente, necessária, de modo a potenciar um incremento do conhecimento dos alunos, pelo conhecimento (e práticas) dos professores. Sendo os números racionais um dos tópicos problemáticos para os alunos, é fundamental identificar quais as situações matematicamente (mais) críticas para os professores de modo que, pela formação facultada, possam deixar de o ser. Neste artigo, tendo por foco o conhecimento matemático do professor e as suas especificidades, discutimos alguns aspetos desse conhecimento de futuros professores sobre números racionais, em concreto o sentido de número racional, identificando as suas componentes mais problemáticas e equacionando alguns dos porquês em que se sustentam. Terminamos com algumas considerações sobre implicações para a formação de professores e responsabilidade dos seus formadores.; Abstract: In recent years, teachers’ knowledge has come to be recognized as one of the core aspects in and for the development of students’ mathematical knowledge. This being the case...
Rational numbers with purely periodic $\beta$-expansion
Adamczewski, Boris; Frougny, Christiane; Siegel, Anne; Steiner, Wolfgang
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
46.29%
We study real numbers $\beta$ with the curious property that the $\beta$-expansion of all sufficiently small positive rational numbers is purely periodic. It is known that such real numbers have to be Pisot numbers which are units of the number field they generate. We complete known results due to Akiyama to characterize algebraic numbers of degree 3 that enjoy this property. This extends results previously obtained in the case of degree 2 by Schmidt, Hama and Imahashi. Let $\gamma(\beta)$ denote the supremum of the real numbers $c$ in $(0,1)$ such that all positive rational numbers less than $c$ have a purely periodic $\beta$-expansion. We prove that $\gamma(\beta)$ is irrational for a class of cubic Pisot units that contains the smallest Pisot number $\eta$. This result is motivated by the observation of Akiyama and Scheicher that $\gamma(\eta)=0.666 666 666 086 ...$ is surprisingly close to 2/3.
A negative result on algebraic specifications of the meadow of rational numbers
Bergstra, Jan A.; Bethke, Inge
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
46.17%
$\mathbb{Q}_0$ - the involutive meadow of the rational numbers - is the field of the rational numbers where the multiplicative inverse operation is made total by imposing $0^{-1}=0$. In this note, we prove that $\mathbb{Q}_0$ cannot be specified by the usual axioms for meadows augmented by a finite set of axioms of the form $(1+ \cdots +1+x^2)\cdot (1+ \cdots +1 +x^2)^{-1}=1$.; Comment: 5 pages, 2 tables
Representation, simplification and display of fractional powers of rational numbers in computer algebra
Rich, Albert D.; Stoutemyer, David R.
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
46.18%
Simplification of fractional powers of positive rational numbers and of sums, products and powers of such numbers is taught in beginning algebra. Such numbers can often be expressed in many ways, as this article discusses in some detail. Since they are such a restricted subset of algebraic numbers, it might seem that good simplification of them must already be implemented in all widely used computer algebra systems. However, the algorithm taught in beginning algebra uses integer factorization, which can consume unacceptable time for the large numbers that often arise within computer algebra. Therefore some systems apparently use various ad hoc techniques that can return an incorrect result because of not simplifying to 0 the difference between two equivalent such expressions. Even systems that avoid this flaw often do not return the same result for all equivalent such input forms, or return an unnecessarily bulky result that does not have any other compensating useful property. This article identifies some of these deficiencies, then describes the advantages and disadvantages of various alternative forms and how to overcome the deficiencies without costly integer factorization.; Comment: 23 pages, 1 figure, 4 tables
Free monoids and forests of rational numbers
Nathanson, Melvyn B.
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
46.17%
The Calkin-Wilf tree is an infinite binary tree whose vertices are the positive rational numbers. Each such number occurs in the tree exactly once and in the form $a/b$, where are $a$ and $b$ are relatively prime positive integers. This tree is associated with the matrices $L_1 = \left( \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right)$ and $R_1 = \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)$, which freely generate the monoid $SL_2(\mathbf{N}_0)$ of $2 \times 2$ matrices with determinant 1 and nonnegative integral coordinates. For other pairs of matrices $L_u$ and $R_v$ that freely generate submonoids of $GL_2(\mathbf{N}_0)$, there are forests of infinitely many rooted infinite binary trees that partition the set of positive rational numbers, and possess a remarkable symmetry property.; Comment: 10 pages
Properties of proper rational numbers
Zelator, Konstantine
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
46.28%
This short article is aimed at educators and teachers of mathematics.Its goal is simple and direct:to explore some of the basic/elementary properties of proper rational numbers.A proper rational number is a rational which is not an integer. A proper rational r can be written in standard form: r=c/b,where c and b are relatively prime integers; and with b greater than or equal to 2. There are seven theorems, one proposition, and one lemma; Lemma1, in this paper. Lemma1 is a very well known result, commonly known as Euclid's lemma.It is used repeatedly throughout this paper, and its proof can be found in reference[1]. Theorem4(i) gives precise conditions for the sum of two proper rationals to be an integer.Theorem5(a) gives exact conditions for the product to be an integer. Theorem7 states that there exist no two proper rationals both of whose sum and product are integers.This follows from Theorem6 which states that if two rational numbers have a sum being an integer; and a product being an integer;then these two rationals must both be in fact integers.; Comment: 6 pages
Beta-expansions of rational numbers in quadratic Pisot bases
Hejda, Tomáš; Steiner, Wolfgang
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
46.21%
We study rational numbers with purely periodic R\'enyi $\beta$-expansions. For bases $\beta$ satisfying $\beta^2=a\beta+b$ with $b$ dividing $a$, we give a necessary and sufficient condition for $\gamma(\beta)=1$, i.e., that all rational numbers $p/q\in[0,1)$ with $\gcd(q,b)=1$ have a purely periodic $\beta$-expansion. A simple algorithm for determining the value of $\gamma(\beta)$ for all quadratic Pisot numbers $\beta$ is described.; Comment: 12 pages, 3 figures, 2 tables
Fields of quantum reference frames based on different representations of rational numbers as states of qubit strings
Benioff, Paul
Tipo: Artigo de Revista Científica | 4,776 | 19,840 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-34 | longest | en | 0.595993 |
https://www.studypool.com/discuss/304885/can-someone-help-please-20?free | 1,481,360,557,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543030.95/warc/CC-MAIN-20161202170903-00154-ip-10-31-129-80.ec2.internal.warc.gz | 1,017,171,870 | 14,277 | Chemistry Tutor: None Selected Time limit: 1 Day
In an experiment 6.910 g of magnesium is reacted with 6.734 g of oxygen gas according to the equation:
Mg + O2 ==> MgO.
A. What is the limiting reactant?
B. What mass of the other reactant is in exess?
C. What mass of MgO is produced?
Dec 11th, 2014
Divide by molar mass to find # of moles of each. O2 is 32, Mg is 12, MgO is 28.
Balance the equation: 2Mg + O2 = 2MgO
Just offhand, the limiting reactant is O2 since there are less moles of it, and 6.734/32 moles of O2 are consumed in this reaction.
Let me go get a calculator...
6.734/32 = 0.2104 moles of O2 reacted to produce twice as many moles of MgO.
So multiply by 28 and then by 2 to get 11.78 g of MgO produced.
0.4208 moles of Mg is 5.051g, which reacted; the rest did not react.
Thus, the leftover Mg is 6.910-5.051 = 1.859g.
Dec 11th, 2014
...
Dec 11th, 2014
...
Dec 11th, 2014
Dec 10th, 2016
check_circle | 334 | 927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.912621 |
https://search.r-project.org/CRAN/refmans/CompRandFld/html/WlsInit.html | 1,624,032,462,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487637721.34/warc/CC-MAIN-20210618134943-20210618164943-00581.warc.gz | 461,142,885 | 2,681 | WlsInit {CompRandFld} R Documentation
## Computes Starting Values based on Weighted Least Squares
### Description
Subroutine called by FitComposite. The function returns opportune starting values for the composite-likelihood fitting procedure based on weigthed least squares.
### Usage
```WlsInit(coordx, coordy, coordt, corrmodel, data, distance, fcall, fixed,
grid, likelihood, margins, maxdist, maxtime, model, numblock,
param, parscale, paramrange, replicates, start, taper, tapsep,
threshold, type, varest, vartype, weighted, winconst, winstp)
```
### Arguments
`coordx` A numeric (d x 2)-matrix (where `d` is the number of points) assigning 2-dimensions of coordinates or a numeric vector assigning 1-dimension of coordinates. `coordy` A numeric vector assigning 1-dimension of coordinates; `coordy` is interpreted only if `coordx` is a numeric vector otherwise it will be ignored. `coordt` A numeric vector assigning 1-dimension of temporal coordinates. `corrmodel` String; the name of a correlation model, for the description. `data` A numeric vector or a (n x d)-matrix or (d x d x n)-matrix of observations. `distance` String; the name of the spatial distance. The default is `Eucl`, the euclidean distance. See the Section Details. `fcall` String; "fitting" to call the fitting procedure and "simulation" to call the simulation procedure. `fixed` A named list giving the values of the parameters that will be considered as known values. `grid` Logical; if `FALSE` (the default) the data are interpreted as a vector or a (n x d)-matrix, instead if `TRUE` then (d x d x n)-matrix is considered. `likelihood` String; the configuration of the composite likelihood. `margins` String; the type of the marginal distribution of the max-stable field. `maxdist` Numeric; an optional positive value indicating the maximum spatial distance considered in the composite-likelihood computation. `maxtime` Numeric; an optional positive value indicating the maximum temporal separation considered in the composite-likelihood computation. `model` String; the name of the model. Here the default is `NULL`. `numblock` Numeric; the observation size of the underlying random field. Only in case of max-stable random fields and in the simulation. `param` A numeric vector of parameter values required in the simulation procedure of random fields. `parscale` A numeric vector with scaling values for improving the maximisation routine. `paramrange` A numeric vector with the range of the parameter space. `replicates` Logical; if `FALSE` (the default) one spatial random field is considered, instead if `TRUE` the data are considered as iid replicates of a field. `start` A numeric vector with starting values. `taper` String; the name of the type of covariance matrix. It can be `Standard` (the default value) or `Tapering` for taperd covariance matrix. `tapsep` Numeric; an optional value indicating the separabe parameter in the space time quasi taper (see Details). `threshold` Numeric; a value indicating a threshold for the binary random field. `type` String; the type of estimation method. `varest` Logical; if `TRUE` the estimates' variances and standard errors are returned. `FALSE` is the default. `vartype` String; the type of estimation method for computing the estimate variances, see the Section Details. `weighted` Logical; if `TRUE` the likelihood objects are weighted, see `FitComposite`. `winconst` Numeric; a positive real value indicating the window size used from the sub-sampling method for the estimation of the parameters variances.
.
`winstp` Numeric; a value in (0,1] for defining the window step. See `FitComposite`.
### Author(s)
`FitComposite`, `WLeastSquare`. | 859 | 3,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-25 | latest | en | 0.68466 |
http://www.xpmath.com/forums/showpost.php?s=a3568d9aa663a4845efc1e842cec1805&p=24150&postcount=1 | 1,606,767,547,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00333.warc.gz | 156,742,626 | 3,516 | XP Math - Forums - View Single Post - Homework help
Thread: Homework help View Single Post
04-14-2008 #1 Tess213 Guest Posts: n/a Homework help The numbers in brackets are supposed to be in subscript so V(2) would be V subscript 2 and the number in front is just multiply, all the variables are vectors so they have a arrow on top. Determine if, W(1) = 2V(1) + 3V(2) , W(2) = V(2) + 2V(3) and W(3) = -V(1) - 3V(3) are coplanar Please help | 145 | 442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-50 | latest | en | 0.880141 |
https://www.daniweb.com/programming/software-development/threads/397230/magic-square-array-4x4-help | 1,722,887,993,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640454712.22/warc/CC-MAIN-20240805180114-20240805210114-00756.warc.gz | 578,846,578 | 17,705 | Problem 1:
Magic Squares:
An n x n array, that is filled with integers 1, 2, 3, …,n2 is a magic square if the sum of the elements in each row, in each column, and in the two diagonals is the same value and each value in the array is unique.
16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1
Write a program that reads in n2 integer values from the user and tests whether they form a magic square when put into an n x n array. You need to test two features:
• does each of the numbers 1 to n2 occur in the user input?
• when the numbers are put into a square, are the sums of the rows, columns and diagonals equal to each other?
Input: n and n2 integer numbers: if the user enters 3, the array has 3 rows and 3 columns and the integers in the array range from 1 to 9. As data validation, be sure the input is in the correct range and that each value is entered only once. (All elements in the array must be unique).
Output: a printout of the n x n array with each row of data on a separate line and a message of whether it is a magic square or not.
That is the problem of my assignment. What I have so far is:
``````#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
int n;
cout<< "Please enter an odd integer: ";
cin>>n;
int MagicSquare[n][n];
int newRow,
newCol;
// Set the indices for the middle of the bottom i
int i =0 ;
int j= n / 2;
// Fill each element of the array using the magic array
for ( int value = 1; value <= n*n; value++ )
{
MagicSquare[i][j] = value;
// Find the next cell, wrapping around if necessary.
newRow = (i + 1) % n;
newCol = (j + 1) % n;
// If the cell is empty, remember those indices for the
// next assignment.
if ( MagicSquare[newRow][newCol] == 0 )
{
i = newRow;
j = newCol;
}
elsen2 integer values
{
// The cell was full. Use the cell above the previous one.
i = (i - 1 + n) % n;
}
}
for(int x=0; x<n; x++)
{
for(int y=0; y<n; y++)
cout << MagicSquare[x][y]<<" ";
cout << endl;
}
}``````
The outcome of this is not like the numbers that the assignment have. Getting really frustrated! Please help if you know!!
Thank you!!!
Thank you. I need this program to be in c++ .... I was wondering if people know my errors, because I don't know my errors.. The outcome of my codes is junk numbers.
There is not one error mentioned in the original post other than a vague "output doesn't look right". Without any details, what can we say?
the magic constant is probably a standard 34 since the numbers are 1..16.
I just updated the solver and moved the solver code after the form (so it works better - form fields are loaded at time script is executed), and I added a function and button to call it which calculates the magic constant based on what is in the magic square.
the numbers of a magic square are supposed to add up to a magic constant. that magic constant is the sum of all the cells / 4.
you are welcome to use the code, since it is imperfect, unless it is for an assignment. in that case, use the properties list and design your own solution. I would have to see your code. and it's probably awful long like mine is, pages and pages with diagrams so you don't lose track of what code does what...
javascript is very close to C++ in most of its syntax. where you see it mostly differ is with creation of variables and variable types, definition of function, and how input and output is done. in my case, I am accessing form controls.
Thank you. I need this program to be in c++ .... I was wondering if people know my errors, because I don't know my errors.. The outcome of my codes is junk numbers.
1. Of course you don't know your errors. In fact you don't seem to have understood a thing about what the programm needs to do (if this is your piece of code)
2. The outcome is junk because the code is junk!
what was that? who are you aiming that comment at?
magic squares are a source of continual study and leisure. there is usually somebody coming up with new patterns for various sizes. there is a google group (newsgroup) dedicated to this subject.
completely "solving" for a magic square is not a fully defined task. you can find different magic square properties on different web sites (and then there's the newsgroup). I suppose this is finite, but you wonder sometimes.
the task of solving for one instance of a pattern as a check of validity would usually be adding up the numbers in that pattern and seeing if they add up to the magic constant.
I am not sure about whether this is a true property or not, but I found some interesting dependent symmetry patterns (2 patterns whose sums are not the magic constant but depend upon one another, but no matter what numbers you put in, the symmetries are true). I am not sure if the magic square folk would accept this however, but it was interesting to find. I did it for the challenge and because I thought I could do this with a program.
It is aimed to itzcarol as the quote implies.
If you want to see the javascript code, then do a view source on the web page.
I don't want to post it here because it's over 1000 lines long of tedious stuff.
http://jesusnjim.com/fun/4x4-magic-square-solver.html
If you want to see the javascript code, then do a view source on the web page.
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We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge. | 1,332 | 5,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-33 | latest | en | 0.738913 |
http://stackoverflow.com/questions/7521754/php-round-array-to-nearest-hundredth-and-seperate-evenly | 1,464,261,483,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049275835.98/warc/CC-MAIN-20160524002115-00243-ip-10-185-217-139.ec2.internal.warc.gz | 284,734,048 | 20,481 | # PHP Round array to nearest hundredth and seperate evenly
I am pulling 10 integers from a MySQL DB:
I want to be able to determine the largest value then round up to the nearest hundredth. With this I want to then be able to create a list of 10 numbers that are evenly separated down to zero.
In Example : { 232, 10, 0, 55, 130, 423, 102, 22, 98, 3 }
Take : 423 => Round up to 500
Then with the number 500 come back with equally separated numbers such as :
500 , 450 , 400, 350, 300, 250, 200, 150, 100, 50, 0
Thank you for the help in advance!
Chris
-
What exactly is the problem? – Kerrek SB Sep 22 '11 at 21:32
I have a bunch of integers and I need to properly format a graph. Therefor on the side I want to find the MAX then evenly show numbers on the side axis. So for example if I had up to 5000, it would be 5000 all the way down to 0. Evenly across 10 seperate numbers. – infinititech Sep 22 '11 at 21:35
I understand the setup. I just don't understand what you're having difficulties with. – Kerrek SB Sep 22 '11 at 21:36
Taking my maximum number, lets say 423 -> becomes 500. Now taking that 500 and echoing back 10 even numbers.. 500...450...400..350..etc – infinititech Sep 22 '11 at 21:37
This should do it:
``````<?php
\$array = array(232, 10, 0, 55, 130, 423, 102, 22, 98, 3);
\$max = max(\$array);
\$max_rounded = ceil(\$max/100)*100;
\$diff = (\$max_rounded/10);
while(\$max_rounded >= 0){
echo \$max_rounded.' ';
\$max_rounded -= \$diff;
}
?>
``````
-
Awesome! One thing, I see the 50 is set. How could we figure that out on the fly as in if we had a massive number 35,000. If it counted at 50s it would be huge. What if we can set it as no matter way we need 10 numbers always. – infinititech Sep 22 '11 at 21:41
This will only work if the number rounds to 500. The difference needs to be calculated. – Timothy Allyn Drake Sep 22 '11 at 21:41
Sorry, didn't payed attention to the 10 numbers part, let me see. – derp Sep 22 '11 at 21:47
I added : \$count = 10; \$diff = (\$max_rounded/\$count); – infinititech Sep 22 '11 at 21:59
Well I just edited it to account for a few things, hope it works for you. – derp Sep 22 '11 at 22:02
``````\$maximal = max(\$array);
function roundNearestHundredUp(\$number)
{
return ceil( \$number / 100 ) * 100;
}
\$counting = roundNearestHundredUp(\$maximal);
while(\$counting >= 0){
echo \$counting;
\$counting -= 10;
}
``````
-
Thank you for responding. I see that this will get us the max value awesome. However how could I take that number and then evenly segment (10) numbers from that so lets say 5000 seperated evenly all the way to 0. – infinititech Sep 22 '11 at 21:36
`for(\$i = 5000; \$i =< 0;\$i-5){ /* do stuff */ }` – genesis Sep 22 '11 at 21:41
something like `\$segments=range(\$max, 0, ceil(\$max/\$num_of_segments))` – Jonathan Kuhn Sep 22 '11 at 21:43
Looping is verbose. PHP's `range()` works the same, but more concisely. – John Kary Sep 22 '11 at 21:49
You can get the maximum number of your results set by using MySQL's `MAX()` function. No sense in returning results you don't need:
``````SELECT MAX(your_column) FROM some_table;
``````
Then use PHP to do the math on the number:
``````function getChoices(\$start, \$count)
{
\$start = ceil(\$start / 100) * 100;
return range(\$start, 0, (\$start / \$count));
}
\$choices = getChoices(460, 10);
``````
If you don't want to use MySQL's `MAX()`: `\$choices = getChoices(max(\$results), 10);`
-
Thank you for your response. I'm having a bit of trouble implementing this, will this count evenly? – infinititech Sep 22 '11 at 21:53
I don't know what you mean by "count evenly". The range() function creates a range between x --> y with the third argument being the number of evenly spaced increments. Read about `range()` here: php.net/manual/en/function.range.php . You can use the function by thinking of it like this: `getChoices(\$maximumNumberOfYourResults, \$howManyIncrementsYouWant)` – John Kary Sep 22 '11 at 22:20 | 1,220 | 3,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2016-22 | latest | en | 0.88193 |
https://www.physicsforums.com/threads/understanding-distances-derivatives-and-integrals-help.839024/ | 1,719,179,640,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864850.31/warc/CC-MAIN-20240623194302-20240623224302-00647.warc.gz | 821,324,251 | 20,178 | # Understanding distances, derivatives, and integrals help?
• Monkey D. Luffy
In summary, two trains, A and B, are moving at constant speeds of 50 km/h and 60 km/h, respectively, on two intersecting railroad tracks at station O. Train A leaves the station at 10AM and one hour later, train B passes through the station. The problem is to find the minimum distance between the two trains. To solve this, we set up equations for the position of each train, with x being the position of A and y being the position of B. We use the Pythagorean theorem to represent the distance between the trains, denoted as s. To find the minimum distance, we need to find the critical point where the derivative of s is equal to 0. This is found
Monkey D. Luffy
## Homework Statement
Two railroad tracks intersect at right angles at station O. At 10AM the train A, moving west with constant speed of 50 km/h, leaves the station O. One hour later train B, moving south with the constant speed of 60 km/h, passes through the station O. Find minimum distance between these trains
dx/dt
dy/dt
## The Attempt at a Solution
I got up to the point where i formulated two equations for the trains:
Xa= 50t
Yb= 60t -60
From here i was lost as to how to figure out the minimum distance because today before i attempted the problem, in a tutorial class we learned that minimum distance is (Yb-Xa) in this case and i do not understand how to do it that way. So i searched for a way to do it online and got this as an answer:
Let x be the position of train A, let y be the position of train B, and let O be the origin. West will be the positive x direction and south will be the positive y direction. We are given that (dx/dt)=50 and (dy/dt)=60. Train A starts at time t=0, so its position is x=50t. Train B starts at time t=1, so its position is y=60(t-1) when t is greater than or equal to 1. (y should equal 0 when t is 1 hour.) The formula we need to minimize is the distance between the trains given by the pythagorean theorem. We denote the distance between the trains s, and write s^2=x^2+y^2. The mimimum will be at a critical point where the derivative is equal to 0, so we differentiate with regards to t. 2s(ds/dt)=2x(dx/dt)+2y(dy/dt). We are looking for the point at which (ds/dt)=0, so we need to find: 0=2x(dx/dt)+2y(dy/dt). We can divide by 2 and not change a thing (0/2=0). 0=x(dx/dt)+y(dy/dt). Now we substitute our equations for x and y, and our numbers for (dx/dt) and (dy/dt). 0=(50t)(50)+(60t-60)(60). Simplify: 0=2500t+3600t-3600. Add the t terms: 0=6100t-3600. Add 3600 to both sides: 3600=6100t. This gives us t=0.59. Now we know the time at which the shortest distance was at, so now we find the distance. x=50(0.59)=29.51km and y=60(0.59-1)=-24.6km. Now we use s^2=(29.51)^2+(-26.6)^2=39.73km. The shortest distancw between the trains was 39.73km.
I understand a lot of it but i do not get when this person said the derivative is equal to 0, this is the equation:
2x(dx/dt)+2y(dy/dt)
why would it not just be Xa-Yb still? why do we multiply the X and Y equations with their derivatives? i am very confused by the notation used here for derivatives as well as the method and why these steps (after the one i just listed) was taken... Could anyone help me please? Also, i know i am asking a lot but could you also explain the integrals notation? since its a derivative taken backwards, i thought it would be simple enough except i don't get what it means when you "take the integral with respect to something else."
Thank you so much guys and the class is principles of physics so i thought this would be the place to create this forum.
Monkey D. Luffy said:
## Homework Statement
Two railroad tracks intersect at right angles at station O. At 10AM the train A, moving west with constant speed of 50 km/h, leaves the station O. One hour later train B, moving south with the constant speed of 60 km/h, passes through the station O. Find minimum distance between these trains
dx/dt
dy/dt
## The Attempt at a Solution
I got up to the point where i formulated two equations for the trains:
Xa= 50t
Yb= 60t -60
From here i was lost as to how to figure out the minimum distance because today before i attempted the problem, in a tutorial class we learned that minimum distance is (Yb-Xa) in this case and i do not understand how to do it that way. So i searched for a way to do it online and got this as an answer:
Let x be the position of train A, let y be the position of train B, and let O be the origin. West will be the positive x direction and south will be the positive y direction. We are given that (dx/dt)=50 and (dy/dt)=60. Train A starts at time t=0, so its position is x=50t. Train B starts at time t=1, so its position is y=60(t-1) when t is greater than or equal to 1. (y should equal 0 when t is 1 hour.) The formula we need to minimize is the distance between the trains given by the pythagorean theorem. We denote the distance between the trains s, and write s^2=x^2+y^2. The mimimum will be at a critical point where the derivative is equal to 0, so we differentiate with regards to t. 2s(ds/dt)=2x(dx/dt)+2y(dy/dt). We are looking for the point at which (ds/dt)=0, so we need to find: 0=2x(dx/dt)+2y(dy/dt). We can divide by 2 and not change a thing (0/2=0). 0=x(dx/dt)+y(dy/dt). Now we substitute our equations for x and y, and our numbers for (dx/dt) and (dy/dt). 0=(50t)(50)+(60t-60)(60). Simplify: 0=2500t+3600t-3600. Add the t terms: 0=6100t-3600. Add 3600 to both sides: 3600=6100t. This gives us t=0.59. Now we know the time at which the shortest distance was at, so now we find the distance. x=50(0.59)=29.51km and y=60(0.59-1)=-24.6km. Now we use s^2=(29.51)^2+(-26.6)^2=39.73km. The shortest distancw between the trains was 39.73km.
I understand a lot of it but i do not get when this person said the derivative is equal to 0, this is the equation:
2x(dx/dt)+2y(dy/dt)
why would it not just be Xa-Yb still? why do we multiply the X and Y equations with their derivatives? i am very confused by the notation used here for derivatives as well as the method and why these steps (after the one i just listed) was taken... Could anyone help me please? Also, i know i am asking a lot but could you also explain the integrals notation? since its a derivative taken backwards, i thought it would be simple enough except i don't get what it means when you "take the integral with respect to something else."
Thank you so much guys and the class is principles of physics so i thought this would be the place to create this forum.
Whoever did the analysis was doing it the hard way. It is easier to evaluate ##s^2## before taking the derivative: ##s^2 = (50 t)^2 + (60 t - 60)^2 = 6100\, t^2 - 7200\, t + 3600##. Now ## ds^2/dt = 12200 \, t - 7200##. This equals 0 when ##t = 36/61 \doteq 0.590## (hours). At that time the squared distance between the trains is obtained by substituting ##t = 36/61## into the expression for ##s^2##, and that simplifies to ##s^2_{\min} = 90 000/61## (km^2). The minimum distance is ##s_{\min} = \sqrt{90 000/61} \doteq 38.41## km. This is a bit different from the answer you gave above. The differences are likely due to excessive roundoff that the author above performed. It is better to keep more figures during the calculation, then round off the answer after all the work is finished. (In this case it makes quite a difference--- more than 1 km in the final answer.)
Ray Vickson said:
Whoever did the analysis was doing it the hard way. It is easier to evaluate ##s^2## before taking the derivative: ##s^2 = (50 t)^2 + (60 t - 60)^2 = 6100\, t^2 - 7200\, t + 3600##. Now ## ds^2/dt = 12200 \, t - 7200##. This equals 0 when ##t = 36/61 \doteq 0.590## (hours). At that time the squared distance between the trains is obtained by substituting ##t = 36/61## into the expression for ##s^2##, and that simplifies to ##s^2_{\min} = 90 000/61## (km^2). The minimum distance is ##s_{\min} = \sqrt{90 000/61} \doteq 38.41## km. This is a bit different from the answer you gave above. The differences are likely due to excessive roundoff that the author above performed. It is better to keep more figures during the calculation, then round off the answer after all the work is finished. (In this case it makes quite a difference--- more than 1 km in the final answer.)
OHHH i see! thanks i see your way of doing it and it makes a lot more sense now! my only remaining question is why we take the derivative and set it to 0, like what does setting it to 0 have to do with anything? i have done optimization before but it was usually for a dog running on a beach then swimming to get a stick in the water or something along those lines. in that situation, optimization gave us the fastest time to fetch the stick. why does it now give us "closest distance"?
CWatters said:
I'm not sure if that's where you are stuck but it should be covered in your course work. For example see..
http://www.themathpage.com/acalc/max.htm
https://www.mathsisfun.com/calculus/maxima-minima.html
ah see i understand derivatives pretty decently and your links brought back some ideas i remember practicing in Grade 12 Calculus, but now its integrals and the notation that confuses me. for example let's say we have -pdV/T what does that notation mean? (Im just using an example we had in class, one that confused me then as well because this is the first time i am doing integrals)
## 1. What is the importance of understanding distances, derivatives, and integrals?
Understanding distances, derivatives, and integrals is crucial in many fields of science, as it allows us to model and analyze various physical phenomena such as motion, growth, and change. It also provides the foundation for more advanced mathematical concepts and tools used in scientific research and problem-solving.
## 2. How do distances, derivatives, and integrals relate to each other?
Distances, derivatives, and integrals are all interconnected concepts that build upon each other. Distances are the basic building blocks of derivatives, which describe the rate of change of a function. Integrals, on the other hand, are the reverse process of derivatives and are used to calculate the total distance or area under a curve.
## 3. How can understanding these concepts help in real-world scenarios?
Understanding distances, derivatives, and integrals can be applied in various real-world situations, such as predicting the trajectory of a moving object, optimizing processes in engineering and economics, and analyzing data in various fields of science. It also helps in developing critical thinking and problem-solving skills.
## 4. What are some common misconceptions about distances, derivatives, and integrals?
One common misconception is that these concepts are only relevant in mathematics and have no practical applications. Another misconception is that they are only applicable in specific fields and not relevant to other areas of science. In reality, understanding these concepts can be beneficial in many different fields and industries.
## 5. How can one improve their understanding of distances, derivatives, and integrals?
To improve understanding, it is important to first have a strong foundation in basic mathematical concepts such as algebra and geometry. It is also helpful to practice solving problems and applying these concepts in various scenarios. Seeking out additional resources, such as textbooks or online tutorials, can also aid in improving understanding.
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745 | 3,172 | 12,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.94268 |
https://forum.daffodilvarsity.edu.bd/index.php/topic,40999.0.html | 1,606,272,518,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00529.warc.gz | 292,406,732 | 6,872 | Amazing Wizard Tricks You Can Do With Basic Math
### Author Topic: Amazing Wizard Tricks You Can Do With Basic Math (Read 741 times)
#### Masuma Parvin
• Sr. Member
• Posts: 323
##### Amazing Wizard Tricks You Can Do With Basic Math
« on: April 17, 2017, 12:18:38 PM »
Here are a few easy but impressive tricks that you can use to intimidate your friends and amaze your enemies.
1.Perfectly Sort Coins Without Looking at Them:
The Magic is :
Let's start with a trick a moron could do. Let's say we're using a handful of coins. First you close your eyes (or get blindfolded) and tell your spectators to shake up the coins and throw them on the table. All you need is one piece of information: how many of the coins are facing heads-up.Then, without ever peeking at the coins, you sort them into two piles, blindly flipping and shuffling them as if your hands were being guided by the spirit world. You magically wind up with the exact same number of heads in each pile. Every time.
The Math:
Let's say your spectators tell you that there are six heads-up coins in the pile. All you need to do is grab that number of coins and flip them. Just any six random coins. Take the ones you flipped and move them to their own pile, which we'll call Pile #1. The remaining coins are Pile #2. Both piles will contain the same number of heads.
#### Masuma Parvin
• Sr. Member
• Posts: 323
##### Re: Amazing Wizard Tricks You Can Do With Basic Math
« Reply #1 on: April 17, 2017, 12:26:44 PM »
Quickly Calculate the Weekday for Anyone's Birthday
The Math:
You don't have to be very smart to calculate the day of the week for any date -- not when eminent mathematician John Conway already devised a clever shortcut for that, which he (somewhat overdramatically) named the doomsday rule. Crack open a 2014 calendar and you'll see that 4/4, 6/6, 8/8, 10/10, and 12/12 all land on the same weekday: Friday. The same goes for other easy-to-remember dates like 5/9 and 9/5, 7/11 and 11/7, the last day of February, Pi Day, July 4, Halloween, and Michael Jackson's birthday (August 29, as you know). Again, all Fridays.
In 2013, those were all Thursdays:In 2012, they were all Wednesdays. Starting to notice a pattern? Since 2012 was a leap year, in 2011 the "doomsday" jumped to a Monday:And so on. So, let's say you're trying to find out the day of the week for July 9, 1987. First you have to figure out the doomsday for that year, by using an important world event as a point of reference, for example. As you probably learned in history class, Captain EO came out on Friday September 12, 1986, so from there it's easy to calculate that 9/5 was a Friday, too. If 1986's doomsday was a Friday, then 1987's was a Saturday, which makes July 4 of that year a Saturday as well. Therefore, a quick finger count tells us July 9, 1987 was ... a Thursday. | 754 | 2,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | latest | en | 0.922679 |
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The scale of a map is 1: 500 000
1. 1 cm on the map represents how many cm on land?
2. 1 cm on the map represents how many km on land?
3. The map distance between two towns is 2.5 cm. How many km are they apart on land?
4. If the two towns are 187.5 km apart, how far will they be apart on the map.
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1. 500 000cm
2. 500 000/100 000 = 5km
3. (2.5 x 500 000)/100 000 = 12.5km
4. (187.5 x 100 000)/500 000 = 37.5cm
by Diamond (42,256 points)
1 like 0 dislike | 275 | 764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-17 | latest | en | 0.820818 |
https://documen.tv/question/how-many-liters-of-c3h6o-are-present-in-a-sample-weighing-25-6-grams-23855938-96/ | 1,660,611,522,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00629.warc.gz | 227,800,351 | 15,810 | How many liters of C3H6O are present in a sample weighing 25.6 grams?
Question
How many liters of C3H6O are present in a sample weighing 25.6 grams?
in progress 0
6 months 2021-07-30T04:31:42+00:00 1 Answers 3 views 0
1. To Find :
Number of moles of C₃H₆O present in a sample weighing 25.6 grams.
Solution :
Molecular mass of C₃H₆O is :
M = (6×12) + (6×1) + (16×1) grams
M = 94 grams/mol
We know, number of moles of 25.6 grams of C₃H₆O is :
Hence, this is the required solution. | 178 | 489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.913423 |
https://programmer.ink/think/622463a3adcb7.html | 1,656,619,256,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103877410.46/warc/CC-MAIN-20220630183616-20220630213616-00383.warc.gz | 511,889,009 | 7,523 | # LeetCode question brushing record
Posted by skroks609 on Sun, 06 Mar 2022 08:32:08 +0100
# Sword finger Offer 30 Stack containing min function
Use the auxiliary value s2 to store the minimum value. If there is one smaller than s2, let the smaller value enter s2. When the minimum value in the original s1 comes out of the stack, judge that if it is equivalent to the s2 value, it means that the minimum value needs to be updated, then the s2 stack pushes out of the stack
```class MinStack {
public:
stack<int>s1;
stack<int>s2;
/** initialize your data structure here. */
MinStack() {
s2.push(INT_MAX);
}
void push(int x) {
s1.push(x);
if(x<=s2.top()){
s2.push(x);
}
}
void pop() {
if(s1.top()==s2.top()) s2.pop(); //s1 out of stack element is equal to the minimum value saved at the top of s2 stack
s1.pop();
}
int top() {
return s1.top();
}
int min() {
return s2.top(); //The stack top of s2 always keeps the minimum value
}
};
```
# Interview question 32 - I. print binary tree from top to bottom
The hierarchical traversal of binary tree 3, 9, 20, 15 and 7 join the team in turn
```class Solution {
public:
vector<int> levelOrder(TreeNode* root) {
vector<int>v;
if(root == NULL) return v;
queue<TreeNode*>q; //Initialize queue
q.push(root); //Root node join
while(!q.empty()){
TreeNode * node = q.front(); //Team first team
q.pop();
v.push_back(node->val);
if(node->left) q.push(node->left); //The left child of this node joins the team
if(node->right) q.push(node->right); //Right child joins the team
}
return v;
}
};
```
# Sword finger Offer 32 - II Print binary tree II from top to bottom
It is almost as like as two peas. It is more stratified output. How to divide into groups: root node joining, queue size() 1, root node children joining the team, queue size() 2, assuming two full tree, then down, size() 4. All the time, there are several nodes in each layer, i=q.size(), and then i -- output each node
```class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
// Hierarchical traversal problem
vector<vector<int>> v;
if(root==NULL){
return v;
}
queue<TreeNode *>q;
q.push(root);
while(!q.empty()){ //When the queue is not empty
vector<int>temp;
for(int i=q.size();i>0;i--) //Key layered approach
{
TreeNode * node = q.front();
q.pop();
temp.push_back(node->val);
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
v.push_back(temp); //Save the results of this layer
}
return v;
}
};
```
# Sword finger Offer 32 - III. print binary tree III from top to bottom![image-20220306113134809]
##### Method 1: the same as the previous question, but use a number to record the number of layers, and then the odd layers are normal, and the even layers can be reversed
```class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
// Hierarchical traversal problem
vector<vector<int>> v;
if(root==NULL){
return v;
}
queue<TreeNode *>q;
q.push(root);
int deep = 1;
while(!q.empty()){//When the queue is not empty
vector<int>temp;
for(int i=q.size();i>0;i--)//Key layered approach
{
TreeNode * node = q.front();
q.pop();
temp.push_back(node->val);
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
if(deep%2 ==0){
reverse(temp.begin(),temp.end());
}
v.push_back(temp);
deep++;
}
return v;
}
};
```
##### Double ended queues can also be used
```// Use double ended queue (even layer of tree): tail in (first left child node and then right child node) and head out; Odd layer of tree: head in (first right child node and then left child node) tail out)
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
if (root == nullptr) {
return {};
}
vector<vector<int>> vec;
deque<TreeNode*>dqe;
int level = 0;
dqe.push_back(root); // The root node is in layer 0 (even layer), so it enters from the end of the queue
while (!dqe.empty()) {
int level_nodes = dqe.size();
vec.push_back({});
if (level % 2 != 0) { // Odd layer: enter from the head of the queue (first the right child node and then the left child node), and exit from the end of the queue
while (level_nodes) {
TreeNode* p_node = dqe.back();
if (p_node->right != nullptr) dqe.push_front(p_node->right);
if (p_node->left != nullptr) dqe.push_front(p_node->left);
vec[level].push_back(p_node->val);
dqe.pop_back();
--level_nodes;
}
++level;
}
else { // Even layer: enter from the end of the queue (first the left child node and then the right child node), and exit from the head of the queue
while (level_nodes) {
TreeNode* p_node = dqe.front();
if (p_node->left != nullptr) dqe.push_back(p_node->left);
if (p_node->right != nullptr) dqe.push_back(p_node->right);
vec[level].push_back(p_node->val);
dqe.pop_front();
--level_nodes;
}
++level;
}
}
return vec;
}
};
```
# 88. Merge two ordered arrays
That is, give two non decreasing arrays, and then let you combine the first n bits of the first array and the first m bits of the second array into one array 1 (the length of array 1 is M+N)
Reverse order double pointer (the idea of double pointer is commonly used)
```class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int i = nums1.size() - 1;
m--; //The subscript starts at 0, so subtract 1 first
n--;
while (n >= 0) {
while (m >= 0 && nums1[m] > nums2[n]) {
nums1[i--]=nums1[m--];
}
nums1[i--]=nums2[n--];
}
}
```
# Sword finger Offer 39 A number that appears more than half the time in the array
This question is the original question that appeared in 408. The best thing to think of is to make statistics and sort, but this question uses the voting method
Because the mode must appear. Every time the mode appears, it will be + 1, not - 1. Finally, it must be greater than 0
```class Solution {
public:
int majorityElement(vector<int>& nums) {
int res = nums[0]; //Let nums[0] be the mode first
int count = 1;
for(int i = 1;i<nums.size();i++){
if(nums[i]==res){ //Equal value++
count++;
}else{ //Further judgment of unequal values
if(count==1){ //If there is only one time left, update the mode
res = nums[i];
}else{
count--; //Number of occurrences--
}
}
}
return res;
}
};
```
The writing method of the boss: this question has particularity, that is, it is not necessary to judge whether it is a mode. It is best to add a judgment
```class Solution {
public:
int majorityElement(vector<int>& nums) {
int x = 0, votes = 0, count = 0;
for(int num : nums){
if(votes == 0) x = num;
votes += num == x ? 1 : -1;
}
// Verify that x is mode
for(int num : nums)
if(num == x) count++;
return count > nums.size() / 2 ? x : 0; // Returns 0 when there is no mode
}
};
```
# Sword finger Offer 40 Minimum number of k
Four solutions to spike TopK (fast row / heap / binary search tree / count sort) ❤️ - Minimum k number - LeetCode (LeetCode CN. Com)
Sword finger Offer 40 Minimum number of k (array division based on quick sorting, clear illustration) - minimum number of k - LeetCode (leetcode-cn.com)
1. Fast selection algorithm
Optimized quick sort, because you only need to find the smallest k, and don't care about their order
As with the fast platoon, it is used frequently
```class Solution {
public:
vector<int> getLeastNumbers(vector<int>& arr, int k) {
if(k>=arr.size()) return arr;
return quick_select(arr, k, 0, arr.size() - 1);
}
vector<int>quick_select(vector<int>&arr,int k,int low,int high){
int i = low;
int j = high;
int pivot = arr[i];//Select pivot
while(i<j){
while(i<j && arr[j]>=pivot) --j;
arr[i] = arr[j] ;
while(i<j && arr[i]<=pivot) ++i;
arr[j] = arr[i];
}
arr[i] = pivot; //Pivot return
// Then judge whether to continue. If I > k, the range is large and the range is reduced
//If I < K represents that our scope is small, it needs to be divided again
if (i > k) return quick_select(arr, k, low, i - 1);
if (i < k) return quick_select(arr, k, i + 1, high);
vector<int> res;
return res;
}
};
```
# 414. The third largest number
##### Method 1: use the set (default sorting) and save only 3 numbers
```class Solution {
public:
int thirdMax(vector<int> &nums) {
set<int> s;
for (int num : nums) {
s.insert(num);
if (s.size() > 3) { //If there are more than 3 elements, delete the smallest one
s.erase(s.begin());
}
}
return s.size() == 3 ? *s.begin() : *s.rbegin(); //If the number is less than 3, the maximum number will be returned (in the case of example 2)
}
};
```
##### Solution 2: with three pointers, you can find it with only one round of scanning
```class Solution {
public:
int thirdMax(vector<int> &nums) {
int *a = nullptr, *b = nullptr, *c = nullptr;
for (int &num : nums) {
if (a == nullptr || num > *a) {
c = b;
b = a;
a = #
} else if (*a > num && (b == nullptr || num > *b)) {
c = b;
b = #
} else if (b != nullptr && *b > num && (c == nullptr || num > *c)) {
c = #
}
}
return c == nullptr ? *a : *c;
}
};
```
# 215. The kth largest element in the array (this condition is more stringent than the above)
##### Solution 2: quick selection
```class Solution {
public:
int quickSelect(vector<int>& a, int l, int r, int index) {
int q = randomPartition(a, l, r);
if (q == index) {
return a[q];
} else {
return q < index ? quickSelect(a, q + 1, r, index) : quickSelect(a, l, q - 1, index);
}
}
inline int randomPartition(vector<int>& a, int l, int r) {
int i = rand() % (r - l + 1) + l;//Introduce randomization The time complexity of fast scheduling depends on the partition
swap(a[i], a[r]);
return partition(a, l, r);
}
inline int partition(vector<int>& a, int l, int r) {//divide
int x = a[r], i = l - 1;
for (int j = l; j < r; ++j) {
if (a[j] <= x) {
swap(a[++i], a[j]);
}
}
swap(a[i + 1], a[r]);
return i + 1;
}
int findKthLargest(vector<int>& nums, int k) {
srand(time(0));
return quickSelect(nums, 0, nums.size() - 1, nums.size() - k);
}
};
```
##### Solution 3: using large root pile
```class Solution {
public:
void maxHeapify(vector<int>& a, int i, int heapSize) {
int l = i * 2 + 1, r = i * 2 + 2, largest = i;
if (l < heapSize && a[l] > a[largest]) {
largest = l;
}
if (r < heapSize && a[r] > a[largest]) {
largest = r;
}
if (largest != i) {
swap(a[i], a[largest]);
maxHeapify(a, largest, heapSize);
}
}
void buildMaxHeap(vector<int>& a, int heapSize) { //Reactor building operation
for (int i = heapSize / 2; i >= 0; --i) {
maxHeapify(a, i, heapSize);
}
}
int findKthLargest(vector<int>& nums, int k) {
int heapSize = nums.size();
buildMaxHeap(nums, heapSize);
for (int i = nums.size() - 1; i >= nums.size() - k + 1; --i) {
swap(nums[0], nums[i]);
--heapSize;
maxHeapify(nums, 0, heapSize);
}
return nums[0];
}
};
```
Topics: C++ leetcode | 3,100 | 10,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-27 | longest | en | 0.674913 |
https://blog.vedicmathsindia.org/amazing-benefits-of-rapid-mental-ma/ | 1,571,195,471,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986661296.12/warc/CC-MAIN-20191016014439-20191016041939-00231.warc.gz | 410,541,964 | 31,407 | SHARE
Today you can define mental math in various different ways. Some would say, memorizing times table and remembering the solutions can form the part of mental mathematics. Some would say ability to perform simple calculations in your head can be mental mathematics. The web dictionary defines mental mathematics as “Computing an exact answer without using pencil and paper or other physical aids.” Today there are five methods available to learn and practice mental mathematics.
Let’s begin with the first one called ‘Learning by Heart’ or better known as the rote memorizing method where your teachers ask you to mug up boring multiplication tables. It not only kills the interest of the child in mathematics but also makes sure that he develops hatred towards the subject for the rest of the years he studies it. This system gives its ardent devotee some degree of success initially as he is able to answer easy problems but then when the supposedly bigger application problems come the steam is almost over. The second one gives you a good degree of success and I would highly recommend it to the younger lot out there. It hails from China and is popular by the name of The Abacus (also known as the Soroban in Japan). An abacus is a calculating tool, often constructed as a wooden frame with beads sliding on wires. With the use of this tool one can perform calculations relating to addition, subtraction, multiplication and division with ease. Gradually one practices with the tool in one’s hand and later on when experienced he learns to do it without the tool. This tool is then fitted into the mind mentally and he can then add, subtract multiply and divide in seconds. This tool also enhances a child’s concentration levels. The main drawback of this system is that it focuses only on the 4 mathematical operations. Concepts beyond these operations such as Algebra, Square Roots, Cubes, Squares, Calculus, and Geometry etc cannot be solved using it at all. Also one needs a longer time to be able to fully get a grasp of the system hence you see courses in the abacus stretching to over 2 years which leads the child to boredom and then quitting from the course. Another Chinese system mainly collected from the book The Nine Chapters on the Mathematical Art lays out an approach to mathematics that centers on finding the most general methods of solving problems. Entries in the book usually take the form of a statement of a problem, followed by the statement of the solution, and an explanation of the procedure that led to the solution. The methods explained in this system can hardly be termed mental and they lack speed to top it all. The Chinese were definitely the most advanced of the civilization thanks to the Yangtze and Yellow Rivers but if I were to choose out of the two methods given by this culture It would be the abacus. If wars have a 99.99% downside, sometimes they can have an upside too for they give birth to stories of hope and creativity. The next mental math system was developed during the Second World War in the Nazi Concentration Camp by a Ukrainian Mathematician Jakow Trachtenberg to keep his mind occupied. What resulted is now known as the Trachtenberg Speed System of Mathematics and consists of Rapid Mental Methods of doing Mathematics. The system consists of a number of readily memorized patterns that allow one to perform arithmetic computations very quickly. It has wider applications than the Abacus and apart from the four basic operation methods it covers Squares and Square Roots. The method focuses mostly on Multiplication and it even gives patterns for multiplication by particular number say 5,6,7 and even 11 and 12. It then gives a general method for rapid multiplication and a special two finger method. After practicing the method myself I realized that the multiplication was a very applicable mental method but the other methods covered to solve division and square roots were not very friendly and were impossible to be done mentally. I was in search of a much better wholesome method where I could easily perform other operations also. Another drawback of this system was that it too like the abacus failed to have a wider scope i.e to encompass other fields like Algebra, Calculus, Trignometry, Cube Roots etc A Recommendation by a friend of mine from America introduced me to what is known as the Kumon Math Method. It was founded by a Japanese educator Toru Kumon in 1950s and as of 2007 over 4 million children were studying under the Kumon Method in over 43 different countries. Students do not work together as a class but progress through the curriculum at their own pace, moving on to the next level when they have achieved mastery of the previous level. This sometimes involves repeating the same set of worksheets until the student achieves a satisfactory score within a specified time limit. In North American Kumon Centers, the mathematics program starts with very basic skills, such as pattern recognition and counting, and progresses to increasingly challenging subjects, such as calculus, probability and statistics. The Kumon Method does not cover geometry as a separate topic but provides sufficient geometry practice to meet the prerequisites for trigonometry, which is covered within the Kumon math program.
Kumon Worksheets: Not Mental at all ; Repeating till you get Right.
I was much impressed with the glamour around Kumon but a glimpse of its curriculum deeply disappointed me. It is not mental at all. It does not offer any special methods to do mathematics and one does not improve one’s speed by doing Kumon Math. There is a set curriculum of worksheets which one does till one achieves mastery in the subject. So say for example a sheet on Divison- one would continue to do division by the conventional method till he gets a satisfactory score and then he moves on to a higher level. This certainly doesn’t make division any faster and the process is certainly not mental. A deep thought on the reason of its tremendous popularity in America led me to conclude was the lack of a franchisee business model of the abacus and the Trachtenberg speed system in the 1950s. The franchisee model was essential in taking the course from country to country. This is where Toru Kumon thrived. Dissapointed with other cultures in the world, my search made me look in my own Indian culture. What I found astonished and amazed me so much that I fell in love with the system and started coaching neighbourhood students in it. This is easily the World’s Fastest Mental Mathematics System called High Speed Vedic Mathematics. It has its roots in Ancient Indian Scriptures called the Vedas meaning ‘the fountain head of knowledge’. With it not only you can add, subtract, multiply or divide which is the limiting factor of the abacus but you can also solve complex mathematics such as algebra, geometry, Calculus, and Trigonometry. Some of the most advanced, complex and arduous problems can be solved using the Vedic Maths method with extreme ease.And all this with just 16 word formulas written in Sanskrit. High Speed Vedic Mathematics was founded by Swami Sri Bharati Krishna Tirthaji Maharaja who was the Sankaracharya (Monk of the Highest Order) of Govardhan Matha in Puri between 1911 and 1918. They are called “Vedic” as because the sutras are contained in the Atharva Veda – a branch of mathematics and engineering in the Ancient Indian Scriptures. High Speed Vedic Mathematics is far more systematic, simplified and unified than the conventional system. It is a mental tool for calculation that encourages the development and use of intuition and innovation, while giving the student a lot of flexibility, fun and satisfaction . For your child, it means giving them a competitive edge, a way to optimize their performance and gives them an edge in mathematics and logic that will help them to shine in the classroom and beyond. Therefore it’s direct and easy to implement in schools – a reason behind its enormous popularity among academicians and students. It complements the Mathematics curriculum conventionally taught in schools by acting as a powerful checking tool and goes to save precious time in examinations. The Trachtenberg Method is often compared to Vedic Mathematics. Infact even some of the multiplication methods are strikingly similar. The Trachtenberg system comes the closest to the Vedic System in comparison and ease of the methods. But the ease and mental solvability of the other method especially division, square roots, cube roots, Algebraic Equations, Trigonometry, Calculus etc clearly gives the Vedic System an edge. Even NASA is said to be using some of this methods applications in the field of artificial intelligence. There are just 16 Vedic Math sutras or word formulas which one needs to practice in order to be efficient in Vedic Math system. Sutras or Word Math Formulas such as the Vertically and Crosswise, All from Nine and Last from ten helps to solve complex problems with ease and also a single formula can be applied in two or more fields at the same time. The Vertically and Crosswise formula is one such gem by which one can multiply, find squares, solve simultaneous equations and find the determinant of a matrix all at the same time. If either of these methods is learned at an early age, a student aged 14 can perform lightening fast calculations easily during his examinations and ace through them.
Vedic Mathematics
is fast gaining popularity in this millennium. It is being considered as the only mental math system suited for a child as it helps to develop his numerical as well as mental abilities. The methods are new and practical and teach only Mental Rapid Mathematics. The system does not focus on learning by repetition as in the Kumon Method. The system focuses on improving intelligence by teaching fundamentals and alternate methods. The purpose is not limited to improving performance in the school or tests, but on providing a broader outlook resulting in improved mathematical intelligence and mental sharpness. To know more about the Vedic Mathematics Sutras – The World’s Fastest Mental Math System you can visit https://vedicmathsindia.org
This Article is by Gaurav Tekriwal,, The President of the Vedic Maths Forum India who has been conducting High Speed Vedic Math Workshops for the last five years and has trained over seven thousand students across the world in the field. He is the author of the best selling DVD on the subject which contains over 10 hours on the subject. He is an expert in the field and revolutionizes the way children learn math.
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Previous articleThe Vedic Numerical Codified Language
Gaurav Tekriwal is the founder President of the Vedic Maths Forum India. Through television programs, workshops, DVDs, and Books he has taken the Vedic Maths System to over 4 million students in India, South Africa, United States, Australia, UAE, Ghana, and Colombia.
#### 6 COMMENTS
1. The simplest benefit of Vedic Mathematics is that it enables us to carry out calculations mentally (though the methods can also be written down). There are many other advantages in using a flexible system. Students can discover their very own methods;, which leads to more imaginative, interested and intelligent students. Research is being carried out in many areas. Researches include studying Vedic Maths effects on children; developing powerful applications of the Vedic Sutras in different fields such as geometry, calculus, computing etc.
But the real charm and usefulness of Vedic Mathematics can be fully treasured only after practicing the system actually.
2. It is rightly said
Vedic Mathematics is fast gaining popularity in this millennium.It is a noteworty effort from some God gifted genius like Gaurav Tekriwal,, The President of the Vedic Maths Forum India.His hardworking effort has helped thousands of people all over the world and to learn the techniques of Quick mental math.Thanks Gaurav and wish you very best for your upcoming journey
3. Its true that vedic maths is one of faster method to solve the maths calculation. It is very useful for competition exam. it is a key of success.
Sanjay Sharma
4. I was looking for a different method of teaching math to my kids other than Abacus. And this article is very helpful for me. Its creative, its effective and fast. | 2,472 | 12,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | longest | en | 0.956108 |
https://www.redhotpawn.com/forum/posers-and-puzzles/net-riddle.23926 | 1,561,021,877,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999200.89/warc/CC-MAIN-20190620085246-20190620111246-00202.warc.gz | 871,940,836 | 9,850 | # Net Riddle
Alcra
Posers and Puzzles 17 May '05 08:51
1. Alcra
Lazy Sod
17 May '05 08:51
Anyone here been to http://nr.wireplay.co.uk/modules.php?name=Net_Riddle
It is a riddle "game" with a cash prize. At the moment, seems most of the planet is stuck on #48. Thought we here at RHP can show them how its done!
If you need help getting to #48 (and some of the puzzles are plain stupid IMHO), go to
http://www.wizardgroup.net/home/
2. Daemon Sin
I'm A Mighty Pirate™
17 May '05 12:24
Originally posted by Alcra
Anyone here been to http://nr.wireplay.co.uk/modules.php?name=Net_Riddle
It is a riddle "game" with a cash prize. At the moment, seems most of the planet is stuck on #48. Thought we here at RHP can show them how its done!
If you need help getting to #48 (and some of the puzzles are plain stupid IMHO), go to
http://www.wizardgroup.net/home/
Looks like a laugh, I'm working my way through it now!
3. Daemon Sin
I'm A Mighty Pirate™
17 May '05 14:10
#48 - The one everyone's stuck on!
A) I am white's first move in the game of kings I have how many choices?
B) Number of ways of winnings 0s and Xs
C) "As I was going to St. Ives, I met a man with seven wives. Every wife had seven sacks, every sack had seven cats, every cat had seven kitts. Kitts, cats, sacks, wives, how many were going to St. Ives?"
D) How many rings not owned by elves?
U = (CD) + (AB)
Please use BODMAS, also spell your answer, for example 2 would become two. And lastly good luck!
******************
There's probably some stupid trick in the question that I haven;t spotted as per usual, but.... fas as I see:
A = 10 possible moves
B = Either 1 (Get 3 in a row) or 8 (possible ways of getting 3 in a row)
C = 1 (Only I am going to St. Ives)
D = 13 Rings out of the 16... I hate LOTR and had to ask many people for that answer.
Where am I going wrong?!
4. Palynka
Upward Spiral
17 May '05 14:30
Originally posted by Daemon Sin
#48 - The one everyone's stuck on!
A) I am white's first move in the game of kings I have how many choices?
B) Number of ways of winnings 0s and Xs
C) "As I was going to St. Ives, I met a man with seven wives. Every wife had seven sacks, every sack had seven cats, every cat had seven kitts. Kitts, cats, sacks, wives, how many were going to St. ...[text shortened]... of the 16... I hate LOTR and had to ask many people for that answer.
Where am I going wrong?!
I think it's something related to the letters
U = CD + AB
You(U) see(C) the(D) plus? (+) ay? (A) be? (B)
Placebo? (+ab)?
U = the answer. Why U? Why the CD first? Why is the strange grammar?
5. Daemon Sin
I'm A Mighty Pirate™
17 May '05 14:51
Originally posted by Palynka
I think it's something related to the letters
U = CD + AB
You(U) see(C) the(D) plus? (+) ay? (A) be? (B)
Placebo? (+ab)?
U = the answer. Why U? Why the CD first? Why is the strange grammar?
All very good points!
Especially the order of the equation.... =/
6. Alcra
Lazy Sod
17 May '05 14:521 edit
Originally posted by Daemon Sin
#48 - The one everyone's stuck on!
A) I am white's first move in the game of kings I have how many choices?
B) Number of ways of winnings 0s and Xs
C) "As I was going to St. Ives, I met a man with seven wives. Every wife had ...[text shortened]... ad to ask many people for that answer.
Where am I going wrong?!
Nice thinking, except A <> 10, A= 20
each pawn can move 1 of two spaces = 16
+ each knight can move to two squares = 4
16 + 4 =20
However, some people have said that white can also resign in move 1, so maybe 21?
EDIT: Tidied up a bit. First go was rushed 😳
7. Alcra
Lazy Sod
17 May '05 14:53
Originally posted by Palynka
I think it's something related to the letters
U = CD + AB
You(U) see(C) the(D) plus? (+) ay? (A) be? (B)
Placebo? (+ab)?
U = the answer. Why U? Why the CD first? Why is the strange grammar?
Nice thinking. I knew if i posted here, we would get some different thoughts.
8. Alcra
Lazy Sod
17 May '05 14:55
Everyone thinks D is LOTR? Why?
Can anyone think of anything that is not LOTR here, might be interesting?
Also, a "clue" was given that states the answer is not numeric? This may help or hinder.
9. Daemon Sin
I'm A Mighty Pirate™
17 May '05 15:01
Originally posted by Alcra
Everyone thinks D is LOTR? Why?
Can anyone think of anything that is not LOTR here, might be interesting?
Also, a "clue" was given that states the answer is not numeric? This may help or hinder.
LOTR is the most generic "Elves" and "Rings" connection. I can't think of any other mainstream connection that everyone else would know...?
10. Daemon Sin
I'm A Mighty Pirate™
17 May '05 17:12
Originally posted by Daemon Sin
#48 - The one everyone's stuck on!
A) I am white's first move in the game of kings I have how many choices?
B) Number of ways of winnings 0s and Xs
C) "As I was going to St. Ives, I met a man with seven wives. Every wife had seven sacks, every sack had seven cats, every cat had seven kitts. Kitts, cats, sacks, wives, how many were going to St. ...[text shortened]... of the 16... I hate LOTR and had to ask many people for that answer.
Where am I going wrong?!
Just to completely throw a spanner in the works...
Is the 'Game of Kings' definately chess?! Thinking about it, I'm sure Backgammon has been referred to as the 'Game of Kings' by my friends before and it could even possibly be about Checkers/Draughts =/
11. AThousandYoung
17 May '05 18:313 edits
Originally posted by Daemon Sin
#48 - The one everyone's stuck on!
A) I am white's first move in the game of kings I have how many choices?
B) Number of ways of winnings 0s and Xs
C) "As I was going to St. Ives, I met a man with seven wives. Every wife had ...[text shortened]... ad to ask many people for that answer.
Where am I going wrong?!
A) Does checkers have a White side? I thought it was Red and Black. If "The Game of Kings" is chess, there are 8x2 Pawn moves = 16, and four possible Knight moves = 20 total. A = 20. Does Backgammon have a White side?
B) 8. Two diagonals, three horizontals, three verticals. Well maybe we should double that because either O or X can win. B = 16 or 8.
C) When I heard this puzzle, I was told no kits, cats, sacks or wives were going to St. Ives; I am going to St Ives, and I met them, so supposedly they must be going the other way. So C=0. However it's possible that they were just going slower than I was, so C would equal 7 wives + 7^2 sacks +7^3 cats + 7^4 kits.
D) This is almost certainly LOTR. Originally, there were the Three, which the Elves always had; the Seven, which the Dwarves had; and the Nine, which the Men had. Sauron made the One Ring to control the others (I guess through the Law of Sympathy, that which seems alike is alike, and so there was a magical connection between the One and the others?). At the time of the LOTR story, the Nine are held by the Ringwraiths; the Elves still have the Three; the Seven were lost and/or destroyed by dragonfire; and of course Frodo has the One. One could interpret this question any number of ways, and the best interpretations would include how many of the Seven were destroyed. However the simple answer would be 1+7+9 = 17 Rings not held by Elves.
Woops! The Ring of Fire is held by Gandalf, so that makes 18. And Elrond is only Half Elven, and he has the Ring of Air...does that mean 18.5 minus any of the Seven that were destroyed?
12. Daemon Sin
I'm A Mighty Pirate™
20 May '05 12:21
There's other riddle/puzzle/quiz games on the rest of the site too, you know. Quite good if you think you're a film boffin
13. 20 May '05 15:35
for the chess game, there are 21 moves!!!! You can resign, you dolts!
14. Daemon Sin
I'm A Mighty Pirate™
20 May '05 16:43
for the chess game, there are 21 moves!!!! You can resign, you dolts!
Check out Alcra's 4th post...
15. The Plumber
Leak-Proof
20 May '05 17:321 edit
Originally posted by Alcra
Anyone here been to http://nr.wireplay.co.uk/modules.php?name=Net_Riddle
It is a riddle "game" with a cash prize. At the moment, seems most of the planet is stuck on #48. Thought we here at RHP can show them how its done!
If you need ...[text shortened]... s are plain stupid IMHO), go to
http://www.wizardgroup.net/home/ | 2,372 | 8,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.954669 |
https://fifthgradewinterhaven.org/2017/03/12/3-6-march-13-17-2017/ | 1,628,123,488,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155268.80/warc/CC-MAIN-20210805000836-20210805030836-00031.warc.gz | 266,491,138 | 21,537 | ## 3.6: March 13-17, 2017
#### Aesop
Important Dates
March
17: Garden Work Party, Food Forest, 3:00-5:00 p.m..
24: Garden Work Party, Native Garden, 3:00-5:00 p.m..
Many hands make light work
27-31: Spring Break- No School
April
14:
MAC Attack Run (Sponsor sheets due April 11)
21:
Earth Day Celebration and Performance, 10:00 a.m. (Morning performance only)
State Testing Schedule for Fifth Grade
May 8, 2017: Science
Week of May 15, 2017: Math
Week of May 22, 2017: Reading
General testing and opting Out Information
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This week in…
さんす(Math)
Grade 5 Module 4: Multiplication and Division of Fractions and Decimal Fractions
Module 4: Topic E: Multiplication of a fraction by a fraction. Supports Lessons 13-20
2. EngageNY
Lesson 13: Objective: Multiply unit fractions by unit fractions. Homework due Tuesday
Lesson 14: Objective: Multiply unit fractions by non-unit fractions.
Homework due Wednesday
Lesson 15: Objective: Multiply non-unit fractions by non-unit fractions.Homework due Thursday
Lesson 16: Objective: Solve word problems using tape diagrams and fraction-by-fraction multiplication. No Math Homework due Friday
3. Systems Thinking Project Teams: Using math (measurement) to increase understanding of the real world and to understand and reduce waste. Ongoing. Telling stories by the numbers.
4. Math Games: Math 24- Multiplying, Dividing, Adding, and Subtracting Fractions; order of operations p.e.m.d.a.s.
Different Mirror
• Chapter 8: From China to Gold Mountain
• Chapter 9: Dealing with the Indians
• Chapter 10: The Japanese and “Money Trees”
• Chapter 11: Jews are Pushed from Russia
Genre- Nonfiction: Informational
Socratic Seminar:
Classroom Law Project
Trial # 2: Criminal Trial
Systems Thinking:
The Case of the Mummified Pigs
Writing
Haiku
-haiku, haibun, and tanka
-history of haiku in Japan
Publish Haiku or Haibun
Earth Day Performance
The Lorax Returns Script due on Friday March 24
Whole Class Activity:
-Students will read our scripts rough draft and develop lines for characters
Small Group work:
-Students will propose essential systems thinking and permaculture principles into the story.
What would Dr. Seuss want us to do? Is the problem the solution?
• Developing a heart or core message then communicate through visual mathematics
• Use Fibonacci pattern, exponential growth snowballs, and diversity to assure resiliency
• Stock and Flow models, BOTG’s, connection circles, causal loops, and mental models
• Integrate the wisdom of Buckminster Fuller, Rachel Carson, Albert Einstein, and George Washington Carver
Dialogue and grammar (pages 13-15 only)
Dialogues in hiragana (dialogue 5 review, dialogue 6 review, dialogue 7 new)
△五 Review
△六 Review
△七 New
Vocabulary
1. koko de – here at this location
2. resutoran – restaurant
3. resutoran de – at the restaurant
4. nani – what?
5. sake – rice wine
6. o-sake – polite way to say “rice wine”
7. hoshiku arimasen – I don’t want
8. hoshii desu – I want
9. Nanbantei Resutoran – name of a famous restaurant in Japan | 810 | 3,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-31 | latest | en | 0.808125 |
https://astarmathsandphysics.com/a-level-maths-notes/d2/3394-the-travelling-salesman-problem-finding-n-upper-bound.html?tmpl=component&print=1&page= | 1,532,035,898,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591296.46/warc/CC-MAIN-20180719203515-20180719223515-00264.warc.gz | 597,435,059 | 2,542 | ## The Travelling Salesman Problem – Finding n Upper Bound
The 'travelling salesman problem' is, for a given network of vertices connected by arcs of given lengths, to find the minimum distance that visits each vertex once and returns to the starting vertex.
For a small network, it is possible to find the minimum distance by examining every possible route, but as the size of the network grows, so does the number of routes that must be examined. It is helpful to find upper and lower bounds for the minimum distance.
The upper bound is easily found. We can construct the minimum spanning tree, using Prim's algorithm for example. The minimum spanning tree is the network of least length that connects every vertex. Each vertex can be visited by tracing forwards and backwards, in some order, each edge in the minimum spanning tree. The length of the path travelled is then twice the total length of the minimum spanning tree.
Suppose we have the network below.
The minimum spanning tree is shown below.
It has length 11, so the upper bound for the travelling salesman problem for this network is 2*11=22. | 227 | 1,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.916832 |
https://thebestpaperwriters.com/answer-in-microeconomics-for-isaac-mensah-179255/ | 1,695,468,622,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.7/warc/CC-MAIN-20230923094750-20230923124750-00130.warc.gz | 632,230,876 | 13,050 | # Answer in Microeconomics for isaac mensah #179255
Given utility maximization problem U= Q1Q2 subject to 10Q1 +2Q2=240 a. Derive the Lagrange function b. Derive the first order conditions c. Use Cramer’s rule to find the critical values of Q1, Q2 and �
i.) Deriving the Lagrange function:
### “Z = Q1Q2+ u03bb(240-10Q1-2Q2)”
ii.) first-order conditions:
“ZQ1 = Q2u2212 u03bb10 = 0\nnn ZQ2 = Q1u2212 u03bb 2 = 0 \nnnZu03bb = 240 u2212 10Q1 u22122 Q2 =0.”
“Zlambda=240-10Q1-2Q2=0”
“ZQ1=Q2- lambda10=0”
“ZQ2=Q1-lambda2=0”
iii.) “begin{bmatrix}n 0 & -10 & -2 \n -10 & 0 &1 \n-2 & 1 & 0nend{bmatrix}” “begin{bmatrix}n lambda \n Q1 \ Q2nend{bmatrix}” = “begin{bmatrix}n -240 \n 0 \ 0nend{bmatrix}”
Q1M“=frac{240}{2[-10]}= -12”
Q2M“=frac{240}{2[-2]}=-60”
“lambda=frac{240}{2[-10.-2]}=6”
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Customers referred by a friend | 1,042 | 3,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-40 | latest | en | 0.813281 |
https://kratos-wiki.cimne.upc.edu/index.php?title=DEM_Application&direction=next&oldid=8015 | 1,660,647,794,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00109.warc.gz | 334,819,817 | 12,170 | # DEM Application
The DEM Kratos Team
## Theory
This application solve the the equations.... Mathematical approach to the problems.
Nothing numerical
### Integration Schemes
Forward Euler Scheme
### Contact Laws
Concept of indentation HMD, LSD
##### Normal Force Laws
###### Linear Repulsive Force
The most simple representation of a repulsive contact force between a sphere and a wall is given by a linear law, where the force acting on the sphere when contacting a plane is a linear function of the indentation, which in turn would bring a quadratic dependence with the contact radius. The next figure shows this simple law:
###### Hertzian Repulsive Force
On the other hand, Hertz solved in 1882 the non-cohesive normal contact between a sphere and a plane. In 1971 Johnson, Kendall and Roberts presented the solution (JKR-Theory) for the same problem in this case adding cohesive behaviour. Not much later, Derjaguin, Müller and Toporov published similar results (DMT-Theory).
Both theories are very close and correct, the main difference being that the JKR theory is oriented to the study of flexible, large spheres, while the DMT theory is specially suited to represent the behaviour of rigid, small ones.
The previous figure shows the standard representation of a Linear or Hertzian contact between a sphere and a wall. The distribution of contact pressures between both bodies follow a parabolic law.
###### JKR Cohesive Force
The preceding capture shows the a representation of a JKR contact between a sphere and a wall. In this case, the distribution of pressures between both bodies is more complex due to the formation of a neck at the boundaries of the contact. A later figure shows a detailed view of the pressures involved.
In the previous graphic, it is very interesting to note the existence of two singular values of contact radius: one for which the total forces acting on the contacting sphere is zero, and another for which the maximum value of adhesion is achieved.
In the previous figure, the blue area represents the distribution of pressures acting on the sphere when contacting a wall if a Hertzian Law is followed. On the other hand, the sum of both green and blue areas represents the JKR distribution. Note the larger values and the existence of adhesive behaviour at both sides of the pressures distribution.
An example of granular simulation without cohesive forces:
The same simulation as before, this time with cohesive forces in both sphere-sphere and sphere-plane contacts.
References:
V. L. Popov. Contact Mechanics and Friction (2010).
(restit. coef)
## Numerical approach (implementation)
Structure of the code (Strategy, Scheme, Element, Node, Utilities, functions frequently used like FastGet,...)
### Search Strategies
The contact search is a very important part of the method in terms of computational cost (range 60%-90% of simulation time in simulations with large number of particles) and it is possibly the most difficult part to treat when dealing with particles that have no spherical/circular shape.
The contact detection basically consists in determining, for every target object, what other objects are in contact with, and then, judge for the correspondent interaction. It is usually not needed to perform a search every time step, which is generally limited for the stability of the explicit integration of the equations of motion. Due to the fact that the search is an expensive step a lower search frequency can be selected without much loss of accuracy.
The most naïve method of search that can be set is the brute search; for every element, the method does a loop for any other element checking for the contact. The order of the number of operations needed is quadratic: n2. Normally, the strategy to avoid such an expensive scheme is to divide the contact search in two basic stages, a global search and a subsequent local resolution of the contact; in this case the computation time of the contact search is proportional to n log n. In the DEMApplication a Grid/Cell based algorithm is used in this purpose.
#### Global Search
In a generic way, there are two types of elements: searcher elements (particles or finite elements) and target elements (particles or finite elements). Hereafter searcher elements will be called S.E. and target elements T.E.
The steps needed to perform contact search are:
a) Build bounding box of S.E. (Figure 2(a)).
b) Build bins cells based on size and position of S.E. (Figure 2(b)).
c) Collocate S.E. in bins and construct hash table with relates coordinates with cells which point to the contacting S.E. (Figure 2(c)).
d) Build bounding box of T.E. (Figure 2(d)).
e) Loop over T.E., detect the intersecting cells to each T.E., check the intersection with the possible found cells and add the entire S.E. contained in the cells intersected by each T.E. (Figure 2(e)).
f) Solve the contact with local resolution (Figure 2(f)).
Note: In the case of FE and DE the FE are selected as the S.E. to construct the Bins and the Spheres are T.E. to be found in that bins.
#### Local Search
Once the possible neighbours are detected, the local resolution check takes place. For the case of two spherical particles, the check is easy; only the sum of the radius has to be compared against the distance between centres. Other geometries may demand a much complicated check. The followed strategy is to mesh all the geometries with a discretization of triangles. In 3D, surface meshes are used for contact detection. Now, the contact detection should be performed between particles and triangles; if no contact is found, particle contact against lines is searched for; and if contact is still not found, contact against points is performed. Figure 3 shows how the local search is performed. Particle i searches contact against element j, then against lines k, l and m and finally against points n, o and p.
This is known as a hierarchical algorithm. For further explanation please refer to the paper: "3D contact algorithms for particle DEM with rigid and deformable solids" - M.Santasusana, J.Irazábal, E.Oñate, J.M.Carbonell. Where the advantges and the drawback of this method and other proposed algoriths are detailed and analysed in complicated situations like multicontact.
Fig. 3 Particle-Face contact detection.
## Benchmarks
The DEM Benchmarks consist of a set of 9 simple tests which are run every night and whose object is to make sure both that the application performs correctly and that the code did not break after the daily changes. They are the following:
### Test1: Elastic normal impact of two identical spheres
Check the evolution of the elastic normal contact force between two spheres with time.
If the coefficient of restitution is 1, the module of the initial and final velocities should be the same. Also, by symmetry, velocities should be equal for both spheres.
_
### Test2: Elastic normal impact of a sphere with a rigid plane
Check the evolution of the elastic normal contact force between a sphere and a plane.
If the coefficient of restitution is equal to 1, the module of the initial and final velocity should remain unchanged.
_
### Test3: Normal contact with different restitution coefficients
Check the effect of different restitution coefficients on the damping ratio.
If total energy is conserved, the restitution coefficient and the damping ratio values should be identical.
_
### Test4: Oblique impact of a sphere with a rigid plane with a constant resultant velocity but at different incident angles
Check the tangential restitution coefficient, final angular velocity and rebound angle of the sphere.
_
### Test5: Oblique impact of a sphere with a rigid plane with a constant normal velocity but at different tangential velocities
Check the final linear and angular velocities of the sphere.
_
### Test6: Impact of a sphere with a rigid plane with a constant normal velocity but at different angular velocities
Check the final linear and angular velocities of the sphere.
_
### Test7: Impact of two identical spheres with a constant normal velocity and varying angular velocities
Check the final linear and angular velocities of both spheres.
By symmetry, the tangential final velocity of both spheres should be zero. Additionally, for a coefficient of restitution of 1, there should be no changes in the modules of both linear and angular velocities and their values should conserve symmetry.
_
### Test8: Impact of two differently sized spheres with a constant normal velocity and varying angular velocities
Check the final linear and angular velocities of both spheres.
In this case, it is interesting to note that, the bigger and/or denser sphere 2 is, the more this test resembles the sphere versus plane simulation.
_
### Test9: Impact of two identical spheres with a constant normal velocity and different restitution coefficients
Check the effect of different restitution coefficients on the damping ratio.
If total energy is conserved, the restitution coefficient and the damping ratio values should be identical.
References:
Y.C.Chung, J.Y.Ooi. Benchmark tests for verifying discrete element modelling codes at particle impact level (2011).
## How to analyse using the current application
### Pre-Process
GUI's & GiD
##### D-DEMPack
D-DEMPack is the package that allows a user to create, run and analyze results of a DEM simulation for discontinuum / granular / little-cohesive materials. It is written for GiD. So in order to use this package, you should install GiD first.
You can read the D-DEMPack manual or follow the D-DEMPack Tutorials for fast learning on how to use the GUI.
##### C-DEMPack
Continuum / Cohesive
Fluid coupling
## Application Dependencies
The Swimming DEM Application depends on the DEM application
FEM-DEM
## Programming Documentation
The source code is accessible through this site.
## Problems!
#### What to do if the Discrete Elements behave strangely
In the case you notice that some discrete elements cross walls, penetrate in them or simply fly away of the domain at high velocity, check the following points:
In the case of excessive penetration:
• Check that the Young Modulus is big enough. A small Young Modulus makes the Elements and the walls behave in a very smooth way. Sometimes they are so soft that total penetration and trespass is possible.
• Check the Density of the material. An excessive density means a big weight and inertia that cannot be stopped by the walls.
• Check the Time Step. If the time step is too big, the Elements can go from one side of the wall to the other with no appearence of a reaction.
• Check the frequency of neighbour search. If the search is not done frequently enough, the new contacts with the walls may not be detected soon enough.
In the case of excessive bounce:
• Check that the Young Modulus is not extremely big. An exaggerated Young Modulus yields extremely large reactions that can make the Elements bounce too fast in just one time step. Also take into account that the stability of explicit methods depends on the Young Modulus (the higher the modulus, the closer to instability).
• Check the Density of the material. A very low density means a very small weight and inertia, so any force exerted by other elements or the walls can induce big accelerations on the element.
• Check the Time Step. If the time step is too big, the method gains more energy, and gets closer to instability.
• Check the restitution coefficient of the material. Explicit integration schemes gain energy noticeably, unless you use a really small time step. In case the time step is chosen to be big (but still stable), use the restitution coefficient to compensate the gain of energy and get more realistic results.
## Contact
-Miguel Angel Celigueta: maceli@cimne.upc.edu
-Guillermo Casas: gcasas@cimne.upc.edu
-Salva Latorre: latorre@cimne.upc.edu
-Miquel Santasusana: msantasusana@cimne.upc.edu
-Ferran Arrufat: farrufat@cimne.upc.edu | 2,501 | 12,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.901805 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-linear-algebra-7th-edition/chapter-2-matrices-2-2-properties-of-matrix-operations-2-2-exercises-page-60/42 | 1,575,814,894,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540510336.29/warc/CC-MAIN-20191208122818-20191208150818-00200.warc.gz | 744,165,574 | 11,961 | ## Elementary Linear Algebra 7th Edition
$(AB)^T=B^TA^T$.
Since we have $$AB=\left[\begin{array}{ccc}{2} &{1} &{-1}\\{0}& {1}&{3}\\{4}&{0}&{2} \end{array}\right] \left[\begin{array}{ccc}{1} &{0} &{-1} \\ {2}& {1} &{-2}\\{0}&{1}&{3} \end{array}\right]= \left[\begin{array}{ccc}{4} &{0}&{-7} \\{2}& {4}&{7} \\{4}&{2}&{2} \end{array}\right]$$ then $$(AB)^T=\left[\begin{array}{ccc}{4} &{2}&{4} \\{0}& {4}&{2}\\{-7}&{7}&{2} \end{array}\right].$$ Now, $$B^TA^T=\left[\begin{array}{ccc}{1} &{2} &{0} \\{0} & {1}&{1} \\ {-1}&{-2}&{3} \end{array}\right] \left[\begin{array}{ccc}{2} &{0}&{4} \\{1}& {1} &{0}\\{-1}&{3}&{2} \end{array}\right]=\left[\begin{array}{ccc}{4} &{2}&{4} \\{0}& {4}&{2}\\{-7}&{7}&{2} \end{array}\right].$$ It is easy to see that $(AB)^T=B^TA^T$. | 374 | 760 | {"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} | 4.5 | 4 | CC-MAIN-2019-51 | latest | en | 0.405692 |
http://cms.itcclassmatestationery.com/sf/8/Mathematics/8RP3 | 1,597,310,126,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738964.20/warc/CC-MAIN-20200813073451-20200813103451-00192.warc.gz | 22,979,074 | 3,297 | # Direct Variationn
Direct variation may be understood by scenarios from our daily life.
For example: An employee who works for hourly wages may be paid according to the number of hours he worked. The two quantities x (the number of hours worked) and y (the amount paid) are related in such a way that when x changes, y changes proportionately such that the ratio remains a constant.
I.e., y varies directly with x. Let us represent the constant by k, i.e.
or y = kx ( where k ≠ 0)
If y varies directly as x, this relation is written as y ∝ and read as y varies as x. The sign “ ∝ ” is read “varies as” and is called the sign of variation.
Example:
If y varies directly as x and given y = 9 when x = 5, find:
• the equation connecting x and y
• the value of y when x = 15
• the value of x when y = 6
Solution:
a) y x i.e. y = kx where k is a constant
Substitute x = 5 and y = 9 into the equation:
y = x
b) Substitute x = 15 into the equation
y = = 27
c) Substitute y = 6 into the equation | 322 | 1,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-34 | latest | en | 0.713843 |
https://people.cs.umass.edu/~barring/cs250s05/exams/finsol.html | 1,669,514,146,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710155.67/warc/CC-MAIN-20221127005113-20221127035113-00162.warc.gz | 493,186,590 | 5,674 | # Final Exam Solutions
#### 17 May 2005
Question text is in black, solutions in blue.
• Question 1 (15): Suppose we throw three identical, fair six-sided dice. (In gaming terms, we "throw 3D6".) The six sides of each die are labeled 1, 2, 3, 4, 5, and 6.
• (a,5) What is the probability that the three numbers of the three dice are all different? Is this probability greater than, equal to, or less than 1/2?
The total number of possible throws, each equally likely, is 63 = 216 (first counting problem). The total number of throws with the three results all different is 63 = 6*5*4 = 120 (second counting problem). The probability is thus 120/216 = 5/9, which is greater than 1/2.
• (b,5) How many possibilities are there for the set of numbers appearing on the three dice? (For example, if all three dice show 5, this set is {5}. If the numbers are 4, 2, and 3, this set is {2,3,4}.)
The possible sets are exactly the subsets of {1,2,3,4,5,6} of size either 1, 2, or 3. There are (6 choose 1) = 6 sets of size 1, (6 choose 2) = 15 sets of size 2, and (6 choose 3) sets of size 3, for a total of 6 + 15 + 20 = 41. (For each individual size, this is an instance of the third counting problem.)
• (c,5) How many possibilities are there for the multiset of numbers appearing on the three dice? (For example, if all three dice show 5, the muliset is {5,5,5}. If all three numbers are different, the multiset is the same as the set.)
By the fourth counting problem, with n=6 and k=3, the number of multisets is (6+3-1 choose 3-1) or (6+3-1 choose 6), which is 8*7 or 56. Another way to see this is that the sets of size 1 or 3 in part (c) are in one-to-one correspondence with multisets, but a set of size 2 such as {2,5} corresponds to two multisets ({2,2,5} and {2,5,5}). So the total number of multisets is 6 + 2*15 + 20 = 56.
• Question 2 (25): Both this question and Question 3 deal with a predicate P(w), where w represents a string over the alphabet {a,b}. The predicate is defined recursively by four rules:
1. P(a) is true.
2. If w is any string and P(w) is true, then P(bbw) is also true.
3. If w is any string and P(w) is true, then P(wb) is also true.
4. If P(w) is not forced to be true by rules 1, 2, and 3, then it is false.
• (a,10) Prove by induction on this definition that for any string w such that P(w) is true, w has exactly one a.
Base case: The string "a" has exactly one a. Inductive cases: Assume that w has exactly one a. Then bbw (rule 2) and wb (rule 3) also have exactly one a. So all strings formed by rules 1, 2, and 3 have exactly one a, and by rule 4 these are all the strings for which P(w) is true.
• (b,5) Let A be the language {w: P(w)}. Give a regular expression denoting A.
(bb)*ab*
• (c,10) Describe or draw a λ-NFA whose language is A. (The λ-NFA made from the regular expression by the construction has nine states. There is a correct λ-NFA that has only three states. Remember that any ordinary NFA is also a &lambda-NFA.)
ASCII art:
``````
From the construction ("L" means "lambda"):
L
__________ L
/ \ ____
L / b b V L a L / b V L
>(1)--->(2)--->(3)--->(4)--->(5)--->(6)--->(7)--->(8)--->((9))
<-----------/ <---/
L L
A simpler, ordinary NFA:
(2)
|A
b||b _
|| / \ b
V| a V \
>(1)---------->((3))/
``````
• Question 3 (25): Let f(n) be the function defined recursively by the following rules:
1. f(0) = 0, f(1) = 1, and f(2) = 1
2. If n ≥ 3, then f(n) = f(n-1) + f(n-2) - f(n-3)
• (a,10) Explain why f(n) is exactly the number of strings of length n in the language A from Question 2. (Hint: Use the result of Question 2 part (a) to explain the base case. For the inductive case, use the Double Counting Rule to count the strings of length n formed by Rule 2 and Rule 3, determining exactly how many strings are formed in both ways.)
For the base case, by (2a) we know that there are 0 strings of length 0 in A. We know that there is at most one string of length 1 by (2a), and by Rule 1 there is exactly 1. For n=2 (2a) rules out all strings except ab and ba. We see that ab is in by Rules 1 and 3, and that ba is out because we cannot form it by either rule. So the three base values of f(n) for n=0, n=1, and n=2 match the number of strings in A of those lengths.
For the inductive case, let n ≥ 3 be arbitrary. A string in A of length n must be formed either by Rule 2 or by Rule 3. Since Rule 2 can form a string in A from any string of length n-2 in A, and Rule 3 can do so from any string of length n-1 in A, we have that the strings of length n are the union of two sets, whose size (by the strong IH) are f(n-2) and f(n-1) respectively. To find the size of this union we must subtract out the number of strings in both sets, i.e., those that can be formed by either Rule 2 or Rule 3. These are exactly the strings bbub for every string u in A of size n-3, so there are exactly f(n-3) of these by the strong IH, and our total number of strings of length n is f(n-1) + f(n-2) - f(n-3) as claimed.
• (b,15) Compute f(n) for n up to at least 5, and determine a formula or rule that gives the value of f(n) for any n. (You may find the Java integer division operator useful.) Prove by induction that your rule is correct. (A good way to do this is by strong induction, with separate cases for odd and even n in the inductive step.)
We have f(0)=0, f(1)=1, f(2)=1, f(3) = 1 + 1 - 0 = 2, f(4) = 2 + 1 - 1 = 2, f(5) = 2 + 2 - 1 = 3, f(6) = 3 + 2 - 2 = 3, f(7) = 3 + 3 - 2 = 4, and f(8) = 4 + 3 - 3 = 4. The rule is that f(n) = (n+1)/2, where the "/" is Java integer division. That is, if n = 2k then f(n) = k, and if n = 2k+1 then f(n) = k+1.
Let g(n) be (n+1)/2. We must prove by strong induction for all naturals n that f(n) = g(n):
• For the base cases n=0, n=1, and n=2, we need only verify that g(0) = (0+1)/2 = 0, g(1) = (1+1)/2 = 1, and g(2) = (2+1)/2 = 1.
• For the inductive case, assume that f(i) = g(i) for all i with i ≤ n. We must prove that f(n+1) = g(n+1).
• If n is even, that is n = 2k, then g(n+1) = k+1 and f(n+1) = f(n) + f(n-1) - f(n-2) = (by the strong IH) g(n) + g(n-1) - g(n-2) = k + k - (k-1) = k+1.
• If n is odd, that is n = 2k+1, then g(n+1) = k+1 and f(n+1) = f(n) + f(n-1) - f(n-2) = (by the strong IH) g(n) + g(n-1) - g(n-2) = (k+1) + k - k = k+1.
• So in each case we have proved f(n+1) = g(n+1) and we have completed the strong induction.
• Question 4 (25+10): Define the following four predicates on strings over the alphabet {0,1}:
• C(x) means "x is a valid description of a Turing machine"
• H(x,y) means "C(x) is true, and x halts when run on input y"
• A(x,y) means "H(x,y) is true, and x outputs "yes" when run on input y"
• T(x) means "∀y:H(x,y)"
(If you are unsure about Turing machines, don't panic. Parts (a) and (b) of this question can be answered by the rules of logic without specific knowledge about Turing machines.)
• (a,5) Prove that for any string x, T(x) → C(x). You may use either symbolic language or clear English.
Let x be an arbitrary string and assume that T(x) is true. By the definition of T(x), H(x,y) is true for all y so it is true for y = λ. (We could pick any specific string for y but we must specify a particular y to continue the reasoning.) By the definition of H, H(x,λ) means that C(x) and something else is true, so by left separation C(x) is true. We have completed a direct proof of T(x) → C(x).
• (b,15) Using quantifier rules carefully, prove the statement
¬∃u:∀v: C(u) ∧ [H(u,v) ↔ ¬H(v,v)]
(Hint: Assume the negation of this statement and derive a contradiction using quantifier rules.)
• Assume the negation of the given statement, which is
∃u:∀v: C(u) ∧ [H(u,v) ↔ ¬H(v,v)].
• By Instantition (∃-elim), let a be a string making
∀v: C(a) ∧ [H(a,v) ↔ ¬H(v,v)] true.
• By Specification (∀-elim) with v = a, conclude
C(a) ∧ [H(a,a) ↔ ¬H(a,a)].
• By right separation, conclude
H(a,a) ↔ ¬H(a,a).
• Since this is of the form "p ↔ ¬p", it is a contradiction.
• Since we derived a contradiction from the negation of the original statement, we have proved that original statement.
• (c,5) Explain the meaning of the statement of part (b) in English. What does it assert to be impossible?
There is no string u that denotes a Turing machine such that for any string v, u halts on v if and only if v does not halt on v. That is, it is impossible to design a Turing machine that halts on exactly those inputs that are not Turing machines that halt on themselves. In other words, the set of descriptions of TM's that do not halt on themselves (called in lecture the "Barber of Seville language") is not Turing recognizable.
• (d,10) Explain the meaning of the statement
∀x:[T(x) → ∃z: C(z) ∧ ∀w: [H(z,w) ↔ A(x,w)]]
Explain informally why it is true.
If a string x denotes a TM that halts on every input, there is another string z that denotes a TM that halts on any input if and only if x says yes on it. That is, any Turing decidable language {w: x says yes on w} is also a Turing recognizable language {w: z halts on w}.
This statement is true because we can take any TM x and modify it so that when it was going to halt, it actually halts if it was going to say yes and it runs forever if it was going to say no.
• Question 5 (35): Define the following λ-NFA M: The state set is {1,2}, the start state is 1, the final state set is {1}, and the transition relation Δ is {(1,a,1), (1,λ,2), (2,a,1), (2,b,2)}.
• (a,5) Draw a diagram for M. Remember that you should have one arrow for each triple in Δ.
``````
_ a _ b
| V L | V
>((1))------->(2) Again, "L" means "lambda".
<--------/
a
``````
• (b,10) Using the "killing λ-moves" construction, build an ordinary NFA N that is equivalent to M.
The state set and start state are the same, and the final state set is the same because the start state is already final. The transition (1,a,1) gives us itself and (1,a,2). The transition (2,a,1) gives us itself, (1,a,1), (1,a,2), and (2,a,1). The transition (2,b,2) gives us itself and (1,b,2). We thus have six distinct transitions in the ordinary NFA:
``````
_ a _ a,b
| V a,b | V
>((1))------->(2)
<--------/
a
``````
• (c,10) Use the subset construction to build a DFA D that is equivalent to M and N. (Partial credit for any correct DFA for this language. If you have trouble with (b), make sure you give some answer here so that you can do part (d).)
The initial state is {1}, a final state. It has an a-move to {1,2}, a final state, and a b-move to {2}, a non-final state. The state {1,2} has an a-move to itself and a b-move to {2}. The state {2} has an a-more to {1,2} and a b-move to itself. The DFA can be drawn as follows:
``````
_____
| \ a
a V /
>(({1}))----------->(({1,2}))/
| | A
| b b| | a
V | |
->({2})<--------------- |
| | \_________________/
\__|b
``````
• (d,10) Apply the state minimization construction to your D to get a minimal DFA (which may or may not be D itself).
We divide the states into two sets F = {{1}, {1,2}} and N = {{2}}. Since N contains only one state, which is reachable, we are finished with it. Both the states in F have a-arrows into F and b-arrows into N, so we may merge these two states into a single state to get the following minimal DFA:
``````
_ a _ b
| V b | V
>((F))------->(N)
<--------/
a
`````` | 3,623 | 11,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.914232 |
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