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http://stackoverflow.com/questions/24868681/apply-divinff-to-discontinuous-data-frame-in-r | 1,444,172,657,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736679281.15/warc/CC-MAIN-20151001215759-00142-ip-10-137-6-227.ec2.internal.warc.gz | 294,473,494 | 18,639 | # Apply divinff to discontinuous data frame in R
I have a set of position measurements that I have to pass through a linear regression in order to obtain slope value.
I do some subsetting first and I get this data frame (showing subset).
`````` us[1:30,1:5]
3087 3 618.2 16561 15861 16313
3088 3 618.4 16561 15861 16313
3089 3 618.6 16561 15861 16313
3090 3 618.8 16561 15861 16313
3091 3 619.0 16561 15861 16313
3092 3 619.2 16561 15861 16313
3093 3 619.4 16561 15861 16313
3094 3 619.6 16561 15861 16313
3095 3 619.8 16561 15861 16313
3096 3 620.0 16561 15861 16313
3097 3 620.2 16561 15861 16313
3098 3 620.4 16561 15861 16313
3099 3 620.6 16561 15861 16313
3100 3 620.8 16561 15861 16313
3101 3 621.0 16561 15861 16313
3102 3 621.2 16561 15861 16313
3103 3 621.4 16561 15861 16313
3104 3 621.6 16561 15861 16313
3105 3 621.8 16561 15860 16313
3106 3 622.0 16561 15860 16313
3107 3 622.2 16561 15860 16313
3108 3 622.4 16561 15859 16313
3109 3 622.6 16561 15859 16313
3110 3 622.8 16561 15859 16313
3111 3 623.0 16561 15859 16313
3112 3 623.2 16561 15859 16312
3113 3 623.4 16561 15859 16310
3114 3 623.6 16561 15859 16309
3115 3 623.8 16561 15859 16308
3116 3 624.0 16561 15859 16307
``````
I have to set the first value of each time start to zero (you can see that "Tiempo" column is not continuous, it jumps values in sets of 45 rows) and relativize the next values within each set to that initial position. The idea behind that is to obtain a set of increasing values for each column (as the position varies) and plot that against "Tiempo" variable to get the slope of each column later.
If I use
``````veltraining<-cbind(us\$Tiempo,diffinv(abs(diff(as.matrix(us[,3:length(us)])))))
``````
The discontinuos jump ruins the job.
`````` veltraining[1:30,1:5]
col1 col2 col3 col4 col5
[1,] 618.2 0 0 0 0
[2,] 618.4 0 0 0 0
[3,] 618.6 0 0 0 0
[4,] 618.8 0 0 0 0
[5,] 619.0 0 0 0 0
[6,] 619.2 0 0 0 10
[7,] 619.4 0 0 0 19
[8,] 619.6 0 0 0 25
[9,] 619.8 0 0 0 33
[10,] 620.0 0 0 0 39
[11,] 620.2 0 0 0 42
[12,] 620.4 0 0 0 42
[13,] 620.6 0 0 0 42
[14,] 620.8 0 0 0 43
[15,] 621.0 0 0 0 44
[16,] 621.2 0 0 0 44
[17,] 621.4 0 0 0 45
[18,] 621.6 0 0 0 47
[19,] 621.8 0 1 0 49
[20,] 622.0 0 1 0 51
[21,] 622.2 0 1 0 53
[22,] 622.4 0 2 0 55
[23,] 622.6 0 2 0 56
[24,] 622.8 0 2 0 58
[25,] 623.0 0 2 0 58
[26,] 623.2 0 2 1 72
[27,] 623.4 0 2 3 80
[28,] 623.6 0 2 4 80
[29,] 623.8 0 2 5 83
[30,] 624.0 0 2 6 92
``````
The expected output should be a data frame like this (subset). Don't mind the column names, it's just I named us columns after I did this post, the order is the same.
``````primer[1:10,1:7]
Tiempo UT TR UT.CHX TR.CHX TR.CHX.1 UT.CHX.1
1 618.2 0 0 0 0 0 0
2 618.4 0 0 0 0 0 0
3 618.6 0 0 0 0 0 0
4 618.8 0 0 0 0 0 0
5 619.0 0 0 0 0 9 0
6 619.2 0 0 0 10 14 0
7 619.4 0 0 0 19 15 0
8 619.6 0 0 0 25 18 0
9 619.8 0 0 0 33 39 0
10 620.0 0 0 0 39 64 0
``````
I don't know how to split the data frame simply - meaning without a lot of subsetting, naming and stuff like that - and I don't know if the "diffinv(abs(diff..." strategy is the best.
Thank you
-
I'm having a little trouble understanding what you are trying to do - could you possibly explain it in a different way / in more detail? Or maybe just take a very small chunk of your supplied data, say 10 or 15 rows of `us`, and manually create an example of what that small `data.frame` should look like after it has been appropriately transformed? – nrussell Jul 21 '14 at 15:35
@nrussell I put the expected output in my post with the position built from zero and rising. You can notice that the dput veltraining is the same but the whole thing with the position jumps when time jumps. – Matias Andina Jul 21 '14 at 15:42
I also cannot understand what you are trying to do. The fact that the sample data doesn't match up to the sample results is a quite confusing as well. you really haven't clearly identified where the problem is. Also, the sample `us` values seem unnecessarily large. Minimal, reproducible examples are best. – MrFlick Jul 21 '14 at 16:01 | 2,100 | 4,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2015-40 | latest | en | 0.484883 |
https://web2.0calc.com/questions/help_50640 | 1,591,410,750,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348509264.96/warc/CC-MAIN-20200606000537-20200606030537-00586.warc.gz | 600,670,273 | 5,783 | +0
# HELP
-1
54
1
+475
Find the height of a pine tree that casts a 86 foot shadow on the ground in the angle of elevation of the sun is 29 degrees 25'
Apr 27, 2020
#1
+23710
+2
29o 25' = 29.4167 o
tan (29.4167) = opposite / adjacent = height / 86
86 tan (29.4167) = height
Apr 27, 2020 | 117 | 296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.724393 |
https://us.metamath.org/ileuni/preq2i.html | 1,725,866,638,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00691.warc.gz | 570,355,894 | 3,543 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > preq2i GIF version
Theorem preq2i 3636
Description: Equality inference for unordered pairs. (Contributed by NM, 19-Oct-2012.)
Hypothesis
Ref Expression
preq1i.1 𝐴 = 𝐵
Assertion
Ref Expression
preq2i {𝐶, 𝐴} = {𝐶, 𝐵}
Proof of Theorem preq2i
StepHypRef Expression
1 preq1i.1 . 2 𝐴 = 𝐵
2 preq2 3633 . 2 (𝐴 = 𝐵 → {𝐶, 𝐴} = {𝐶, 𝐵})
31, 2ax-mp 5 1 {𝐶, 𝐴} = {𝐶, 𝐵}
Colors of variables: wff set class Syntax hints: = wceq 1332 {cpr 3557 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-io 699 ax-5 1424 ax-7 1425 ax-gen 1426 ax-ie1 1470 ax-ie2 1471 ax-8 1481 ax-10 1482 ax-11 1483 ax-i12 1484 ax-bndl 1486 ax-4 1487 ax-17 1503 ax-i9 1507 ax-ial 1511 ax-i5r 1512 ax-ext 2136 This theorem depends on definitions: df-bi 116 df-tru 1335 df-nf 1438 df-sb 1740 df-clab 2141 df-cleq 2147 df-clel 2150 df-nfc 2285 df-v 2711 df-un 3102 df-sn 3562 df-pr 3563 This theorem is referenced by: opid 3755 funopg 5197 df2o2 6368 fzprval 9962 fzo0to2pr 10095 fzo0to42pr 10097 2strstr1g 12232
Copyright terms: Public domain W3C validator | 628 | 1,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.186 |
https://ok-em.com/help-make-a-parabola-45 | 1,674,841,815,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495001.99/warc/CC-MAIN-20230127164242-20230127194242-00218.warc.gz | 428,773,057 | 5,660 | ## How to Graph Parabolas
The vertex form of a parabola is: f(x) = a(x - h) 2 + k. The a in the vertex form of a parabola corresponds to the a in standard form. If a is positive, the parabola will open upwards. If a is negative, the parabola will open downwards. In vertex
## How to Graph a Parabola: 13 Steps (with Pictures)
Simply draw the cube first, then use the angles of the cube to make the curves. 2 Make it using yarn or string. This is even cooler than doing it on
## How to graph parabolas — Krista King Math
5 = a (1) + 3. 2 = a. To finish, we rewrite the pattern with h, k, and a: 2. Find the equation of the parabola: This is a vertical parabola, so we are using the pattern. Our vertex is (-4, -1), so we will substitute those numbers in for h and k: Now
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Get calculation assistance online | 258 | 924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.844557 |
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# 3.2 Logarithmic Functions and Their Graphs PowerPoint PPT Presentation
3.2 Logarithmic Functions and Their Graphs. Definition of Logarithmic Function. 2 3 = 8. Ex. 3 = log 2 8. Ex.log 2 8. = x. 2 x = 8. x = 3. Properties of Logarithms and Natural Logarithms. log a 1 = 0 log a a = 1 log a a x = x. ln 1 = 0 ln e = 1 ln e x = x. Ex.
3.2 Logarithmic Functions and Their Graphs
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#### Presentation Transcript
3.2
Logarithmic Functions
and Their Graphs
Definition of Logarithmic Function
23 = 8
Ex. 3 = log2 8
Ex.log28
= x
2x = 8
x = 3
Properties of Logarithms and Natural Logarithms
• loga 1 = 0
• loga a = 1
• loga ax = x
• ln 1 = 0
• ln e = 1
• ln ex = x
Ex.
Use the definition of logarithm to write in
logarithmic form.
Ex.
4x = 16
log4 16 = x
e2 = x
ln x = 2
Graph and find the domain of the following functions.
y = ln x
x y
-2
-1
0
1
2
3
4
.5
cannot take
the ln of a (-)
number or 0
0
ln 2 = .693
ln 3 = 1.098
ln 4 = 1.386
D: x > 0
ln .5 = -.693
Graph y = 2x
y = x
x y
-2
-1
0
1
2
2-2 =
2-1 =
1
2
4
The graph of y = log2 x
is the inverse of y = 2x.
The domain of y = b +/- loga (bx + c), a > 1 consists
of all x such that bx + c > 0, and the V.A. occurs when
bx + c = 0. The x-intercept occurs when bx + c = 1.
Ex. Find all of the above for y = log3 (x – 2). Sketch.
D: x – 2 > 0
D: x > 2
V.A. @ x = 2
x-int. x – 2 = 1
x = 3
(3,0) | 710 | 1,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-13 | longest | en | 0.835714 |
https://equationbalancer.com/al-no3-3 | 1,685,938,982,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650620.66/warc/CC-MAIN-20230605021141-20230605051141-00416.warc.gz | 250,329,888 | 13,557 | # Al(no3)3
Aluminum nitrate appears as a white, crystalline solid. Noncombustible but can accelerate the burning of combustible materials.
Sarah Taylor-
Published on
## Introduction
Chemical equations are used to describe the reactants and products in a chemical reaction. However, the equation needs to be balanced to ensure that the number of atoms on each side of the equation is equal. In this article, we'll show you how to balance the equation for aluminum nitrate and water using algebraic equations.
## Word Equation
Aluminum nitrate + water → nitric acid + aluminum hydroxide
## Input Interpretation
Al(NO3)3 + 3H2O → 3HNO3 + Al(OH)3
Step 1
Write the Balanced Equation Algebraically To balance the equation, we need to add stoichiometric coefficients to the reactants and products. Let's call these coefficients c1, c2, c3, and c4:
c1Al(NO3)3 + c2H2O → c3 3HNO3 + c4Al(OH)3
Step 2
Equate the Number of Atoms Next, we need to set the number of atoms in the reactants equal to the number of atoms in the products for each element. For aluminum (Al), we have:
c1 = c4
For nitrogen (N), we have:
3c1 = c3
For oxygen (O), we have:
9c1 + c2 = 3c3 + 3c4
For hydrogen (H), we have:
2c2 = c3 + 3c4
Step 3
Solve for the Coefficients Since there are four unknowns and only four equations, we can solve for the coefficients by choosing an arbitrary value for one of them. Let's set c1 = 1:
c1 = 1 c2 = 3 c3 = 3 c4 = 1
Step 4
Substitute the Coefficients Now we can substitute the coefficients back into the balanced equation:
Al(NO3)3 + 3H2O → 3HNO3 + Al(OH)3 | 461 | 1,582 | {"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} | 4.5625 | 5 | CC-MAIN-2023-23 | latest | en | 0.857153 |
https://nl.mathworks.com/matlabcentral/cody/problems/44633-sum-of-diagonals/solutions/1520224 | 1,563,616,950,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526506.44/warc/CC-MAIN-20190720091347-20190720113347-00485.warc.gz | 483,007,099 | 14,972 | Cody
# Problem 44633. Sum of diagonals
Solution 1520224
Submitted on 7 May 2018 by J-G van der Toorn
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
a = 1; s_correct = 1; assert(isequal(maxDiagonalSum(a),s_correct))
2 Pass
a = [1 2 3; 2 3 4; 4 5 10]; s_correct = 14; assert(isequal(maxDiagonalSum(a),s_correct))
3 Pass
a = [1 2 ; 4 3 ]; s_correct = 6; assert(isequal(maxDiagonalSum(a),s_correct)) | 180 | 525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.657868 |
https://www.coursehero.com/file/5971164/Quiz1-Generic-without-solution-2001/ | 1,490,838,373,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191444.45/warc/CC-MAIN-20170322212951-00112-ip-10-233-31-227.ec2.internal.warc.gz | 859,980,645 | 253,467 | Quiz1 Generic without solution 2001
# Quiz1 Generic - 2 c Suppose that the forward rate that you observe in the market is lower than the one you computed in part(b How would you take
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FINANCE 100 Corporate Finance Professor Roberts Sample Quiz #1 NAME: SECTION: Question Maximum Student Score Q u e s t i o n 1 7 0 Question 2 30 TOTAL 100 Instructions:
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Question 1: (60 points) Consider the following prices for zero coupon bonds with \$100 face value: Maturity Price 1 \$95.2381 2 \$88.9996 3 \$82.7849 a) What is the yield to maturity of these bonds if the yield is compounded annually? (15 points) Maturity Yield 1 2 3 b) What is the forward rate for a one-year loan, one year from today? (15 points)
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Unformatted text preview: 2 c) Suppose that the forward rate that you observe in the market is lower than the one you computed in part (b). How would you take advantage of this discrepancy? Explain in detail the transactions that you would conduct and the corresponding cashflows. (40 points). 3 4 Question 2: (30 points) Find the price of a 10% Government coupon bond that pays semiannual coupons, matures in exactly 3 years and 6 months, and face value of \$100. The yield to maturity is 8% p.a. compounded semiannually . The bond has just paid its semi-annual coupon, hence you need to price it assuming that the first coupon that you will receive is due in exactly 6 months....
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## This note was uploaded on 10/06/2010 for the course FNCE 100 taught by Professor Jaffe during the Spring '10 term at UNC Asheville.
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Quiz1 Generic - 2 c Suppose that the forward rate that you observe in the market is lower than the one you computed in part(b How would you take
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Ask a homework question - tutors are online | 527 | 2,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-13 | longest | en | 0.935458 |
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### Can you solve this genius math puzzle? Number Series Puzzle – Crack the Code: Decode the Mathematical Puzzle! 🧮🔍
🌟 Challenge your math skills with this mind-boggling puzzle! Each equation seems to follow a unique pattern. Can you unveil the logic behind it? Let’s dive into the numbers and solve the mystery!
🤔 Puzzle Question:
1. 5+3=28
2. 9+1=810
3. 8+6=214
4. 5+4=19
🔢 Now, it’s your turn: 7+3=??
## Genius Math Puzzle Image:
💡 Solution Insight: (7−3=4)
(7+3=10) | 197 | 680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.812308 |
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Pictire riddles - Our online picture riddles are the best for kids and adults. We offer more than 100 picture puzzles to train your mind.
##### Divide in Equal Shape
Can you divide the below-given shape into four equal and identical pieces?
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In the Chess Board picture below white army is arranged. You need to add a black army on the board such that no piece is under any threat.
Note: Army comprised of 1 king, 1 queen, 2 rooks, 2 bishops, 2 knights, and 8 pawns.
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##### Missing Number
Can you find out the missing number in the grid?
Asked by Mehmat on 12 Nov 2023
##### Dots on Lines
There are nine dots in the picture that has been attached with this question. Can you join all the dots by drawing four straight lines without picking up your pen?
Asked by Vinod on 11 Nov 2023
##### Cross the Gate
You need to complete the maze by entering from the entrance marked below in the figure near the yellow circle, bottom left and leaving from the exit point near the green circle, bottom middle.
Rule of Game: You can move only by exchanging green and yellow circles.
Asked by Sachin on 21 Oct 2023
##### Triangles Count
Count the number of triangles in the below picture?
Asked by Sachin on 12 Oct 2023
##### Matchsticks Square
Move four matchsticks to the below square to form three squares?
Asked by Sachin on 09 Oct 2023
##### Odd one out
Out of four images which one is the odd one out?
Asked by Sachin on 02 Oct 2023
##### English Clockwise
If you read clockwise, you can form a word by inserting three missing letters in the picture given below. Can you do it?
Asked by Sachin on 10 Sep 2023
##### Maths Picture Puzzle
Can you solve the maths in the below-given picture equation?
Asked by Sachin on 06 Sep 2023
### Amazing Facts
###### Gambling
In Canada, a mathematical puzzle must be solved in order to win the lottery to classify it as a “game of skill” not gambling. | 490 | 2,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | longest | en | 0.933269 |
http://math.stackexchange.com/questions/148009/average-of-all-numbers-from-1-to-100-that-end-in-2/148010 | 1,469,702,555,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828010.65/warc/CC-MAIN-20160723071028-00137-ip-10-185-27-174.ec2.internal.warc.gz | 165,817,164 | 17,306 | # Average of all numbers from $1$ to $100$ that end in $2$
Can anyone please tell me what would be the fastest way to find the average of all numbers from $1$ to $100$ that end in $2$?
-
Evidently, the fastest way is to ask on this website. 2 answers so far in under 4 minutes. – Gerry Myerson May 22 '12 at 1:40
## 2 Answers
The numbers are of the form $10k+2$ where $k \in \{0,1,2,\ldots,9\}$. Hence, the average is $$\frac{\displaystyle \sum_{k=0}^{9} (10k+2)}{\displaystyle \sum_{k=0}^{9} 1} = 47$$ A quick way is to notice that the numbers are in arithmetic progression and hence the average is $$\dfrac{\text{First term}+\text{Last term}}{2} = \dfrac{2+92}{2} = 47$$
-
Thanks that helped. Forgot about AP formula – Rajeshwar May 22 '12 at 1:42
Group $2$ with $92$, $12$ with $82$, $22$ with $72$, $32$ with $62$, and $42$ with $52$. Each group can be further "split" into $47$ and $47$ (for example $2+92=47+47$). So the average is $47$.
- | 332 | 952 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2016-30 | latest | en | 0.903575 |
https://es.mathworks.com/help/stats/linearmodel.coefci.html | 1,675,306,819,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499954.21/warc/CC-MAIN-20230202003408-20230202033408-00496.warc.gz | 255,008,458 | 21,967 | # coefCI
Confidence intervals of coefficient estimates of linear regression model
## Syntax
``ci = coefCI(mdl)``
``ci = coefCI(mdl,alpha)``
## Description
example
````ci = coefCI(mdl)` returns 95% confidence intervals for the coefficients in `mdl`.```
example
````ci = coefCI(mdl,alpha)` returns confidence intervals using the confidence level ```1 – alpha```.```
## Examples
collapse all
Fit a linear regression model and obtain the default 95% confidence intervals for the resulting model coefficients.
Load the `carbig` data set and create a table in which the `Origin` predictor is categorical.
```load carbig Origin = categorical(cellstr(Origin)); tbl = table(Horsepower,Weight,MPG,Origin);```
Fit a linear regression model. Specify `Horsepower`, `Weight`, and `Origin` as predictor variables, and specify `MPG` as the response variable.
```modelspec = 'MPG ~ 1 + Horsepower + Weight + Origin'; mdl = fitlm(tbl,modelspec);```
View the names of the coefficients.
`mdl.CoefficientNames`
```ans = 1x9 cell Columns 1 through 4 {'(Intercept)'} {'Horsepower'} {'Weight'} {'Origin_France'} Columns 5 through 7 {'Origin_Germany'} {'Origin_Italy'} {'Origin_Japan'} Columns 8 through 9 {'Origin_Sweden'} {'Origin_USA'} ```
Find confidence intervals for the coefficients of the model.
`ci = coefCI(mdl)`
```ci = 9×2 43.3611 59.9390 -0.0748 -0.0315 -0.0059 -0.0037 -17.3623 -0.3477 -15.7503 0.7434 -17.2091 0.0613 -14.5106 1.8738 -18.5820 -1.5036 -17.3114 -0.9642 ```
Fit a linear regression model and obtain the confidence intervals for the resulting model coefficients using a specified confidence level.
Load the `carbig` data set and create a table in which the `Origin` predictor is categorical.
```load carbig Origin = categorical(cellstr(Origin)); tbl = table(Horsepower,Weight,MPG,Origin);```
Fit a linear regression model. Specify `Horsepower`, `Weight`, and `Origin` as predictor variables, and specify `MPG` as the response variable.
```modelspec = 'MPG ~ 1 + Horsepower + Weight + Origin'; mdl = fitlm(tbl,modelspec);```
Find 99% confidence intervals for the coefficients.
`ci = coefCI(mdl,.01)`
```ci = 9×2 40.7365 62.5635 -0.0816 -0.0246 -0.0062 -0.0034 -20.0560 2.3459 -18.3615 3.3546 -19.9433 2.7955 -17.1045 4.4676 -21.2858 1.2002 -19.8995 1.6238 ```
The confidence intervals are wider than the default 95% confidence intervals in Find Confidence Intervals for Model Coefficients.
## Input Arguments
collapse all
Linear regression model object, specified as a `LinearModel` object created by using `fitlm` or `stepwiselm`, or a `CompactLinearModel` object created by using `compact`.
Significance level for the confidence interval, specified as a numeric value in the range [0,1]. The confidence level of `ci` is equal to 100(1 – `alpha`)%. `alpha` is the probability that the confidence interval does not contain the true value.
Example: `0.01`
Data Types: `single` | `double`
## Output Arguments
collapse all
Confidence intervals, returned as a k-by-2 numeric matrix, where k is the number of coefficients. The jth row of `ci` is the confidence interval of the jth coefficient of `mdl`. The name of coefficient j is stored in the `CoefficientNames` property of `mdl`.
Data Types: `single` | `double`
collapse all
### Confidence Interval
The coefficient confidence intervals provide a measure of precision for regression coefficient estimates.
A 100(1 – α)% confidence interval gives the range that the corresponding regression coefficient will be in with 100(1 – α)% confidence, meaning that 100(1 – α)% of the intervals resulting from repeated experimentation will contain the true value of the coefficient.
The software finds confidence intervals using the Wald method. The 100*(1 – α)% confidence intervals for regression coefficients are
`${b}_{i}±{t}_{\left(1-\alpha /2,n-p\right)}SE\left({b}_{i}\right),$`
where bi is the coefficient estimate, SE(bi) is the standard error of the coefficient estimate, and t(1–α/2,np) is the 100(1 – α/2) percentile of t-distribution with n – p degrees of freedom. n is the number of observations and p is the number of regression coefficients.
## Version History
Introduced in R2012a | 1,142 | 4,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-06 | longest | en | 0.557387 |
https://www.enotes.com/homework-help/6-1-kg-object-hangs-one-end-rope-thatis-attached-294558 | 1,670,317,649,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00820.warc.gz | 792,305,617 | 18,928 | # A 6.1 kg object hangs at one end of a rope that is attached to a support on a railroad boxcar. When the car accelerates to the right, the rope makes an angle of 39 with the vertical. Find the acceleration of the car. The acceleration due to gravity is 9.8 m/s^2.
The 6.1 kg object hangs at one end of a rope that is attached to a support on a railroad boxcar.
There is a force of gravitation that acts on the object and pulls it in a direction vertically downwards. The magnitude of the force of gravity acting on the...
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The 6.1 kg object hangs at one end of a rope that is attached to a support on a railroad boxcar.
There is a force of gravitation that acts on the object and pulls it in a direction vertically downwards. The magnitude of the force of gravity acting on the object is 6.1*9.8 N.
When the railroad boxcar accelerates to the right there is another force exerted on the object. This is in a horizontal direction and has a magnitude of 6.1*A where A denotes the acceleration of the car.
The resultant force of gravity and that due to the acceleration of the boxcar is making the direction of the rope from which the object is hanged inclined to the vertical at an angle of 39 degrees.
tan 39 = A/9.8
=> A = 9.8* tan 39
=> A = 7.935 m/s^2
The rate at which the car accelerates is 7.935 m/s^2 | 359 | 1,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.940996 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=151&t=19676&view=print | 1,601,293,865,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401600771.78/warc/CC-MAIN-20200928104328-20200928134328-00608.warc.gz | 420,069,643 | 1,852 | Page 1 of 1
### 15.61 equation
Posted: Sun Feb 19, 2017 10:54 pm
In book problem 15.61 why does the solution manual use the equation
ln(k'/k) = (Ea/R) x ((T' - T) / (T'T))
ln(k'/k) = (Ea/R) x ((1/T) - (1-T')) ?
The solution manual says the two equations are equal but I am confused how the second one is derived / why it is used / why it produces a different answer than the first.
### Re: 15.61 equation
Posted: Mon Feb 20, 2017 12:01 am
Those two equations are the same if you find a common factor of the denominator.
### Re: 15.61 equation
Posted: Fri Mar 15, 2019 8:42 pm
is the second equation necessary to use? would the first equation produce the same answer?
### Re: 15.61 equation
Posted: Fri Mar 15, 2019 9:16 pm
Madelyn Cearlock wrote:is the second equation necessary to use? would the first equation produce the same answer?
Yes, they produce the same answer. These two equations are the same thing, just in different formats. | 280 | 947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.903565 |
http://www.ee.nmt.edu/~thomas/ee321_f05/grading-hw.html | 1,513,346,957,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948572676.65/warc/CC-MAIN-20171215133912-20171215155912-00027.warc.gz | 351,997,424 | 2,066 | • First do the homework and copy just the short answers onto another paper and turn it in on Mondays.
• I will post solutions after class on Monday. The solutions will be in the 2nd floor study EE area and on reserve at the library.
• Look at the solutions and grade it as right or wrong - no partial credit. Next study the solutions and rework incorrect problems using the solutions. If it doesn't make sense ask for help. Get a grade sheet off the web or draw one on your paper and record your grades and total them.
• Scoring - Each problem is worth 10 points. To make things easy and uniform give your self credit only for correct answers. If there is only one answer the score could be 0 or 10. If there are 4 answers the score could be 0, 3, 5, 8, 10 (round fractions in your favor).
• Wednesday, hand in the grade sheet, your graded homework (the original set) and the reworked solutions.
• I will record the two grades from the grade sheet for each home work set.
• The first grade is for answers handed in on Mondays - NO late credit.
• The second grade is for the reworked solutions - This one should be perfect. Late solutions are reduced 10% per period.
• Both grades will be averaged for the final homework grade (along with the pop quizzes and reading summaries).
Here is a sample grade sheet. There should be one on the web for each homework set, if not make one like this.
Home work set # 101
Prob. first score 0-10 check here if you did not redo the problem second score 0-10 1 - - - 2 - - - 3 - - - 4 - - - 5 - - - 6 - - - 7 - - - 8 - - - 9 - - - Total - - -
If you did the problem correctly the first time and got 10 points, check 'did not redo' and put 10 in for the second score.
If you elect not to redo a problem check 'did not redo' and put the same score in for the second score.
If you didn't work the problems and just copied the solutions put zero everywhere.
FAQ: Why do i need to rework homework for this class, I don't have to for my other classes?
Answer: On a job in the real world when you are given a problem, you are expected to work on it until it is solved, not just one good try. It will help you learn the material better.
FAQ: What should I do if I find an error in the your solution?
Answer: This really happens. Send me e-mail explaining the error. The first one to notify me will receive 20 extra points. I will let everyone know by e-mail. If there really wasn't an error there is no penalty. | 612 | 2,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-51 | latest | en | 0.922599 |
https://gmplib.org/list-archives/gmp-devel/2009-December/001290.html | 1,708,697,480,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474412.46/warc/CC-MAIN-20240223121413-20240223151413-00434.warc.gz | 274,315,328 | 2,233 | # gcdext_1 questions
Paul Zimmermann Paul.Zimmermann at loria.fr
Fri Dec 4 13:39:01 CET 2009
``` Niels,
> > Thus one iteration is equivalent to Stein's binary algorithm, except you
> > replace the subtraction by an addmul_si (and doing that you remove more bits).
>
> A slightly different way to look at it is as follows.
> We have u and v odd, and update
>
> u <-- (u + q v) / 2^{j+number of extra trailing zeros}
>
> where q = - u v^{-1} mod 2^j (represented in some symmetric or
> assymmetric range).
yes exactly, except q = - u v^{-1} mod 2^{j+1} in our algorithm.
> In the binary algorithm as described in Stehlé's and your paper, j is
> determined by the number of "extra trailing bits" from the previous
> iteration. But from a practical point of view, the important thing is
> that u and q v are of the same bitsize, so that u + q*v is not much
> larger than u, growing at most by a carry bit.
>
> So I think one could do a little better by using count_leading_zeros, to
> take into account precisely what happens to the bits also in the most
> significant end (instead of just relying on the growth there being
> bounded as a fibonacci sequence), we may by luck have a bit or two
> cancelling there too.
you mean you want to determine the best q modulo 2^{j+1}, i.e., the value
so that |u + q v| is minimal?
> Like,
>
> while (u != v)
> {
> if (u >v)
> {
> j = #u - #v + 1; // I let # denote bitsize
> q = -u v^{-1} mod 2^j;
> u += q v;
> u >>= (trailing zeros, >= j);
> }
> else { ... Analogous update v += q * v ... }
> }
>
> where there then are several options for how to handle signs of u, v and
> q.
I'm not sure to understand your code, since j is not defined that way in our
algorithm. Do you mean a variant of our algorithm?
Paul
``` | 525 | 1,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-10 | latest | en | 0.899359 |
https://www.tutorialspoint.com/the-perimeter-of-a-triangle-with-vertices-0-4-0-0-and-3-0-is-a-5-b-12-c-11-d-7plus-sqrt-5 | 1,669,698,079,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710685.0/warc/CC-MAIN-20221129031912-20221129061912-00115.warc.gz | 1,128,769,624 | 9,608 | # The perimeter of a triangle with vertices $(0,4),(0,0)$ and $(3,0)$ is(A) 5(B) 12(C) 11(D) $7+\sqrt{5}$
#### Complete Python Prime Pack
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Given:
The vertices of a triangle are $(0,4),(0,0)$ and $(3,0)$.
To do:
We have to find the perimeter of the triangle,
Solution:
Let the vertices of the triangle be $A(0,4),B(0,0)$ and $C(3,0)$.
We know that,
Perimeter of a triangle $=$ Sum of the lengths of the sides of the triangle
This implies,
Perimeter $=AB+BC+CA$
Using the distance formula,
$d=\sqrt{( x_2-x_1)^2+( y_2-y_1)^2}$
Perimeter $=\sqrt{(0-0)^{2}+(0-4)^{2}}+\sqrt{(3-0)^{2}+(0-0)^{2}}+\sqrt{(3-0)^{2}+(0-4)^{2}}$
$=\sqrt{0+16}+\sqrt{9+0}+\sqrt{9=16}$
$=4+3+\sqrt{25}$
$=7+5$
$=12$
The perimeter of the triangle is $12$.
Updated on 10-Oct-2022 13:28:26 | 349 | 930 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-49 | latest | en | 0.479919 |
https://inginious.org/course/ProblMath/log-q1 | 1,653,236,755,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545875.39/warc/CC-MAIN-20220522160113-20220522190113-00143.warc.gz | 372,601,645 | 5,193 | ### Information
Author(s) Philippe Delsarte, Manon Oreins Deadline No deadline Submission limit No limitation Category tags Logarithme/exponentielle
### Tags
Logarithme/exponentielle
## Exponentielle et logarithme - Q1
Résoudre les équations suivantes
##### Question 1:
$$2^x + 2^{x+1} =3$$
##### Question 2:
$$5 \cdot 3^{x-1} - 2 \cdot 3^{1-x} = 3$$
##### Question 3:
$$30 \cdot 3^x - 9^x - 81 = 0$$
##### Question 4:
$$4^{x^2} = 2^{5x-2}$$
##### Question 5:
$$4^x - 3^{x + \frac{1}{2}} = 3^{x - \frac{1}{2}} - 2^{2x}$$ | 212 | 536 | {"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} | 3.28125 | 3 | CC-MAIN-2022-21 | latest | en | 0.212837 |
https://nl.msx.org/nl/node/52814?page=3 | 1,670,394,417,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711150.61/warc/CC-MAIN-20221207053157-20221207083157-00777.warc.gz | 445,097,025 | 8,642 | # 3D rasterisation
Pagina 4/4
1 | 2 | 3 |
Right — if you're ray casting then doing conventional perspective is actually a little more expensive, at one extra multiply per column. If a pet peeve stemming from all raycasters that I have ever seen isn't too off-topic: the thing people usually get wrong there is thinking that if you want to cover an x degree field of view in u pixels then angles should be picked through a linear scaling of x to u. When people have made that mistake you usually see them pick a small field of view, because the error is smaller, and still have things be slightly sickeningly-distorted towards the edges. The correct thing to do is to think about the right-angled triangle each cast ray makes from viewer to screen, and get the angle from that. So it's an arctan thing. And then just a lookup table. No extra processing cost for getting it right.
My more helpful reason for returning is that I realised my previous comment about an odd-shaped lookup table for polygon edge scanning (or line rasterisation for an MSX1, I guess) is hokum because you've two values to store and each table is triangle-shaped. So you can just flip one and push them together.
Suppose you're limited to a 192x192 drawing area. Then the lookup table is 36kb, and when |x| > |y| the algorithm looks something like:
address = (|x| * 192) + |y| // with shifting rather than an actual multiplication, obviously
x1 += (quotient >> 1)
error = y
while(y1 != y2) {
output(x1, y1)
error -= remainder
if(error < 0) {
error += y
x1++
}
x1 += quotient
y1++
}
Typed extemporaneously. Check before use. You'd probably have four variations for the four arrangements of incrementing and decrementing x and y, and have ended up with |x| and |y| in picking one.
The |y| > |x| version instead outputs either quotient or quotient+1 locations downward from (x1, y1) before incrementing x1 (and have four arrangements of that).
If you're on an MSX1 or any other machine without hardware line drawing then to draw a line, don't seed x1 to halfway across a span, and draw all the spans. Make sure you handle x = y, x = 0 and y = 0 as special cases.
Nice port of the wireframe 3D intro of Silpheed to MSX: https://youtu.be/fYyeU7QZhFE
Using 512x212 resolution! Not sure whether it’s screen 6 or 7, but if it’s drawn with LINE VDP commands then there wouldn’t be much of a performance difference anyway.
(For comparison, the PC version: https://youtu.be/3n4nOiURgzg)
Very impressive. Doesn't look like it uses more than 4 colors, though. I imagine it uses page-flipping to avoid flickering and tearing. Maybe it uses the extra available pages in SCREEN 6 to store the background graphics?
The first one looks a lot smoother… Does anybody know why?
Google translate of the description of the first video:
Quote:
Hey, it seems that Silpheed was scheduled to be released on MSX! This is that demo!
… It's April Fool's Well! That's too bad! Huh…? Already in June …?
T Sato was created, “I tried playing the wireframe of Silpheed OP with BASIC's one screen program.”
Looking at, I made it because I thought “How fast do you get out of MSX?”
Except for wire frame data and fonts, it is an eye / ear copy.
To be a bad thing, it will not work unless the primary mapper is 512 KB or more. There are 250 KB wire frame data anyway.
Since it does not discriminate the model, it operates even though “extended memory of MSX 2 or more + DOS 2 + 512 KB or more”, but the speed does not come out.
Because A1ST has only 256 KB as the primary mapper, it does not work even if you stick an extended memory (it probably works if you extend and remodel the main memory)
Real (It is dedicated largely due to the program) It is for A1GT only.
Why would it be such a thing, if you ask someone familiar with MSX it will be fun to teach you about 30 minutes.
Among them, you may make making and commentary videos.
For users of MZ-700 / MZ-1500 / MZ-2500 / SMC-777 / MB-S1 / Pasopia 7, the next is your turn! What?
Created by T Sato - X1 turbo version
Created by nori6001 - PC6601SR version
Google translate of the description of the second video:
Quote:
Zakary, you look a little pale?
It's a palette problem. dont 'worry.
…?
Don’t worry.
Haa …
The AV version wireframe demo is skipped one frame of 88 version.
I think that I wanted on-memory instead of processing speed problem.
Thanks to this, memory is enough even for non-remodeled ST. I did it!
At the same time, it became SCREEN 5 and the frame rate also dropped, so the speed at which MSX 2 (Necessary DOS 2) can finally be seen also comes out.
So it seems the second video halved the number of frames to reduce the memory usage for the sake of the turboR ST, as well as make it run on MSX2. For the ST though, too bad the quality was reduced instead of just supporting an external mapper (which is trivial)…
By the way, looks like all the 3D math has been preprocessed, and the data is essentially a wireframe FMV? May be a useful idea. Takes more memory than I would expect though…
wireframe FMV? As in, for each frame there's a list of start and end coordinates to draw lines between?
Is what I think, indeed.
Metalion posted an interesting optimisation approach here:
Grauw wrote:
Metalion wrote:
3) As I'm drawing in fact a polygon, I'm needing 2 of those triangles and a rectangle in between. The rectangle is done with a HMMV, and I'm using the time needed by the VDP to draw it to make computation for a next frame. If I draw larger triangles because of the HMMV command, it reduces the size of the rectangle and therefore the time available for computations. Of course, it's a balance between the 2, and the gain on drawing speed could also be used. But that needs to be checked.
Aha, interesting! I was already wondering what you needed right triangles for :). I would implement scan line algorithm with two Bresenham functions, one for the left edge and one for the right of a flat-base or flat-top triangle.
But if I understand correctly you’re aiming to exploit larger block fills to optimise the performance, by rendering the triangle in groups of lines at a time (say 8), and then filling the inner area that doesn’t have any edges with HMMV while using something slower for the corners. Or differently put, rendering with HMMV as it were at a lower resolution.
Although in that case to maximise CPU-VDP parallelism my first thought would be to render the corners completely by the CPU, so that it can be done in parallel with the HMMV execution.
Grauw wrote:
By the way, looks like all the 3D math has been preprocessed, and the data is essentially a wireframe FMV? May be a useful idea. Takes more memory than I would expect though…
If that was the case, why the author didn't do hidden line removal? With 3D math predone, it would not cost anything, on the contrary it would save bandwidth.
Pagina 4/4
1 | 2 | 3 | | 1,708 | 6,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.931372 |
https://www.physicsforums.com/threads/pfr-vs-cstr-reactor-size-for-various-order-reactions.768935/ | 1,720,819,161,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514452.62/warc/CC-MAIN-20240712192937-20240712222937-00463.warc.gz | 740,050,842 | 22,229 | PFR vs. CSTR reactor size for various order reactions
• gfd43tg
In summary, the conversation between two individuals revolves around a problem in Fogler's Chemical Reaction Engineering, specifically parts (a) through (c) which have been completed. The equations used for this problem were the design equations for PFR and CSTR, and the individual also went beyond the scope of the problem by using Matlab to plot reactor size vs. input stream flow rate. One question that was asked was why the concentration of a species in the exit stream is used in the reaction rate equation rather than an instantaneous concentration. The answer is that this is only true for a Continuous Stirred Tank Reactor (CSTR) due to the intense mixing that occurs. In a CSTR, the concentration of reactant is perfectly uniform due to the
gfd43tg
Gold Member
Hello,
I have been working on a very interesting problem out of Fogler's Chemical Reaction Engineering. I have completed the problem (parts (a) through (c) which is what i'll do because I've been working on this problem for a while and I'm too tired to do part (d) right now), but I want to share my results and ask for some help in the interpretation of the results.
The equations used were the design equations for PFR and CSTR, which are
$$V_{CSTR} = \frac {F_{out} - F_{in}}{r_{A}}$$
$$\frac {dF_{A}}{dV_{PFR}} = r_{A}$$
where ##V## is the volume of the reactor, ##F## is the molar flow rate of the species, and ##r_{A}## is the reaction rate of the species.
I went beyond the scope of the problem and did some work in Matlab to write some code to plot reactor size vs. input stream flow rate, for fun and to gain a deeper understanding for this as well as keep up with my MATLAB skills! For the sake of context, I will post the problem as well as share my m-file and plots.
One preliminary detail I want to ask about is the following. Why is it that when you have the reaction rate for species A ##r_{A} = -kC_{A}##, ##C_{A}## is the concentration of A in the exit stream, and not some sort of instantaneous concentration?
So now for the meat of the interpretation.
So to begin with the zeroth order reaction, why is it that they require the same volume for a given input feed? As I worked and refined through this problem, it became very apparent to me that the reaction order makes a great deal of difference.
For the first and second order reactions, why is it that the CSTR requires such a larger volume to have the same conversion than the PFR? The difference between zeroth and first order is extremely obvious, but even for the second order reaction, it appears that the relative difference in size requirement decreases more than for the first order. My prediction for a third order reaction would be that the line for the PFR would become even flatter compared to the CSTR. What is the physical reason for this?
My observation is that as the order of the reaction increases, the relative difference in reactor volume between the PFR and CSTR increases. It seems that the PFR begins to flatten out, and the PFR continues to slope up. I haven't done the test, but I would suppose at some ##n^{th}## order reaction, the line for the PFR would eventually have zero slope.
Why does the relative difference between the two seem to decrease as the order of the reaction increases?
You can see that the actual value of the reactor size increases by orders of magnitude, from ##10^3## to ##10^4## to ##10^5## liters for zeroth, first, and second order reactions, respectively. Why does the order of a reaction affect the actual size of the reactor necessary to convert a input the same amount, so much as an increase of a magnitude of order? I am looking for some sort of physical reasoning.
Lastly, to follow up with this problem I was wondering if there are any other relations I should look at and plot to gain physical insight into sizing of a chemical reactor?
Thank you
Attachments
• CSTR vs. PFR reactor size.jpg
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• Fogler chp 1 prob 15.m
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• zeroth order reaction.jpg
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• first order reaction.jpg
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• second order reaction.jpg
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Last edited:
I've been very busy with family matters today, but I'll try to address your questions soon.
Chet
I had the wrong titles for my graphs!
Attachments
• first order reaction.jpg
17.4 KB · Views: 1,893
• second order reaction.jpg
16.7 KB · Views: 2,307
Maylis said:
Hello,
I have been working on a very interesting problem out of Fogler's Chemical Reaction Engineering. I have completed the problem (parts (a) through (c) which is what i'll do because I've been working on this problem for a while and I'm too tired to do part (d) right now), but I want to share my results and ask for some help in the interpretation of the results.
The equations used were the design equations for PFR and CSTR, which are
$$V_{CSTR} = \frac {F_{out} - F_{in}}{r_{A}}$$
$$\frac {dF_{A}}{dV_{PFR}} = r_{A}$$
where ##V## is the volume of the reactor, ##F## is the molar flow rate of the species, and ##r_{A}## is the reaction rate of the species.
I went beyond the scope of the problem and did some work in Matlab to write some code to plot reactor size vs. input stream flow rate, for fun and to gain a deeper understanding for this as well as keep up with my MATLAB skills! For the sake of context, I will post the problem as well as share my m-file and plots.
One preliminary detail I want to ask about is the following. Why is it that when you have the reaction rate for species A ##r_{A} = -kC_{A}##, ##C_{A}## is the concentration of A in the exit stream, and not some sort of instantaneous concentration?
This is true only for a Continuous Stirred Tank Reactor (CSTR), not for a Plug Flow Reactor (PFR).
CSTR: In an ideal CSTR, the mixing is so intense that there are no concentration variations within the reactor. Each parcel of fluid within the tank has the same number of molecules of the reactant; and, therefore, the stream coming out of the reactor has the same concentration of reactant as within the reactor.
Mixing is a process that involves a combination of convection and diffusion. The mixing convection creates very thin striations of fluid adjacent to one another. Even if these striations have quite different concentrations, because the striations are very thin, the process of diffusion rapidly removes these concentration differences. In the limit of infinite mixing rate, the concentration in the CSTR is perfectly uniform.
If you examine any parcel of liquid within an ideal CSTR, it contains molecules of reactant having all different ages (relative to when the molecules entered the tank). It contains molecules that entered the tank just a short time ago, and molecules that entered the tank a long time ago. This is what the high degree of mixing accomplishes. However, the total number of reactant molecules per unit volume (of all ages) in every parcel will be exactly the same.
This is why, in an ideal CSTR, there is only one reactant concentration value (i.e., uniform concentration) in the tank, and why it is also the concentration of reactant in the exit stream.
PFR: In an ideal PFR, the situation is reversed. There is absolutely no mixing taking place, and the concentration within the PFR varies with axial position along the pipe. Furthermore, we neglect the fact that there is a no-slip boundary condition at the wall, and assume that the velocity profile in the pipe is perfectly flat. We also neglect axial diffusion resulting from axial concentration gradients. Under these circumstances, at steady state, the differential mass balance equation for the PRF becomes:
$$v\frac{dC}{dx}=r$$
where v is the axial velocity and x is axial position in the pipe. If we multiply numerator and denominator of the rhs of this equation by the cross sectional area of the pipe, we obtain:
$$F\frac{dC}{dV}=r$$
where F is the volumetric flow rate in the pipe, and V is the differential pipe volume between x and x + dx. These equations can also be written in another form:
$$\frac{dC}{dt}=r$$
where t =x/v is the cumulative residence time from the inlet to the reactor x = 0 to some arbitrary location x. In this form, you can recognize the equation as the same as for a batch reactor. If you situated yourself on a little parcel of fluid as it passes through a PFR, your parcel would essentially be a little batch reactor, and you would measure the same concentration at time t as in a batch reactor at time t. This is what happens in analyzing a PFR reactor when you adopt a so called Lagrangian (material) frame of reference.
So now for the meat of the interpretation.
So to begin with the zeroth order reaction, why is it that they require the same volume for a given input feed? As I worked and refined through this problem, it became very apparent to me that the reaction order makes a great deal of difference.
If the reaction rate doesn't depend on the concentration of the reactant (zero order reaction), the reaction rate at every location in a PFR will be exactly the same as the reaction rate at every location in a CSTR. So the only thing that matters is the average amount of time that the fluid elements spend in the reactor. This is equal to the volume of the reactor divided by the volumetric flow rate.
Gotta go now. I'll get back to your other questions later.
Chet
1 person
I see where you sent me a thanks for this post. Do you still want me to try to field your other questions, or are you OK with all this now?
Chet
Oh, well I still was wondering about the other parts too thank you
I was thanking you for what you've already written
Last edited:
Let's look at a first order reaction, and let's assume that the exit concentration from the CSTR is the same as the exit concentration from the PFR, for the same feed rate. In the CSTR, the concentration of reactant throughout the reactor is Cfinal, and this is the concentration in the exit stream. In the PFR, the concentration of reactant throughout the reactor is higher than the exit concentration Cfinal (it is decreasing with axial location along the reactor). Since the rate of reaction is kC, the average rate of reaction in the PFR reactor is higher than the average rate of reaction in the CSTR. So the volume of the PFR can be less.
If the reaction is second order, the reaction rate goes as the square of the concentration, rather than the concentration to the first power. So, for a second order reaction, the ratio of the average rate of reaction in a CSTR to the average rate of reaction in the PFR is even higher than for a first order reaction.
Chet
But why does the PFR volume not increase so significantly for an increasing molar flow rate, as compared to the CSTR?
Maylis said:
But why does the PFR volume not increase so significantly for an increasing molar flow rate, as compared to the CSTR?
Who says? Are we talking about absolute increase or relative increase? Far a first order reaction, the percentage increases in volume should be the same.
Chet
Look at my graph!
Maylis said:
Look at my graph!
Well, consider this:
PFR: ##C_f=C_0e^{-\frac{kV}{F}}##
CSTR: ##C_f=\frac{C_0}{1+\frac{kV}{F}}##
Note that V/F occurs in combination. So, if you double F, you have to double V. How do the above equations compare with your results in the graph? Your graph shows that, in both cases, V
is proportional to F. So a 10% increase in F will be accompanied by a 10% increase in V in each case.
Chet
1 person
1. What is the difference between PFR and CSTR reactors?
PFR (plug flow reactor) and CSTR (continuous stirred tank reactor) are two types of reactors commonly used in chemical reactions. The main difference between them is the way they mix the reactants. In a PFR, the reactants flow through a tube without any mixing, while in a CSTR, the reactants are continuously stirred to ensure uniform mixing.
2. How does the order of a reaction affect the reactor size for PFR and CSTR?
The order of a reaction, which is determined by the rate law, directly affects the reactor size for both PFR and CSTR. For a first-order reaction, the reactor size for a PFR will be smaller than that of a CSTR, while for a second-order reaction, the opposite is true.
3. What is the ideal reactor size for a first-order reaction in a PFR vs. a CSTR?
The ideal reactor size for a first-order reaction is different for a PFR and a CSTR. In a PFR, the ideal reactor size is the smallest possible, as the reactants are continuously flowing and mixing is not necessary. In a CSTR, the ideal reactor size is larger than that of a PFR, as the reactants need to be continuously stirred for uniform mixing.
4. How do the conversion and selectivity of a reaction affect the reactor size for PFR and CSTR?
The conversion and selectivity of a reaction also play a role in determining the reactor size for PFR and CSTR. Higher conversion and selectivity require a larger reactor size, as more reactants need to be converted to reach the desired product. This applies to both PFR and CSTR, but the effect may be more significant in a PFR due to the absence of mixing.
5. Are there any limitations to using PFR or CSTR for certain order reactions?
Both PFR and CSTR have their limitations when it comes to certain order reactions. For example, a PFR may not be suitable for second-order reactions, as the reactants may not have enough time to mix and react efficiently. Similarly, a CSTR may not be ideal for first-order reactions, as the continuous mixing may cause unwanted side reactions.
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2K | 3,328 | 14,202 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.970917 |
https://en.wikipedia.org/wiki/Piecewise-deterministic_Markov_process | 1,680,203,352,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00724.warc.gz | 266,457,838 | 17,516 | # Piecewise-deterministic Markov process
In probability theory, a piecewise-deterministic Markov process (PDMP) is a process whose behaviour is governed by random jumps at points in time, but whose evolution is deterministically governed by an ordinary differential equation between those times. The class of models is "wide enough to include as special cases virtually all the non-diffusion models of applied probability."[1] The process is defined by three quantities: the flow, the jump rate, and the transition measure.[2]
The model was first introduced in a paper by Mark H. A. Davis in 1984.[1]
## Examples
Piecewise linear models such as Markov chains, continuous-time Markov chains, the M/G/1 queue, the GI/G/1 queue and the fluid queue can be encapsulated as PDMPs with simple differential equations.[1]
## Applications
PDMPs have been shown useful in ruin theory,[3] queueing theory,[4][5] for modelling biochemical processes such as DNA replication in eukaryotes and subtilin production by the organism B. subtilis,[6] and for modelling earthquakes.[7] Moreover, this class of processes has been shown to be appropriate for biophysical neuron models with stochastic ion channels.[8]
## Properties
Löpker and Palmowski have shown conditions under which a time reversed PDMP is a PDMP.[9] General conditions are known for PDMPs to be stable.[10]
Galtier and Al.[11] studied the law of the trajectories of PDPM and provided a reference measure in order to express a density of a trajectory of the PDMP. Their work opens the way to any application using densities of trajectory. (For instance, they used the density of a trajectories to perform importance sampling, this work was further developed by Chennetier and Al.[12] to estimate the reliability of industrial systems.)
• Jump diffusion, a generalization of piecewise-deterministic Markov processes
• Hybrid system (in the context of dynamical systems), a broad class of dynamical systems that includes all jump diffusions (and hence all piecewise-deterministic Markov processes)
## References
1. ^ a b c Davis, M. H. A. (1984). "Piecewise-Deterministic Markov Processes: A General Class of Non-Diffusion Stochastic Models". Journal of the Royal Statistical Society. Series B (Methodological). 46 (3): 353–388. doi:10.1111/j.2517-6161.1984.tb01308.x. JSTOR 2345677.
2. ^ Costa, O. L. V.; Dufour, F. (2010). "Average Continuous Control of Piecewise Deterministic Markov Processes". SIAM Journal on Control and Optimization. 48 (7): 4262. arXiv:0809.0477. doi:10.1137/080718541.
3. ^ Embrechts, P.; Schmidli, H. (1994). "Ruin Estimation for a General Insurance Risk Model". Advances in Applied Probability. 26 (2): 404–422. doi:10.2307/1427443. JSTOR 1427443.
4. ^ Browne, Sid; Sigman, Karl (1992). "Work-Modulated Queues with Applications to Storage Processes". Journal of Applied Probability. 29 (3): 699–712. doi:10.2307/3214906. JSTOR 3214906.
5. ^ Boxma, O.; Kaspi, H.; Kella, O.; Perry, D. (2005). "On/off Storage Systems with State-Dependent Input, Output, and Switching Rates". Probability in the Engineering and Informational Sciences. 19. CiteSeerX 10.1.1.556.6718. doi:10.1017/S0269964805050011.
6. ^ Cassandras, Christos G.; Lygeros, John (2007). "Chapter 9. Stochastic Hybrid Modeling of Biochemical Processes" (PDF). Stochastic Hybrid Systems. CRC Press. ISBN 9780849390838.
7. ^ Ogata, Y.; Vere-Jones, D. (1984). "Inference for earthquake models: A self-correcting model". Stochastic Processes and Their Applications. 17 (2): 337. doi:10.1016/0304-4149(84)90009-7.
8. ^ Pakdaman, K.; Thieullen, M.; Wainrib, G. (September 2010). "Fluid limit theorems for stochastic hybrid systems with application to neuron models". Advances in Applied Probability. 42 (3): 761–794. arXiv:1001.2474. doi:10.1239/aap/1282924062.
9. ^ Löpker, A.; Palmowski, Z. (2013). "On time reversal of piecewise deterministic Markov processes". Electronic Journal of Probability. 18. arXiv:1110.3813. doi:10.1214/EJP.v18-1958.
10. ^ Costa, O. L. V.; Dufour, F. (2008). "Stability and Ergodicity of Piecewise Deterministic Markov Processes" (PDF). SIAM Journal on Control and Optimization. 47 (2): 1053. doi:10.1137/060670109.
11. ^ Galtier, T. (2019). "On the optimal importance process for piecewise deterministic Markov process". ESAIM: PS. 23. doi:10.1051/ps/2019015.
12. ^ Chennetier, G. (2022). "Adaptive importance sampling based on fault tree analysis for piecewise deterministic Markov process" (PDF). Preprint. | 1,251 | 4,477 | {"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} | 2.8125 | 3 | CC-MAIN-2023-14 | latest | en | 0.926545 |
https://www.stata.com/statalist/archive/2012-02/msg00879.html | 1,586,289,956,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371805747.72/warc/CC-MAIN-20200407183818-20200407214318-00538.warc.gz | 1,158,505,570 | 5,562 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: Interval regression: Categories with 100% censoring
From Gillian.Frost@hsl.gov.uk To statalist@hsphsun2.harvard.edu Subject Re: st: Interval regression: Categories with 100% censoring Date Mon, 20 Feb 2012 09:57:46 +0000
```Hello Stas,
Gillian
From: Stas Kolenikov <skolenik@gmail.com>
To: statalist@hsphsun2.harvard.edu
Date: 20/02/2012 03:33
Subject: Re: st: Interval regression: Categories with 100%
censoring
Sent by: owner-statalist@hsphsun2.harvard.edu
In this case, the regression parameter for the region may not be
identified. The numeric maximization algorithm pushes this parameter
to the right until the probability of being censored becomes as little
as possible, i.e., computer zero (or c(epsdouble), if you like).
Hence, the probability being above the point of left censoring becomes
computer 1-c(epsdouble). If this is the case, then (1) you would see
absurdly large standard errors on the shift parameter for this region;
(2) the point estimate would likely be a few sigmas (e(sigma)) away
from where it should have been:
. display abs( invnorm( c(epsfloat) ) )
5.1665781
Simple demonstration:
clear
set obs 20
gen byte k = mod(_n,4)
set seed 12345
gen e1 = rnormal()
gen e2 = rnormal()
gen y1 = min( k + e1, k + e2 )
gen y2 = max( k+e1, k+e2 )
replace y1 = . in 15/20
replace y2 = . in 1/7
replace y2 = . if k==1
Then we have
. intreg y1 y2 ibn.k, nocons nolog
Interval regression Number of obs = 19
Wald chi2(4) = 207.14
Log likelihood = -4.4521754 Prob > chi2 = 0.0000
------------------------------------------------------------------------------
| Coef. Std. Err. z P>|z| [95% Conf.
Interval]
-------------+----------------------------------------------------------------
k |
0 | .1423818 .2701057 0.53 0.598 -.3870157
.6717792
1 | 2.570719 233.2821 0.01 0.991 -454.6537
459.7952
2 | 2.460023 .9607074 2.56 0.010 .5770706
4.342974
3 | 3.053332 .2163009 14.12 0.000 2.62939
3.477274
-------------+----------------------------------------------------------------
/lnsigma | -1.227765 .5440687 -2.26 0.024 -2.29412
-.1614097
-------------+----------------------------------------------------------------
sigma | .2929467 .1593831 .1008501
.8509434
------------------------------------------------------------------------------
Observation summary: 5 left-censored observations
0 uncensored observations
9 right-censored observations
5 interval observations
. display (_b[1.k]-1)/e(sigma)
5.3617906
On Sun, Feb 19, 2012 at 1:49 PM, <Gillian.Frost@hsl.gov.uk> wrote:
> Hello all,
>
>
> A number of water samples have been collected, and a microbiological
> examination undertaken to assess the number of colony forming units per
> 100ml (CFU). Some observations are right censored, some are left
> censored, and the censoring point is not always the same. I have
> therefore been using interval regression (-intreg-) to look for regional
> differences in the organism levels.
>
> Based on previous advice from Statalist (thank you!), my dependent
> variable is log10 colony forming units and my independent variable is
the
> categorical variable of region. Some samples were collected from the
same
> location. My command is as follows:
>
> intreg depvar1 depvar2 i.region, vce(cluster location)
>
> This seems to be working quite nicely and the results seem sensible.
> However, when I have a region where all observations are, say, right
> censored, then the predicted log10 CFU for this region is substantially
> higher than the other regions (statistically significantly so). But its
> value does not seem sensible - for example, all regions generally
predict
> around 3-4 CFU, whereas the region with all censored values has a
> predicted value of around 7 CFU!
>
> I am guessing that this is happening because all of the observations are
> right censored and so the model doesn't really have sufficient
information
> to estimate a reliable coefficient. But I cannot find this written
> anywhere.
>
> Do you think it is justifiable to say that "results can be unreliable
when
> all observations are censored, and so regions where this happens were
> excluded from the analysis", or something along those lines?
>
> I would be grateful for any advice.
>
> Many thanks,
>
> Gillian
>
>
>
>
>
>
>
>
>
>
>
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``` | 1,801 | 7,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.666784 |
https://forums.dungeon-quest.com/t/calculating-battle-arena-power/17727?u=mr_scooty | 1,670,033,557,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710918.58/warc/CC-MAIN-20221203011523-20221203041523-00375.warc.gz | 286,824,597 | 10,531 | # Calculating Battle Arena "Power"
When I did my firsts steps on battle arena, I was wondering what “power” on arena was supposed to be. At first glance, I thought it was how much “power points” I had invested on my character, but facing players with higher power values against my fully invested power character made me to discard this hypothesis. After it, I just figured out it was a potential power measure, that I could compare against my opponent. Although, I have found it was not absolute: Facing and defeating players with higher power values and also losing for some with less power than me just encouraged to see battle arena power as a “magic number”. I’ve ignored it until recently, when I found the below post of @marwinberna on forum:
I thought it was such an absurd, cause I never saw a non-hacker player with this value. It made me to question how “power” is really calculated. So I did some empirical tests and I have found a formula that is pretty accurate (also that what Bekhor said is partial true).
## 1 Basics
If you're familiar with damage calculation, just jump for the next section.
### 1. 1 Full Weapon Damage (FWD) (Main Hand Values)
FWD = [ BWD * (1 + IQ) * (1 + WD%) + WD+ + 5 * 75 + floor(PP * 19 / 98) * 75 ] * (1 + ED%) + ED+
Where:
• BWD is MH Weapon’s Base. Every weapon usually have a different BWD, but since it affects in a minimal way on PVP, you can use the 144 (average value for many weapons). For detailed info, look at my spreadsheet here.
• IQ is MH item quality. It is a percentage, so 25% means 0,25.
• WD% is sum of all WD% on MH.
• WD+: Flat Weapon Damage on MH (only).
• 5 * 75: Basic damage, you always have 5 power points.
• PP: Hero PVE Power Points. “floor” means round down function.
• ED%: Elemental Damage% across all gear. Percentage.
• ED+: Flat Elemental Damage across all gear.
### 1.2 Level Conversion Formula
If you have an affix with value X on level 100, then Y is its value at level 20, which can be found using following formula: Y = Xmin + 0.2 * (X - Xmin)
Where:
Xmin is minimal affix value. When crafting items, you will notice that affixes come with ranges, e.g. [10-50] Glasscannon, [10-75] Luck, etc. Value on left bracket is minimal value of it. Items which can’t be crafted, such PTL, you can take similar affixes to deduce its value, e. g. PTL is similar to Glasscannon, so you can assume it has same affix range (it really has). Also, that formula only works for non-eternal item affixes. In order to use it, just divide its value by 2 e. g. 86% Eternal PTL becomes 43%.
Examples:
50% Glasscannon turns: 10% + 0.2 * (50% - 10%) = 18%
45% Push the Limit turns: 10% + 0.2 * (45% - 10%) = 17%
74% Total HP% turns: 5% + 0.2 * (74% - 5%) = 18,8%
etc.
## 2 Evaluating Power
### 2.1 Power Affixes
Unlike your real power (damage output), there are few affixes which affects your battle arena "power". Full Weapon Damage affixes and following affects your battle arena power: - Push the Limit - Glasscannon (and Empower Talent) - Barbarian - Equality - Momentum - Pathfinder - Epiphany Bonus - Haunting Bonus
Not a complete list, I didn’t test every affix in this game, but many to deduce some things as I comment below.
Someone:
• “Wow, I won’t read it anymore, cause you miss Defiant, then you must be wrong.”
It was a big surprise to me that Defiant doesn’t insta boost power, because its description say that there is a guaranteed 25% boost, but you guys will notice that most of “conditional boost affixes” didn’t enter in this list either.
By conditional boosts, I mean boosts that only apply under very specific conditions. Examples: Angelic only boost skills with CD less than 1 sec, Masochism only when your HP is below 75%, Identity only when skill isn’t of your class, Skilled only for primary weapons and others. Even if you satisfy it for one skill, it won’t increases your battle arena power, because it counts only what increases all skills. This is somewhat the case of Defiant: While it always increases 25% damage, it has a conditional parcel which comes from your current HP. This shows that battle arena power is very limited and can be far different from your real power, which comes from conditional boosts and also DPS.
You can also deduce that things such Multi Attack, Extra Attack Chance, Attack SpD, Elemental Crit Chance/DMG, Crit Chance/DMG, Procs and others don’t boost your arena power, since I didn’t mention on my list (LOL) and tested them all.
### 2.2 The Formula
Power = 0.05 * FWD * (1 + MOD1%) * (1 + MOD2%) * (...) * (1 + MODn%)
Where:
-FWD is Full Weapon Damage
-MOD% is a power affix from my list, e. g. you have 2 PTL, then they become 36% (section 1.2), so (1 + MODPTL%) becomes 1 + 0.36 = 1.36
-Product of them all, because different affixes are multiplicative.
-0.05 comes from 95% damage reduction on arena.
### 2.3 Examples
0 Stats build (Quartz Everything) with Gauntlet (BWD = 384) at 25% Quality, 0 PP FWD = (384 * (1 + 0.25) * (1 + 0) + 5 * 75 + floor(0 * 19 / 98) * 75) * (1 + 0) + 0 = 855 Power = 855 * 0.05 = 42.75
0 Stats build (Quartz Everything) with Gauntlet (BWD = 384) at 25% Quality, 98 PP
FWD = (384 * (1 + 0.25) * (1 + 0) + 5 * 75 + floor(98 * 19 / 98) * 75 ) * (1 + 0) + 0 = 2280
Power = 2280 * 0.05 = 114.0
0 Stats build (Quartz Everything) with Gauntlet (BWD = 384) at 25% Quality, 98 PP, 2 +2 All Sets and 1 Equality Set Affix [Since with no HP/MP Mods, thus HP = MP]
FWD = (384 * (1 + 0.25) * (1 + 0) + 5 * 75 + floor(98 * 19 / 98) * 75 ) * (1 + 0) + 0 = 2280
Power = 2280 * (1 + 3 * 0.075) * 0.05 = 139.65
Hypothetical Bekhor’s Build (Solved: True)
Wand has 162 BWD and 50% IQ (Eternal)
3 Eternal ED+ and 4 Epic ED+ = 3 * 2080 + 4 * 1040 = 10400
2x PTL legend + 1x Eternal PTL = 0.18 * 2 + 0.36 = 0.72, but it is capped at 0.4, then it is just 0.4
2x GC + 1x Eternal PTL = 0.18 * 2 + 0.36 = 0.72 (see this update)
4 Empower Talent = 0.025 * 4 = 0.1
FWD = (162 * (1 + 0.5) * (1 + 0) + 5 * 75 + floor(98 * 19 / 98) * 75 ) * (1 + 0) + 10400 = 12443
Power = 12443 * (1 + 0.4) * (1 + 0.72) * (1 + 0.1) * 0.05 = 1648
Not 1700, but very near. Just add one barbarian and you can have more than 1700:
Power = 12443 * (1 + 0.4) * (1 + 0.72) * (1 + 0.1) * (1 + 0.18) * 0.05 = 1945
## Final Commentaries
As you guys could see, battle arena power isn't that great, as it only tells you about a portion of your real power, since it doesn't consider many affix on its own calculation and consider only your MH weapon stats. But if you're still interested on it, this guide can help you with your curiosity. Another use sure is checking classic offensive affixes on a build that you see on arena and "reverse" his damage affixes (homework).
I also recommend making an excel spreadsheet to automatize calculation process, if you are really interested in power, or even if you aren’t, almost everything I said here applies in real power calculation.
For BWD values for all weapons, check my spreadsheet: https://tinyurl.com/DQBWD30
Special Thanks: Bekhor for causing my doubts and Midlumer for his Damage Maximization Guide, from where I got an explained version of damage formula.
Sorry for my bad english wherever you have found it.
@update 06/09/2017: Glasscannon cap is 100%, not 40% like it is supposed to be. Same applies for Barbarian. Not sure about PTL yet, but it’s probably working properly.
@update 07/09/2017: Even WD+ on MH not affecting OH damage, it still has a contribution on Battle Arena Power.
13 Likes
Good Job @luisfsk. Thank you for taking out the time to provide this level of detail and understanding of the PvP power number for all players.
2 Likes
I’m trying to rebuild the said build I told you before. I just need sometime to farm Eternal Pet (which is dream to me)
1 Like
Thank you for this well though out guide.
1 Like
Thank you, it is pleasure understanding this game mechanics, so no problem about spending some free time xD. After all info that I got here, It is a minimal favor I could give back.[quote=“CuzegSpiked, post:4, topic:17727, full:true”]
Thank you for this well though out guide.
[/quote]
Thank you, but I wish I could have a better english to fully clear ideas.
Just in your dreams
1 Like
Your English is good .
1 Like
did you find it? it was a multiplication by 0
Sorry. I get it now.
Are you certain that defiant don’ boost your power in arena? Just my thought, it say that it boost your dmg though. The conditional dmg is when you’re HP becomes low.
I think defiant Increase damage but I can’t tell that easily or if it gives DMG boost as you lose HP. It definitely works at damage Reduction though so no issues there. Lets test.
It boosts damage on arena, I’m sure about it, but doesn’t boost shown battle arena’s power. I mean, there is a conditional part on defiant and no conditional bonus increases shown battle arena’s power.
1 Like
Thank you.
@luisfsk is correct. Defiant does not show in power display but it’s damage boost works in your attacks.
Absoloutely
It seems like flat WD is not part of thr formula. Why is that?
1 Like
Just a minor oversight. Flat weapon damage is added after the weapon quality.
1 Like
Because I forgot. I tested and it also increases power, even it only affecting MH damage. I already did correction on guide, thank you.
1 Like
By any chance you know if AI in practise match have a fix affixes on them?
Probably not, cause sometimes I notice they’re recovering, sometimes reflecting DMG, etc. so I doubt too much.
Is “damage to elites” good to have, or is it pointless…?
It seems pointless although it is free DMG, only to magic+ enemies. It’s only 25% though and only found on certain gar. Whether it’s worth it or not depends on you but I just seen better days to spend affix slots.
In PvP , it’s only 9% instead of 18%. Halved the value of Glasscannon and Barbarian but the difference is that no cost for 25% increased DMG, the only difference is reliance on elites. More powerful with things like Frozen though but not often used in general.
It’s not on you whether it’s worth it or not. I didn’t find it worth the affix slots and because of such conditions. I preferred Execute DMG 100% or Demonic for eg and other forms of DMG that were more easier to achieve without reliance on elites.
2 Likes | 2,925 | 10,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.957416 |
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A 2-Digit number is squared. When this 2-digit number is reversed and squared, the difference between the squares is also a square. What is the 2-digit number? | 214 | 990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-47 | latest | en | 0.894128 |
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Thinking in 3D is sometimes harder than it appears.
Each of the three openings (circle, square, triangle) in this gauge corresponds to an orthogonal projection view of a simple solid.
Three challenges:
1. Provide an alternate description of the solid using two words.
2. Describe a procedure for fabricating the solid.
3. Describe a procedure for modelling the solid in Inventor.
The circle has one inch diameter.
The square has one inch side.
The triangle has one inch base and one inch height.
1: chisel tip?
2: Cut a 1" length of 1" rod. Mark a vertical diameter on one end, and a horizontal diameter on the other. Grind off material such that the ground faces become half-ellipses, so that the curved “end” touches the tip of the horizontal diamter and the straight “end” is the vertical diameter.
(It might be easier and/or safer to do the grinding first, then the cutting. It is also possible to cut the diagonals rather than to grind them off.)
3: Extrude a 1" circle, a 1" square, and the 1"x1" triangle. Intersect them all at right angles (e.g. cylinder along the x axis, square prism along the y axis, and triangular prism along the z axis).
This is a very cool site that contains drawings a ton of polygons. It also has 3D color pictures that can be rotated of polyhedrons of varying complexity, I found one that looks like the Epcot center.
On to the goods
1.) Omni Cork Plug
2.) Put a screw in the bottom of a 1x1 cylinder and hold it verticly ^ (when its done) sorta like that drawn and mill out the triangle shape.
3.) Extrude a 1in. Dia. Cylinder, with a 60 deg. taper, 1in.
Looking at this site i described above i noticed that on the square face the piece removed by the triangle is a perfect parabola.
Actually, that would have to be a half ellipse, since the two extremes are parallel. The sides of a parabola never reach parallel.
Chinese Screwdriver?!
Epcot’s Spaceship Earth is the world’s largest complete geodesic sphere. There are many other geodesic domes scattered around the world and a sphere or two, as well.
If you were to extend the length on the main cylindrical feature the extremes would be parallel, but if you to expand the height and width so that the poece would always be a perfect square the sides of the parabola would just be intersecting with the corners of the square shape and would continue in a parabolic “line”
Good guess.
That’s the one I was looking for.
This one is the probably the best. Note that the mathworld link says there is an infinite family of shapes that satisfy the challenge. The one with minimum volume has triangular cross sections.
I think the mimimum volume cork plug can be (approximately) modelled in Inventor as a loft from a thin rectangle one inch long by 1/1000th inch wide to a one inch diameter circle. The rectangle is sketched on a plane one inch from the circle and the long centerline of the rectangle projects onto the circle’s diameter. I used a model made this way to generate an STP file from which we made a “minimum volume” cork plug on a Viper SLA machine at Emerson.
Of course, Alan’s procedure is more practical. Just round the elliptical edges a bit and you end up pretty close to the minimum volume.
GW is right. The curved edge is elliptical, not parabolic. Referring again to the the mathworld site, an ellipse is a section through a cylinder (a circle is the special case in which the section plane is perpendicular to the cylinder’s axis), while a parabola is a conic section (section through a cone).
Correct. Specifically, a parabola is the special case where the section plane is parallel to one “edge” of the cone, so that the curve never closes on itself: Conic Section -- from Wolfram MathWorld
If you make this like the way everyone else is wouldn’t there me more lines on the triangle? They wouldnt be hidden lines because they are outside of the projection of the triangle face. | 904 | 3,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.911944 |
https://www.techwhiff.com/issue/all-coronado-industries-produces-a-product-requiring--418572 | 1,659,989,508,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570871.10/warc/CC-MAIN-20220808183040-20220808213040-00097.warc.gz | 897,654,994 | 12,271 | All Coronado Industries produces a product requiring 4 pounds of material costing $3.50 per pound. During December, All Coronado purchased 4700 pounds of material for$15792 and used the material to produce 500 products. What was the materials price variance for December
Question:
All Coronado Industries produces a product requiring 4 pounds of material costing $3.50 per pound. During December, All Coronado purchased 4700 pounds of material for$15792 and used the material to produce 500 products. What was the materials price variance for December
50 points. View attached file.
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The doors _____ slowly.
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Ashanti worked hard to make the end of her speech memorable and to leave her audience with an intense feeling that compelled them to clap enthusiastically. This is known as the __________ of a speech.
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1. At 7 a.m., the science classroom had a temperature of 68°F. By 3 p.m., the temperature had increased to 75°F. Calculate the rate of temperature change from 7 a.m. to 3 p.m.
1. At 7 a.m., the science classroom had a temperature of 68°F. By 3 p.m., the temperature had increased to 75°F. Calculate the rate of temperature change from 7 a.m. to 3 p.m....
If there are 6 randomly selected digits in an automobile license tag, and each digit must be one of the 10 integers (0-9), then there are 106 possible license tags.A. TrueB. False
If there are 6 randomly selected digits in an automobile license tag, and each digit must be one of the 10 integers (0-9), then there are 106 possible license tags.A. TrueB. False...
If z = 4.0 and y = 5.6, what is the value of x? z=y/x Express your answer to the tenths place (i.e., one digit after the decimal point).
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What was the name of the first puritian settlement
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A graph is shown below
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Turn the sentences from active to passive.1 The lead actor directs the film.2 Critics gave the film a good review.3 They made the book into a blockbuster4 Computers created the special effects.5 Thousands of fans watched The Hobbit.6 The famous actress plays the lead role.
Turn the sentences from active to passive.1 The lead actor directs the film.2 Critics gave the film a good review.3 They made the book into a blockbuster4 Computers created the special effects.5 Thousands of fans watched The Hobbit.6 The famous actress plays the lead role....
If a human or a prey animal is out and about at night, they are more at risk of being eaten. if the prey are in a safe place sleeping and conserving energy, they are more likely to remain unharmed. this reasoning is consistent with the ________ of sleeping.
If a human or a prey animal is out and about at night, they are more at risk of being eaten. if the prey are in a safe place sleeping and conserving energy, they are more likely to remain unharmed. this reasoning is consistent with the ________ of sleeping....
SOH CAH TOA Find the missing side. Round to the nearest tenth.
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A line has a slope of 2 and contains the points (1, 3) and (2, y). The value of y is -1 1 5 please hurry!
A line has a slope of 2 and contains the points (1, 3) and (2, y). The value of y is -1 1 5 please hurry!...
Which type of matter is likely to absorb the most sound waves? A. Metal door B. Loudspeaker C. Hot air O D. Foam wall
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What is the significance of the Paleolithic era in world history
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What was the result of cesar chavez going on a hunger strike to protest conditions for grape pickers?
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Which group has three unpaired electrons in its p orbital
which group has three unpaired electrons in its p orbital... | 1,040 | 4,371 | {"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} | 2.859375 | 3 | CC-MAIN-2022-33 | latest | en | 0.919638 |
https://math.stackexchange.com/questions/3417937/in-2-norm-case-if-l-is-unitary-frac-l-2-u-2-a-2-leq-n | 1,713,886,421,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818711.23/warc/CC-MAIN-20240423130552-20240423160552-00820.warc.gz | 354,097,091 | 34,758 | # In 2-norm case, if L is unitary, $\frac{\||L|\|_2\||U|\|_2}{\||A|\|_2} \leq n$
Let A, L, U be $$n \times n$$ non-singular matrices such that A=LU.
$$\gamma_1 = \frac{\||L|\|_2\||U|\|_2}{\|A\|_2}$$
$$\gamma_2 = \frac{\|L\|_2\|U\|_2}{\|A\|_2}$$
Show that in 2-norm case, if L is unitary,
$$\gamma_2 = 1, \gamma_1 \leq n$$
for $$\gamma_2 = 1$$, this is my idea:
$$\gamma_2 = \frac{\|L\|_2\|U\|_2}{\|A\|_2} \leq \frac{\|L\|_2\|L^{-1}A\|_2}{\|A\|_2} \leq \frac{\|L\|_2\|L^{-1}\|_2\|A\|_2}{\|A\|_2} = \|L\|_2\|L^{-1}\|_2 = 1$$
$$\gamma_2 = \frac{\|L\|_2\|U\|_2}{\|A\|_2} \geq \frac{\|LU\|_2}{\|A\|_2} = \frac{\|A\|_2}{\|A\|_2} = 1$$
So, $$\gamma_2$$ = 1
But I have no idea for $$\gamma_1 \leq n$$
By the way, there are some facts:
(1) $$\gamma_2 \leq min(\kappa(L),\kappa(U))$$
(2) $$\gamma_1 \leq \gamma_2$$ when 1-norm or $$\infty$$-norm is used
Can any one help me?
• I took out $\gamma$ since it was a bit superfluous and also if you want $n \times n$, use \times, not x. Nov 1, 2019 at 15:34
• Thank you ! \times looks much prettier than x ! Nov 2, 2019 at 0:55 | 552 | 1,076 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-18 | latest | en | 0.569944 |
https://maker.pro/forums/threads/light-meter-problem.276041/ | 1,701,188,649,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099892.46/warc/CC-MAIN-20231128151412-20231128181412-00498.warc.gz | 435,091,826 | 62,970 | # Light Meter Problem
#### muashr
Feb 11, 2014
5
Hi,
Has anyone got experience with this circuit? or can help me out with this?
http://circuitswiring.com/bar-graph-light-meter-by-ic-741/
The LEDs do not light up as the intenstiy of light increases. The voltage tried ranges from (10- 15)V. My questions are:
What type of phototransistor should be used (two legs or three legs and wavelength)?
What type of LEDs should be used (green or red or yellow etc.)?
Why the potentiometer is needed between pins 1 & 5?
What potentiometer's specification (power) work?
I tried to make this circuit work by applying voltages ranging from (10-15)V and adjusting the potentiometer but the it only results in heating up the IC and potentiometer.
Any ideas
Thanks
#### Martaine2005
May 12, 2015
4,776
Hello,
Are you using a dual power supply?
The pot is a 10K.
LEDs can be any colour.
Martin
#### GPG
Sep 18, 2015
452
So many things wrong its hard to know where to start The pot wiper should be connected to -9 for starters
#### Colin Mitchell
Aug 31, 2014
1,416
Have you connected a 9v battery from the 0v rail to the +9v rail and another battery from the 0v rail to the -9v rail?
Now remove the transistor and put a 1k then 10k then 47k then 100k then 220k then 330k in place of the transistor.
If the LEDs don't change, the circuit is faulty.
#### GPG
Sep 18, 2015
452
If the LEDs don't change, the circuit is faulty.
The fact that the pot and IC got hot............
#### CDRIVE
##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
Talk about horrible design! That thing takes the cake. It's also EXTREMELY power hungry! That poor little 9V battery is screaming "You're Killing Me"!
Chris
#### GPG
Sep 18, 2015
452
The question is: do you need a light meter ?
If so there are better ways.
#### CDRIVE
##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
The question is: do you need a light meter ?
If so there are better ways.
Ha, I spiced that circuit for the hell of it. I think you were being too kind when stating there are "better" ways. Any way is better than that way.
Cheers,
Chris
#### CDRIVE
##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
If you want to stay with an LED type display I would recommend something like the LM3914(Dot-Bar) Display Driver. You may be able to directly control it with the proper LDR.
Another option and a whole lot of fun to learn is Picaxe
Cheers,
Chris
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### Comment on
Need Help??
A few weeks after posting this node, I noticed that Korea's leading golfer, "leonid", had jumped from 118 strokes to 111 strokes, thus snatching the Ruby lead from flagitious and me.
To respond to this provocation, I began by staring at my 112 stroker:
```r=*\$<;'XXXXXXXXXXXX'.each_byte{|i|\$><<"%#{327%i/9}c"%
'ohmx'[1>>(i-r.to_i)%12^2&2<<i%12-r[3,2].to_i/5]+\$/*(i/85)}
[download]```
The most obvious candidate for improvement here is the ungainly expression in 'ohmx'[expr], namely:
```1>>(i-r.to_i)%12^2&2<<i%12-r[3,2].to_i/5
[download]```
which can be clarified somewhat by writing as:
```1>>(i-h)%12^2&2<<i%12-m
[download]```
where h represents hours (0..23) and m minutes (0..11). How to shorten?
Though relying on inspiration in cases like these might work on a good day, a more dependable approach is to write a program to search for shorter solutions. Accordingly, I wrote the following Ruby brute force searcher:
```# Search for Ruby solutions by building expressions from the sym
# and numsymop arrays below, then evaluating them (via eval).
# 0, 11, 1, 10, 2, 9, 3, 8, 4, 7, 5, 6
i_arr = [ 120, 47, 253, 22, 194, 21, 183, 44, 196, 55, 125, 246 ]
sym = [ '*', '/', '%', '+', '-', '>>', '<<', '&', '^', '|' ]
numsymop = (-12..12).to_a
numsymop.unshift('i'); numsymop.unshift('-i'); numsymop.unshift('~i')
numsymop.unshift('m'); numsymop.unshift('-m'); numsymop.unshift('~m')
for op2 in numsymop
for s2 in sym
\$stderr.print "#{op2} #{s2}\n"
for op3 in numsymop
for s3 in sym
for op4 in numsymop
for s4 in sym
for op5 in numsymop
expr = op2.to_s + s2 + op3.to_s + s3 + op4.to_s + s4 + o
+p5.to_s
i = 120; m = 0 # (d = 0)
c_expr = "i=#{i};m=#{m};" + expr
veq1 = -4242
begin
veq1 = eval(c_expr)
rescue
end
next if veq1 == -4242
i = 47; m = 11 # (d = 11)
c_expr = "i=#{i};m=#{m};" + expr
veq2 = -4242
begin
veq2 = eval(c_expr)
rescue
end
next if veq2 == -4242
next if veq1 != veq2
i = 120; m = 11 # (d = 0)
c_expr = "i=#{i};m=#{m};" + expr
vneq1 = -4242
begin
vneq1 = eval(c_expr)
rescue
end
next if vneq1 == -4242
next if vneq1 == veq1
i = 47; m = 10 # (d = 11)
c_expr = "i=#{i};m=#{m};" + expr
vneq2 = -4242
begin
vneq2 = eval(c_expr)
rescue
end
next if vneq2 == -4242
next if vneq1 != vneq2
good = 1
for i in i_arr
d = i % 12
m = d
c_expr = "i=#{i};m=#{m};" + expr
veq2 = -4242
begin
veq2 = eval(c_expr)
rescue
end
if veq2 == -4242
good = 0
break
end
if veq1 != veq2
good = 0
break
end
end
next if good == 0
for i in i_arr
d = i % 12
for m in 0..11
next if m == d
c_expr = "i=#{i};m=#{m};" + expr
vneq2 = -4242
begin
vneq2 = eval(c_expr)
rescue
end
if vneq2 == -4242
good = 0
break
end
if vneq1 != vneq2
good = 0
break
end
end
break if good == 0
end
next if good == 0
print "woohoo: #{expr}: veq=#{veq1} vneq=#{vneq1}\n"
\$stdout.flush
end
end
end
end
end
end
end
[download]```
As an aside, writing "eval" search programs like these seems to be a good way to unearth bugs in dusty corners of a language; at least, I had to remove '**' from the sym array above to workaround various random crashes and hangs of the Ruby interpreter.
Anyway, after dutifully chugging away for several hours, the above program did find a lucky hit, namely:
```woohoo: ~m%i%12/11: veq=1 vneq=0
[download]```
Further simple variations of this search program found new solutions for the "hours" portion of the expression also, such as:
```(i-h)%12/~i
[download]```
Combining these two new expressions enabled me to shorten the overall expression as shown below:
```1>>(i-h)%12^2&2<<i%12-m # original
(i-h)%12/~i^~m%i%12/11 # new and improved
[download]```
and thus match leonid's 111 strokes via:
```r=*\$<;'XXXXXXXXXXXX'.each_byte{|i|\$><<"%#{327%i/9}c"%
'hxmo'[(i-r.to_i)%12/~i^~r[3,2].to_i/5%i%12/11]+\$/*(i/85)}
[download]```
Of course, I have no idea if this is the same one stroke improvement that leonid found. If not, and he reads this post, I expect shortly to watch him move ahead to 110 strokes. :)
My main point is that relying on inspiration alone, I doubt I would ever have thought of this new shortened solution. That is, writing and running little search programs like these is part and parcel of being a serious golfer.
Update: Some time later, I discovered an easier way to get to 111 by changing the way hours and minutes are extracted from stdin, as shown below:
```r=*\$<; r r[3,2] # original
gets;~/:/; \$_ \$' # new
[download]```
By way of explanation, note that the Ruby gets function automatically sets \$_. And, like Perl, ~/:/ automatically sets \$'. Unlike Perl though, ~/:/ returns the (zero-based) index of the match. In this game therefore, ~/:/ always returns two (the index of : in, for example, 12:34).
At first glance, the new method looks like a step backwards because it consumes one more stroke than the original. As is often the case in golf though, a tactical trick is available to shorten it. By pure luck, you see, the original 2&2<<i%12-m expression contains two twos! So we can replace one of them with the return value of ~/:/ thus transforming the new approach from one stroke longer to one stroke shorter:
```gets;'XXXXXXXXXXXX'.each_byte{|i|\$><<"%#{327%i/9}c"%'ohmx'[1>>(i-\$_.to
+_i)%12^~/:/&2<<i%12-\$'.to_i/5]+\$/*(i/85)}
[download]```
Alas, this new tactical trick cannot be similarly applied to my shortened ~m%i%12/11 expression because, unluckily, it does not contain a two.
Python Update
Inspired by this new Ruby magic formula, I was later able to shave a stroke from my 127-stroke Python solution by shortening the last part of the magic formula from:
```(i-int(r[:2]))%-12/42
[download]```
to:
```(i-int(r[:2]))%-12/i
[download]```
This (cheating) 126-stroke Python "solution" is not correct for all possible inputs but has a good chance of passing the (poor) set of test data provided for this game. To clarify, I chose 42 in the original magic formula for artistic effect only; any number greater than eleven will do. Replacing 42 with i therefore works for eleven of the twelve string ord values, namely 48, 23, 85, 22, 86, 87, 20, 88, 31, 77, 78 ... but fails for the single ord value that is less than eleven, namely the 9 that forms the string's sixth character.
Hallvabo was kind enough to send me his two best solutions to this game. Here is his (second-placed) 128 stroke Python entry:
```_=raw_input()
for c in'XXXXXXXXXXXX':n=ord(c);print 3459%n/9*' '+'ohmx'[n%13==int(_[
+:2])%12::2][n%13==int(_[3:])/5]+n%3*'\n',
[download]```
where XXXXXXXXXXXX above is a string with ord values 130, 180, 92, 75, 197, 48, 185, 138, 134, 228, 148, 188. And here is an alternate hallvabo 128-stroker:
```_=raw_input()
for c in'XXXXXXXXXXXX':n=ord(c);print 247%n/7*' '+'ohmx'[n%12==int(_[:
+2])%12::2][n%12==int(_[3:])/5]+n/72*'\n',
[download]```
where XXXXXXXXXXXX above is a string with ord values 72, 71, 205, 34, 158, 9, 147, 20, 160, 43, 101, 186. Note that both these solutions require a leading 3-character BOM (0xef, 0xbb, 0xbf). Note too that hallvabo employed his favourite "slice and dice" trick once again, just as he did in 99 Bottles of Beer.
It never occurred to me to use "slice and dice" in this game. As it turns out, the two different techniques produce similar lengths, as indicated below:
```[(i%12==int(r[3:])/5)^(i-int(r[:2]))%-12/42]
[-(i%12==int(r[3:])/5)^1>>(i-int(r[:2]))%12]
[i%12==int(r[3:])/5::2][i%12==int(r[:2])%12]
[i%12==int(r[3:])/5::2][1>>(i-int(r[:2]))%12]
[download]```
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Notices? | 3,010 | 9,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.730008 |
https://nctbsolution.com/nctb-class-8-math-chapter-ten-exercise-10-3-solution/ | 1,709,017,931,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474671.63/warc/CC-MAIN-20240227053544-20240227083544-00130.warc.gz | 430,631,226 | 20,042 | # NCTB Class 8 Math Chapter Ten Exercise 10.3 Solution
NCTB Class 8 Math Chapter Ten Exercise 10.3 Solutions by Math Expert. Bangladesh Board Class 8 Math Solution Chapter Ten Circle 10.3 Solution.
Board NCTB Class 8 Subject Mathematics Chapter 10 Chapter Name Circle Exercise 10.3 Solution
Exercise 10.3
1> On a plane
(i) Innumerable circles can be drawn with two particular points.
(ii) the three points are not on one straight line. Hence, only one circle can be drawn.
(iii) A straight line can intersect at more than two points in a circle. Which one of the following is correct.
Ans: (A)
2> In a circle with radius 2r-
(i)’Circumference is 4 π r unit.
(ii) Diameter is 4r unit.
(iii) Area is 2 π2r2 sq. unit.
Which one of the following is correct?
Ans: (A)
3> For a circle with radius 3 cm, What will be the length of the chord 6 cm from the centre in cm?
Ans: (A)
4> What will be the area of a circle with unit radius?
Ans: (D)
5> What will be the length of a radius of a circle with circumference 23 cm?
Ans:(B)
6> What will be the area in between the space of the two uni centered circles with radius 3 cm and 2 cm?
Ans:(D)
7> The diametre of a wheel of a vehicle is 38 cm. What will be the distance covered by two complete round?
Ans:
Answer questions 8, 9 and 10 on the basis of the following figure:
8> In the figure, 0 is the centre of the circle? what will be the lengths of CD in cm?
Ans:
9> AB=CD, OB =3 cm. What will be the radius of the circle in cm?
Ans:
10> If AB>CD, which of the following will be correct?
Ans:
12> Find the circumference of the circles with the following radius:
Ans:
13> Find the area of the circles given below :
Ans:
14> If the circumference of a circular sheet is 154 cm, find its radius. Also, find the area of the sheet.
Ans:
15> A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase if he makes 2 rounds of the fence. Also, find the cost of the rope if it costs Tk. 18 per metre. 14mm
Ans: Circumference = 2 πr
= 2 x 22/7 x 21
= 132m.
Rope needs= (132 x 2)
= 264m.
Therefore, 1m. rope cost 18 tk
264m. rope cost wil be (18 x 264) = 4752 tk.
16> Find the perimeter of the given shape.
Ans:
18> The height of a right circular cylinder of radius 5.5 cm is 8 cm. Find the area of the whole surfaces of the cylinder (T = 3.14)
Ans:
Updated: March 31, 2021 — 3:20 pm | 693 | 2,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-10 | longest | en | 0.864575 |
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# F(X) = X3 $-$ 1 on R . - CBSE (Arts) Class 12 - Mathematics
#### Question
f(x) = x$-$ 1 on R .
#### Solution
We can observe that f(x) increases when the values of x increase and f(x) decreases when the values of x decrease.
Also, f(x) can be reduced by giving smaller values of x.
Similarly, f(x) can be enlarged by giving larger values of x.
So, f(x) does not have a minimum or maximum value.
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution F(X) = X3 $-$ 1 on R . Concept: Graph of Maxima and Minima.
S | 163 | 564 | {"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} | 3.625 | 4 | CC-MAIN-2019-43 | latest | en | 0.774792 |
https://cs.stackexchange.com/questions/128799/sat-satisfaction-with-10-variables | 1,720,786,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514387.30/warc/CC-MAIN-20240712094214-20240712124214-00160.warc.gz | 119,825,285 | 39,326 | SAT satisfaction with 10 variables
I am trying to prove that the next problem is NPC:
$$A = \{ \langle\phi\rangle \ \big| \ \phi \ \text{is CNF and has sat. assignment where exactly 10 vars are TRUE} \}$$
I am trying to find polynomial mapping reduction from SAT but I can't find a way to force exactly 10 variables to get TRUE assignment. My idea was to create new formula, with 10 clauses, each clause is the intersection of a new variable $$x_i$$ with the old formula, but I don't see how my idea helpful.
I would appreciate help.
• this problem is in P with with an algorithm of $n^{10}$ Commented Jul 29, 2020 at 8:14
• Thanks. I just checked that and I understand. thank you:)
– Ella
Commented Jul 29, 2020 at 8:28
The problem you mentioned is in $$P$$ so it is not NP-complete. We know that $$|\phi| = n$$ so the number of variables is less than $$n$$ and we know that members of $$A$$ exactly have 10 True assignments. So by a brute force algorithm check every possible assignment to variables. Choose 10 variables from n, $$(^{n}_{10})$$ and set them to $$True$$ and set other variables to $$False$$ and check if this is a satisfying assignment. The run time of this algorithm is $$(^{n}_{10})*n = O(n^{11})$$. | 329 | 1,225 | {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-30 | latest | en | 0.943466 |
http://www.appszoom.com/android_developer/educomp-solutions-ltd_fmkdl.html | 1,387,684,976,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345777159/warc/CC-MAIN-20131218054937-00027-ip-10-33-133-15.ec2.internal.warc.gz | 262,488,058 | 8,425 | # Educomp Solutions Ltd.
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FREE | 503 | 2,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.843767 |
https://dice.mpi-inf.mpg.de/fact/horseshoe/is-junction | 1,679,857,967,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00358.warc.gz | 254,916,219 | 3,628 | # horseshoe: is junction
from Quasimodo
## Related concepts
Parents gameWeight: 0.59, picnic tableWeight: 0.52, symbolWeight: 0.51, swimming poolWeight: 0.51, thingWeight: 0.49 Siblings table tennisWeight: 0.33, soccer ballWeight: 0.32, chess gameWeight: 0.32, baseball gameWeight: 0.32, tennis gameWeight: 0.32
## Related properties
Similarity Property is junction 1.00 junction 1.00
## Clauses
### Salient implies Plausible
0.24
Rule weight: 0.28
Evidence weight: 0.86
Similarity weight: 1.00
Evidence: 0.76
Plausible(horseshoe, is junction)
Evidence: 0.59
¬ Salient(horseshoe, is junction)
### Similarity expansion
0.78
Rule weight: 0.85
Evidence weight: 0.91
Similarity weight: 1.00
Evidence: 0.88
Remarkable(horseshoe, is junction)
Evidence: 0.74
¬ Remarkable(horseshoe, junction)
0.74
Rule weight: 0.85
Evidence weight: 0.87
Similarity weight: 1.00
Evidence: 0.11
Typical(horseshoe, is junction)
Evidence: 0.15
¬ Typical(horseshoe, junction)
0.70
Rule weight: 0.85
Evidence weight: 0.83
Similarity weight: 1.00
Evidence: 0.76
Plausible(horseshoe, is junction)
Evidence: 0.71
¬ Plausible(horseshoe, junction)
0.70
Rule weight: 0.85
Evidence weight: 0.83
Similarity weight: 1.00
Evidence: 0.59
Salient(horseshoe, is junction)
Evidence: 0.42
¬ Salient(horseshoe, junction)
### Typical and Remarkable implies Salient
0.13
Rule weight: 0.14
Evidence weight: 0.96
Similarity weight: 1.00
Evidence: 0.59
Salient(horseshoe, is junction)
Evidence: 0.11
¬ Typical(horseshoe, is junction)
Evidence: 0.88
¬ Remarkable(horseshoe, is junction)
### Typical implies Plausible
0.47
Rule weight: 0.48
Evidence weight: 0.97
Similarity weight: 1.00
Evidence: 0.76
Plausible(horseshoe, is junction)
Evidence: 0.11
¬ Typical(horseshoe, is junction) | 617 | 1,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-14 | latest | en | 0.798197 |
https://daytradetowin.com/blog/understanding-the-24-hour-time-format-on-day-trading-charts/ | 1,716,312,314,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058504.35/warc/CC-MAIN-20240521153045-20240521183045-00470.warc.gz | 158,309,628 | 19,776 | # Understanding the 24-hour Time Format on Day Trading Charts
In the U.S. at least, the everyday person does not often encounter the 24-hour time format or “military time” as it’s sometimes called. New traders often encounter the 24-hour format when looking at charts. To assist with understanding, we’ve written this little guide.
Unfortunately, many trading platforms (NinjaTrader included) do not provide charts with regular time. You’re stuck looking at the 24-hour format.
Therefore, to make sense of a chart, you have to do some mental math and memorization until you automatically know what you’re looking at. Here are some tips for understanding the 24-hour time format:
• From 1 a.m. to noon regular time, the 24-hour time format is the same, so no conversion is needed. For example, 1:00 a.m. is 01:00 on a NinjaTrader chart. Likewise, 9:00 a.m. is 09:00 and 12:00 p.m. is 12:00.
• Now, your brainpower is needed to make sense of the following. Once 1:00 p.m. occurs, the time is 13:00. See what we did there? We did +1 from 12:00 p.m., the last “normal” representation of time. Continuing, if we needed to figure out what two hours after 12 p.m. is (2 p.m.), we would do 12 p.m. + 2 = 14:00. The minutes stay the same, by the way.
• Well, what about figuring out the normal time from 24-hour time? Well, you would only need to labor your brain with this if the time is after 12 p.m. and before 1 a.m., right? So, to continue, you would subtract 12 from the 24-hour time to get the regular time. For example, 13:00 – 12 = 1:00 p.m. regular time. Another example: 23:30 – 12 = 11:30 p.m. regular time. Remember, the minutes stay the same.
• On some 24-hour clocks and perhaps NinjaTrader charts, you may see either 00:00 or 24:00 representing midnight. However, once a second rolls past the midnight hour, the time will start with “00”. For example, 00:00:01 is 12:00:01 a.m. regular time.
• Again, minutes (and seconds) behave the same in both time formats. Count up to to 59 seconds, and a new minute begins with the next second.
• If you need to convert a time zone to local time, you will still add or subtract the time zone difference between your time zone to the other geographical region. Not much changes. After your time zone conversion math is done, use the concepts described above to change the format to 24-hour or regular time.
If you want to get used to the 24-hour time faster, consider using a wristwatch that also has 24-hour times listed or adjust your smartphone’s time display. This will help reinforce your understanding, as you are sort of forcing your brain to adapt. | 672 | 2,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.943491 |
https://wizardofvegas.com/forum/gambling/craps/4984-the-mathematics-of-real-world-craps/ | 1,718,584,371,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861674.39/warc/CC-MAIN-20240616233956-20240617023956-00863.warc.gz | 548,270,956 | 13,023 | FleaStiff
• Posts: 14484
Joined: Oct 19, 2009
April 7th, 2011 at 6:41:31 AM permalink
Okay. Several different concepts in one thread here. Sorry about that but I'd like to discuss "a" real world craps betting system. Perhaps its similar to yours, perhaps not. I'd like to know the math related to a certain betting style and a certain time at table rather than a per roll basis.
Here is what the style of play is:
One PassLine bet with 2x odds asap. Odds once up are never off or down.
Two ComeBets with 2x odds asap. Odds once up are never off or down.
Place the Inside Numbers and leave up for FIVE rolls, using any interim place bet winnings to press that placebet. I believe it would be customary that any such place bets would be moved to the "sister number" so as to avoid placing the point number, but its up to the math whizzes to figure out if they want to handle such things or not.
What is a house edge for something like this? Its clear that there is a certain expectation of winning some of those place bets, but the money will be pressed and will not be taken as "winnings".
Is this appreciably different from considering just the textbook house edge? If I resolve to never remove the odds and to always press any winning place bets unless five rolls have gone by, does it make any difference?
TIMSPEED
• Posts: 1246
Joined: Aug 11, 2010
April 7th, 2011 at 1:38:10 PM permalink
When playing the passline, why would you want to make a come bet, because the advantage to both bets (pass and come) is that the 7 is a winner initially. So on your come bet, your advantage becomes the disadvantage because it'll lose your passline+odds bet.
Gambling calls to me...like this ~> http://www.youtube.com/watch?v=4Nap37mNSmQ
odiousgambit
• Posts: 9618
Joined: Nov 9, 2009
April 8th, 2011 at 2:58:46 AM permalink
Quote: FleaStiff
Okay. Several different concepts in one thread here. Sorry about that but I'd like to discuss "a" real world craps betting system. Perhaps its similar to yours, perhaps not. I'd like to know the math related to a certain betting style and a certain time at table rather than a per roll basis.
Here is what the style of play is:
One PassLine bet with 2x odds asap. Odds once up are never off or down.
Two ComeBets with 2x odds asap. Odds once up are never off or down.
Place the Inside Numbers and leave up for FIVE rolls, using any interim place bet winnings to press that placebet. I believe it would be customary that any such place bets would be moved to the "sister number" so as to avoid placing the point number, but its up to the math whizzes to figure out if they want to handle such things or not.
What is a house edge for something like this? Its clear that there is a certain expectation of winning some of those place bets, but the money will be pressed and will not be taken as "winnings".
the math would wear me out and probably be wrong [g], but I would expect the amount you are betting on your place bets to work on your bankroll eventually; hopefully you have some good wins before that.
Quote:
Is this appreciably different from considering just the textbook house edge? If I resolve to never remove the odds and to always press any winning place bets unless five rolls have gone by, does it make any difference?
anytime your system limits your betting, it preserves your bankroll in theory. Otherwise, it is pure superstition to think that some event is now "due" or whatever. But you knew that, right?
Quote: TIMSPEED
When playing the passline, why would you want to make a come bet, because the advantage to both bets (pass and come) is that the 7 is a winner initially. So on your come bet, your advantage becomes the disadvantage because it'll lose your passline+odds bet.
It is a temporary hedge, but more often than not becomes a point to make. I asked the WoO here once about this and might try to find that; essentially he feels the hedging is a non-factor because it is in effect only on the first roll and then becomes an independent action. In the long run your results should be the same if you made all those bets in series or in parallel action. The effect of the 7-out on all those multiple bets that typically start riding simultaneously is devastating to a guy's morale though.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
DrEntropy
• Posts: 199
Joined: Nov 13, 2009
April 8th, 2011 at 8:08:28 AM permalink
Quote: FleaStiff
Okay. Several different concepts in one thread here. Sorry about that but I'd like to discuss "a" real world craps betting system. Perhaps its similar to yours, perhaps not. I'd like to know the math related to a certain betting style and a certain time at table rather than a per roll basis.
Whatever strategy I come up with I like to code up in "WinCraps". It can figure out the house edge, and it also helps you to clearly define your strategy. There are some things that are not so well defined in your specification. What happens after the five rolls? When do put the place numbers back up? But if you code it up yourself, you can specify all this.
The weakest part of your strategy, on the face of it, is placing the 5 and 9, which has a high house edge.
"Mathematical expectation has nothing to do with results." (Sklansky, Theory of Poker).
FleaStiff
• Posts: 14484
Joined: Oct 19, 2009
May 21st, 2011 at 4:43:36 AM permalink
Let us consider a ten dollar table.
Its going to be 30.00 each for the PassLine and ComeBets... so that is 90.00 "invested".
Its going to be 44.00 Inside for the 5, 6, 8, 9 Place Bets.
Worst case would then be: \$134.00 "invested" only to have no interim hits and to then have a Seven-Out that wipes away my entire "investment".
Best case would be that those place bets hit and be pressed higher and higher so that after my mandated Five Rolls, every successful roll after that means higher money on those Inside Numbers.
But if I am committing myself to 134 dollars invested and five rolls of risk being involved, its not really some sort of 1.414 percent house edge game, is it?
MathExtremist
• Posts: 6526
Joined: Aug 31, 2010
May 21st, 2011 at 7:57:31 AM permalink
Quote: FleaStiff
Okay. Several different concepts in one thread here. Sorry about that but I'd like to discuss "a" real world craps betting system. Perhaps its similar to yours, perhaps not. I'd like to know the math related to a certain betting style and a certain time at table rather than a per roll basis.
Here is what the style of play is:
One PassLine bet with 2x odds asap. Odds once up are never off or down.
Two ComeBets with 2x odds asap. Odds once up are never off or down.
Place the Inside Numbers and leave up for FIVE rolls, using any interim place bet winnings to press that placebet. I believe it would be customary that any such place bets would be moved to the "sister number" so as to avoid placing the point number, but its up to the math whizzes to figure out if they want to handle such things or not.
What is a house edge for something like this? Its clear that there is a certain expectation of winning some of those place bets, but the money will be pressed and will not be taken as "winnings".
Is this appreciably different from considering just the textbook house edge? If I resolve to never remove the odds and to always press any winning place bets unless five rolls have gone by, does it make any difference?
You haven't been specific enough -- when do you make the place bets, etc? Is it after or before the come bets? Before the come-out roll - and are they working?
The house edge is simply the house edge of the various bets you're making times the weighted average of how often they're working. The second come bet will happen less than 100% of the time, for example, since sometimes the shooter does point-7 or point-winner and you can't make the bet.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
FleaStiff
• Posts: 14484
Joined: Oct 19, 2009
May 21st, 2011 at 8:31:28 AM permalink
I was contemplating Placing the Inside numbers as soon as possible and working whenever possible.
I do realize that between the two extremes, there are a variety of possibilities wherein some of those place bets pay off and some of those rolls are naturals, some are craps and some are point numbers.
I just want to focus on the fact that I'm committing myself to not take any winnings but to press bets.
An alternative would be to just make a line bet and an odds bet that total about the same amount, 134.00. I just don't know if there is a way to analyze the wisdom of having separate modest bets for five rolls or one bet, with odds, that either Passes or fails.
Its just that I can't see this 1.414 percent edge as what the casino is really making its money on. The casino keeps 20 percent of most people's buy-ins and 100 percent of my buy-ins, so this 1.414 percent sure ain't the proper figure.
MathExtremist
• Posts: 6526
Joined: Aug 31, 2010
May 21st, 2011 at 9:10:33 AM permalink
Quote: FleaStiff
Its just that I can't see this 1.414 percent edge as what the casino is really making its money on. The casino keeps 20 percent of most people's buy-ins and 100 percent of my buy-ins, so this 1.414 percent sure ain't the proper figure.
First, it's not the passline where the casino makes its money. A \$10 line bet is worth 14c to the casino. At the standard rate of 35 come-outs per hour, that's about \$5/hour. They make their money on the prop bets and outside place/buy bets. And the field.
Second, don't confuse buy-in with action. If Joe buys in with \$100 and leaves with \$20, he's lost 20% of his buy-in. If he gave the table \$1000 in action, he's only lost 2% of his action. The EV of a wager is related to win/action, not win/buy-in (e.g. win/drop).
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
thecesspit
• Posts: 5936
Joined: Apr 19, 2010
May 21st, 2011 at 9:12:32 AM permalink
Remember, it's 1.41% of every pass line bet made till resolution. So the house takes it chunk out of buy ins over several bets made (the 20% drop). If you place the six, then push each time it hits, your paying a bigger and bigger actual EV time as you raise it up.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
SanchoPanza
• Posts: 3502
Joined: May 10, 2010
May 21st, 2011 at 9:53:59 AM permalink
Quote: thecesspit
Remember, it's 1.41% of every pass line bet made till resolution.
Not if the resolution is a win for the player.
thecesspit
• Posts: 5936
Joined: Apr 19, 2010
May 21st, 2011 at 3:28:48 PM permalink
Quote: SanchoPanza
Not if the resolution is a win for the player.
At that point, the expected value becomes an actual value...
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
SanchoPanza
• Posts: 3502
Joined: May 10, 2010
May 21st, 2011 at 3:48:49 PM permalink
Quote: thecesspit
At that point, the expected value becomes an actual value...
Isn't it quite a bit more? Because in calculating expected value, isn't the likelihood or unlikelihood of a loss is factored in?
thecesspit | 3,026 | 11,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.957251 |
https://numberworld.info/12371517 | 1,590,419,502,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00443.warc.gz | 448,475,354 | 3,870 | Number 12371517
Properties of number 12371517
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
bcc63d
Base 32:
bphht
sin(12371517)
0.50414104578283
cos(12371517)
0.86362133250458
tan(12371517)
0.58375242343861
ln(12371517)
16.330907372258
lg(12371517)
7.0924229562447
sqrt(12371517)
3517.3167329656
Square(12371517)
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12371517 which is pronounced (twelve million three hundred seventy-one thousand five hundred seventeen) is a great number. The cross sum of 12371517 is 27. If you factorisate the figure 12371517 you will get these result 3 * 3 * 1374613. The number 12371517 has 6 divisors ( 1, 3, 9, 1374613, 4123839, 12371517 ) whith a sum of 17869982. The number 12371517 is not a prime number. The figure 12371517 is not a fibonacci number. The figure 12371517 is not a Bell Number. 12371517 is not a Catalan Number. The convertion of 12371517 to base 2 (Binary) is 101111001100011000111101. The convertion of 12371517 to base 3 (Ternary) is 212021112112100. The convertion of 12371517 to base 4 (Quaternary) is 233030120331. The convertion of 12371517 to base 5 (Quintal) is 11131342032. The convertion of 12371517 to base 8 (Octal) is 57143075. The convertion of 12371517 to base 16 (Hexadecimal) is bcc63d. The convertion of 12371517 to base 32 is bphht. The sine of the number 12371517 is 0.50414104578283. The cosine of 12371517 is 0.86362133250458. The tangent of the figure 12371517 is 0.58375242343861. The square root of 12371517 is 3517.3167329656.
If you square 12371517 you will get the following result 153054432881289. The natural logarithm of 12371517 is 16.330907372258 and the decimal logarithm is 7.0924229562447. You should now know that 12371517 is very great number! | 657 | 1,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-24 | latest | en | 0.751865 |
https://en.wikipedia.org/wiki/Repeated_squaring | 1,713,316,224,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817112.71/warc/CC-MAIN-20240416222403-20240417012403-00763.warc.gz | 212,762,145 | 32,860 | # Exponentiation by squaring
(Redirected from Repeated squaring)
In mathematics and computer programming, exponentiating by squaring is a general method for fast computation of large positive integer powers of a number, or more generally of an element of a semigroup, like a polynomial or a square matrix. Some variants are commonly referred to as square-and-multiply algorithms or binary exponentiation. These can be of quite general use, for example in modular arithmetic or powering of matrices. For semigroups for which additive notation is commonly used, like elliptic curves used in cryptography, this method is also referred to as double-and-add.
## Basic method
### Recursive version
The method is based on the observation that, for any integer ${\displaystyle n>0}$, one has:
${\displaystyle x^{n}={\begin{cases}x\,(x^{2})^{(n-1)/2},&{\mbox{if }}n{\mbox{ is odd}}\\(x^{2})^{n/2},&{\mbox{if }}n{\mbox{ is even}}\end{cases}}}$
If the exponent is zero then the answer is 1 and if the exponent is negative then we can reuse the previous formula by rewriting the value using a positive exponent. That is,
${\displaystyle x^{-n}=\left({\frac {1}{x}}\right)^{n}\,.}$
Together, these may be implemented directly as the following recursive algorithm:
In: an integer x; an integer n
Out: xn
exp_by_squaring(x, n)
if n < 0 then
return 1 / exp_by_squaring(x, -n);
else if n = 0 then
return 1;
else if n is even then
return exp_by_squaring(x * x, n / 2);
else if n is odd then
return x * exp_by_squaring(x * x, (n - 1) / 2);
end function
In each recursive call, the least significant digits of the binary representation of n is removed. It follows that the number of recursive calls is ${\displaystyle \lceil \log _{2}n\rceil ,}$ the number of bits of the binary representation of n. So this algorithm computes this number of squares and a lower number of multiplication, which is equal to the number of 1 in the binary representation of n. This logarithmic number of operations is to be compared with the trivial algorithm which requires n − 1 multiplications.
This algorithm is not tail-recursive. This implies that it requires an amount of auxiliary memory that is roughly proportional to the number of recursive calls -- or perhaps higher if the amount of data per iteration is increasing.
The algorithms of the next section use a different approach, and the resulting algorithms needs the same number of operations, but use an auxiliary memory that is roughly the same as the memory required to store the result.
### With constant auxiliary memory
The variants described in this section are based on the formula
${\displaystyle yx^{n}={\begin{cases}(yx)\,(x^{2})^{(n-1)/2},&{\mbox{if }}n{\mbox{ is odd}}\\y\,(x^{2})^{n/2},&{\mbox{if }}n{\mbox{ is even}}.\end{cases}}}$
If one applies recursively this formula, by starting with y = 1, one gets eventually an exponent equal to 0, and the desired result is then the left factor.
This may be implemented as a tail-recursive function:
Function exp_by_squaring(x, n)
return exp_by_squaring2(1, x, n)
Function exp_by_squaring2(y, x, n)
if n < 0 then return exp_by_squaring2(y, 1 / x, -n);
else if n = 0 then return y;
else if n is even then return exp_by_squaring2(y, x * x, n / 2);
else if n is odd then return exp_by_squaring2(x * y, x * x, (n - 1) / 2).
The iterative version of the algorithm also uses a bounded auxiliary space, and is given by
Function exp_by_squaring_iterative(x, n)
if n < 0 then
x := 1 / x;
n := -n;
if n = 0 then return 1
y := 1;
while n > 1 do
if n is odd then
y := x * y;
n := n - 1;
x := x * x;
n := n / 2;
return x * y
The correctness of the algorithm results from the fact that ${\displaystyle yx^{n}}$ is invariant during the computation; it is ${\displaystyle 1\cdot x^{n}=x^{n}}$ at the beginning; and it is ${\displaystyle yx^{1}=xy}$ at the end.
These algorithms use exactly the same number of operations as the algorithm of the preceding section, but the multiplications are done in a different order.
## Computational complexity
A brief analysis shows that such an algorithm uses ${\displaystyle \lfloor \log _{2}n\rfloor }$ squarings and at most ${\displaystyle \lfloor \log _{2}n\rfloor }$ multiplications, where ${\displaystyle \lfloor \;\rfloor }$ denotes the floor function. More precisely, the number of multiplications is one less than the number of ones present in the binary expansion of n. For n greater than about 4 this is computationally more efficient than naively multiplying the base with itself repeatedly.
Each squaring results in approximately double the number of digits of the previous, and so, if multiplication of two d-digit numbers is implemented in O(dk) operations for some fixed k, then the complexity of computing xn is given by
${\displaystyle \sum \limits _{i=0}^{O(\log n)}{\big (}2^{i}O(\log x){\big )}^{k}=O{\big (}(n\log x)^{k}{\big )}.}$
## 2k-ary method
This algorithm calculates the value of xn after expanding the exponent in base 2k. It was first proposed by Brauer in 1939. In the algorithm below we make use of the following function f(0) = (k, 0) and f(m) = (s, u), where m = u·2s with u odd.
Algorithm:
Input
An element x of G, a parameter k > 0, a non-negative integer n = (nl−1, nl−2, ..., n0)2k and the precomputed values ${\displaystyle x^{3},x^{5},...,x^{2^{k}-1}}$.
Output
The element xn in G
y := 1; i := l - 1
while i ≥ 0 do
(s, u) := f(ni)
for j := 1 to k - s do
y := y2
y := y * xu
for j := 1 to s do
y := y2
i := i - 1
return y
For optimal efficiency, k should be the smallest integer satisfying[1]
${\displaystyle \lg n<{\frac {k(k+1)\cdot 2^{2k}}{2^{k+1}-k-2}}+1.}$
## Sliding-window method
This method is an efficient variant of the 2k-ary method. For example, to calculate the exponent 398, which has binary expansion (110 001 110)2, we take a window of length 3 using the 2k-ary method algorithm and calculate 1, x3, x6, x12, x24, x48, x49, x98, x99, x198, x199, x398. But, we can also compute 1, x3, x6, x12, x24, x48, x96, x192, x199, x398, which saves one multiplication and amounts to evaluating (110 001 110)2
Here is the general algorithm:
Algorithm:
Input
An element x of G, a non negative integer n=(nl−1, nl−2, ..., n0)2, a parameter k > 0 and the pre-computed values ${\displaystyle x^{3},x^{5},...,x^{2^{k}-1}}$.
Output
The element xnG.
Algorithm:
y := 1; i := l - 1
while i > -1 do
if ni = 0 then
y := y2' i := i - 1
else
s := max{i - k + 1, 0}
while ns = 0 do
s := s + 1[notes 1]
for h := 1 to i - s + 1 do
y := y2
u := (ni, ni-1, ..., ns)2
y := y * xu
i := s - 1
return y
Many algorithms for exponentiation do not provide defence against side-channel attacks. Namely, an attacker observing the sequence of squarings and multiplications can (partially) recover the exponent involved in the computation. This is a problem if the exponent should remain secret, as with many public-key cryptosystems. A technique called "Montgomery's ladder"[2] addresses this concern.
Given the binary expansion of a positive, non-zero integer n = (nk−1...n0)2 with nk−1 = 1, we can compute xn as follows:
x1 = x; x2 = x2
for i = k - 2 to 0 do
if ni = 0 then
x2 = x1 * x2; x1 = x12
else
x1 = x1 * x2; x2 = x22
return x1
The algorithm performs a fixed sequence of operations (up to log n): a multiplication and squaring takes place for each bit in the exponent, regardless of the bit's specific value. A similar algorithm for multiplication by doubling exists.
This specific implementation of Montgomery's ladder is not yet protected against cache timing attacks: memory access latencies might still be observable to an attacker, as different variables are accessed depending on the value of bits of the secret exponent. Modern cryptographic implementations use a "scatter" technique to make sure the processor always misses the faster cache.[3]
## Fixed-base exponent
There are several methods which can be employed to calculate xn when the base is fixed and the exponent varies. As one can see, precomputations play a key role in these algorithms.
### Yao's method
Yao's method is orthogonal to the 2k-ary method where the exponent is expanded in radix b = 2k and the computation is as performed in the algorithm above. Let n, ni, b, and bi be integers.
Let the exponent n be written as
${\displaystyle n=\sum _{i=0}^{w-1}n_{i}b_{i},}$
where ${\displaystyle 0\leqslant n_{i} for all ${\displaystyle i\in [0,w-1]}$.
Let xi = xbi.
Then the algorithm uses the equality
${\displaystyle x^{n}=\prod _{i=0}^{w-1}x_{i}^{n_{i}}=\prod _{j=1}^{h-1}{\bigg [}\prod _{n_{i}=j}x_{i}{\bigg ]}^{j}.}$
Given the element x of G, and the exponent n written in the above form, along with the precomputed values xb0...xbw−1, the element xn is calculated using the algorithm below:
y = 1, u = 1, j = h - 1
while j > 0 do
for i = 0 to w - 1 do
if ni = j then
u = u × xbi
y = y × u
j = j - 1
return y
If we set h = 2k and bi = hi, then the ni values are simply the digits of n in base h. Yao's method collects in u first those xi that appear to the highest power ${\displaystyle h-1}$; in the next round those with power ${\displaystyle h-2}$ are collected in u as well etc. The variable y is multiplied ${\displaystyle h-1}$ times with the initial u, ${\displaystyle h-2}$ times with the next highest powers, and so on. The algorithm uses ${\displaystyle w+h-2}$ multiplications, and ${\displaystyle w+1}$ elements must be stored to compute xn.[1]
### Euclidean method
The Euclidean method was first introduced in Efficient exponentiation using precomputation and vector addition chains by P.D Rooij.
This method for computing ${\displaystyle x^{n}}$ in group G, where n is a natural integer, whose algorithm is given below, is using the following equality recursively:
${\displaystyle x_{0}^{n_{0}}\cdot x_{1}^{n_{1}}=\left(x_{0}\cdot x_{1}^{q}\right)^{n_{0}}\cdot x_{1}^{n_{1}\mod n_{0}},}$
where ${\displaystyle q=\left\lfloor {\frac {n_{1}}{n_{0}}}\right\rfloor }$. In other words, a Euclidean division of the exponent n1 by n0 is used to return a quotient q and a rest n1 mod n0.
Given the base element x in group G, and the exponent ${\displaystyle n}$ written as in Yao's method, the element ${\displaystyle x^{n}}$ is calculated using ${\displaystyle l}$ precomputed values ${\displaystyle x^{b_{0}},...,x^{b_{l_{i}}}}$ and then the algorithm below.
Begin loop
Find ${\displaystyle M\in [0,l-1]}$, such that ${\displaystyle \forall i\in [0,l-1],n_{M}\geq n_{i}}$.
Find ${\displaystyle N\in {\big (}[0,l-1]-M{\big )}}$, such that ${\displaystyle \forall i\in {\big (}[0,l-1]-M{\big )},n_{N}\geq n_{i}}$.
Break loop if ${\displaystyle n_{N}=0}$.
Let ${\displaystyle q=\lfloor n_{M}/n_{N}\rfloor }$, and then let ${\displaystyle n_{N}=(n_{M}{\bmod {n}}_{N})}$.
Compute recursively ${\displaystyle x_{M}^{q}}$, and then let ${\displaystyle x_{N}=x_{N}\cdot x_{M}^{q}}$.
End loop;
Return ${\displaystyle x^{n}=x_{M}^{n_{M}}}$.
The algorithm first finds the largest value among the ni and then the supremum within the set of { ni \ iM }. Then it raises xM to the power q, multiplies this value with xN, and then assigns xN the result of this computation and nM the value nM modulo nN.
## Further applications
The approach also works with semigroups that are not of characteristic zero, for example allowing fast computation of large exponents modulo a number. Especially in cryptography, it is useful to compute powers in a ring of integers modulo q. For example, the evaluation of
13789722341 (mod 2345) = 2029
would take a very long time and much storage space if the naïve method of computing 13789722341 and then taking the remainder when divided by 2345 were used. Even using a more effective method will take a long time: square 13789, take the remainder when divided by 2345, multiply the result by 13789, and so on.
Applying above exp-by-squaring algorithm, with "*" interpreted as x * y = xy mod 2345 (that is, a multiplication followed by a division with remainder) leads to only 27 multiplications and divisions of integers, which may all be stored in a single machine word. Generally, any of these approaches will take fewer than 2log2(722340) ≤ 40 modular multiplications.
The approach can also be used to compute integer powers in a group, using either of the rules
Power(x, −n) = Power(x−1, n),
Power(x, −n) = (Power(x, n))−1.
The approach also works in non-commutative semigroups and is often used to compute powers of matrices.
More generally, the approach works with positive integer exponents in every magma for which the binary operation is power associative.
## Signed-digit recoding
In certain computations it may be more efficient to allow negative coefficients and hence use the inverse of the base, provided inversion in G is "fast" or has been precomputed. For example, when computing x2k−1, the binary method requires k−1 multiplications and k−1 squarings. However, one could perform k squarings to get x2k and then multiply by x−1 to obtain x2k−1.
To this end we define the signed-digit representation of an integer n in radix b as
${\displaystyle n=\sum _{i=0}^{l-1}n_{i}b^{i}{\text{ with }}|n_{i}|
Signed binary representation corresponds to the particular choice b = 2 and ${\displaystyle n_{i}\in \{-1,0,1\}}$. It is denoted by ${\displaystyle (n_{l-1}\dots n_{0})_{s}}$. There are several methods for computing this representation. The representation is not unique. For example, take n = 478: two distinct signed-binary representations are given by ${\displaystyle (10{\bar {1}}1100{\bar {1}}10)_{s}}$ and ${\displaystyle (100{\bar {1}}1000{\bar {1}}0)_{s}}$, where ${\displaystyle {\bar {1}}}$ is used to denote −1. Since the binary method computes a multiplication for every non-zero entry in the base-2 representation of n, we are interested in finding the signed-binary representation with the smallest number of non-zero entries, that is, the one with minimal Hamming weight. One method of doing this is to compute the representation in non-adjacent form, or NAF for short, which is one that satisfies ${\displaystyle n_{i}n_{i+1}=0{\text{ for all }}i\geqslant 0}$ and denoted by ${\displaystyle (n_{l-1}\dots n_{0})_{\text{NAF}}}$. For example, the NAF representation of 478 is ${\displaystyle (1000{\bar {1}}000{\bar {1}}0)_{\text{NAF}}}$. This representation always has minimal Hamming weight. A simple algorithm to compute the NAF representation of a given integer ${\displaystyle n=(n_{l}n_{l-1}\dots n_{0})_{2}}$ with ${\displaystyle n_{l}=n_{l-1}=0}$ is the following:
${\displaystyle c_{0}=0}$
for i = 0 to l − 1 do
${\displaystyle c_{i+1}=\left\lfloor {\frac {1}{2}}(c_{i}+n_{i}+n_{i+1})\right\rfloor }$
${\displaystyle n_{i}'=c_{i}+n_{i}-2c_{i+1}}$
return ${\displaystyle (n_{l-1}'\dots n_{0}')_{\text{NAF}}}$
Another algorithm by Koyama and Tsuruoka does not require the condition that ${\displaystyle n_{i}=n_{i+1}=0}$; it still minimizes the Hamming weight.
## Alternatives and generalizations
Exponentiation by squaring can be viewed as a suboptimal addition-chain exponentiation algorithm: it computes the exponent by an addition chain consisting of repeated exponent doublings (squarings) and/or incrementing exponents by one (multiplying by x) only. More generally, if one allows any previously computed exponents to be summed (by multiplying those powers of x), one can sometimes perform the exponentiation using fewer multiplications (but typically using more memory). The smallest power where this occurs is for n = 15:
${\displaystyle x^{15}=x\times (x\times [x\times x^{2}]^{2})^{2}}$ (squaring, 6 multiplies),
${\displaystyle x^{15}=x^{3}\times ([x^{3}]^{2})^{2}}$ (optimal addition chain, 5 multiplies if x3 is re-used).
In general, finding the optimal addition chain for a given exponent is a hard problem, for which no efficient algorithms are known, so optimal chains are typically used for small exponents only (e.g. in compilers where the chains for small powers have been pre-tabulated). However, there are a number of heuristic algorithms that, while not being optimal, have fewer multiplications than exponentiation by squaring at the cost of additional bookkeeping work and memory usage. Regardless, the number of multiplications never grows more slowly than Θ(log n), so these algorithms improve asymptotically upon exponentiation by squaring by only a constant factor at best.
1. ^ In this line, the loop finds the longest string of length less than or equal to k ending in a non-zero value. Not all odd powers of 2 up to ${\displaystyle x^{2^{k}-1}}$ need be computed, and only those specifically involved in the computation need be considered. | 4,758 | 16,622 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 61, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.84164 |
https://www.studypool.com/discuss/489937/recording-entry-accounting?free | 1,506,166,156,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689624.87/warc/CC-MAIN-20170923104407-20170923124407-00101.warc.gz | 860,237,475 | 14,188 | ##### recording entry - accounting
label Accounting
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Derby University sells 4,400 season basketball tickets at \$280 each for its 11-game home schedule.
Give the entry to record (a) the sale of the season tickets and (b) the revenue recognized by playing the first home game.
Apr 22nd, 2015
(a)Dr. Cash \$12,32000
Cr. Unearned basketball ticket revenue \$12,32000
(To record sale of 4,400 season tickets)
(b)\$12,32000 / 11 = \$1,12,000
Dr. Unearned basketball ticket revenue \$1,12,000
Apr 22nd, 2015
Thank you so much!
Apr 22nd, 2015
u welcome
Apr 22nd, 2015
A)Beth's pay for the period is:
(40 X \$22) + (10 X \$33) = \$1210
Entries (based on the facts given) would be:
Debit Salary Expense \$1210
Debit FICA Expense \$92.57 (Employer's portion calculated as \$1210 X 7.65%)
Credit Payroll Taxes Payable \$189.57
Credit Salaries Payable \$1113
The entries under B would then be:
Debit Payroll Taxes Payable \$189.57
Debit Salaries Payable \$1113
Credit Cash \$1302.57
Hope u will be happy
Apr 22nd, 2015
...
Apr 22nd, 2015
...
Apr 22nd, 2015
Sep 23rd, 2017
check_circle | 367 | 1,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | latest | en | 0.840181 |
https://vypros.com/what-happens-to-the-pressure-of-a-gas-if-its-volume-is-decreased-by-half-its-original-size-at-constant-temperature/ | 1,686,360,365,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656869.87/warc/CC-MAIN-20230609233952-20230610023952-00181.warc.gz | 672,492,370 | 22,808 | # What happens to the pressure of a gas if its volume is decreased by half its original size at constant temperature?
W
## What happens to the pressure of a gas if its volume is decreased by half its original size at constant temperature?
Another way of thinking about this law is that the values of pressure and volume are inversely proportional; if one goes up, the other must decrease by the same factor. If you trap gas in a cylinder, and then reduce the internal volume of the cylinder to half its original value, the pressure will double.
## What would happen to the volume of a gas if the pressure on it were decreased and then the gas’s temperature were increased?
What effect does increasing pressure have on gas particles? when the pressure of a gas at constant temperature is increased, the volume of the gas decreases. when the pressure is decreased, the volume increases. the volume of a gas is directly proportional to its temperature(in kelvin) at constant pressure.
Which of the following will happen if the volume of a gas is decreased while the temperature remains constant?
Which of the following will happen if the volume of a gas is decreased while the temperature remains constant? Thus the pressure of the gas will increases as the volume of the gas decreases at constant temperature.
### What happens to the pressure of a gas if you decrease the volume of its container by 1 4?
Boyle’s law Because the volume has decreased, the particles will collide more frequently with the walls of the container. Each time they collide with the walls they exert a force on them. More collisions mean more force, so the pressure will increase. When the volume decreases, the pressure increases.
### What happens to the pressure if the volume is reduced to half and the temperature is doubled?
The law itself can be stated as follows: for a fixed amount of an ideal gas kept at a fixed temperature, P (pressure) and V (volume) are inversely proportional—that is, when one doubles, the other is reduced by half. The moving wall converts the effect of molecular collisions into pressure and acts as a pressure gauge.
What would happen to the volume of a gas if the pressure on that gas were doubled and then the absolute temperature of the gas were doubled?
Doubling the absolute temperature of a gas also doubles its volume, if the pressure is constant, and vice versa.
## What happens to a gas if the pressure is increased?
Boyle found that when the pressure of gas at a constant temperature is increased, the volume of the gas decreases. this relationship between pressure and volume is called Boyle’s law. So, at constant temperature, the answer to your answer is: the volume decreases in the same ratio as the ratio of pressure increases.
## What happens to the pressure of a gas if the temperature is decreased?
the lower the temperature, the lower the kinetic energy of a gas will be, and it will be easier to compress the gas. Using the ideal gas equation PV=nRT, if the volume is constant, a decrease in temperature will cause a decrease in the pressure of the gas.
What will happen to the pressure of a gas if its volume triples increases by a factor of three as the amount of gas and the temperature are held constant?
Explanation: Boyle’s law states that the pressure of a gas and the volume it occupies are inversely proportional. Therefore, if the pressure increases by a factor of 3 (tripled), then at constant temperature, we expect the volume to decrease by a factor of 13 (“cut in third”).
### What happens to the pressure if the gas volume is cut in half while N and T are held constant?
If volume increases, then pressure decreases and vice versa, when the temperature is held constant. Therefore, when the volume is halved, the pressure is doubled; and if the volume is doubled, the pressure is halved.
### What happens when the volume of a gas decreases?
Because the volume has decreased, the particles will collide more frequently with the walls of the container. Each time they collide with the walls they exert a force on them. More collisions mean more force, so the pressure will increase. When the volume decreases, the pressure increases.
What happens when the pressure and temperature of a fixed amount of gas?
If you decrease both the pressure and temperature of a fixed amount of gas, any changes you observe will be in the volume of the gas. Temperature is directly related to volume, and pressure is inversely related to volume.
## How is the pressure of a gas inversely proportional to its volume?
When the volume decreases, the pressure increases. This shows that the pressure of a gas is inversely proportional to its volume. This is shown by the following equation – which is often called Boyle’s law.
## How is the volume of gas affected by charles’law?
Calculations using Charles’ Law involve the change in either temperature (T2) or volume (V2) from a known starting amount of each (V1and T1): Boyle’s Law -states that the volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are constant. | 1,039 | 5,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-23 | latest | en | 0.948687 |
http://projectivegeometricalgebra.org/wiki/index.php?title=Motor&diff=157&oldid=71 | 1,652,740,710,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00490.warc.gz | 48,837,574 | 7,975 | Difference between revisions of "Motor"
A motor is an operator that performs a proper isometry in Euclidean space through the sandwich product. Such isometries encompass all possible combinations of any number of rotations and translations. The name motor is a portmanteau of motion operator.
In the 4D projective geometric algebra $$\mathcal G_{3,0,1}$$, a motor $$\mathbf Q$$ has the general form
$$\mathbf Q = r_x \mathbf e_{41} + r_y \mathbf e_{42} + r_z \mathbf e_{43} + r_w {\large\unicode{x1d7d9}} + u_x \mathbf e_{23} + u_y \mathbf e_{31} + u_z \mathbf e_{12} + u_w$$ .
To possess the geometric property, the components of $$\mathbf Q$$ must satisfy the equation
$$r_x u_x + r_y u_y + r_z u_z + r_w u_w = 0$$ .
Norm
The bulk norm of a motor $$\mathbf Q$$ is given by
$$\left\Vert\mathbf Q\right\Vert_\unicode{x25CF} = \sqrt{\mathbf Q \mathbin{\unicode{x27D1}} \mathbf{\tilde Q}} = \sqrt{u_x^2 + u_y^2 + u_z^2 + u_w^2}$$ ,
and its weight norm is given by
$$\left\Vert\mathbf Q\right\Vert_\unicode{x25CB} = \sqrt{\mathbf Q \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}}\vphantom{\mathbf{\tilde Q}}} = {\large\unicode{x1D7D9}}\sqrt{r_x^2 + r_y^2 + r_z^2 + r_w^2}$$ .
The geometric norm of a motor $$\mathbf Q$$ is thus
$$\left\Vert\mathbf Q\right\Vert = \sqrt{\dfrac{u_x^2 + u_y^2 + u_z^2 + u_w^2}{r_x^2 + r_y^2 + r_z^2 + r_w^2}}$$ ,
and this is equal to half the distance that the origin is moved by the operator.
Exponential Form
A motor $$\mathbf Q$$ can be expressed as the exponential of a line $$\mathbf L$$ multiplied by the dual number $$d + \phi{\large\unicode{x1D7D9}}$$, where $$\phi$$ is half the angle of rotation about the line $$\mathbf L$$, and $$d$$ is half the displacement distance along the line $$\mathbf L$$. This gives us
$$\mathbf Q = \exp_\unicode{x27C7}((d + \phi{\large\unicode{x1D7D9}}) \mathbin{\unicode{x27C7}} \mathbf L)$$ .
This expands to
$$\mathbf Q = \mathbf L\sin\phi + (d \mathbin{\unicode{x27C7}} \mathbf L)\cos\phi - d\sin\phi + {\large\unicode{x1D7D9}}\cos\phi$$ .
Square Root
The square root of a motor $$\mathbf Q$$ is given by
$$\sqrt{\mathbf Q} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\sqrt{2 + 2Q_\smash{\large\unicode{x1D7D9}}}} \mathbin{\unicode{x27C7}} \left({\large\unicode{x1D7D9}} - \dfrac{Q_\mathbf 1}{2 + 2Q_{\large\unicode{x1D7D9}}}\right)$$ .
If $$\mathbf Q$$ is a simple motor, then $$Q_{\mathbf 1} = 0$$, and this reduces to
$$\sqrt{\mathbf Q} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\sqrt{2 + 2Q_\smash{\large\unicode{x1D7D9}}}} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\left\Vert\mathbf Q + \smash{\large\unicode{x1D7D9}}\right\Vert_\unicode{x25CB}}$$ .
Conversion from Motor to Matrix
Given a specific unitized motor $$\mathbf Q$$, define the matrices
$$\mathbf A = \begin{bmatrix}1 - 2(r_y^2 + r_z^2) & 2r_xr_y & 2r_zr_x & 2(r_yu_z - r_zu_y) \\ 2r_xr_y & 1 - 2(r_z^2 + r_x^2) & 2r_yr_z & 2(r_zu_x - r_xu_z) \\ 2r_zr_x & 2r_yr_z & 1 - 2(r_x^2 + r_y^2) & 2(r_xu_y - r_yu_x) \\ 0 & 0 & 0 & 1\end{bmatrix}$$
and
$$\mathbf B = \begin{bmatrix}0 & -2r_zr_w & 2r_yr_w & 2(r_wu_x - r_xu_w) \\ 2r_zr_w & 0 & -2r_xr_w & 2(r_wu_y - r_yu_w) \\ -2r_yr_w & 2r_xr_w & 0 & 2(r_wu_z - r_zu_w) \\ 0 & 0 & 0 & 0\end{bmatrix}$$ .
Then the corresponding 4×4 matrix $$\mathbf M$$ that transforms a point $$\mathbf p$$, regarded as a column matrix, as $$\mathbf p' = \mathbf{Mp}$$ is given by
$$\mathbf M = \mathbf A + \mathbf B$$ .
The inverse of $$\mathbf M$$, which transforms a plane $$\mathbf f$$, regarded as a row matrix, as $$\mathbf f' = \mathbf{fM^{-1}}$$ is given by
$$\mathbf M^{-1} = \mathbf A - \mathbf B$$ .
Conversion from Matrix to Motor
Let $$\mathbf M$$ be an orthogonal 4×4 matrix with determinant +1 having the form
$$\mathbf M = \begin{bmatrix} M_{00} & M_{01} & M_{02} & M_{03} \\ M_{10} & M_{11} & M_{12} & M_{13} \\ M_{20} & M_{21} & M_{22} & M_{23} \\ 0 & 0 & 0 & 1\end{bmatrix}$$ .
Then, by equating the entries of $$\mathbf M$$ to the entries of $$\mathbf A + \mathbf B$$ from above, we have the following four relationships based on the diagonal entries of $$\mathbf M$$:
$$M_{00} - M_{11} - M_{22} + 1 = 4r_x^2$$
$$M_{11} - M_{22} - M_{00} + 1 = 4r_y^2$$
$$M_{22} - M_{00} - M_{11} + 1 = 4r_z^2$$
$$M_{00} + M_{11} + M_{22} + 1 = 4(1 - r_x^2 - r_y^2 - r_z^2) = 4r_w^2$$
And we have the following six relationships based on the off-diagonal entries of $$\mathbf M$$:
$$M_{21} + M_{12} = 4r_yr_z$$
$$M_{02} + M_{20} = 4r_zr_x$$
$$M_{10} + M_{01} = 4r_xr_y$$
$$M_{21} - M_{12} = 4r_xr_w$$
$$M_{02} - M_{20} = 4r_yr_w$$
$$M_{10} - M_{01} = 4r_zr_w$$
If $$M_{00} + M_{11} + M_{22} \geq 0$$, then we calculate
$$r_w = \pm \dfrac{1}{2}\sqrt{M_{00} + M_{11} + M_{22} + 1}$$ ,
where either sign can be chosen. In this case, we know $$|r_w|$$ is at least $$1/2$$, so we can safely divide by $$4r_w$$ in the last three off-diagonal relationships above to solve for $$r_x$$, $$r_y$$, and $$r_z$$. Otherwise, if $$M_{00} + M_{11} + M_{22} < 0$$, then we select one of the first three diagonal relationships based on the largest diagonal entry $$M_{00}$$, $$M_{11}$$, or $$M_{22}$$. After calculating $$r_x$$, $$r_y$$, or $$r_z$$, we plug its value into two of the first three off-diagonal relationships to solve for the other two values of $$r_x$$, $$r_y$$, and $$r_z$$. Finally, we plug it into one of the last three off-diagonal relationships to solve for $$r_w$$.
Setting $$t_x = M_{03}$$, $$t_y = M_{13}$$, and $$t_z = M_{23}$$, the values of $$u_x$$, $$u_y$$, $$u_z$$, and $$u_w$$ are given by
$$u_x = \dfrac{1}{2}(r_wt_x + r_zt_y - r_yt_z)$$
$$u_y = \dfrac{1}{2}(r_wt_y + r_xt_z - r_zt_x)$$
$$u_z = \dfrac{1}{2}(r_wt_z + r_yt_x - r_xt_y)$$
$$u_w = \dfrac{1}{2}(-r_xt_x - r_yt_y - r_zt_z)$$ . | 2,310 | 5,753 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-21 | latest | en | 0.631385 |
https://origin.geeksforgeeks.org/inorder-tree-traversal-without-recursion-and-without-stack/?ref=lbp | 1,675,340,104,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00385.warc.gz | 458,444,873 | 37,649 | # Inorder Tree Traversal without recursion and without stack!
• Difficulty Level : Hard
• Last Updated : 20 Jun, 2022
Using Morris Traversal, we can traverse the tree without using stack and recursion. The idea of Morris Traversal is based on Threaded Binary Tree. In this traversal, we first create links to Inorder successor and print the data using these links, and finally revert the changes to restore original tree.
1. Initialize current as root
2. While current is not NULL
If the current does not have left child
a) Print current’s data
b) Go to the right, i.e., current = current->right
Else
a) Find rightmost node in current left subtree OR
node whose right child == current.
If we found right child == current
a) Update the right child as NULL of that node whose right child is current
b) Print current’s data
c) Go to the right, i.e. current = current->right
Else
a) Make current as the right child of that rightmost
node we found; and
b) Go to this left child, i.e., current = current->left
Although the tree is modified through the traversal, it is reverted back to its original shape after the completion. Unlike Stack based traversal, no extra space is required for this traversal.
## C++
#include using namespace std; /* A binary tree tNode has data, a pointer to left child and a pointer to right child */ struct tNode { int data; struct tNode* left; struct tNode* right; }; /* Function to traverse the binary tree without recursion and without stack */ void MorrisTraversal(struct tNode* root) { struct tNode *current, *pre; if (root == NULL) return; current = root; while (current != NULL) { if (current->left == NULL) { cout << current->data << " "; current = current->right; } else { /* Find the inorder predecessor of current */ pre = current->left; while (pre->right != NULL && pre->right != current) pre = pre->right; /* Make current as the right child of its inorder predecessor */ if (pre->right == NULL) { pre->right = current; current = current->left; } /* Revert the changes made in the 'if' part to restore the original tree i.e., fix the right child of predecessor */ else { pre->right = NULL; cout << current->data << " "; current = current->right; } /* End of if condition pre->right == NULL */ } /* End of if condition current->left == NULL*/ } /* End of while */ } /* UTILITY FUNCTIONS */ /* Helper function that allocates a new tNode with the given data and NULL left and right pointers. */ struct tNode* newtNode(int data) { struct tNode* node = new tNode; node->data = data; node->left = NULL; node->right = NULL; return (node); } /* Driver program to test above functions*/ int main() { /* Constructed binary tree is 1 / \ 2 3 / \ 4 5 */ struct tNode* root = newtNode(1); root->left = newtNode(2); root->right = newtNode(3); root->left->left = newtNode(4); root->left->right = newtNode(5); MorrisTraversal(root); return 0; } // This code is contributed by Sania Kumari Gupta (kriSania804)
## C
#include #include /* A binary tree tNode has data, a pointer to left child and a pointer to right child */ typedef struct tNode { int data; struct tNode* left; struct tNode* right; }tNode; /* Function to traverse the binary tree without recursion and without stack */ void MorrisTraversal(tNode* root) { tNode *current, *pre; if (root == NULL) return; current = root; while (current != NULL) { if (current->left == NULL) { printf("%d ", current->data); current = current->right; } else { /* Find the inorder predecessor of current */ pre = current->left; while (pre->right != NULL && pre->right != current) pre = pre->right; /* Make current as the right child of its inorder predecessor */ if (pre->right == NULL) { pre->right = current; current = current->left; } /* Revert the changes made in the 'if' part to restore the original tree i.e., fix the right child of predecessor */ else { pre->right = NULL; printf("%d ", current->data); current = current->right; } /* End of if condition pre->right == NULL */ } /* End of if condition current->left == NULL*/ } /* End of while */ } /* UTILITY FUNCTIONS */ /* Helper function that allocates a new tNode with the given data and NULL left and right pointers. */ tNode* newtNode(int data) { tNode* node = (tNode *)malloc(sizeof(tNode)); node->data = data; node->left = NULL; node->right = NULL; return (node); } /* Driver program to test above functions*/ int main() { /* Constructed binary tree is 1 / \ 2 3 / \ 4 5 */ tNode* root = newtNode(1); root->left = newtNode(2); root->right = newtNode(3); root->left->left = newtNode(4); root->left->right = newtNode(5); MorrisTraversal(root); return 0; } // This code is contributed by Sania Kumari Gupta (kriSania804)
## Java
// Java program to print inorder // traversal without recursion // and stack /* A binary tree tNode has data, a pointer to left child and a pointer to right child */ class tNode { int data; tNode left, right; tNode(int item) { data = item; left = right = null; } } class BinaryTree { tNode root; /* Function to traverse a binary tree without recursion and without stack */ void MorrisTraversal(tNode root) { tNode current, pre; if (root == null) return; current = root; while (current != null) { if (current.left == null) { System.out.print(current.data + " "); current = current.right; } else { /* Find the inorder predecessor of current */ pre = current.left; while (pre.right != null && pre.right != current) pre = pre.right; /* Make current as right child of its * inorder predecessor */ if (pre.right == null) { pre.right = current; current = current.left; } /* Revert the changes made in the 'if' part to restore the original tree i.e., fix the right child of predecessor*/ else { pre.right = null; System.out.print(current.data + " "); current = current.right; } /* End of if condition pre->right == NULL */ } /* End of if condition current->left == NULL*/ } /* End of while */ } // Driver Code public static void main(String args[]) { /* Constructed binary tree is 1 / \ 2 3 / \ 4 5 */ BinaryTree tree = new BinaryTree(); tree.root = new tNode(1); tree.root.left = new tNode(2); tree.root.right = new tNode(3); tree.root.left.left = new tNode(4); tree.root.left.right = new tNode(5); tree.MorrisTraversal(tree.root); } } // This code has been contributed by Mayank // Jaiswal(mayank_24)
## Python 3
# Python program to do Morris inOrder Traversal: # inorder traversal without recursion and without stack class Node: """A binary tree node""" def __init__(self, data, left=None, right=None): self.data = data self.left = left self.right = right def morris_traversal(root): """Generator function for iterative inorder tree traversal""" current = root while current is not None: if current.left is None: yield current.data current = current.right else: # Find the inorder # predecessor of current pre = current.left while pre.right is not None and pre.right is not current: pre = pre.right if pre.right is None: # Make current as right # child of its inorder predecessor pre.right = current current = current.left else: # Revert the changes made # in the 'if' part to restore the # original tree. i.e., fix # the right child of predecessor pre.right = None yield current.data current = current.right # Driver code """ Constructed binary tree is 1 / \ 2 3 / \ 4 5 """ root = Node(1, right=Node(3), left=Node(2, left=Node(4), right=Node(5) ) ) for v in morris_traversal(root): print(v, end=' ') # This code is contributed by Naveen Aili # updated by Elazar Gershuni
## C#
// C# program to print inorder traversal // without recursion and stack using System; /* A binary tree tNode has data, pointer to left child and a pointer to right child */ class BinaryTree { tNode root; public class tNode { public int data; public tNode left, right; public tNode(int item) { data = item; left = right = null; } } /* Function to traverse binary tree without recursion and without stack */ void MorrisTraversal(tNode root) { tNode current, pre; if (root == null) return; current = root; while (current != null) { if (current.left == null) { Console.Write(current.data + " "); current = current.right; } else { /* Find the inorder predecessor of current */ pre = current.left; while (pre.right != null && pre.right != current) pre = pre.right; /* Make current as right child of its inorder predecessor */ if (pre.right == null) { pre.right = current; current = current.left; } /* Revert the changes made in if part to restore the original tree i.e., fix the right child of predecessor*/ else { pre.right = null; Console.Write(current.data + " "); current = current.right; } /* End of if condition pre->right == NULL */ } /* End of if condition current->left == NULL*/ } /* End of while */ } // Driver code public static void Main(String[] args) { /* Constructed binary tree is 1 / \ 2 3 / \ 4 5 */ BinaryTree tree = new BinaryTree(); tree.root = new tNode(1); tree.root.left = new tNode(2); tree.root.right = new tNode(3); tree.root.left.left = new tNode(4); tree.root.left.right = new tNode(5); tree.MorrisTraversal(tree.root); } } // This code has been contributed // by Arnab Kundu
## Javascript
Output
4 2 5 1 3
Time Complexity : O(n) If we take a closer look, we can notice that every edge of the tree is traversed at most three times. And in the worst case, the same number of extra edges (as input tree) are created and removed.
Auxiliary Space: O(1) since using only constant variables
References:
www.liacs.nl/~deutz/DS/september28.pdf
www.scss.tcd.ie/disciplines/software_systems/…/HughGibbonsSlides.pdf | 3,426 | 12,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.768241 |
http://mathhelpforum.com/discrete-math/198829-propositional-logic-prove-disprove-print.html | 1,527,184,438,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866733.77/warc/CC-MAIN-20180524170605-20180524190605-00489.warc.gz | 184,983,340 | 2,920 | # propositional logic: prove or disprove
Printable View
• May 14th 2012, 08:59 PM
nehal234
propositional logic: prove or disprove
i'm supposed to prove (p → q) ∧ (¬p → r) ⇒ (q ∨ r) with a truth table and im stuck. i've created most of the truth table by hand. but i do no know how to prove or disprove it.
any help is much appreciated.
• May 14th 2012, 10:17 PM
Soroban
Re: propositional logic: prove or disprove
Hello, nehal234!
I don't understand your difficulty.
Quote:
i'm supposed to prove (p → q) ∧ (¬p → r) ⇒ (q ∨ r) with a truth table and im stuck.
i've created most of the truth table by hand. . Most if it? . . . What does that mean?
But i do no know how to prove or disprove it. . Do you understand what a truth table does?
. . $\displaystyle \begin{array}{|c|c|c||cccccccccccc|} p & q & r & [(p & \to & q) & \wedge & (\sim p & \to & r)] & \to & (q & \vee & r) \\ \hline T&T&T & T&T&T&T&F&T&T&{\color{red}T}&T&T&T \\T&T&F & T&F&T&F&F&T&F&{\color{red}T}&T&T&F \\ T&F&T & T&F&F&F&F&T&T&{\color{red}T}&F&T&T \\ T&F&F & T&F&F&F&F&T&F&{\color{red}T}&F&F&F \\ F&T&T & F&T&T&T&T&T&T&{\color{red}T}&T&T&T \\ F&T&F & F&T&T&F&T&F&F&{\color{red}T}&T&T&F \\ F&F&T & F&T&F&T&T&T&T&{\color{red}T}&F&T&T \\ F&F&F & F&T&F&F&T&F&F&{\color{red}T}&F&F&F \\ \hline &&& 1&2&1&3&1&2&1&4&1&2&1 \end{array}$
• May 15th 2012, 12:05 AM
nehal234
Re: propositional logic: prove or disprove
i see how you have used ⇒ and → to be the same. this makes it clear for me.
thanks | 601 | 1,461 | {"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} | 3.625 | 4 | CC-MAIN-2018-22 | latest | en | 0.757058 |
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Mathematics Education Section Curriculum Development Institute Education Bureau
2009 (with updates in August 2018)
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### Contents
Page
Foreword i
Foundation Knowledge 1
Learning Unit 1 Odd and even functions 2
Learning Unit 2 Mathematical induction 3
Learning Unit 3 The binomial theorem 5
Learning Unit 4 More about trigonometric functions 7
Learning Unit 5 Introduction to e 9
Calculus 11
Learning Unit 6 Limits 12
Learning Unit 7 Differentiation 14
Learning Unit 8 Applications of differentiation 17
Learning Unit 9 Indefinite integration and its applications 19
Learning Unit 10 Definite integration 22
Learning Unit 11 Applications of definite integration 24
Algebra 25
Learning Unit 12 Determinants 26
Learning Unit 13 Matrices 28
Learning Unit 14 Systems of linear equations 30
Learning Unit 15 Introduction to vectors 32
Learning Unit 16 Scalar product and vector product 34
Learning Unit 17 Applications of vectors 36
Further Learning Unit
Learning Unit 18 Inquiry and investigation 37
Acknowledgements 38
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### Foreword
The Mathematics Curriculum and Assessment Guide (Secondary 4 – 6) (abbreviated as “C&A Guide” in this booklet) was updated in December 2017 to keep abreast of the ongoing renewal of the school curriculum. The Senior Secondary Mathematics Curriculum consists of a Compulsory Part and an Extended Part. The Extended Part has two optional modules, namely Module 1 (Calculus and Statistics) and Module 2 (Algebra and Calculus).
In the C&A Guide, the Learning Objectives of Module 2 are grouped under different learning units in the form of a table. The notes in the “Remarks” column of the table in the C&A Guide provide supplementary information about the Learning Objectives. The explanatory notes in this booklet aim at further explicating:
1. the requirements of the Learning Objectives of Module 2;
2. the strategies suggested for the teaching of Module 2;
3. the connections and structures among different learning units of Module 2; and 4. the curriculum articulation between the Compulsory Part and Module 2.
Teachers may refer to the “Remarks” column and the suggested lesson time of each Learning Unit in the C&A Guide, with the explanatory notes in this booklet being a supplementary reference, for planning the breadth and depth of treatment in learning and teaching. Teachers are advised to teach the contents of the Compulsory Part and Module 2 as a connected body of mathematical knowledge and develop in students the capability to use mathematics to solve problems, reason and communicate. Furthermore, it should be noted that the ordering of the Learning Units and Learning Objectives in the C&A Guide does not represent a prescribed sequence of learning and teaching. Teachers may arrange the learning content in any logical sequence that takes account of the needs of their students.
Comments and suggestions on this booklet are most welcomed. They should be sent to:
Chief Curriculum Development Officer (Mathematics) Curriculum Development Institute
Education Bureau
4/F, Kowloon Government Offices 405 Nathan Road, Kowloon Fax: 3426 9265
E-mail: ccdoma@edb.gov.hk
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1
Foundation Knowledge
The content of Foundation Knowledge comprises five Learning Units and is considered as the pre-requisite knowledge for Calculus and Algebra of Module 2. These Learning Units serve to bridge the gap between the Compulsory Part and Module 2. Therefore, it should be noted that complicated treatment of topics in Foundation Knowledge is not the objective of the Curriculum.
Learning Unit “Odd and even functions” provides the background knowledge to help students understand the properties of definite integrals involving odd and even functions. Learning Unit
“The binomial theorem” forms the basis of the proofs of some rules in Learning Unit
“Differentiation”. Students should be able to prove propositions by applying Mathematical induction. Learning Unit “More about trigonometric functions” introduces the radian measure of angles, the three new trigonometric functions and some trigonometric formulae commonly used in the learning of Calculus. Students are required to understand the importance of the radian measure in Calculus. Learning Unit “Introduction to e” helps students understand that e and the natural logarithm are important concepts in mathematics and play crucial roles in differentiation and integration in Calculus.
As there is a strong connection between Foundation Knowledge, Calculus and Algebra, teachers should arrange suitable teaching sequences to suit their students’ needs. For example, teachers may integrate Learning Unit “Introduction to e ” into Learning Unit “Limits ” when teaching the definition of e to form a coherent set of learning contents.
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2
Learning Unit Learning Objective Time
Foundation Knowledge 1 Odd and even
functions
1.1 recongise odd and even functions and their graphs 2
Explanatory Notes:
In Learning Unit “Functions and graphs” of the Compulsory Part, students learnt the concept of functions. In this Learning Unit, students should recognise the definitions of odd function and even function and their graphs. In Learning Objective 10.2, some properties of definite integrals involve the concepts of odd and even functions. However, the properties of odd and even functions, for example, ‘odd function + odd function = odd function’ and ‘even function + even function = even function’, etc. are not required in the curriculum.
Students are required to recognise the definition of absolute value function yx and its graph, and that it is an example of even functions. Besides, the formula 1dx ln x
x C
###
and
its proof in Learning Objective 9.2 also involve the concept of the absolute value function.
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3
Learning Unit Learning Objective Tim
e Foundation Knowledge
2. Mathematical induction
2.1 understand the principle of mathematical induction 3
Explanatory Notes:
Mathematical induction is an important tool in proving mathematical propositions. In this Learning Unit, students should be able to use mathematical induction to prove propositions related to the summation of a finite sequence. Students are required to understand the principle of mathematical induction, follows the procedures of mathematical induction and use mathematical induction to solve problems. Using mathematical induction to prove propositions involving inequalities and divisibility are not required in the curriculum.
Teachers may guide students to guess the formulae for some summations of finite sequences and ask them to examine their guesses. Teachers should point out that even though we know that a proposition is true for some positive integers, it was still not sufficient to guarantee that the proposition is true for all positive integers, e.g
1 3 5 ... (2 n 1) n2 (n 1)(n2)(n3)(n4)(n holds when n = 1, 2, 3, 4 and 5 5) but it is not true for other positive integers.
In order to prove mathematical propositions P(n) that are true for all positive integers n by mathematical induction, students are required to notice that the following two steps in the principle of mathematical induction are crucial:
(1) Prove that P(1) is true.
(2) Prove that for any positive integer k, if P(k) is true, then P(k + 1) is also true.
Teachers may use counter-examples to illustrate that if one of the two above steps is incomplete, we cannot prove that P(n) is true for all positive integers n, for example:
(a) For any positive integer n, 1 2 3 1 2 n n
.
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4
(b) For any positive integer n, 1 2 3 ( 1) 2 2 n n n
.
Example (a) shows that only step 1 but not step 2 can be completed for the statement. As a result, the statement “ P(n) is true for all positive integers n” cannot be proved by mathematical induction.
Example (b) shows that only step 2 but not step 1 can be completed for the statement. As a result, the statement “P(n) is true for all positive integers n” cannot be proved by mathematical induction.
In Learning Unit “Arithmetic and geometric sequences and their summations” of the Compulsory Part, students learnt the formulae for the summation of the arithmetic sequence and that of the geometric sequence. Students may try to prove the related formulae by mathematical induction.
Students should be able to apply mathematical induction to prove the binomial theorem in Learning Unit 3.
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5
Learning Unit Learning Objective Time
Foundation Knowledge 3. The binomial
theorem
3.1 expand binomials with positive integral indices using the binomial theorem
3
Explanatory Notes:
At Key Stage 3, students learnt the laws of integral indices, the operations of polynomials and the identity,
ab
###
2 a22ab b 2. To introduce the binomial theorem in this Learning Unit, teachers may let students recognise that using the methods they have learnt at Key Stage 3 to expand
a b
###
n will become very complicated when n is very large.
Students should be able to prove the binomial theorem by mathematical induction learnt in Learning Unit 2.
To present the binomial expansion in a more concise form, students are required to recognise the summation notation ( ):
(a + b)n= C0nan+C1nan–1b +C2nan–2b2+...+Cn–1n abn–1+Cnnbn =
###
r=0 n
Crnan–rbr where n is a positive integer.
Students are also required to recognise the relationships:
1 n
i
a na
and
1 1 1
n n n
r r r r
r r r
ax by a x b y
###
, where a, b are constants.
As the binomial theorem belongs to a learning unit in Foundation Knowledge, the problems and examples involved should be simple and straightforward. In this connection, the following contents are not required in the curriculum:
expansion of trinomials
the greatest coefficient, the greatest term, and the properties of binomial coefficients
applications to numerical approximation
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In Learning Unit 7, students should be able to use the binomial theorem to prove the formula ( n) n 1
d x nx dx
, where n is a positive integer, from first principles.
Besides, teachers may introduce the following historical facts to students:
The arrangement of the binomial coefficients in a triangle is named after Blaise Pascal as he included this triangle with many of its application in his treatise, Traité du triangle arithmétique (1654). In fact, in the 13th century, Chinese mathematician Yang Hui (楊輝) presented the triangle in his book 《詳解九章算法》(1261) and pointed out that Jia Xian (賈憲) had used the triangle to solve problems. Thus, the triangle is also named Yang Hui’s Triangle (楊輝三角) or Jia Xian’s Triangle (賈憲三角).
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7
Learning Unit Learning Objective Time
trigonometric functions
4.1 understand the concept of radian measure
4.2 understand the functions cosecant, secant and cotangent 4.3 understand compound angle formulae and double angle
formulae for the functions sine, cosine and tangent, and product-to-sum and sum-to-product formulae for the functions sine and cosine
15
Explanatory Notes:
In Module 2, students should be able to use radian to express the magnitude of an angle and perform the conversion between radian and degree. In Learning Unit 6 and 7, teachers may explain the significance of learning radian measure in deriving the formula
0
limsin 1
( is in radians) and finding the derivatives of trigonometric functions.
In Learning Unit “More about trigonometry” of the Compulsory Part, students learnt the trigonometric functions sine, cosine and tangent, and their graphs and properties (including maximum and minimum values and periodicity). In this Learning Unit, students are required to understand the other three trigonometric functions cosecant, secant and cotangent, including their definitions and the two related identities: 1 + tan2 = sec2 and 1 + cot2 = cosec2 .
Students should also be able to use these identities to simplify others trigonometric expressions.
Students are required to understand the following formulae:
sin(A B )sin cosA Bcos sinA B
cos(A B )cos cosA B sin sinA B
tan tan
tan( )
1 tan tan
A B
A B A B
sin 2A2sin cosA A
cos 2Acos2Asin2A 1 2sin2A2cos2A 1
2
2 tan tan 2
1 tan A A
A
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8
sin2 1
1 cos 2
A 2 A
cos2 1
1 cos 2
###
A2 A
2sin cosA Bsin(A B ) sin( A B )
2cos cosA Bcos(A B ) cos( A B )
2sin sinA Bcos(A B ) cos( A B )
sin sin 2sin cos
2 2
A B A B
A B
sin sin 2 cos sin
2 2
A B A B
AB
cos cos 2 cos cos
2 2
A B A B
A B
cos cos 2sin sin
2 2
A B A B
AB
Besides, teachers may introduce some connections between the compound angle formulae and the Chord table constructed by Claudius Ptolemy (around AD 100 – 170) of Alexandria and the theorem mainly used in the Chord table, called “Ptolemy’s Theorem”. This theorem can be a topic for investigation in Further Learning Unit of the Compulsory Part.
Students will find that sin2 1
1 cos 2
###
A 2 A and cos2 1
1 cos 2
###
A2 A together with the product-to-sum and sum-to-product formulae are important tools in finding integrals.
In Learning Unit “More about trigonometry” of the Compulsory Part, students learnt to solve simple trigonometric equations with solutions from 0 to 360 only. In this regard, students should also be able to solve trigonometric equations with solutions from 0 to 2 only and this content can be applied to solve optimisation problems in Learning Objective 8.4.
Subsidiary angle form is not required in the curriculum.
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Learning Unit Learning Objective Time
Foundation Knowledge
5. Introduction to e 5.1 recognise the definitions and notations of e and the natural logarithm
2
Explanatory Notes:
Students will find that e and the natural logarithm that are learnt in this Learning Unit will have a significance for the study of Calculus. In Learning Unit “Exponential and logarithmic functions” of the Compulsory Part, students learnt the exponential and logarithmic functions and their graphs. In this Learning Unit, students are required to understand the exponential function
e and the natural logarithmic function ln x. x
Teachers may use different methods to introduce e. For example,
(1) lim(1 1)n
n
en
(2)
2 3
1 2! 3!
x x x
e x
Students are required to recognise that 1 1
n
n
will tend to a number, that is e if the value of n increases. After learning the concept of limits, students should recognise that lim(1 1)n
n
en . However, the proof of the existence of lim(1 1)n
nn is not required in the curriculum.
Teachers may ask students to use calculators or spreadsheets to get the approximate value of lim(1 1)n
nn . Besides, teachers may use dynamic mathematics software to plot the graph of (1 1)n
y n to help students observe the trend of 1 1
n
n
as n increases, and estimate the value of lim(1 1)n
n n .
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Students are required to recognise that
2 3
1 2! 3!
x x x
e x , and can find the approximate value of e by putting x = 1 into this expression.
Students are required to recognise that the natural logarithmic function possesses all the properties of logarithm functions in Learning Unit “Exponential and logarithmic functions” of the Compulsory Part. In differentiation of Calculus, the formula for the change of base is important in finding derivatives of logarithmic functions of different bases .
Since this Learning Unit may involve the concept of limits, the teaching of this Learning Unit may be arranged before that of Learning Objective 6.1.
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Calculus
The content of Calculus comprises six Learning Units related to limits, differentiation, integration and their applications.
Students are required to master the concepts of functions, their graphs and properties before studying limits and differentiation. The limit of a function is an important component of Calculus. With the knowledge of the limit of a function, students could understand the concept of the derivative of a function and the related rules of differentiation. In the applications of differentiation, students should be able to solve problems related to rate of change, maximum and minimum.
The indefinite integral and differentiation are related as reverse processes to each other. The Fundamental Theorem of Calculus links up the two apparently different concepts. At this stage, the applications of the definite integral focus on finding the areas of plane figures and the volumes of solids of revolution. Students can also appreciate how to apply the definite integral to calculate the areas of non-rectilinear figures, for example, the area of a circle, etc.
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Learning Unit Learning Objective Time
Calculus
6. Limits 6.1 understand the intuitive concept of the limit of a function
6.2 find the limit of a function
3
Explanatory Notes:
Students learnt the concepts of various functions and their graphs in Learning Units “Functions and graphs” and “More about graphs of functions” of the Compulsory Part. Dynamic mathematics software is very useful for the exploration of the graphs of functions. The limit of a function is an important component of Calculus. Students are required to use algebraic method and graphs of functions to understand the intuitive concept of the limit of a function. Teachers may use dynamic mathematics software to assist students to grasp the related concept. It should be noted that the rigorous definition of the limit of a function is not required in the curriculum.
Students are required to recognise that the limit of f x does not exist for some functions ( ) ( )
f x when x tends to a , such as the limit of the function 1 ( )
f x does not exit when x x tends to 0.
To distinguish between continuous functions and discontinuous functions from their graphs is not required in the curriculum.
Students are required to recognise the theorems related to the limits of the sum, difference, product, quotient and scalar multiple of functions, and the limits of composite functions, but the proofs are not required in the curriculum. Students should recognise the required conditions for these theorems. For example, students are required to recognise that the truth of the theorem
###
lim ( ) ( ) lim ( ) lim ( )
x a f x g x x a f x x ag x
has assumed that both lim ( )
x a f x
and lim ( )
x ag x
exist.
On the other hand, teachers may ask students to give examples in which lim
( ) ( )
###
x a f x g x
lim ( ) lim ( )
x a f x x ag x
does not hold.
In this Learning Unit, teachers should introduce the concept of composite functions.
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Students should be able to perform the conversion between the expressions such as, 1
2
x x and 12
x 2 x
###
. In this Learning unit, teachers may introduce the above method of conversion when computing the limits such as
lim0
2 2
x
x
x ,lim0 3 3
5 5
x
x
x
and 0
7 7
limx
x
x
, etc.
Students should be able to use two important formulae
0
limsin 1
( is in radians) and
0
lim 1 1
x
x
e
x
to find the derivatives of trigonometric functions and the derivatives of
exponential functions.
Teachers may use different methods, for example, using diagrams or dynamic mathematics software to explain the reasons why the above two formulae hold.
Besides, students should be able to find the limit of a rational function at infinity.
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14
Learning Unit Learning Objective Time
Calculus
7. Differentiation 7.1 understand the concept of the derivative of a function 7.2 understand the addition rule, product rule, quotient rule
and chain rule of differentiation
7.3 find the derivatives of functions involving algebraic functions, trigonometric functions, exponential functions and logarithmic functions
7.4 find derivatives by implicit differentiation 7.5 find the second derivative of an explicit function
13
Explanatory Notes:
In Learning Unit 6 , students learnt the concept of the limit of a function. In this Learning Unit, students are required to understand: Given a functionyf x( ), if
0
( ) ( )
lim
x
f x x f x
x
exists, this limit is defined as the derivative of a function yf x( ) at x . In addition, students should be able to use the graph of the function yf x( ) to explain the derivative of the function
( )
yf x at x as the limit of the slope of the secant line passing through ( , ( ))x f x and (x x f x, ( x)) when x tends to 0. Teachers should also introduce the concept of the tangent to a curve.
Students should be able to find, from first principles, the derivatives of elementary functions, e.g. constant functions, xn (n is a positive integer),
### x
, sin x, cos x, ex and ln x. They should
also be able to apply method such as the conversion between the expressions
### x xx
and x
x x x
to find the derivative of the function 1
x from first principles. Students should be able to use the binomial theorem to prove d ( n) n 1
x nx dx
, where n is a positive integer, from first principles. Students may also prove this formula by mathematical induction.
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15
Students are required to recognise the notations: y', f '(x)and dy
dx for derivatives.
Testing differentiability of functions is not required in the curriculum.
Students are required to understand the addition rule, product rule, quotient rule and chain rule, and should be able to use these rules to find the derivatives of functions.
The rules include:
Addition rule: d (u v) du dv dx dxdx
Product rule: d (uv) vdu udv dx dx dx
Quotient rule: ( ) 2
du dv
v u
d u dx dx
dx v v
Chain rule: dy dy du dxdu dx
Teachers may choose suitable examples such as
2
2 (sin ) (sin )
(sin ) 2sin cos
(sin )
d d x d x
x x x
dxd xdx
to help students understand the chain rule.
Students are required to understand the following formulae:
( )C 0
(xn) nxn1
(sin )x cosx
(cos )x sinx
(tan )x sec2x
( )ex ex
(ln )x 1
x
Students should be able to use the above rules and formulae to find the derivatives of functions
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involving algebraic functions, trigonometric functions, exponential functions, and logarithmic functions. The following algebraic functions are required:
polynomial functions
rational functions
power functions x
functions formed from the above functions through addition, subtraction, multiplication, division and composition, such as x 2 1
To find the derivatives of logarithmic functions with base not equal to e such as ylog2x, students should be able to use the formula for the change of base learnt in Learning Unit
“Exponential and logarithmic functions” of the Compulsory Part:
(log2 ) dy d
dx dx x ln 1 1
(ln )
ln 2 ln 2 ln 2
d x d
dx dx x x
.
Students should be able to use implicit differentiation in finding derivatives of functions.
Equations such as x33xyy3 and 3 x y y2are examples for illustrating the use of implicit differentiation to find dy
dx. It is not easy or impossible to express y in terms of x for some equations. If the purpose is to find the derivative only, it is not necessary for students to express y in terms of x.
Students should be able to use the technique of logarithmic differentiation to find the derivatives of functions such as y(x22)(3x2) (42 x5)6 and
2 1 4
2 1
y x x
, etc.
Students should be able to find the second derivatives of explicit functions and recognise the notations: y , '' f ''( )x and
2 2
d y
dx . Students should be able to apply the second derivative of a function
### f x
to judge the concavity of its graph in a x b. The second derivatives are useful in determining the concavity of functions and finding the extrema of functions in Learning Objective 8.2.
Third and higher order derivatives are not required in the curriculum.
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Learning Unit Learning Objective Time
Calculus
8. Applications of differentiation
8.1 find the equations of tangents to a curve
8.2 find the maximum and minimum values of a function 8.3 sketch curves of polynomial functions and rational
functions
8.4 solve the problems relating to rate of change, maximum and minimum
14
Explanatory Notes:
Students learnt to find the equations of straight lines in the Compulsory Part. In Learning Objective 8.1, students should be able to find not only the equations of tangents passing through a given point on a given curve, but also the equations of the tangents passing through an external point.
In Learning Unit “Functions and graphs” of the Compulsory Part, students learnt to use the graphical method and the algebraic method to find the maximum and the minimum value of a quadratic function. In this Learning Unit, students should be able to apply differentiation to find the maximum and the minimum values of other functions.
Students are required to understand the concepts of increasing, decreasing and concavity of functions, and should be able to apply the related concepts to find the maximum and the minimum values of functions.
Students should be able to use the first derivative and the second derivative to determine whether a turning point of a function is a maximum point or a minimum point, and to find local extrema (local maximum and minimum values) and global extrema (global maximum and minimum values) of functions. If f( )x0 , the second derivative is not applicable to determine the 0 extremum at x . In this case, students have to use the first derivative to find the extrema of x0 the function.
Students should be able to sketch curves of polynomial functions and rational functions.
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18
Students are required to consider the following points in curve sketching:
symmetry of the curve
limitations on the values of x and y
intercepts with the axes
maximum and minimum points
points of inflexion
vertical, horizontal and oblique asymptotes to the curve
Students should be able to use the second derivative to determine the concavity of a function and use these properties to find the points of inflexion of the curve. Students may use dynamic mathematics software to explore whether the tangent to a curve at a point of inflexion may be horizontal or oblique. For more able students, teachers may further discuss whether the tangent to a curve at a point of inflexion may be vertical.
Students should note that it is not necessary for them to consider all of these features when sketching the curve of a function.
Finding the equation of the oblique asymptote of a rational function may involve long division.
Before teaching this Learning Objective, teachers may consolidate students’ knowledge related to the division of polynomials in Learning Unit “More about polynomials” of the Compulsory Part should be consolidated in this Learning Unit.
Teachers should note that students are required to solve the problems relating to rate of change, maximum and minimum in Learning Objective 8.4, and the problems involving displacement, velocity and acceleration are required.
If the problems involve terms from other disciplines, the definitions of these terms should be provided in the problems unless they are “displacement”, “velocity” and “acceleration” in Learning Objective 8.4.
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Learning Unit Learning Objective Time
Calculus 9. Indefinite
integration and its applications
9.1 recognise the concept of indefinite integration
9.2 understand the properties of indefinite integrals and use the integration formulae of algebraic functions,
trigonometric functions and exponential functions to find indefinite integrals
9.3 understand the applications of indefinite integrals in mathematical contexts
9.4 use integration by substitution to find indefinite integrals 9.5 use trigonometric substitutions to find the indefinite
integrals involving a2x2 ,
2 2
1
ax or 2 2
1 x a
9.6 use integration by parts to find indefinite integrals
15
Explanatory Notes:
Students are required to recognise that indefinite integration is the reverse process of differentiation.
Students are required to recognise the notation of indefinite integral:
###
f x dx( ) and the relation
( ) ( )
f x dxF xC
###
and understand the meaning of the constant of integration C in this relation. Students are required to recognise the terms “integrand”, “primitive function” and
“constant of integration”, etc. Students are required to recognise that different methods of indefinite integration may lead to answers which look different, such as
2 2 3 2
1
( 1) ( 2 1) 1
x dx x x dx3x x x C
###
and
2 2 3
2
( 1) ( 1) ( 1) 1( 1) x dx x d x 3 x C
###
.
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Students are required to understand the following properties of indefinite integrals:
###
kf x dx( ) k f x dx
###
( ) where k is a constant
f x( )g x dx( )
f x dx( )
###
g x dx( )
Students are required to understand and should be able to use the following formulae to find indefinite integrals:
###
k dxkxC where k and C are constants.
1
1
n
n x
x dx C
n
where n 1.
1
ln x xdx C
e dxxexC
sin
x dx cosx C
cos
x dxsinx C
###
sec2x dxtanx C
Students are required to understand the applications of indefinite integrals in mathematical contexts such as geometry. If the problems involve terms from other disciplines, the definitions of these terms should be provided in the problems unless they are “displacement”, “velocity”
and “acceleration” in Learning Objective 8.4.
Students should be able to use integration by substitutions for finding the indefinite integrals.
Students should be able to use trigonometric substitutions to find the indefinite integrals involving the forms a2x2 ,
2 2
1
ax or 2 1 2
xa , and are required to recognise the notations: sin1 x, cos1 x and tan1 x and the concepts of their principal values. The integrands containing sin1 x, cos1 x and tan1 x are not required in the curriculum.
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Students should be able to use integration by parts to find indefinite integrals. Teachers can use ln x dx
###
as an example to illustrate the method of integration by parts. It should be noted that in Module 2, the use of integration by parts is limited to at most two times in finding an integral.
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Learning Unit Learning Objective Time
Calculus 10. Definite
integration
10.1 recognise the concept of definite integration 10.2 understand the properties of definite integrals 10.3 find definite integrals of algebraic functions,
trigonometric functions and exponential functions 10.4 use integration by substitution to find definite integrals 10.5 use integration by parts to find definite integrals
10
Explanatory Notes:
Students are required to recognise the definite integral as the limit of a sum and find a definite integral from the definition. Students are required to recognise the notation: b ( )
a f x dx
###
and the
concept of dummy variables, e.g. b ( ) b ( )
a f x dxa f t dt
###
. Using definite integration to find the sum to infinity of a sequence is not required in the curriculum.
Students are required to understand the following properties of definite integrals:
b ( ) a ( )
a f x dx b f x dx
a ( ) 0
a f x dx
###
b ( ) c ( ) b ( )
a f x dxa f x dxc f x dx
###
b ( ) b ( )
akf x dxk a f x dx
###
where k is a constant.
ab
f x( )g x dx( )
ab f x dx( )
abg x dx( )
a ( ) 0
a f x dx
###
if f x is an odd function. ( )
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23
a ( ) 2 0a ( )
a f x dx f x dx
###
if f x is an even function. ( )
Teachers may discuss with students the geometric meanings of the above properties of definite integrals.
For the definite integral a ( )
a f x dx
###
, where f x involves absolute values and is an odd or ( ) even function, students should be able to apply the properties of definite integrals of odd and even functions to have the result such as:
3
3x x dx 0
###
for yx x is an odd function.
Finding other definite integrals whose integrands involve absolute values is not required in the curriculum.
When using integration by substitution to find definite integrals, students should be able to change the upper limit and the lower limit of the definite integral correspondingly.
Students are required to recognise the Fundamental Theorem of Calculus and through the theorem to recognise the relationship between definite integral and indefinite integral:
( ) ( ) ( )
b
a f x dxF bF a
###
, where dxd F x( ) f x( ).
Teachers may also introduce the proof of the Fundamental Theorem of Calculus.
Students should be able to use integration by parts to find definite integrals but the use of integration by parts is limited to at most two times in finding an integral.
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Learning Unit Learning Objective Time
Calculus
11. Applications of definite
integration
11.1 understand the application of definite integrals in finding the area of a plane figure
11.2 understand the application of definite integrals in finding the volume of a solid of revolution about a coordinate axis or a line parallel to a coordinate axis
4
Explanatory Notes:
In this Learning Unit, the applications of definite integration only confine to the calculations of areas of plane figures and volumes of solids of revolution. Teachers may give geometric demonstration on the relationship between the definite integral and the area of a plane figure.
Students are required to understand and should be able to use disc method to find the volumes of solids of revolution such as finding the volume by revolving the region about the y-axis if the region is bounded by the curves
2
2
y x and yex2, where 1 x 2. The shell method is not required in the curriculum.
For the appreciation of the applications of definite integration, teachers may guide students to derive the formulae of the area of circle, the volume of right circular cone and the volume of sphere by using definite integration.
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Algebra
Algebra consists of Determinants, Matrices, Systems of Linear Equations, and Vectors.
Students are required to understand the concepts, operations and properties of matrices, the existence of inverse matrices and the determinants. Determinants are important tools to investigate the properties of matrices.
Students learnt to solve the simultaneous linear equations in two unknowns by algebraic and graphical method at Key Stage 3. In Module 2, students are required to recognise the concepts of consistency and inconsistency and to further explore the conditions of consistency or inconsistency in a system of linear equations. They should be able to use Cramer’s rule, inverse matrices and Gaussian elimination to solve systems of linear equations. Teachers should guide students to know the strengths and weaknesses of each method and how to choose appropriate methods to solve problems.
In order to extend students’ knowledge in Algebra, the concepts, operations and properties of vectors should be included. The scalar product and the vector product are two useful tools to investigate the geometric properties of vectors including parallelism and orthogonality. In addition, students should be able to use the vector method to find the angle between two vectors and the area of a triangle or a parallelogram, etc.
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Learning Unit Learning Objective Time
Algebra
12. Determinants 12.1 recognise the concept of determinants of order 2 and order 3
2
Explanatory Notes:
Determinant is the important pre-requisite knowledge for the learning of matrices and systems of linear equations in the subsequent two Learning Units.
Teachers should emphasise that in Module 2 determinant is mainly used to find the inverse of a square matrix and to solve system of linear equations.
Students are required to recognise the definitions of determinants of order 2 and order 3, such as:
11 12 11 22 12 21
21 22
a a
a a a a
a a
11 12 13
22 23 21 23 21 22
21 22 23 11 12 13
32 33 31 33 31 32
31 32 33
a a a
a a a a a a
a a a a a a
a a a a a a
a a a
11 12 13
22 23 12 13 12 13
21 22 23 11 21 31
32 33 32 33 22 23
31 32 33
a a a
a a a a a a
a a a a a a
a a a a a a
a a a
Use
11 12 13 11 12
21 22 23 21 22
31 32 33 31 32
a a a a a
a a a a a
a a a a a
to denote the determinant of order 3 as follows:
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11 12 13
21 22 23 11 22 33 12 23 31 13 21 32 13 22 31 11 23 32 12 21 33
31 32 33
a a a
a a a a a a a a a a a a a a a a a a a a a
a a a
Teachers may explain to students that the above 3 definitions of a determinant of order 3 are the same.
Students are required to recognise that both |A| and det A are two common notations of the determinant of the matrix A.
Teachers may introduce some geometric uses of determinants, for example:
In the figure, OAB is a triangle where O is the origin, A=(a,b), B=(c,d) and O, A and B are arranged in anticlockwise direction.
Area of the triangle OAB =1 2
a b c d .
The properties of determinants are not required in the curriculum.
y
x B(c,d)
A(a,b) O
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Learning Unit Learning Objective Time
Algebra
13. Matrices 13.1 understand the concept, operations and properties of matrices
13.2 understand the concept, operations and properties of inverses of square matrices of order 2 and order 3
10
Explanatory Notes:
Students are required to understand the general form of a matrix with m rows and n columns, namely “m n matrix”. Students should be able to perform addition, subtraction, scalar multiplication and multiplication of matrices and understand the following properties:
A B B A
A(B C )(A B ) C
( )AAA
(A B )AB
A BC( )(AB C)
A B C( ) ABAC
(A B C ) ACBC
(A)(B)()AB
ABA B
Students are required to understand that the commutative property does not hold for matrix multiplication , i.e. AB is not necessarily equal to BA.
The general proof of ABA B where A and B are square matrices of order n is not required in the curriculum. However, teachers may have more in-depth discussions on this property with students for determinants of order 2 since the proofs are simpler.
Studens are required to recognise the terms, “zero matrix”, “identity matrix”, “transpose of a matrix” and “square matrix” and to understand the concepts, operations and the following
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properties of inverse of square matrices of order 2 and order 3:
the inverse of A is unique
(A 1) 1A
(A)11A1
(An)1(A1)n
(AT)1(A1)T
A1A1
(AB)1B A1 1
where A and B are invertible matrices and is a non-zero scalar.
Students should be able to determine whether a square matrix is invertible and to find the inverse of an invertible matrix, e.g. by using the adjoint matrix and using elementary row operations. In addition, in some circumstances, students may need to use mathematical induction to prove propositions involving matrices.
In order to determine whether a 2 2 matrix a b c d
is invertible, students may consider to
solve the matrix equation 1 0
0 1 a b x y
c d z w
for unknowns x, y, z and w.
(36)
30
Learning Unit Learning Objective Time
Algebra
14. Systems of linear equations
14.1 solve the systems of linear equations in two and three variables by Cramer’s rule, inverse matrices and Gaussian elimination
6
Explanatory Notes:
At Key Stage 3, students learnt using the algebraic and graphical methods to solve linear equations in two unknowns. In this Learning Unit, students should be able to use Cramer’s rule, inverse matrices and Gaussian elimination to solve systems of linear equations in two and three variables, and are required to recognise the terms “homogeneous”, “non-homogeneous”,
“consistency” and “inconsistency”.
Cramer’s rule is an important topic of determinants. Students have to recognise that by Cramer’s rule, for the system of linear equations Ax = b, if is the determinant of the coefficient matrix and 0, the system has a unique solution. If = 0, Cramer’s rule cannot be used. Teachers may discuss with students the logical relation between Ax = b and
x = x , y = y and z = z (*).
For example:
Should any solution in Ax = b be a solution in (*)?
Should any solution in (*) be a solution in Ax = b?
x is the determinant obtained by replacing the first column of the coefficient matrix by the column vector b, y is the determinant obtained by replacing the second column of the coefficient matrix by the column vector b and z is the determinant obtained by replacing the third column of the coefficient matrix by the column vector b. Besides, students are required to recognise the following conclusions:
Case Condition Conclusion
1 0 The system has a unique solution.
2 0 and at least one of x , y or z
0 The system has no solutions.
3 0 and x = y = z = 0 The system has no solutions or infinitely many solutions.
(37)
31
In Case 1, the system has a unique solution and x x,y y,z z
.
In Case 2, as the given condition contradicts (*), the system has no solutions.
In Case 3, teachers may use the following examples to illustrate that the systems have no solutions or infinitely many solutions.
1 2 3 x y z x y z x y z
(no solutions)
3
2 2 2 6
3 3 3 9
x y z
x y z
x y z
(infinitely many solutions)
Matrix is another important tool for solving systems of linear equations. With the knowledge of Learning Unit 13, students should be able to rewrite a system of linear equations in matrix form.
If the inverse of the coefficient matrix exists, the system can be solved by using the inverse matrix. Students are required to recognise that this method becomes invalid if the inverse matrix does not exist.
Students should also be able to solve systems of linear equations by using Gaussian elimination.
By setting up the augmented matrix, elementary row operations can be applied to solve systems of linear equations.
Teachers may demonstrate the linkage between matrices, determinants and elementary row operations in solving systems of linear equations.
Students are required to understand the theorem: a system of homogeneous linear equations has nontrivial solutions if and only if the coefficient matrix is singular. Teachers can use some simple systems of homogeneous linear equations in two variables to guide students to discover this theorem. Students are also required to understand that the systems of homogeneous linear equations in two and three variables are always consistent and know the way to find their nontrivial solution if the coefficient matrices are singular.
(38)
32
Learning Unit Learning Objective Time
Algebra
15. Introduction to vectors
15.1 understand the concepts of vectors and scalars 15.2 understand the operations and properties of vectors 15.3 understand the representation of a vector in the
rectangular coordinate system
5
Explanatory Notes:
In this Learning Unit, teachers should emphasise that the magnitude and direction are two key concepts of vectors. Teachers should explain to students the difference between vectors and scalars. All vectors are restricted to R2 or R3 in the discussion of the vector properties. Students are required to understand the concepts of zero vector, unit vectors, equal vectors and negative vectors.
Students are required to recognise some common notations of vectors in printed form (including a and AB ) and in written form (including a , AB and a ) ; and some notations for magnitude (including a and a ).
Students are required to understand the concepts of addition, subtraction and scalar multiplication of vectors, and the following properties of vectors:
a b b a
a (b c) (a b) c
a 0 a
0 a 0
( a) ( )a
( )aaa
(a b ) ab
If ab1a1b (a and b are non-zero and are not parallel to each other), then
1 and 1.
(39)
33
Teachers may use the representation of vectors in the rectangular coordinate system to discuss the above properties of vectors with students.
Teachers should introduce the vectors i, j and k representing the unit vectors in the directions of the positive x-, y- and z-axis respectively. Student should be able to use the form
xiyj and xiyjzk to express any vector in R2 and R3 respectively.
Students are required to understand the following formulae:
1. |OP| x2y2 when OP xi yj in R2.
2. sin 2y 2
x y
and
2 2
cos x
x y
when is the angle that a non-zero vector OP
makes with the positive x-axis and OP xi yj . 3. |OP| x2y2z2 when OP xi yj zk in R3.
The concept of direction cosines is not required in the curriculum.
Now, nearly all of the current flows through wire S since it has a much lower resistance than the light bulb. The light bulb does not glow because the current flowing through it
- promoting discussion before writing to equip students with more ideas and vocabulary to use in their writing and to enable students to learn how to work in discussion groups and
• Is the school able to make reference to different sources of assessment data and provide timely and effective feedback to students according to their performance in order
Students should also be able to appreciate the interrelation between bonding, structures and properties of substances by learning the properties of metals, giant ionic
In Learning Objective 23.1, students are required to understand and prove the properties of parallelograms, including opposite sides equal, opposite angles equal, and two
Students in this Learning Unit should recognise the concepts of sample statistics and population parameters and their relationships:. Population Parameter
In this Learning Unit, students are required to solve compound linear inequalities in one unknown involving logical connectives “and” or “or”, quadratic inequalities in one
In practice, ρ is usually of order 10 for partial pivot | 11,937 | 46,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-26 | latest | en | 0.809282 |
http://stackoverflow.com/questions/14669563/build-efficient-array-integer-incrementer-with-different-caps-per-number | 1,394,826,044,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678694619/warc/CC-MAIN-20140313024454-00055-ip-10-183-142-35.ec2.internal.warc.gz | 130,475,946 | 16,225 | # Build efficient array integer incrementer with different caps per number
I want to program a counter which is represented by an array of numbers, starting with:
``````[0, 0, 0]
``````
The constraint here is, that each position has a different cap, so it's not necessarily 9 or something else, but it is given. For instance:
``````[4, 2, 1]
``````
Which would lead to the following incrementation sequence:
``````[0, 0, 0]
[0, 0, 1]
[0, 1, 0]
[0, 1, 1]
[0, 2, 0]
[0, 2, 1]
[1, 0, 0]
.
.
.
``````
Of course I can think of a solution using modulo and adding each carryover onto the next position. But has someone an idea how to implement this efficiently, respectively with nice Ruby syntax without cluttering it too much?
That is my naive implementation:
``````max = [10, 1, 1, 1, 10]
counter = [0, 0, 0, 0, 0]
i = counter.length-1
while counter != max do
counter[i] = counter[i] + 1
while counter[i] > max[i]
counter[i] = 0
i = i - 1
counter[i] = counter[i] + 1
end
i = counter.length-1
end
``````
-
Create an array for each cap, with values from 0 upto cap. Take the first array and calculate the Cartesian product with the rest of the arrays.
``````caps = [4, 2, 1]
arrs = caps.map{|cap| (0..cap).to_a} #=>[[0, 1, 2, 3, 4], [0, 1, 2], [0, 1]]
p arrs.shift.product(*arrs)
# =>[[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 1, 1], [0, 2, 0], [0, 2, 1], ...
``````
If you don't want a memory-consuming array with the results, then provide a block. `product` will yield each element to it, one by one.
``````arrs = caps.map{|cap| (0..cap).to_a}
arrs.shift.product(*arrs){|el| puts el.join} #no resulting array
#000
#001
#010
#011
#...
``````
-
I'm not sure about efficiency but here's my shot at it:
``````start = [0, 0, 0]
cap = [4, 2, 1]
start.zip(cap).map{ |i, c| (i..c).to_a }.reduce(&:product).map &:flatten
``````
Produces something like:
``````[[0, 0, 0],
[0, 0, 1],
[0, 1, 0],
[0, 1, 1],
[0, 2, 0],
[0, 2, 1],
[1, 0, 0],
[1, 0, 1],
[1, 1, 0],
[1, 1, 1],
[1, 2, 0],
[1, 2, 1],
[2, 0, 0],
[2, 0, 1]...]
``````
-
Edit: I was writing this before you made your edit. It seemed like you wanted a counter object, not just to output a list.
1) I would recommend specifying not the limits but (limit+1) of each of the digits. For example, for a [second, minute, hour, day, year] counter it makes more sense (to me) to write [60, 60, 24, 365] instead of [59,59,23,364].
2) You'll have to figure out what to do if your counter overflows the last limit of your array. I added an extra position that counts to infinity.
3) I would also recommend reversing the order of the array, at least in the internal representation to avoid inverting subscripts. If you don't want it like that, you can `.reverse` the `bases` in `initialize` and `@digits` in `to_s`
``````class MyCounter
def initialize bases
@bases = bases
@bases << 1.0/0 # Infinity
@digits = Array.new(bases.size, 0)
prod = 1
@digit_values = [1] + @bases[0..-2].map { |b| prod *= b }
end
def to_s
@digits
end
def increment(digit=0)
v = @digits[digit] + 1
if v < @bases[digit]
@digits[digit] = v
else
@digits[digit] = 0
increment(digit+1)
end
self
end
def +(integer)
(@digits.size - 1).step(0,-1).each do |i|
@digits[i] += integer / @digit_values[i]
integer = integer % @digit_values[i]
end
self
end
end
c1 = MyCounter.new [2,3,5]
20.times { c1.increment; p c1 }
c2 = MyCounter.new [2,3,5]
c2 += 20
p c2
``````
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http://www.mywordsolution.com/question/determine-the-costs-per-equivalent-unit-for/918423 | 1,487,964,594,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171629.92/warc/CC-MAIN-20170219104611-00230-ip-10-171-10-108.ec2.internal.warc.gz | 527,420,156 | 8,819 | +1-415-315-9853
info@mywordsolution.com
## Accounting
Accounting Basics Cost Accounting Financial Accounting Managerial Accounting Auditing Taxation
Q1) Bell Computers, Ltd., located in Liverpool, England, assembles standardized personal computer from parts it buys from different suppliers. Production process comprises of many steps, starting with assembly of \"mother\" circuit board, that contains central processing unit. This assembly takes place in CPU Assembly Department. The company recently hired new accountant who prepared following report for department for May using weighted-average method:
Quantity Schedule Units to be accounted for: Work in process, May 1 (materials 90% complete; conversion 80% compete) 5,000 Started into production 29,000 Total units 34,000 Units to be accounted for as follows: Trasferred to next department 30 Work in process, May 31 (materials 75% complete; conversion 50% compete) 4,000 Total units 34,000
Cost Reconciliation Cost to be accounted for: Work in process, May 1 £ 13,400 Cost added in the department 87,800 Total cost to be accounted for £101,200 Cost accounted for as follows: Work in process, May 31 £ 8,200 Trasferred to next department 93,000 Total cost accounted for £101,200
Company\'s management would like some extra information about May\'s operation in CPU Assembly Department. (Currency in England is pound, which is denoted by symbol £.)
Determine the costs per equivalent unit for May?
Accounting Basics, Accounting
• Category:- Accounting Basics
• Reference No.:- M918423
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Practical mapping and data visualization in Graph Builder
### Episode 1: The Economics of Snow Plow Services
Every now and again, we JMP Systems Engineers run into interesting questions that would fall somewhat outside the normal range of JMP usage. These applications are generally clever and bring home how using data isn’t just for business or technical problems.
Here is an example of a question I received from a user (also named Mike) in New York state that he and I thought would be fun to share.
### The Question
Mike [JMP],
I’m playing around with JMP using data I found for snowfall totals last year. I moved to a new house this summer, and I now have a 387-foot-long driveway that will need to be plowed. I was recently talking with a snowplow contractor, and he was trying to talk me into a seasonal contract. Normally when he charges per push, it’s done anytime we have 3 or more inches of snow at \$75 per push. He offered a price of \$2,000 where he plows anytime we have snow. Last year, he told me, that we had 3” or more of snow about 20 times, so if this happened again this year I would end up paying \$1,500 (\$75 x 20). So I would only need to pay \$500 more to have him plow every time it snowed.
As you can see from my data, there were only 5 times last year with 3” or more of snow. Of course, I told this contractor I was not interested. I ended up hiring another guy that will do it for \$50 a push. My question for you is why does JMP put the 15” snow total on the graph between 2.5” and 3.0”? Since the totals are in descending order, why isn’t the 15” all the way to the right?
Thanks,
Mike [User]
### My Response
Hi, Mike!
In answer to your question, I think you’ll find that the column “Greenfield Center 1.6 W” has the Data Type set as “Character.” Beyond the obvious alphabetical order the ordering gets convoluted with character data. When you pair that with the Count-descending option in Graph Builder it gets more challenging to predict.
Here’s what I think is happening: Graph Builder is ordering by “Count Descending” first. So, the categories with the most rows will appear on the left and decrease as you move to the right. When you have multiple groups with the same number of data points, i.e., 1 & 2, 1.5 & 2.5, and 15 – 9.5, Graph Builder will use normal ordering rules. Since the column is character, that means that it will follow a pseudo-alphabetical order. This means that the ordering for the last group would be 15, 3, 5.5, 6, 9.5, because alphabetical order is by character, left to right. Using a more controlled example, how JMP orders the numbers between 1 and 20 is different depending on if the data type is character or numeric. Numerically, it would be 1,2,3, …, 20, like you’d expect. Alphabetically, it would be 1, 10, 11,12, 13, …, 19, 2, 20, 3, 4, 5, 6, 7, 8, 9.
A way to override this issue is to use a Value Order column property. You can define the order you want JMP to use and it overrides the default. With a user defined Value Order the hierarchy used in Graph Builder would be Count Descending, then Value Order, then the default for anything not defined in the Value Order column property.
Best,
M
### More Thoughts
After thinking about this, I realized there were some additional ways to approach this and some more visualizations that could be used to prove the point. First, let’s deal with some data clean-up. The data Mike used comes from the NOAA’s website. The site is set up to download a month at a time, so it requires some concatenation work. Looking at the data, there are some character codes used in the otherwise numerical precipitation data for each station. M and T are used for “missing” and “trace” respectively. This means two things. First, we have Ghost Data.
Second, we have some threshold below which reporting of specific values is not deemed necessary by the NOAA. We’ll have to deal with these somehow. Also, the data is provided with a day-month format that JMP considers Character and is missing a year. This means that if we were to generate a time series scatterplot, the ordering would be incorrect. This will also need to be corrected.
First, correcting the date is a series of Find & Replace operations using “-<mon>” as the "find" and “/mm/yy” as the "replace." In practice, this looks like this:
After the dates have been corrected, it’s probably a good idea to stack the data. This gives more options for analysis. With the data stacked, I can now assign a missing value column property to have JMP consider the “M” as missing. I can also use recode to change the “T” to a numerical value. The NOAA's FAQ says it considers trace to be “a small amount of precipitation that will wet a rain gauge but is less than the 0.01-inch measuring limit.” In this case, I’m just going to recode it to 0.01.
The last thing I’m going to do is create a new column, “Precipitation Groups.” I actually used Brady Brady’s Interactive Binning (V2) Add-In, but the logic is pretty straightforward to do in Formula Editor. If “Precipitation” is greater than 3 inches, then it returns “Plow”; less than 3 inches is “No Plow”; and if the data is missing, it returns “Missing.”
After I set this column up, I hid and excluded all the missing values.
There are three types of visualizations that come to mind in this case. A Distribution, a bar chart, and a bubble plot. Since the data is stacked, each of these is going to use a Local Data Filter to select the station of interest. The Distribution, built on the indicator column I created earlier, shows very clearly that plowing would have been required five times in the 2016-17 season.
A bar chart, created in Graph Builder, using a reference line and coloring by whether plowing was required, shows a similar story.
Bubble plots are just fun, and animating the story is a great way to improve the impact of the message. There are several ways of approaching the visualization, depending on taste and message. The bubble plot can also be exported to interactive HTML5 (export by running the scripts in the attached data table and then using the correct export option for your OS).
The first bubble plot is an animated time plot. When animated, it traces the daily precipitation and colors the line based on the how close the precipitation level is to the critical 3-inch level for plowing.
The second example shows all the stations in the data set, again with the daily recorded precipitation on the y-axis. In this example, I included the precipitation value as a sizing variable so the markers get bigger as the climb the y-axis.
The map example was created by locating the NOAA station coordinate list on the same site the data came from. I created a look-up table and used a virtual join to bring in the latitude and longitude for each station based on station name. Like the second bubble plot, the markers are sized by the daily recorded precipitation at a given station. The map isn’t particularly important for the original question, but when animated it provides an interesting visual for when, where, and how severe the snow storms in New York were during the 2016-17 winter season. For a little fun, see if you can locate the storm that allowed people to snowboard in Times Square in New York City (hint: it was in 2017).
Do you have fun questions or data sets that you’d like to have a systems engineer look at? Leave me a comment or message. Maybe we’ll turn this into a running thing.
### A Note From The Author
For those of you reading this again, no you’re not crazy, the title did change a little. I received a request to update the title so that it would be easier for our search engines to find. I can only guess this is because some search engines are modern incarnations of 19th-century reference librarians with no sense of humor. Apologies for any inconvenience. -M
Editor's note: This is the first in a series called Engineering Mailbag written by JMP Systems Engineers, who field all kinds of technical questions about JMP.
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• October 2019
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MEAM 535
Matlab, Numerical Integration, and Simulation n
Matlab tutorial n n n
n
Numerical integration n n n
n
Basic programming skills Visualization Ways to look for help Integration methods: explicit, implicit; one-step, multi-step Accuracy and numerical stability Stiff systems
Programming examples n n
Solutions to HW0 using Matlab Mass-spring-damper system University of Pennsylvania
Peng Song
GRASP
1
MEAM 535
MATLAB n
MATLAB (MATrix LABoratory) is an interpretative (interactive) programming language n
n
MATLAB working environment. n
n
n
2D and 3D data visualization, image processing, animation, and presentation graphics
MATLAB mathematical function library MATLAB API n
Peng Song
tools for developing, managing, debugging, and profiling M-files
MATLAB graphics system n
n
control flow statements, functions, data structures, input/output, and object-oriented programming features
interact with C and Fortran programs University of Pennsylvania
GRASP
2
1
MEAM 535
Basic Matrix-Vector Operations n
Enter a matrix: >>A=[ 3 2; 2 4] n All matrices are enclosed in a pair of bracket [ ]. n Case-sensitive: matrix A and matrix a are different.
n
Enter a vector >>b = [4;3]; b is a 2 x 1 column vector. n The semicolon sign at the end of the line indicates that the interpreter will not echo that line on the screen. n
n
Matrix operations >>c >>AA >>bb >>x
= = = =
Basic Operators: + - * / \ ^ ’ .+ .- .* ./ .\ .^ .’
A*b A^2 b.^2 A\b
University of Pennsylvania
Peng Song
GRASP
3
MEAM 535
M-file Programming n n
n
Use Matlab Editor create a program and save it as a m-file with a file-extension [.m] Simply enter the file name of the m-file (without the file extension) in the command window, the m-file will get executed. Example: use Matlab Editor to create a file called test.m. In test.m, enter the following lines A c A x
= = = =
[3 2; 2 4]; b=[3; 2]; A*b; A^2; b = b.^2; A\b
Then in the command window, enter >>test; Peng Song
University of Pennsylvania
GRASP
4
2
MEAM 535
Function n n
n
n
A function is a special m-file. A general syntax of a function: function [outData1, outData2] = myfunction(inData1, inData2, inData3) Example: function [c,x] = test_function(A,b) Workspace variables are not directly accessible to functions. They are normally passed to functions as arguments. All variable defined within a function are local variables. These variables will be erased after the execution of the function. University of Pennsylvania
Peng Song
GRASP
5
MEAM 535
Visualization 2-D plot Plot 2-D data with linear scales for both axes loglog Plot with logarithmic scales for both axes
3-D n
Lines Plot3 Plot 3-D data with linear scales for all axes contour Plot contour lines quiver Plot vector fields
n
Surfaces mesh, surf: surface plot meshc, surfc: surface plot with contour plot beneath it surfl: surface plot illuminated from specified direction surface: low-level function for creating surface graphics objects
Peng Song
University of Pennsylvania
GRASP
6
3
MEAM 535
Examples Line plots
Surface plots
t = 0:pi/50:10*pi; plot3(sin(t),cos(t),t) axis square; grid on
[X,Y] = meshgrid(-8:.5:8); R = sqrt(X.^2 + Y.^2) + eps; Z = sin(R)./R; mesh(Z)
Peng Song
University of Pennsylvania
GRASP
7
MEAM 535
More Examples contour plots [X,Y,Z] = peaks; contour(X,Y,Z,20);
Peng Song
Vector fields n = –2.0:.2:2.0; [X,Y,Z] = peaks(n); contour(X,Y,Z,10); [U,V] = gradient(Z,.2); hold on quiver(X,Y,U,V)
University of Pennsylvania
GRASP
8
4
MEAM 535
Help n n
Help on the toolbar Command line >>help plot >>helpwin plot >>doc plot
n
Demos >>demo
n
Tutorial from mathworks: http://www.mathworks.com/products/education/student_version/tutorials/laun chpad.shtml University of Pennsylvania
Peng Song
GRASP
9
MEAM 535
Numerical Integration of ODEs dy = f(x , y) dx n
Initial value problem: Given the initial state at y0=y(x0), to compute the whole trajectory y(x)
y
y ' = 1 - 2y , y (0) = 0
x Explicit Euler Solution Peng Song
University of Pennsylvania
GRASP
10
5
MEAM 535
Euler’s method n
Explicit: evaluate derivative using values at the beginning of the time step
y i + 1 = y i + h ⋅ f ( xi , yi ) + O( h2 ) n
n
Not very accurate (global accuracy O(h)) & requires small time steps for stability
Implicit: Evaluate derivative using values at the end of the time step
y i + 1 = yi + h ⋅ f ( xi +1 , y i + 1 ) + O (h2 ) n
n n
May require iteration since the answer depends upon what is calculated at the end. Still not very accurate (global accuracy O(h)). Unconditionally stable for all time step sizes (why?). University of Pennsylvania
Peng Song
GRASP
11
MEAM 535
Truncation errors
Local truncation error
Global truncation error
y
y
y i+1 yi o
Peng Song
y i+1 yi xi
x i+1
x
o
xi
University of Pennsylvania
x i+1
x i+2
x
GRASP
12
6
MEAM 535
Stability n
n n
n
A numerical method is stable if errors occurring at one stage of the process do not tend to be magnified at later stages. A numerical method is unstable if errors occurring at one stage of the process tend to be magnified at later stages. In general, the stability of a numerical scheme depends on the step size. Usually, large step sizes lead to unstable solutions. Implicit methods are in general more stable than explicit methods.
University of Pennsylvania
Peng Song
GRASP
13
MEAM 535
Stability Characteristics of Euler Methods Model:
dy = − λ y; dx
n
Error:
dε = − λε dx
n
Explicit Euler Method is conditionally stable:
n
y( 0) = 1
y ( x ) = 1 − e −λx
λ > 0
ε i +1 = ε i + h( − λ εi ) = ε i (1 − λh) n
ε i+1 = 1 − λh ≤ 1 εi
0≤ h≤
2 λ
Implicit Euler Method is unconditionally stable: ε i +1 = ε i + h( −λ εi +1 ) ⇒ ε i +1 =
εi 1 + λh
ε i +1 1 = ≤ 1 satisfied for all h ≥ 0 εi 1 + λh Peng Song
University of Pennsylvania
GRASP
14
7
MEAM 535
Second-order Runge-Kutta (midpoint method)
n
n
n
Second-order accuracy is obtained by using the initial derivative at each step to find a midpoint halfway across the interval, then using the midpoint derivative across the full width of the interval. In the above figure, filled dots represent final function values, open dots represent function values that are discarded once their derivatives have been calculated and used. A method is called nth order if its error term is O(hn+1) University of Pennsylvania
Peng Song
GRASP
15
MEAM 535
Classic 4th-order R-K method h ( k1 + 2 k 2 + 2 k 3 + k 4 ) 6 k1 = f ( x i , yi )
y i +1 = y i +
h h , yi + k1 ) 2 2 h h k 3 = f ( x i + , yi + k 2 ) 2 2 k 4 = f ( x i + h , y i + hk 3 ) k2 = f ( xi +
where h = x i + 1 − x i is the step size. Local error is of O ( h 4 ). Global error is of O ( h 3 ). Peng Song
University of Pennsylvania
GRASP
16
8
MEAM 535
Runge-Kutta Methods ν
y( x + h) = y( x ) + ∑ w i k i i =1
i −1 k i = hf x + c i h , y ( x ) + ∑ a i , j k j j=1
n
c, a, and w are numerical coefficients chosen to satisfy certain conditions.
n
Peng Song
? is the number of terms. For 4th order R-K method, ? = 4 University of Pennsylvania
GRASP
17
MEAM 535
Runge-Kutta Methods
4,5
3 6 2 1
Peng Song
University of Pennsylvania
GRASP
18
9
MEAM 535
Runge-Kutta Methods 1. Make a first (tentative) step with the Euler method 2. Evaluate slope at intermediate point. 3. Use the adjusted slope and make a second (also tentative) step from the initial point. 4. Evaluate function at additional points and use this information to further adjust the slope to be used at the start 5. Evaluate function at as may other points as required and make further adjustments to the slope to be used at the start. 6. Combine all the estimates to make the actual step. University of Pennsylvania
Peng Song
GRASP
19
MEAM 535
Multi-step Methods n
Runge-Kutta-methods are one step methods, only the current state is used to calculate the next state.
n
Multi-step methods calculate the next value based on a time series of values, that might be from the past, or from the future. n n
Explicit (b0 = 0) & implicit methods. # of the previous steps.
y i + 1 = a 1 y i + a 2 y i − 1 + .. + h ⋅ [ b0 f ( x i + 1 , y i + 1 ) + b1 f ( x i , y i ) + b2 f ( x i − 1 , y i − 1 ) + ... ] n n n
Peng Song
Adams-Bashforth Method (explicit b0 = 0) Admas-Moulton Method (implicit) Predictor-Corrector Methods University of Pennsylvania
GRASP
20
10
MEAM 535
Adams-Bashforth (A-B) Methods (Explicit) h [23 f ( xi , y i ) − 16 f ( xi − 1 , yi −1 ) + 5 f ( xi − 2 , y i − 2 ) ] 12 for i = 2, 3, ........, n - 1. yi +1 = yi +
a1 = 1, a 2 = 0, b0 = 0, b1 = 23 / 12, b2 = −16 / 12 & b3 = 5 / 12.
At the beginning, while y0 is known, y1 & y2 are calculted using one - step method. The local error is of O( h4 ). University of Pennsylvania
Peng Song
GRASP
21
MEAM 535
Adams-Moulton (A-M) Methods (Implicit) h [5 f ( xi +1 , y i +1 ) + 8 f ( xi , y i ) − f ( x i −1 , y i −1 ) ] 12 for i = 1, 2, 3, ........, n-1. y i +1 = y i +
a1 = 1, a2 = 0, b0 = 5 / 12, b1 = 8 / 12, and b2 = −1 / 12. y 0 is known, y1 is calculted using one - step method. The local error is O( h4 ).
Peng Song
University of Pennsylvania
GRASP
22
11
MEAM 535
Predictor-Corrector Methods Adams 3rd-order Predictor-Corrector Methods n
Predictor: using 3rd-order A-B three-step method to predict
yiP+1
for i = 2, 3, ........, n - 1 . y iP+ 1 = y i + n
h [23 f ( x i , y i ) − 16 f ( x i −1 , y i − 1 ) + 5 f ( x i − 2 , y i − 2 ) ] 12
Corrector: using 3rd-order A-M two-step method to compute
y iC+1 = y i +
yiC+1
h [5 f ( x i +1 , y iP+ 1 ) + 8 f ( x i , y i ) − f ( x i −1 , y i − 1 ) ] 12 University of Pennsylvania
Peng Song
GRASP
23
MEAM 535
Stiff ODEs n
Peng Song
Example
University of Pennsylvania
GRASP
24
12
MEAM 535
Stiff ODEs n
A practical example
University of Pennsylvania
Peng Song
GRASP
25
MEAM 535
Stiff ODEs n
n
n
Stiff systems are characterized by some system components which combine very fast and very slow behavior. Requires efficient step size control that adapt the step size dynamically, as only in certain phases they require very small step sizes. Implicit method is the cure! n n n n
Peng Song
Nonlinear systems: solving implicit models by linearization (semiimplicit methods) Rosenbrock – generalizations of RK method Bader-Deuflhard – semi-implicit method Predictor-corrector methods – descendants of Gear’s backwoard differentiation method University of Pennsylvania
GRASP
26
13
MEAM 535
Higher Order ODEs n n
Higher order ODEs can be turned into systems of 1st order ODEs: For example: d3y d2y dy + a +b + cy = 2 sin( x ) 3 2 dx dx dx y2 y1 d ⇒ y3 y2 = dx y 2 sin( x ) − ay − by − cy 3 3 2 1 dy d 2y where y1 = y y 2 = y3 = dx dx 2 University of Pennsylvania
Peng Song
GRASP
27
MEAM 535
Mass-Spring-Damper System Free-body diagram f(t)
kx
f(t) m
c x& m &x& + c x& + kx = f ( t ) y1 = x
y& 1 = y2
y2 = x&
K C f (t ) y& 2 = − y1 − y2 + m m m
Peng Song
0 y& = − K m
University of Pennsylvania
1 0 y + f ( t ) − C m m GRASP
28
14
MEAM 535
Dynamic Simulation Example Given y0=(x0, 0), solve for y(t) for t=[0, T]. Initial conditions
ODE solver
differential equations
(ode45)
(diff_msd.m)
output
(demo_msd.m)
University of Pennsylvania
Peng Song
GRASP
29
MEAM 535
Simulation Results m= 1.0 kg C = 1.0 N*sec/m K = 100.0 N/m
Displacement
0.5
0
-0.5 0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5 Time (s)
6
7
8
9
10
4
Velocity
2 0 -2 -4 -6
Peng Song
University of Pennsylvania
GRASP
30
15
MEAM 535
Matlab Solvers for Nonstiff Problems n
ode45 n n n
n
ode23 n n n
n
Explicit Runge-Kutta (4,5) formula One-step solver Best function to apply as a "first try" for most problems Explicit Runge-Kutta (2,3) One-step solver May be more efficient than ode45 at crude tolerances and in the presence of mild stiffness.
ode113 n n n
Variable order Adams-Bashforth-Moulton PECE solver Multistep solver It may be more efficient than ode45 at stringent tolerances and when the ODE function is particularly expensive to evaluate. University of Pennsylvania
Peng Song
GRASP
31
MEAM 535
Matlab Solvers for Stiff Problems n
ode15s n
n n
n
ode23s n n n
n
Based on a modified Rosenbrock formula of order 2 One-step solver May be more efficient than ode15s at crude tolerances
ode23t n n
Peng Song
Variable-order solver based on the numerical differentiation formulas (NDFs). Optionally it uses the backward differentiation formulas (BDFs). Multistep solver. If you suspect that a problem is stiff or if ode45 failed or was very inefficient, try ode15s first.
An implementation of the trapezoidal rule using a "free" interpolant Use this solver if the problem is only moderately stiff and you need a solution without numerical damping University of Pennsylvania
GRASP
32
16
MEAM 535
References n
Numerical Initial Value Problems in Ordinary Differential Equations, Gear, C.W., Englewood Cliffs, NJ: Prentice-Hall,1971.
n
Numerical Recipes in C : The Art of Scientific Computing William H. Press, Brian P. Flannery, Saul A. Teukolsky, William T. Vetterling, Cambridge University Press, 1992. (online at http://www.library.cornell.edu/nr/bookcpdf.html)
Peng Song
University of Pennsylvania
GRASP
33
17
February 2020 53
December 2019 21
December 2019 32
February 2020 35
December 2019 21
November 2019 31
October 2019 19
December 2019 33
December 2019 56
December 2019 42
December 2019 32
October 2019 35 | 4,357 | 13,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.81853 |
https://numbermatics.com/n/74/ | 1,563,527,399,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526153.35/warc/CC-MAIN-20190719074137-20190719100137-00391.warc.gz | 496,179,940 | 8,712 | # 74
## 74 is an even composite number composed of two prime numbers multiplied together.
74 is an even composite number. It is composed of two distinct prime numbers multiplied together. It has a total of four divisors.
## Prime factorization of 74:
### 2 × 37
See below for interesting mathematical facts about the number 74 from the Numbermatics database.
Quick links:
### Names of 74
• Cardinal: 74 can be written as Seventy-four.
### Scientific notation
• Scientific notation: 7.4 × 101
### Factors of 74
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 2
• Sum of prime factors: 39
### Divisors of 74
• Number of divisors d(n): 4
• Complete list of divisors:
• Sum of all divisors σ(n): 114
• Sum of proper divisors (its aliquot sum) s(n): 40
• 74 is a deficient number, because the sum of its proper divisors (40) is less than itself. Its deficiency is 34
### Bases of 74
• Binary: 1001010 2
• Hexadecimal: 0x4A
• Base-36: 22
### Squares and roots of 74
• 74 squared (742) is 5476
• 74 cubed (743) is 405224
• The square root of 74 is 8.6023252671
• The cube root of 74 is 4.1983364539
### Scales and comparisons
How big is 74?
• 74 seconds is equal to 1 minute, 14 seconds.
• To count from 1 to 74 would take you about thirty-seven seconds.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000.
Note: we do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!
• A cube with a volume of 74 cubic inches would be around 0.3 feet tall.
### Recreational maths with 74
• 74 backwards is 47
• The number of decimal digits it has is: 2
• The sum of 74's digits is 11
• More coming soon!
The information we have on file for 74 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 74, math, Factors of 74, curriculum, school, college, exams, university, STEM, science, technology, engineering, physics, economics, calculator. | 640 | 2,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.885889 |
https://www.itsnotaboutme.tv/how-to-uiyopf/99a2e3-kilogram-definition-chemistry | 1,618,743,065,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038476606.60/warc/CC-MAIN-20210418103545-20210418133545-00093.warc.gz | 928,418,614 | 7,050 | ## kilogram definition chemistry
The SI unit for molality is mol/kg. For almost a century and a half, much of the world used a highly polished, golf ball-sized cylinder of a platinum alloy as the international standard for mass measurement — a kilogram. One kilogram is equal to 1,000 grams. Each of the seven base units of the SI (second, metre, kilogram, ampere, kelvin, mole, and candela) is defined by an agreed reference that has to be readily available and easily realized experimentally by anyone anywhere at any time with sufficient precision (sufficiently low uncertainty) for our needs. From what I understand, the kilogram was initially defined in terms of the mass of a specific volume of water at a specific temperature. The kilogram (kg) is an accepted SI unit of measurement for the mass of an object. Those calculations are always made in kelvins. 1800, illustrating new standards. However, it is now (and has been for a while) defined as exactly the mass of the International Prototype Kilogram (IPK). Las unidades adoptarán su nueva definición el 20 de mayo de 2019, el Día Mundial de la Metrología. The new definition of the kilogram is based on the Planck constant. The mole is the unit of amount of substance and was defined in 1960 by the number of atoms in 0.012 kg of 12C. The term solution is commonly applied to the liquid state of matter, but solutions of gases and solids are possible. The following is a brief summary of the changes that are being adopted in this new SI. This picture shows the Kibble balance – the new way to measure the kilogram, as … When converted to English standard units it is equivalent to about 2.2 pounds. The challenge now though is to explain these new definitions to people — especially non-scientists — so they understand. Definition: A milligram is a unit of weight and mass that is based on the SI (International System of Units) base unit of mass, the kilogram. Kilogram, basic unit of mass in the metric system. As an SI unit, it uses the "milli" SI prefix to denote that it is a submultiple of the base unit. Did You Know? Molality is a property of a solution and is defined as the number of moles of solute per kilogram of solvent. Woodcut ca. Redefining the kilogram will make precision measurement widely available in labs. It is equal to 1/1,000 grams, or 1/1,000,000 kilograms. The kilogram has been transformed as new definition takes hold. It is equivalent to 1.0567 U.S. liquid quarts and is equal to the volume of one kilogram of distilled water at 4°C. Search results for 'kilogram'. Universally accepted units of measurement for the weight or mass of an object are kilograms, grams and pounds. Abbreviation: l See more. Kilogram definition is - the base unit of mass in the International System of Units that is equal to the mass of a prototype agreed upon by international convention and that is nearly equal to the mass of 1000 cubic centimeters of water at the temperature of its maximum density. 1. Long Version. ... Chemistry and physics require many calculations involving temperature. image caption There are now discrepancies between copies of the kilogram and the master version held in a French vault Scientists have changed the way the kilogram is defined. For more than a century, the kilogram (kg) — the fundamental unit of mass in the International System of Units (SI) — was defined as exactly equal to the mass of a small polished cylinder, cast in 1879 of platinum and iridium.. Under the new BIPM proposal, therefore, as the kilogram definition changes the mass of carbon-12 will also change, probably multiple times. Visit BYJU’S to learn more about it. These definitions were revised in May 2019. Comparison of temperature scales: Temperatures of some common events and substances in different units. The kilogram is likely to be redifined in the near future as a function of the number of atoms of a sphere of ultrapure \$\ce{^{28}Si}\$, as per the answer to this Chemistry SE question. The new kilogram definition fixes the relativistic mass of a photon at the Cs hyperfine frequency as m ph = 6.777 265 × 10 −41 kg. Over the years its mass has changed for reasons that are not entirely clear, but which may involve surface chemistry or the leaching of nitrogen or oxygen trapped during its manufacture. Comparing a kilogram to a metal block is easy. Solution, in chemistry, a homogenous mixture of two or more substances in relative amounts that can be varied continuously up to what is called the limit of solubility. To appreciate the whole story, we need to … This modern form of the Metric system is … From today the kilogram is defined using the Planck constant, something that doesn’t change from quantum physics. For redefining the kilogram on the basis of physical constants, two approaches are considered worldwide: the Avogadro experiment, where in … Definition of the Kg ( kilogram) in chemistry Get the answers you need, now! A solution with a molality of 3 mol/kg is often described as “3 molal” or “3 m.” However, following the SI system of units, mol/kg or a related SI unit is now preferred. A kilogram was equal to the heft of this aging hunk of metal, and this cylinder, by definition, weighed exactly a kilogram. Thus, why not keep a mass unit in the mole definition and maintain the present definition. The kilogram is still defined by the mass of a Pt–Ir cylinder conserved at the International Bureau of Weights and Measures. If the cylinder changed, even a little bit, then the entire global system of measurement had to change, too. The Chemistry Glossary contains basic information about basic terms in chemistry, physical quantities, measuring units, classes of compounds and materials and important theories and laws. Technically a kilogram (kg) is now defined: From left to right: the liter, the gram, and the meter. Kept in a triple-locked vault on the outskirts of Paris, the platinum-iridium cylinder was officially called the International Prototype of the Kilogram (IPK). That means that the standard will be the Avogadro number, and the mass will be a derived unit. History/origin: The milligram is based on the SI unit of weight and mass, the kilogram. In other words, a kilogram mass exerts about 9.8 newtons of force. The International System of Units (SI) is system of units of measurements that is widely used all over the world. The definition of the kilogram, the unit of mass in the International System of Units (SI), has not changed in more than 125 years. Since the Planck constant is a fundamental physical constant of nature, it will never change and neither will the value of the kilogram. It is defined by taking the fixed numerical value of the Planck constant h to be 6.626 070 15 x 10 –34 when expressed in the unit J s, which is equal to kg m 2 s –1, where the metre and the second are defined in terms of c and Cs. Definition agreed by the 26th CGPM (November 2018), implemented 20 May 2019: The kilogram, symbol kg, is the SI unit of mass. Liter definition, a unit of capacity redefined in 1964 by a reduction of 28 parts in a million to be exactly equal to one cubic decimeter. The kilogram is the only unit still based on a physical object. The kilogram (kg) was originally defined as the mass of a liter (i.e., of one thousandth of a cubic meter). The original kilogram is defined by the international prototype, a cylinder of platinum-iridium that was cast in 1879.
kilogram definition chemistry 2021 | 1,694 | 7,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-17 | longest | en | 0.914767 |
https://ai.stackexchange.com/tags/fuzzy-logic/hot | 1,713,793,555,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818293.64/warc/CC-MAIN-20240422113340-20240422143340-00530.warc.gz | 71,534,119 | 32,817 | # Tag Info
Accepted
### What is fuzzy logic?
As complexity rises, precise statements lose meaning and meaningful statements lose precision. ( Lofti Zadeh ). Fuzzy logic deals with reasoning that is approximate rather than fixed and exact. This ...
• 3,004
### What is fuzzy logic?
Fuzzy logic is based on regular boolean logic. Boolean logic means you are working with truth values of either true or false (or 1 or 0 if you prefer). Fuzzy logic is the same apart from you can have ...
Accepted
### How can fuzzy logic be used in creating AI?
A classical example of fuzzy logic in an AI is the expert system Mycin. Fuzzy logic can be used to deal with probabilities and uncertainties. If one looks at, for example, predicate logic, then ...
### What is fuzzy logic?
It's analogous to analogue versus digital, or the many shades of gray in between black and white: when evaluating the truthiness of a result, in binary boolean it's either true or false (0 or 1), but ...
• 339
### What is the relationship between fuzzy logic and objective bayesian probability?
This was a somewhat hotly debated question in the 1980s. The debate was more-or-less ended with papers like Cheeseman's In Defense of Probability. The short answer is that Fuzzy Logic does not just ...
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### Why has statistics-based AI become more popular than other forms of AI?
The availability of large data sets. In symbolic/rule-based AI, the 'knowledge' has to be hand-coded, usually by experts. This is expensive and limited to small-scale problems only. In statistical AI/...
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### Are there real applications of fuzzy logic?
You've obviously never heard of fuzzy logic washing machines. ● Typically, fuzzy logic controls the washing process, water intake,water temperature, wash time, rinse performance, and spin speed. ...
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1 vote
### Defining formula for fuzzy equation
After i did a bit more study on this, i found this solution. μsmall[x] 1; -> x≤0 (20-x)/(20-0); -> 0 0; -> x≥20 μmedium[x] 0; -> x≤20 or x≥70 (x-20)/(45-20); -> 20 (70-x)/(70-45); -> 45 μlarge[x] ...
• 39
1 vote
Accepted
### Tuning the b parameter, ANFIS
After re-reading Jang's original (1993?) paper on ANFIS, I learned he recommended simply squaring the b-parameter to deal with the notion that b could be changed to a negative value when using back-...
1 vote
### Are there real applications of fuzzy logic?
Recently, I needed to develop a Fuzzy Logic algorithm to made inferences of any data entrance; the real case applied was in Oil and Gas Industry, that the code needs to infere Joint Types in Fluid ...
1 vote
### How to determine the probability of an "existence" question
You need some sort of interpretation abstraction before your mathematical reasoning. While the text might read "123", you need to parse this into a literal of type Natural Number or Integer. Similarly,...
• 221
1 vote
### How to determine the probability of an "existence" question
A human has an abstract concept of numbers in mind. So 456 is a unique entity which is by definition unlike any other number because that are other unique entities. If you give ∃x ∈ ℕ: x==123 to your ...
• 318
1 vote
### What is fuzzy logic?
Fuzzy Logic is a way of dealing with uncertainties, which is something that computers don't do naturally which human do very well. The way we instantly think of dealing with things and the way that ...
1 vote
### What is fuzzy logic?
It is making deductions based on probability and statistics, like humans make decisions all the time. We are never 100% sure the decision we have made is the right one but there is always some doubt ...
• 219
1 vote
### What is fuzzy logic?
Why is it useful? Many things we don't know for sure. We estimate and are often uncertain, but nearly never 100% sure. It may seem like a weakness, but because of this fuzzy approach we can function ...
• 111
Only top scored, non community-wiki answers of a minimum length are eligible | 935 | 3,985 | {"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": 1, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-18 | latest | en | 0.933997 |
https://www.edureka.co/community/218326/filter-sumproduct-formula-based-on-array | 1,721,615,500,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00394.warc.gz | 649,542,975 | 24,889 | Using Table 1, I can use the SUMPRODUCT function to determine the total revenue. However, I would prefer to have a straight filtering option in the formula to exclude certain places. When region B is removed from the data using the formula below, the proper result (13,000), it is produced is:
`=SUMPRODUCT(--(Sales[Area]<>Exceptions[Area]);Sales[Quantity];Sales[Price per unit])`
However, when I add another area in Table 2, the formula returns an error. Is it possible to filter out multiple variables (areas) directly in the formula?
Dec 16, 2022 in Others 907 views
## 1 answer to this question.
Use ISERROR(MATCH()):
```=SUMPRODUCT(--(ISERROR(MATCH(Sales[Area];Exceptions[Area];0)));Sales[Quantity];Sales[Price per unit])
```
--(ISERROR(MATCH(Sales[Area];Exceptions[Area];0))) will return 1 if the area is not found in the search area because the MATCH will return an error when it is not found in the list.
answered Dec 16, 2022 by
• 63,720 points
## Excel formula to average selected cells based on certain criteria
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https://whyevolutionistrue.com/2014/05/01/a-mathematical-spider-from-the-african-deserts/ | 1,696,079,483,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510676.40/warc/CC-MAIN-20230930113949-20230930143949-00509.warc.gz | 648,669,571 | 33,123 | # A mathematical (?) spider from the African deserts
May 1, 2014 • 5:37 am
A reader sent me the link to an old scientific article (from 1995) about a “mathematical spider” living in the Namibian desert. It turns out that the adjective “mathematical” is pretty misleading, but since the article was interesting I thought I’d give it a brief shout-out.
The paper, published by G. Costa et al. in the Journal of Arid Environments, (reference below, can join ResearchGate and get free copy, or ask me), is about a new species in the family of tube-dwelling spiders, Segrestriidae; the species is named Ariadna sp. (the “sp.” means “species not identified”, although it may well have been in the last 18 years).
24 specimens of this ground-dwelling spider were studied near Gobabeb, a well-known research station in the desert of Namibia. The spiders dig burrows in the ground from whence they venture to get prey. Here’s a picture of the spider:
Here’s its bleak habitat. Life is nearly everywhere on this planet:
Some gypsum casts of its burrows. There are about 2.5 cm to the inch, so the burrows are about five inches long. The spiders line them with silk.
Now for the “mathematical” part. For reasons yet unknown, the spiders pile stones around the entrance of their burrows. The stones are fairly uniform in size, and there are usually about seven, though the number ranges from five to nine. The centimeter scale is at the top, along with a Namibian five-cent coin for extra scale (useful only to Namibians!)
The interesting thing about the stones is that they are usually placed radially, with the narrowest parts near the burrow, and of fairly uniform size. This makes the burrow look a bit like a flower. Stones are clearly selected for size, as you can see by the surrounding stones. Perhaps the spiders use the biggest stones that they’re able to carry. The photo below shows a typical array of 7 stones.
The “mathematical” part, which the authors make a great deal of, is that the mean number of stones (and the mode) is seven, with other numbers distributed fairly symmetrically around that. Here’s the table showing the percentage of stones falling in each category:
Well, that’s interesting, but hardly mathematical. It doesn’t show at all that the spiders can count, and it’s not surprising in any way that the distribution is a bell-shaped curve. What may be going on here is simply that the spiders heft the largest stones they can carry, that their burrows are of a relatively fixed size because spiders are of a relatively fixed size, and the average number of spider-heft-able stones that can surround a burrow happens to be seven. The spiders do apparently exercise a preference for quartz stones, though again this preference isn’t documented statistically.
What is more interesting to me is that the stones appear to be placed radially (though I’d like to see more photos and measurements), and, especially, that the spiders even bother to highlight their burrow this way. Why? The authors raise three possibilities:
a. Detection or attraction of prey. The authors suggest that “the stone ring could perhaps attract prey or facilitate the detection of prey by the spider waiting inside its burrow.” Well, maybe, though the attraction hypothesis seems more viable than the detection one. At any rate, this could be tested, even in the lab, by removing the stones and seeing fewer prey approach the burrow. But neither of these hypotheses seem really convincing.
b. Strengthening the burrow and making it impervious to sand or debris. This seems more likely to me. The raised stones could keep dirt or sand blowing along the ground from entering and clogging the burrow. Again, this could be tested fairlly easily.
c. Deterrence of predators. Here’s what the authors say:
The stone ring might be a way of reducing predatory risk. The evenly lighted circle around the burrow’s entrance could make this appear like, [sic] a black stone or a shaded area. On the other hand, spider holes may simulate the little black stones that are scattered over the gravel plain. The characteristic alternation of light and dark areas on the gravel ground complicates the detection of real burrows by predators.
That’s possible, but again requires testing. Another possibility, which the authors don’t mention, is that the symmetrical pattern may help the spider find its burrow in a complicated patchwork of stones and ground. In other words, the stones could act as landmarks, much the same way that some ground-dwelling “digger” wasps recognize their burrows by the patterns of debris on the ground nearby. (That work, a classic study of animal behavior by Tingergen and Kruyt, showed that the wasps could be confused by simply moving the landmarks around a burrow.) This “recognition” hypothesis may not be likely, though, if the spiders’ vision is poor.
At any rate, it’s a cute behavior whose significance is not yet determined, but would seem to be tractable to easy experiments.
h/t: Hardy
________
Costa, G., A. Petralia, E. Conti, and C. Hanel. 1995. A mathematical spider living on gravel plains of the Namib Desert. Journal of Arid Environments 29:485-494.
## 37 thoughts on “A mathematical (?) spider from the African deserts”
1. How DARE they give it the name Ariadna – it is my mother’s first name!
🙂
2. Do both sexes do this? It would be a good way for one sex to advertise its quality to the other. Bigger stones = “bigger stones”. Or, “Hey, look at me, I can count!”
1. Good question. The authors don’t mention this, so it’s implied that it’s not sexually dimorphic. Nor do they mention finding burrows without stones. But the short answer is: we don’t know.
3. Surely, if a spider does maths, in would be in octal. So, to the spider’s mind, the number of stones would vary from 5 to 11. 🙂
1. infiniteimprobabilit says:
If the spider really did that, I would expect the number of stones to peak at 8. One per leg, in fact.
Maybe there’s a high accident rate and there are a lot of seven-legged spiders around.
4. Jim Knight says:
It may be that the spider gains some advantage by making it’s burrow look like one of the unusual Lithops-like flowers (rock or stone flowers) native to the Namib, that would protect if from some predator, say birds? A little mimicry in action…?
1. I was wondering whether there’s any chance the flower-like arrangement would attract pollinators, who could be prey, but it would depend on how those pollinators see these stones: not the same way we would.
1. Jim Knight says:
Any feature that would serve more than one purpose in helping to make a living in nature sould certainly be useful, and selected for. The pollinator idea is a good one, and testable, as is the camo idea…
2. Lithops flowers don’t look like that. They have very many thin petals.
5. Dennis Hansen says:
Far out, but hey: A radially symmetric flower-like structure could also lure in flying pollinators.
Perhaps not that far out, since e.g. long-tongued flies in Aouth Africa can be fooled by artificial flowers.
1. Dennis Hansen says:
No worries – great minds & all that. Re: your comment below, one of my students found a (potentially new) species of trapdoor spider on Aldabra – it lives in loose coral sand, and its door looks 100% like the surrounding sand. Don’t ask me how he spotted it! (well, ok, the answer would be that the spider had a wee bit of its black, hairy leg sticking out (a bait to lure in smaller predators?)).
6. Protection from clogging by blowing debris seems most likely right now. It does remind me of an Australian trap door spider that lays out twigs and leaves in a radial pattern at the burrow entrance. In this case the function appears to be for detecting passing prey.
1. infiniteimprobabilit says:
Blimey! Looking at the size of that trapdoor hole – how big is that spider?!
7. I remember reading this paper years ago–very interesting (but show me a spider paper that isn’t!). I don’t have any blinding insights, but, in the scheme of spider evolution, there’s nothing terribly unusual about this behavior. I suspect–but don’t know–that Jerry is right that the “mathematical” nature of the stone collection depends more on the strength of the spider than on discrimination by size. (In fact, to me, when viewed from a spider’s point of view, the stones don’t seem so uniform in size.) It’s not unusual at all for burrow dwelling spiders to incorporate nearby debris into their burrow entrances–see photos of collar- and turret-web spiders, among others. In those cases, there is evidence that prey detection is at work. Radial extension of silk lines from the burrow entrance is seen in many species–I can’t remember whether these stones are connected to the inner lining of the burrow by silk, but that would make them similar to radial lines incorporating surrounding debris. I don’t think they would have anything to do with finding one’s way home, as I think these spiders have poor eyesight and in any case probably leave dragline trails of some sort when they venture forth–and they probably don’t go far at all anyway, being burrow-based predators.
1. gravelinspector-Aidan says:
Further to your point, I picked up on Jerry’s :
What may be going on here is simply that the spiders heft the largest stones they can carry, that their burrows are of a relatively fixed size because spiders are of a relatively fixed size, and the average number of spider-heft-able stones that can surround a burrow happens to be seven.
I would bet that larger spiders have larger diameter burrows, and are stronger, so can move larger stones. Which would correlate stone size and pebble size, and also act to reduce variation in pebble count.
So … if you’ve got acess to the full paper, do they document variations in burrow diameter?
1. Marella says:
I accessed the paper. There is a relationship between diameter of hole and stone size. The mean ratio of size of hole and size of stone is M= 0.87(SD=0.1480)with the regression coefficient alpha= 0.0879 and beta=0.6754 (p<<0.001). The sample was not big enough to do a proper analysis of spider weight and hole size but spider weight and hole size were highly correlated, r=0.88(p=<0.05.
Message me with your email if you'd like a copy.
1. gravelinspector-Aidan says:
2. Jo5ef says:
“Very interesting – but show me a spider paper that isn’t!”
Best comment I have seen on the web for awhile 🙂
I back the burrow protection theory
8. eric says:
What about the possibility of using the stones as a ‘heat shield’ to keep the burrow relatively cool? The symmetry and preference for a type of stone could reflect (pun intended) a selection pressure for formations that better reflect or redsitribute heat away from the burrow mouth.
Just a thought. The ‘dust shield’ idea seems very credible too.
1. I thought of that too.
Also, could the stones keep the opening intact for longer periods of time, by shedding rain?
I also wonder if this particular adroitness for radial symmetry is related to that displayed by web-spinning spiders.
9. Seth says:
Looks like a South African 5c piece to me. FWIW
1. Gregory Kusnick says:
And it looks to be about the same size as a US dime.
10. Sean says:
That is a South African 5c coin, featuring the Blue Crane.
Stunning bird that is the republic’s national bird.
Must know how pretty they are to be selected above the King Fisher’s, Fish Eagles etc to be the National Bird.
11. Taz says:
Any way the stone arrangement could help in collecting moisture?
12. Gregory Kusnick says:
The radial placement seems fairly easy to explain if the spider simply grabs a stone by the small end and drags it to the lip of the burrow.
And whatever advantage is conferred by arranging stones in a circle around the burrow entrance, it seems reasonable to expect that natural selection will discover the most efficient packing (wedge-shaped stones oriented small end in).
1. Marella says:
This is how worms drag leaves down their burrows as I recall.
13. To be fair, a 5 cent coin probably isn’t very useful to Namibians, either.
14. Diane G. says:
Very cool spider observation. 🙂
15. Mark Joseph says:
Great post; interesting comments. It would be fun to experiment and find out which of the option(s) for the explanation of the radially-arranged stones pan out.
I’ve plugged it before, but I’ll do it again (since Leslie comments here); if you’re at all interested in spiders or evolution (who, us?), treat yourself to Leslie Brunetta and Catherine L. Craig’s book Spider Silk, subtitled “Evolution and 400 Million Years of Spinning, Waiting, Snagging, and Mating”. A superb read! | 2,924 | 12,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-40 | latest | en | 0.926407 |
https://lifehacks.stackexchange.com/questions/2591/how-to-work-around-european-size-hardware-measurements/2600 | 1,621,289,348,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991870.70/warc/CC-MAIN-20210517211550-20210518001550-00485.warc.gz | 386,022,499 | 38,832 | # How to work around European size hardware measurements
I have some furniture which was made in Europe. The bolts and what-not are all in metric measurements. Some of the bolts came loose and were misplaced, and some of the pieces necessary for converting stages of the furniture were missing.
I went to the local hardware store, and they did not stock more than a very basic supply of metric sized hardware. I searched online and could not find anywhere to order metric sizes.
How can I, using U.S. sizes of hardware, work with European sized threaded-sockets? Both the lengths and the thicknesses are issues. Is there a way to adapt the sockets so that they will fit standard measurement hardware? Perhaps something that can be stuffed into the sockets, or something like that, so that a smaller size standard piece would fit??
• You mean metric? You went to a bad hardware store then... – J. Musser Jan 6 '15 at 5:03
• @J.Musser Home Depot? Either way, you would think I at least would have found online. – Y e z Jan 6 '15 at 5:04
• Are you trying to match an imperial bolt to a metric nut? – apaul Jan 6 '15 at 17:02
• @close-voters - why is this off topic as per either of the upvoted answers here? I explained what my problem is, why conventional attempts did not work and I need a hack, and defined what kind of answer I am looking for. – Y e z Jan 6 '15 at 19:40
• If you can measure the size precisely (preferably using calipers), you can get the exact size metric fastener pretty cheap from my favorite hardware vendor, Fastenal. They have branches all across the US, Canada, and Mexico. – Mike Jan 8 '15 at 1:17
Saying European measurements is not a proper terminology and that is the reason you had a confused clerk at the hardware store.
There are two standard measurements in the world, metric and imperial. Imperial is miles yards, feet, inches etc. Metric is meters, millimeters, centimeters etc.
if your furniture was made in Europe the hardware is more likely then not metric. Take a bolt or nut to the hardware store with you. Show it to the clerk and tell them what you need.
• I don't think my problem was calling it European - the clerk showed me the 3 metric bolts they had in the store, despite my referring to it as European, he somehow figured out what I meant. – Y e z Jan 6 '15 at 19:30
• @YeZ I think you need to go to a different store. – Jon Jan 6 '15 at 19:32
If you're trying to get an imperial bolt to screw in to metric nut, that's likely pretty hopeless.
Beyond the obvious length and diameter issue you will likely encounter differences in the thread pitch. Imperial bolts typically use the Unified Thread Standard (UTS) while metric bolts typically use ISO standard threads.
While there may be a few that seem to screw together, at best you'll end up with a loose, wobbly, and potentially dangerous connection. Save yourself the the time and aggravation and use one or the other.
If you're just looking for a tool that will work for both metric and imperial you could try a crescent wrench or adjustable spanner.
• There is no hacky way to resize the metric sockets? I was hoping someone would come up with some putty or something that I could stuff into the socket to resize it. – Y e z Jan 6 '15 at 19:32
• @YeZ If it was a one time use thing perhaps you could, but I wouldn't recommend it for something as permanent as furniture. – apaul Jan 6 '15 at 19:59
• @YeZ also did you look for metric parts or did you just ask for European parts? – apaul Jan 6 '15 at 20:00
• I found the metric selection they had - it was very minimal. (Even though I think I asked for European - I don't remember, it was a few months ago) – Y e z Jan 6 '15 at 20:13 | 908 | 3,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | longest | en | 0.957672 |
http://mathhelpforum.com/differential-geometry/184180-help-umbilical-points.html | 1,524,416,757,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945624.76/warc/CC-MAIN-20180422154522-20180422174522-00160.warc.gz | 196,503,423 | 9,772 | # Thread: Help with Umbilical Points
1. ## Help with Umbilical Points
we need your help. We had some time trying this one:
If P is an umbilical point then the second fundamental form is constant
Thanks!
2. ## Re: Help with Umbilical Points
What is the link between the second fundamental form and the normal curvature?
3. ## Re: Help with Umbilical Points
k_n = II(X,X) and k_n is constant. But I need that II(X,Y) constant. or am I wrong?
4. ## Re: Help with Umbilical Points
Originally Posted by kezman
k_n = II(X,X) and k_n is constant. But I need that II(X,Y) constant.
Use the bi-linearity.
5. ## Re: Help with Umbilical Points
kn=II(X,X) =???= II(X,Y). I need something in between to reach the right side???? | 201 | 726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-17 | latest | en | 0.915557 |
https://www.coursehero.com/file/6499715/Business-Calc-Homework-w-answers-Part-10/ | 1,529,667,973,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864391.61/warc/CC-MAIN-20180622104200-20180622124200-00241.warc.gz | 778,161,942 | 56,352 | Business Calc Homework w answers_Part_10
# Business Calc Homework w answers_Part_10 - 46 Section 2.1...
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48. continued (b) lim x 2 1 1 f ( x ) 5 0; lim x 2 1 2 f ( x ) 5 0 (c) Yes. The limit is 0. 49. (a) [ 2 2 p , 2 p ] by [ 2 2, 2] (b) ( 2 2 p , 0) < (0, 2 p ) (c) c 5 2 p (d) c 5 2 2 p 50. (a) [ 2 p , p ] by [ 2 3, 3] (b) 1 2 p , } p 2 } 2 < 1 } p 2 } , p 2 (c) c 5 p (d) c 5 2 p 51. (a) [ 2 2, 4] by [ 2 1, 3] (b) (0, 1) < (1, 2) (c) c 5 2 (d) c 5 0 52. (a) [ 2 4.7, 4.7] by [ 2 3.1, 3.1] (b) ( 2‘ , 2 1) < ( 2 1, 1) < (1, ) (c) None (d) None 53. [ 2 4.7, 4.7] by [ 2 3.1, 3.1] lim x 0 ( x sin x ) 5 0 Confirm using the Sandwich Theorem, with g ( x ) 5 2 ) x ) and h ( x ) 5 ) x ) . ) x sin x ) 5 ) x ) ? ) sin x ) # ) x ) ? 1 5 ) x ) 2 ) x ) # x sin x # ) x ) Because lim x 0 ( 2 ) x ) ) 5 lim x 0 ) x ) 5 0, the Sandwich Theorem gives lim x 0 ( x sin x ) 5 0. 54. [ 2 4.7, 4.7] by [ 2 5, 5] lim x 0 ( x 2 sin x ) 5 0 Confirm using the Sandwich Theorem, with g ( x ) = 2 x 2 and h ( x ) 5 x 2 . ) x 2 sin x ) 5 ) x 2 ) ? ) sin x ) # ) x 2 ) ? 1 5 x 2 . 2 x 2 # x 2 sin x # x 2 Because lim x 0 ( 2 x 2 ) 5 lim x 0 x 2 5 0, the Sandwich Theorem gives lim x 0 ( x 2 sin x ) 5 0 55. [ 2 0.5, 0.5] by [ 2 0.25, 0.25] lim x 0 1 x 2 sin } x 1 2 } 2 5 0 Confirm using the Sandwich Theorem, with g ( x ) 5 2 x 2 and h ( x ) 5 x 2 . ) x 2 sin } x 1 2 } ) 5 ) x 2 ) ? ) sin } x 1 2 } ) # ) x 2 ) ? 1 5 x 2 . 2 x 2 # x 2 sin } x 1 2 } # x 2 Because lim x 0 ( 2 x 2 ) 5 lim x 0 x 2 5 0, the Sandwich Theorem give lim x 0 1 x 2 sin } x 1 2 } 2 5 0. 56. [ 2 0.5, 0.5] by [ 2 0.25, 0.25] lim x 0 1 x 2 cos } x 1 2 } 2 5 0 Confirm using the Sandwich Theorem, with g ( x ) 5 2 x 2 and h ( x ) 5 x 2 . ) x 2 cos } x 1 2 } ) 5 ) x 2 ) ? ) cos } x 1 2 } ) # ) x 2 ) ? 1 5 x 2 . 2 x 2 # x 2 cos } x 1 2 } # x 2 Because lim x 0 ( 2 x 2 ) 5 lim x 0 x 2 5 0, the Sandwich Theorem give lim x 0 1 x 2 cos } x 1 2 } 2 5 0. 57. (a) In three seconds, the ball falls 4.9(3) 2 5 44.1 m, so its average speed is } 44 3 .1 } 5 14.7 m/sec. 46 Section 2.1
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(b) The average speed over the interval from time t 5 3 to time 3 1 h is } D D y t } 5 5 } 4.9(6 h h 1 h 2 ) } 5 29.4 1 4.9 h Since lim h →0 (29.4 1 4.9 h ) 5 29.4, the instantaneous speed is 29.4 m/sec. 58. (a) y 5 gt 2 20 5 g (4 2 ) g 5 } 2 1 0 6 } 5 } 5 4 } or 1.25 (b) Average speed 5 } 2 4 0 } 5 5 m/sec (c) If the rock had not been stopped, its average speed over the interval from time t 5 4 to time t 5 4 1 h is } D D y t } 5 5 } 1.25(8 h h 1 h 2 ) } 5 10 1 1.25 h Since lim h →0 (10 1 1.25 h ) 5 10, the instantaneous speed is 10 m/sec.
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https://www.mathworks.com/matlabcentral/profile/authors/6862552 | 1,618,688,294,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464045.54/warc/CC-MAIN-20210417192821-20210417222821-00535.warc.gz | 987,682,956 | 20,641 | Community Profile
# Ngoc-Tram Le
##### Last seen: 5 months ago
7 total contributions since 2020
#### Ngoc-Tram Le's Badges
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Calculate BMI
Given a matrix |hw| (height and weight) with two columns, calculate BMI using these formulas: * 1 kilogram = 2.2 pounds * 1 ...
12 months ago
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Plot Damped Sinusoid
Given two vectors |t| and |y|, make a plot containing a blue ( |b| ) dashed ( |--| ) line of |y| versus |t|. Mark the minimum...
12 months ago
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Solve a System of Linear Equations
*Example*: If a system of linear equations in _x₁_ and _x₂_ is: 2 _x₁_ + _x₂_ = 2 _x₁...
12 months ago
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Find the Oldest Person in a Room
Given two input vectors: * |name| - user last names * |age| - corresponding age of the person Return the name of the ol...
12 months ago
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Convert from Fahrenheit to Celsius
Given an input vector |F| containing temperature values in Fahrenheit, return an output vector |C| that contains the values in C...
12 months ago
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Calculate Amount of Cake Frosting
Given two input variables |r| and |h|, which stand for the radius and height of a cake, calculate the surface area of the cake y...
12 months ago
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Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
12 months ago | 390 | 1,416 | {"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} | 2.828125 | 3 | CC-MAIN-2021-17 | latest | en | 0.740354 |
http://converged.yt/talks/duke-dsm/talk.html | 1,569,154,248,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575513.97/warc/CC-MAIN-20190922114839-20190922140839-00268.warc.gz | 39,734,374 | 6,069 | Density surface models
Duke University, 13 February 2014
Who is this guy?
• Statistician by training (St Andrews)
• PhD University of Bath, w. Simon Wood
• Postdoc, University of Rhode Island
• Research fellow at CREEM
• Developer of distance sampling software
Spatial modelling
What do we want to do?
• Relate covariates to animal abundance
• Estimate abundance in a spatially explicit way
• Calculate uncertainty
• Interpretability to biologists/ecologists
• Often using mixed/historical data
Some “problems”
• How to model covariate effects?
• Model (term) selection
• Reponse distribution
• Uncertain detection
• Availability
• Autocorrelation
Density surface models
Detection
How do we deal with detectability?
• Distance sampling! – Fit detection functions
• Estimate $$\mathbb{P}(\text{ detection } | \text{ object at distance } x) = g(x)$$
• Calculate average detection probability = $$\frac{1}{w}\int_0^w g(x) \text{ d}x$$ (where $$w$$ is truncation distance)
Distance sampling
• Lots of other stuff going on here!
• Covariates that affect detectability
• Double observer ($$g(0)<1$$)
• Detection function formulations
Distance sampling software
• Distance for Windows (6.2 out soon!)
• Easy to use Windows software
• Len Thomas, Eric Rexstad, Laura Marshall
• Distance R package
• Simple way to fit detection functions
• Me!
• mrds R package
• More complex analyses - double observer surveys
• Jeff Laake, me
Two pages generalized additive models (I)
If we are modelling counts:
$\mathbb{E}(n_j) = \exp \left\{ \beta_0 + \sum_k f_k(z_{jk}) \right\}$
• $$n_j$$ has some count distribution (quasi-Poisson, Tweedie, negative binomial)
• $$f_k$$ are smooth functions (splines $$\Rightarrow f_k(x)=\sum_l \beta_l b_l(x)$$)
• $$f_k$$ can just be fixed effects $$\Rightarrow$$ GLM
• Add-in random effects, correlation structures $$\Rightarrow$$ GAMM
• Wood (2006) is a good intro book
Two pages generalized additive models (II)
Minimise distance between data and model while minimizing:
$\lambda_k \int_\Omega \frac{\partial^2 f_k(z_k)}{\partial z_k^2} \text{ d}z_k$
Two options for response
$$n_j$$ - raw counts per segment
$\mathbb{E}(n_j) = A_j \hat{p}_j \exp \left\{ \beta_0 + \sum_k f_k(z_{jk}) \right\}$
$$\hat{n}_j$$ - H-T estimate per segment
$\mathbb{E}(\hat{n}_j) = A_j \exp \left\{ \beta_0 + \sum_k f_k(z_{jk}) \right\}$
$\hat{n}_j = \sum_{i \text{ in segment } j} \frac{s_i}{\hat{p}_i}$
The dsm package
• Design “inspired by” (“stolen from”) mgcv
• Easy to build simple models, possible to build complex ones
• Syntax example:
model <- dsm(count ~ s(x,k=10) + s(depth,k=6),
detection.function,
segment.data,
observation.data,
family=negbin(theta=0.1))
• Utility functions: variance estimation, plotting, prediction etc
Case study I - Seabirds in RI waters
RI seabirds - Aims
• Wind development in RI/MA waters
• Map of usage
• Estimate uncertainty
• Combine maps (Zonation)
Photo by jackanapes on flickr (CC BY-NC-ND)
RI seabirds - The model
• Availability
• correction factor from previous experimental work
• $$p_j \times \mathbb{P}(\text{available for detection})$$
• Term selection by approximate $$p$$-values
• Covariates are collinear (curvilinear)
• select - extra penalty
• REML - better optimisation objective
From Fig. 1 of Wood (2011)
RI seabirds - Uncertainty
Case study II - black bears in Alaska
Case study II - black bears in AK
• Area of 26,482 km2 (~ size of VT/MA)
• Double observer surveys using Piper Super Cubs
• 1238, 35km transects, 2001-2003
Survey protocol
• Surveys in Spring, bears are there, but not too much foliage
• Generally search uphill
• Double observer (Borchers et al, 2006)
• Curtain between pilot and observer; light system
• Go off transect and circle to ID
Black bears
• Truncate at 22m and 450m, leaving 351 groups (out of ~44,000 segments)
• Group size 1-3 (lone bears, sow w. cubs)
• 1402m elevational cutoff
Final model
• bivariate smooth of location
• smooth of elevation
• bivariate smooth of slope and aspect
Abundance estimate for GMU13E
• MRDS estimate: ~1500 black bears
• DSM estimate: ~1200 black bears (968 - 1635, CV ~13%)
• Not a huge difference, so why bother?
Conclusions
• Flexible spatial models
• GLMs + random effects + smooths + other extras
• autocorrelation can be modelled
• Large areas, makes sense
• Spatial component is v. helpful for managers
• Two-stage models can be useful!
• Estimating temporal trends
References
• Borchers, DL, JL Laake, C Southwell, and CGM Paxton. Accommodating Unmodeled Heterogeneity in Double‐Observer Distance Sampling Surveys. Biometrics 62, no. 2 (2006): 372–378.
• Miller, DL, ML Burt, EA Rexstad and L Thomas. Spatial Models for Distance Sampling Data: Recent Developments and Future Directions. Methods in Ecology and Evolution 4, no. 11 (2013): 1001–1010.
• Winiarski, KJ, ML Burt, Eric Rexstad, DL Miller, CL Trocki, PWC Paton, and SR McWilliams. Integrating Aerial and Ship Surveys of Marine Birds Into a Combined Density Surface Model: a Case Study of Wintering Common Loons. The Condor 116, no. 2 (2014): 149–161.
• Winiarski, KJ, DL Miller, PWC Paton, and SR McWilliams. A Spatial Conservation Prioritization Approach for Protecting Marine Birds Given Proposed Offshore Wind Energy Development. Biological Conservation 169 (2014): 79–88.
Talk available at http://dill.github.com/talks/duke-dsm/talk.html
Thanks
• Rhode Island: Kris Winiarski, Peter Paton, Scott McWilliams
• Alaska: Earl Becker, Becky Strauch, Mike Litzen, Dave Filkill
• Elsewhere: Mark Bravington, Natalie Kelly, Eric Rexstad, Louise Burt, Len Thomas, Steve Buckland
Randomised quantile residuals
• Goodness of fit testing
• Dunn, PK, and GK Smyth. Randomized Quantile Residuals. Journal of Computational and Graphical Statistics 5, no. 3 (1996): 236–244.
• Back transform for exactly Normal residuals
• Less problems with artefacts
• (Thanks to Natalie Kelly at CSIRO for the tip) | 1,668 | 5,931 | {"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": 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-2019-39 | latest | en | 0.737538 |
https://share.cocalc.com/share/5d54f9d642cd3ef1affd88397ab0db616c17e5e0/www/talks/2006-05-09-sage-digipen/tutorial/modeling-3.py?viewer=share | 1,576,289,260,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00183.warc.gz | 528,233,037 | 4,623 | Sharedwww / talks / 2006-05-09-sage-digipen / tutorial / modeling-3.pyOpen in CoCalc
Author: William A. Stein
1# Soya 3D tutorial
2# Copyright (C) 2001-2004 Jean-Baptiste LAMY
3#
4# This program is free software; you can redistribute it and/or modify
6# the Free Software Foundation; either version 2 of the License, or
7# (at your option) any later version.
8#
9# This program is distributed in the hope that it will be useful,
10# but WITHOUT ANY WARRANTY; without even the implied warranty of
11# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
12# GNU General Public License for more details.
13#
14# You should have received a copy of the GNU General Public License
15# along with this program; if not, write to the Free Software
16# Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
17
18
19# modeling-3: Vertex colors : the multicolor pyramid
20
21# Models can be designed in Blender and then imported in Soya (as in lesson basic-1.py),
22# but you can also create them from scratch, using Soya primitive. Learning this is
23# the purpose of the modeling-* tutorial series.
24
25# In this lesson, we'll build a pyramid, made of a quad base and 4 triangles.
26
27
28# Imports and inits Soya.
29
30import sys, os, os.path, soya
31
32soya.init()
33soya.path.append(os.path.join(os.path.dirname(sys.argv[0]), "data"))
34
35# Creates the scene.
36
37scene = soya.World()
38
39# Creates the World that will contain the pyramid. We don't create the pyramid in the
40# scene since we are going to compile the pyramid into a shape.
41
42pyramid_world = soya.World()
43
44# This time, we create the 5 vertices first. We do so to be sure each vertex will have
45# the same color on the different faces it belongs to ; though you can still use the way
46# we use in lesson modeling-1.py.
47# The diffuse attribute of vertex is the color of the vertex ; in Soya, colors are
48# (red, green, blue, alpha) tuples where all components are usually in the range 0.0-1.0.
49# Set diffuse to None to disable vertex color.
50
51# Here, the apex of the pyramid is in white, the base1 vertex is red, base2 is yellow,
52# base3 is green and base4 is blue.
53
54apex = soya.Vertex(pyramid_world, 0.0, 0.5, 0.0, diffuse = (1.0, 1.0, 1.0, 1.0))
55base1 = soya.Vertex(pyramid_world, 0.5, -0.5, 0.5, diffuse = (1.0, 0.0, 0.0, 1.0))
56base2 = soya.Vertex(pyramid_world, -0.5, -0.5, 0.5, diffuse = (1.0, 1.0, 0.0, 1.0))
57base3 = soya.Vertex(pyramid_world, -0.5, -0.5, -0.5, diffuse = (0.0, 1.0, 0.0, 1.0))
58base4 = soya.Vertex(pyramid_world, 0.5, -0.5, -0.5, diffuse = (0.0, 0.0, 1.0, 1.0))
59
60# Now, creates the face, using the vertices we creates above.
61
62soya.Face(pyramid_world, [base1, base2, base3, base4])
63soya.Face(pyramid_world, [base2, base1, apex])
64soya.Face(pyramid_world, [base4, base3, apex])
65soya.Face(pyramid_world, [base1, base4, apex])
66soya.Face(pyramid_world, [base3, base2, apex])
67
68# Compile the pyramid into a shape.
69
70pyramid_shape = pyramid_world.shapify()
71
72# Creates a subclass of Volume that permanently rotates.
74
75class RotatingVolume(soya.Volume):
77 self.rotate_lateral(2.0 * proportion)
78
79# Create a rotating volume in the scene, using the cube shape.
80
81pyramid = RotatingVolume(scene, pyramid_shape)
82pyramid.rotate_vertical(60.0)
83
84# Creates a light.
85
86light = soya.Light(scene)
87light.set_xyz(1.0, -1.0, 2.0)
88
89# Creates a camera.
90
91camera = soya.Camera(scene)
92camera.set_xyz(0.0, 0.0, 2.0)
93soya.set_root_widget(camera)
94
95soya.Idler(scene).idle()
96
97 | 1,158 | 3,565 | {"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} | 2.6875 | 3 | CC-MAIN-2019-51 | longest | en | 0.789133 |
https://nxttime.wordpress.com/2009/12/14/freerover-ultrasonic-sensor-low-pass-filter/ | 1,521,954,476,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651820.82/warc/CC-MAIN-20180325044627-20180325064627-00713.warc.gz | 667,567,572 | 15,284 | Today I decided to enhance US sensor output. Why? Sometimes I had the impression the car changed behaviour. I suspected this had something to do with faulty readings from the Ultrasonic sensor. I decided to make a low pass filter for the US sensor. This should eliminate random faulty readings. But first I had to understand low pass filters. Coding a low pass filter isn’t too hard. There is an explanation on Wikipedia. Basicly it’s just one line of code:
lowPassUS=lowPassUS+alpha*(SensorValue[sonar]-lowPassUS);
The filtered value is calculated from the raw value and the previous filtered value. There is one parameter, alpha, that controls the filter. I wanted to understand how the value of alpha changes the behaviour of the filter. That is why I did some experiments in Excel. Here is an example. The red line represents unfiltered values, the green filtered values. The horizontal axis represents time.
The alpha factor does one thing, it controls how fast the filter adjusts to a new situation. The smaller the alpha, the quicker the filter reacts. In the plot it means that the green line follows the red line more closely. The downside is that less noise is filtered out.
There is one other thing that is important to realise. Alpha doesn’t control the time the filter takes to adjust, it controls the number of steps (iterations) the filter needs to adjust. This means that there is a relation between sample rate, how often the output from the US sensor is read, and alpha. With a higher sample rate and constant alpha, the filter adjusts quicker in real time. But also, if the sample rate is changes one also needs to consider to change the alpha.
The graph above shows four scenario’s. At t=1 the sampling starts, the filter needs time to adjust. One should give the filter time to settle, or, one should initialize the filtered value to the unfiltered value.
Around t=0,3 I tested another scenario. The reading of the Sensor changes because an object is in sight. In this example it takes about 0,1 second for the filter to adjust to the new situation. Is this quick enough, for me it is.
Around t=0,46 there is an faulty reading, the filter dampens the reading but for a short time (0,06 seconds) the objects seems further than it is. In my case, The damping needs to be enough so that the filtered value not exceeds treshhold. I need more real life testing to find out the damping is enough. Also I do know i get faulty 255 values sometimes, but I don’t know if I also get faulty zero values. These will affect behaviour more often, as they can influence collision detection.
Around t=0,6 the last scenario is shown. Here an object at 60 cm dissappears and and a background object at 90 cm appears. But this time the objects doesn’t dissappear at once but during a short time it is sometimes sen and at other times it is not seen. Here the filter really smoothens out the jitter. | 653 | 2,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-13 | latest | en | 0.911217 |
https://mathspanda.com/year7.html | 1,713,152,883,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00764.warc.gz | 341,174,583 | 8,465 | # Year 7 (23-24)
### Past examination material
` `
Since the Year 7 programme for 23-24 is new, there are no past papers.
` `
` `
#### November examination
` `
Theme 1 Midway Assessment
` `
Theme 1 Midway Assessment MS
#### Revision sheets
These sheets are useful for Theme examination preparation.
Answers are at the end of each sheet.
Theme 1 midway revision
` `
Theme 1 revision Theme 2 revision
` `
Theme 3 revision - algebra
` `
Theme 3 revision - algebra ANSWERS
` `
Theme 1 to 3 revision
#### Interleaf sheets
These sheets revise several areas at once. (Theme number in brackets).
Interleaf 1 (1) Interleaf 2 (1)
` `
Interleaf 3 (1) Interleaf 4 (1)
` `
Interleaf 5 (2) Interleaf 6 (2)
` `
Interleaf 7 (3) Interleaf 8 (3)
#### End of Year examination
` `
` `
` `
Many of the PowerPoints and all the booklets below come from the OAT academy website.
We would like to thank them for allowing us to host them on our website.
` `
### Autumn term (Sep-Dec)
` `
#### Theme 1: Representing Number
` `
L1 Place Value: Integers (slides 17-20)
` `
Booklet p4 Place value riddles 5a Place value riddles 5b
` `
Comparing 7 digit numbers 7-digit challenges
` `
8-digit challenges
` `
L2 Place Value: Decimals (slides 39-51)
` `
Booklet p6 Expanded form to 10 million
` `
` `
L3 Place Value: Numbers in words (slide 23)
` `
Booklet p3 (1) & (2) Cross number
` `
` `
L4 Integers/Decimals: Position on number line (slide 28-34)
` `
Booklet p4 (4), p5
` `
` `
L5 Integers/Decimals: Comparing decimals (slides 58-67)
` `
Booklet p8-11 CB: Ordering decimals
` `
Manifest game (slide 35) Millions, billons or trillions
` `
L6 Int./Dec.: Multiply/divide by powers of 10 (slides 78-96)
` `
Booklet p13-16 CB: Multiply by 10, 100,...
` `
CB: Multiplying by powers of 10
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CB: Dividing by powers of 10
` `
L7 Fractions: Representations & bar models (slide 8-27)
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Booklet p3-7 Fractions on number lines 1
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Fractions on numberlines 2
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` `
Labelling fractions on number lines
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L8 Fractions: Improper & mixed fractions (slides 36-50)
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Booklet p9-11
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CB: Converting between improper fractions and mixed numbers
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L9 Fractions: Equivalent fractions (slides 55-79)
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Booklet p12-14 Comparing fractions (bar model)
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Shading equivalent fractions CB: Equivalent fractions
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Equivalent fractions on a number line CB: Ordering fractions
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CB: Simplifying fractions
` `
L10 Fractions: Converting fractions & decimals (slides 7-25)
` `
Booklet p3-5 Related fractions and decimals
` `
CB: Converting decimals to fractions
` `
CB: Converting fractions to decimals
` `
L11 Fractions: Converting fractions, decimals & %ages (slides 39-58)
` `
Booklet p6-9 Equivalent FDP
` `
Whole one search CB: Decimals to percentages
` `
CB: FDP Key Equivalents CB: FDP Mixture
` `
CB: Fractions to percentages CB: Percentages to decimals
` `
CB: Percentages to fractions
` `
L12 Fractions: One number as fraction of another (slides 108-128)
` `
Booklet p18-20
` `
CB: Expressing one number as a fraction of another
` `
L13 Fractions: One number as %age of another (slides 61-64)
` `
Booklet p10-11
` `
CB: Expressing one number as a percentage of another
` `
L14 Fractions: Fraction of amount (slides 102-107)
` `
Booklet p17-18 Fraction of amount (bar model)
` `
Fraction of amount (number lines & bar models)
` `
Fraction of amount (number line & bar model template)
` `
CB: Fraction of amount
` `
L15 FDP: Percentage of amount (slides 66-75)
` `
Booklet p11-13 Percentage of amount using bar models
` `
Shading percentage of amount CB: Percentage of amount
` `
L16 Negatives: Position on number line (slides 71-77)
` `
Booklet p11-12
` `
` `
L17 Negatives: Ordering (slides 14-22)
` `
Booklet p3-5 Ordering negative numbers
` `
CB: Ordering numbers, including negatives
` `
L18 Inequalities: Inequalities on a number line (no lesson)
` `
CB: Inequalities on a number line
` `
` `
L19 Graphs: Plotting co-ordinates in all 4 quadrants (slides 2-7)
` `
` `
Co-ordinates extension (slides 7, 9 etc.)
` `
Coordinate - vertices of shapes CB: Coordinates
` `
L20 Graphs: Lines of the form x = a, y = b & y = x (slides 8-21)
` `
Graphs of x = a and y = a
` `
` `
L21 Graphs: Inequalities with x = a, y = b & y = x (no lesson)
` `
CB: Graphical inequalities
` `
` `
L22 Transformations: Reflection in a line (no lesson)
` `
` `
##### Revision
` `
Place value Ordering decimals Powers of 10
` `
Representing fractions Mixed and improper fractions
` `
Converting between fractions and decimals
` `
Converting between fractions and percentages
` `
Fraction of amount Percentage of amount
` `
Ordering negative numbers Inequalities on a number line
` `
Coordinates Drawing linear graphs
` `
Graphical inequalities (x = and y = only)
` `
Reflections
#### Theme 2: Addition and Subtraction
` `
L1 Decimals: Column methods (slides 10-12, 18-20)
` `
Booklet p3 NCETM Checkpoints Activity C (slide 100)
` `
CB: Adding decimals CB: Subtracting decimals
` `
L2 Integers/decimals: Addition/subtraction strategies (slides 24-30)
` `
Booklet p4-5 WRM Addition/subtraction strategies (slide 14)
` `
NCETM Checkpoints 6-10 Subtractions and number lines
` `
L3 Int./dec.: Fact families & inverse operations (slide 71-82)
` `
Booklet p10-13 Addition and subtraction fact families
` `
` `
` `
Booklet p15-16 Fraction pairs
` `
Egyptian fractions
` `
L5 Fractions: Add/subtract with mixed numbers (slides 95-101)
` `
Booklet p17 Mixed number magic squares
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` `
L6 Fractions: Addition/subtraction problems (no lesson)
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` `
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L7 FDP: Increase/decrease by a fraction or %age (slides 100-107)
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Booklet p18-19
` `
` `
L9 Negatives: Addition with negative numbers (slides 33-49)
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Booklet p7-12 NCETM Checkpoints 1-5
` `
` `
Negative arithmagons and magic squares
` `
L10 Negatives: Subtraction with negative numbers (slides 50-70)
` `
Booklet p13-19 NCETM Checkpoints 1-5
` `
More addition and subtraction with negatives
` `
CB: Addition and subtraction with Negatives
` `
L11 Sequences: Term-to-term rule for linear sequences
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CB: Sequences - term-to-term rule
` `
` `
L12 Graphs: Lines y = x + a, x + y = b & y = b - x (no lesson)
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Shape problems involving add and subtract graphs
` `
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L13 Graphs: Inequalities using y = x + a etc. (no lesson)
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` `
` `
L14 Transformations: Column vectors (no lesson)
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L15 Transformations: Translations (no lesson)
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` `
##### Revision
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` `
` `
Increasing/decreasing by a fraction
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Increasing/decreasing by a percentage
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Term-to-term rule for sequences Column vectors Translations
### Spring term (Jan-Apr)
` `
#### Theme 3: Addition to Multiplication
` `
L1 Order of operations: Four operations (slides 29-42)
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Booklet p5-7 NCETM Checkpoints 29-32
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L2 Order of operations: Powers and brackets (slides 43-71)
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Booklet p7-11
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L3 Algebra: Forming expressions (slide 7-23)
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Booklet p3-4 CB: Forming Expressions
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CB: Multiplying terms
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L5 Algebra: Simplifying additions (slides 36-49)
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Booklet p5-7 Adding to and subtracting from expressions
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Algebraic order of operations Simplifying expressions
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L6 Algebra: Collecting like terms (slides 50-56)
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Booklet p7-8
` `
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L7 Algebra: Substitution into expressions & formulae (slides 57-69)
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Booklet p8-9
` `
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L8 Algebra: Equations involving addition/subtraction (slides 74-79)
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Booklet p10-11 Solving equations involving add & subtract
` `
L9 Algebra: Equations of the form ax + b = c (no lesson)
` `
Booklet p11
` `
##### Revision
` `
Order of operations Forming expressions
` `
Algebraic notation Collecting like terms
` `
Substitution into expressions Solving linear equations
#### Theme 4: Multiplication
` `
L1 Integers: Long multiplication - column method (slides 20-23)
` `
Booklet p6-7 CB: Long multiplication
` `
` `
L2 Integers: Long multiplication - grid method (slides 35-48)
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Booklet p8-12 NCETM Checkpoints Activity G (slide 104)
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` `
L3 Integers: Multiplication as repeated addition (slides 64-73)
` `
Booklet p15-16
` `
` `
L4 Decimals: Multiplying decimals (slides 24-34)
` `
Booklet p7-8 NCETM Checkpoints 21-23
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CB: Multiplying decimals
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L5 Negatives: Multiplying (slide 71-82)
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Booklet p20-22
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L6 Algebra: Expanding single brackets (slide 7-33)
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Booklet p3-7 Algebraic multiplications and factors grid
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L7 Algebra: Expanding including CLT & fractions (slide 34-46)
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Booklet p7-9
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L8 Algebra: Factorising expression into one bracket (slides 47-58)
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Booklet p9-10 Factorisation groups
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L9 Algebra: Expanding double brackets (slides 59-74)
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Booklet p10-13 Playing with double brackets
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L10 Fractions: Multiplying fractions (slides 131-151)
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Booklet p23-25 Multplying fractions using area models
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Multiplying unit fractions match
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L11 Fractions: Multiplying mixed numbers (slides 152-155)
` `
Booklet p25
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L12 Graphs: Lines of the form y = mx (no lesson)
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` `
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L13 Graphs: Lines of the form y = mx + c (no lesson)
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L14 Graphs: Inequalties involving y = mx + c (no lesson)
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L15 Sequences: n-th term of a linear sequence (no lesson)
` `
Visual linear sequences
` `
` `
L16 Proportion: Double number lines & ratio tables (slides 81-88)
` `
Booklet p14-17
` `
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L17 Proportion: Linear proportion (slides 7-28, 51-67)
` `
Booklet p3-6 Proportions using 100
` `
NCETM Exploring multiplicative structure
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L18 Proportion: Linear proportion & recipes (slides 68-73)
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Booklet p14-15
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L19 Proportion: Exchange rates & conversion graphs (slides 51-67)
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Booklet p 12-14 CB: Conversion graphs
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Fractions to percentages conversion graph
` `
Proportion and currencies
` `
##### Revision
` `
Long multiplication Exploring multiplication
` `
Multiplying and dividing decimals Arithmetic with negative numbers
` `
### Summer term (Apr-Jul)
` `
#### Theme 5: Division
` `
L1 Integers: Short division (slides 74-96)
` `
Booklet p17-18
` `
` `
L2 Integers: Decomposing (slides 113-120)
` `
Booklet p21-22 NCETM Checkpoints 19
` `
` `
L3 Integers: Distributivity (slides 126-132)
` `
Booklet p21
` `
` `
L4 Decimals: Division strategies with decimals (slide 130-135)
` `
Booklet p22 NCETM Checkpoints 25-28
` `
Division and cupcakes
` `
L5 Negatives: Division (slide 83-88)
` `
Booklet p23-24
` `
` `
L6 Fractions: Divide a fraction by an integer & vice versa (slides 158-168)
` `
Booklet p26
` `
` `
L7 Fractions: Divide a fraction by a fraction (slides 169-176)
` `
Booklet p27-28
` `
` `
L8 Fractions: Division involving mixed numbers (slides 177-181)
` `
Booklet p28-29
` `
` `
L9 Integers: Multiples (slides 52-63)
` `
Booklet p12-15
` `
` `
` `
L10 Integers: Factors (slides 136-145)
` `
Booklet p22-24 Factors & multiples (sometimes/always/never)
` `
How many factors?
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L11 Integers: Product of prime factors (no lesson)
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` `
Prime factor decomposition logic puzzle
` `
L12 Integers: Uses of prime factorisation (no lesson)
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Factors from prime factorisation
` `
` `
` `
L13 Integers: Divisibility rules (slides 176-183)
` `
Booklet p37-39 NCETM Checkpoints 17-18, 24,
` `
Divisibility problems (ppt) Divisibility_problems (pdf)
` `
` `
#### Theme 6: Mensuration
` `
L1 2-D shapes: Perimeter of rectilinear shapes (slides 99-106)
` `
Booklet p17-19
` `
` `
L2 2-D shapes: Area of rectilinear shapes (slides 162-166)
` `
Booklet p30-32
` `
` `
L3 2-D shapes: Area of parallelograms
` `
Area of parallelogram (too much information)
` `
` `
L4 2-D shapes: Area of triangles (no lesson)
` `
` `
` `
L5 2-D shapes: Area of trapezia (no lesson)
` `
` `
` `
` `
L7 3-D solids: Volume of cuboids (slides 167-175)
` `
Booklet p33-37
` `
` `
L8 3-D solids: Volume of prisms, incl. triangular prisms (no lesson)
` `
` `
` `
L6 3-D solids: Nets of 3-D solids (no lesson)
` `
` `
` `
L7 3-D solids: Surface area of cuboids (no lesson)
` `
` `
` `
L8 3-D solids: Surface area of triangular prisms (no lesson)
` `
` `
` `
##### Revision
` `
Please e-mail admin@mathspanda.com if any of the links don't work or if you can think of a way to improve the website. We cannot guarantee an immediate response as we spend most of the day munching through 16 kg of bamboo each and our paws are a bit big for typing. Our reply might end up in your spam folder - whatever one of those is. | 4,123 | 13,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-18 | latest | en | 0.735712 |
https://www.myfxbook.com/lv/community/general/newbiequestion3ausd2fjpypairmorevolatile3f/1355867,1 | 1,537,348,672,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156096.0/warc/CC-MAIN-20180919083126-20180919103126-00394.warc.gz | 809,242,714 | 32,563 | Newbie Question : USD/JPY Pair more volatile?
Biedrs kopš Jan 14, 2017 2 ieraksti Robsingh Jan 20 2017 at 08:18
Hello,
I am new to forex and I am trying to understand things well before I trade with lvie accounts. I understand how profits/losses are calculated but I don't seem to get why profit/losses with USD/JPY pair differ from other pairs such as EUR/USD.
For example, when testing strategies in Demo Account. A buy of 2500 Size (.25 lots) gives me a a higher profit/loss ~between \$50-\$500 for USD/JPY pair. But, when I test the strategy against EUR/USD with same position size -2500, I get a lower profit loss range (\$5-\$50). Two Questions:
Q1.If profit/loss depends on pips then why so much difference in \$ amount ? Is it that USD/JPY pair is much more volatile and there is a greater opportunity for profits or risk for loss?
Q2. Does entering a Position(long) of 2500 USD/JPY mean I need to have greater than \$2500 in my account? Conversely, does entering a Short position with a size 2500 means selling \$2500 USD to buy JPY? The math would say-
To put it in mathematical terms , here is how I am calculating profit when buying
Buying- 0.25 lots @114.50
Enter Long Calculation : 2500 X 114.50 = 286250 YEN
Selling -.25 Lots @114.70
Exit Long Calculation : 2500 X 114.70 = 286750 YEN
Difference =500 YEN . Applying current conversion rate results in 500/114.70 = \$1 Profit @ 20 Pips
In the demo account however, I get a much higher profit amount ~\$1000. What am I not getting???
Thank you for your help!!
Biedrs kopš Dec 17, 2015 30 ieraksti janettte Jan 20 2017 at 13:44
Robsingh posted:
Q2. Does entering a Position(long) of 2500 USD/JPY mean I need to have greater than \$2500 in my account?
Not necessary, it depends on the required margin / leverage you are using for your account as well as the margin call and stop out levels set by your broker. If you trade at 1:100, to open a position of 2500 USD you need to have more than 25 USD in your account, for example. 2500 is actually .025 lots.
Biedrs kopš Dec 17, 2015 30 ieraksti janettte Jan 20 2017 at 13:48
Robsingh posted:
For example, when testing strategies in Demo Account. A buy of 2500 Size (.25 lots) gives me a a higher profit/loss ~between \$50-\$500 for USD/JPY pair. But, when I test the strategy against EUR/USD with same position size -2500, I get a lower profit loss range (\$5-\$50).
These are 2 separate instruments, 2 separate markets, so the same position (long/short) will certainly result in different outcome, so I am not sure what is not clear in this case. Maybe I couldn't get you point correctly.
Biedrs kopš Jan 14, 2017 2 ieraksti Robsingh Jan 22 2017 at 07:49
janettte posted:
Robsingh posted:
For example, when testing strategies in Demo Account. A buy of 2500 Size (.25 lots) gives me a a higher profit/loss ~between \$50-\$500 for USD/JPY pair. But, when I test the strategy against EUR/USD with same position size -2500, I get a lower profit loss range (\$5-\$50).
These are 2 separate instruments, 2 separate markets, so the same position (long/short) will certainly result in different outcome, so I am not sure what is not clear in this case. Maybe I couldn't get you point correctly.
Thank you for the reply.
I understand that the results will be different. What I don't understand is - when look at Pip value for JPY and another pair , the value is approx the same (image attached)
But JPY has a greater profit/loss potential/risk for the same lot size (backtest image attached)
Pielikumi:
Biedrs kopš Apr 18, 2017 256 ieraksti AmDiab May 08 2017 at 11:05
You can use GBP/JPY, this is the most volatile trading pair in Forex! By the way, USD/JPY is my favorite trading pair! For the reason that, I get more than 90% accuracy on USD/JPY from my personal trading tool! Actually, in my live trading I am always interested on major Forex pairs, these all are reliable to use!
Biedrs kopš Nov 21, 2015 6 ieraksti Gpifund May 08 2017 at 14:57
Robsingh posted:
For example, when testing strategies in Demo Account. A buy of 2500 Size (.25 lots) gives me a a higher profit/loss ~between \$50-\$500 for USD/JPY pair. But, when I test the strategy against EUR/USD with same position size -2500, I get a lower profit loss range (\$5-\$50).
The ting that one particular strategy 'works well' in one pair DOESN'T mean that it will also work on another one...the reasons depends on the strategies you are using everytime ;)
Dream big, trade bigger.
Biedrs kopš Feb 12, 2016 522 ieraksti Baldo (BaldoN) May 09 2017 at 11:26
Robsingh posted:
Hello,
I am new to forex and I am trying to understand things well before I trade with lvie accounts. I understand how profits/losses are calculated but I don't seem to get why profit/losses with USD/JPY pair differ from other pairs such as EUR/USD.
The reason for this is because when you trade (for example 1.0 standard MT4 lot):
* USD/JPY - You are buying or selling 100,000 USD vs JPY (PnL result is in JPY); Point here is the 3rd digit
* EUR/USD - You are buying or selling 100,000 EUR vs USD (PnL result is in USD); Point here is 5th digit
You have different currencies and different exchange rates with different place of decimal point and you will have also different pip value.
Biedrs kopš Apr 18, 2017 256 ieraksti AmDiab Jun 13 2017 at 12:46
Robsingh posted:
Hello,
I am new to forex and I am trying to understand things well before I trade with lvie accounts. I understand how profits/losses are calculated but I don't seem to get why profit/losses with USD/JPY pair differ from other pairs such as EUR/USD.
For example, when testing strategies in Demo Account. A buy of 2500 Size (.25 lots) gives me a a higher profit/loss ~between \$50-\$500 for USD/JPY pair. But, when I test the strategy against EUR/USD with same position size -2500, I get a lower profit loss range (\$5-\$50). Two Questions:
Q1.If profit/loss depends on pips then why so much difference in \$ amount ? Is it that USD/JPY pair is much more volatile and there is a greater opportunity for profits or risk for loss?
Q2. Does entering a Position(long) of 2500 USD/JPY mean I need to have greater than \$2500 in my account? Conversely, does entering a Short position with a size 2500 means selling \$2500 USD to buy JPY? The math would say-
To put it in mathematical terms , here is how I am calculating profit when buying
Buying- 0.25 lots @114.50
Enter Long Calculation : 2500 X 114.50 = 286250 YEN
Selling -.25 Lots @114.70
Exit Long Calculation : 2500 X 114.70 = 286750 YEN
Difference =500 YEN . Applying current conversion rate results in 500/114.70 = \$1 Profit @ 20 Pips
In the demo account however, I get a much higher profit amount ~\$1000. What am I not getting???
Thank you for your help!!
Among JPY pairs, GBP/JPY is the most volatile one! It’s known as dragon pair because of it’s nature! I actually always use volatile currency pair since it produces my daily trading target so quickly! On the other hand, USD/JPY is average volatile as like others major currency pairs!
Lūdzu ienāciet, lai komentētu.
10-y Bond Auction (22 min)
PBOC Governor Yi: Will maintain prudent ...(3 min ago)
EURUSD 1.17032 GBPUSD 1.31874 USDJPY 112.358 USDCAD 1.29508
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# Probability with permutations and combinations
## Problem
Each card in a standard deck of $52$ playing cards is unique and belongs to $1$ of $4$ suits:
• $13$ cards are clubs
• $13$ cards are diamonds
• $13$ cards are hearts
• $13$ cards are spades
Suppose that Luisa randomly draws $4$ cards without replacement.
What is the probability that Luisa gets $2$ diamonds and $2$ hearts (in any order)? | 154 | 622 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 18, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | longest | en | 0.928724 |
https://community.qlik.com/t5/Qlik-Sense-Advanced-Authoring/how-to-redistribute-rows-into-other-rows/td-p/1593118 | 1,606,283,887,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181179.12/warc/CC-MAIN-20201125041943-20201125071943-00082.warc.gz | 255,775,101 | 48,183 | Announcements
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## how to redistribute rows into other rows
Table1:
columnname1 Columnname2
v 1
w 2
x 4
y 8
z 30
Suppose i have Filterpane of Dimension columnname1, if i click on Z or filter Z, then columnname2 corresponding column z will be redistributed to other rows in their respective ratios and add with their respective columns.
eg suppose I filter on Z then result will be like
columnname1 Columnname2 colmnname3
v 1 ( (1/(1+2+4+8))*30)+1=3
w 2 ( (2/(1+2+4+8))*30)+2=6
x 4 ( (4/(1+2+4+8))*30)+4=12
y 8 ( (8/(1+2+4+8))*30)+8=24
remove z if only z is filtered ,if 2 columns are selected then sum ofcolumnname2 corresponding to filterd columns will redistributed into the remaining.
eg:
if i filtered y and z then 38 will reditributed in their remaining rows w.r.t their ratios. | 344 | 1,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | latest | en | 0.516403 |
http://clay6.com/qa/25913/an-electric-field-overrightarrow-2-hat-3-hat-n-c-exists-in-space-if-the-pot | 1,537,398,540,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156311.20/warc/CC-MAIN-20180919220117-20180920000117-00420.warc.gz | 50,846,506 | 28,013 | # An electric field $\;\overrightarrow{E}=(2 \hat{i}+3\hat{j})\;N/C$ exists in space . if the potential at the origin is zero , find the potential at $\;(2 ,2 )\;m$
$(a)\;-100 V\qquad(b)\;10 V\qquad(c)\;- 10V\qquad(d)\;100 V$
Explanation : $\;V_{a}-V_{b}=-\;\int_{b}^{a}\;\overrightarrow{E} . \overrightarrow{dl}$
$V_{(2 , 2)}-0= -\;\int_{(0 ,0)}^{(2 , 2)}\;(2 \hat{i}+3\hat{j})(dx \hat{i}+dy \hat{j})$
$V_{(2 , 2)}= -\;\int_{(0 ,0)}^{(2 , 2)}\;(2 dx +2 dy )$
$=(2\times2+3\times2)$
$V_{2 ,2 }=-10 V\;.$ | 240 | 505 | {"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} | 3.5 | 4 | CC-MAIN-2018-39 | longest | en | 0.193232 |
https://www.physicsforums.com/threads/falling-to-earth.715520/ | 1,660,500,905,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572063.65/warc/CC-MAIN-20220814173832-20220814203832-00695.warc.gz | 813,053,969 | 17,689 | # Falling to Earth
pitchwest
I'd like to graph the function of and object falling to earth from a point high enough that air density has an effect on the rate the object falls due to gravity. I'm not really sure where to start though.
I'm thinking about that guy that jumped from the highest point ever free-falled from and broke the sound barrier.
## Answers and Replies
Staff Emeritus
Homework Helper
First, figure out what function you are talking about. Velocity versus time? distance versus time?
pitchwest
What do you think would best show the change in gravity due to the change in air resistance against an object?
I was thinking that I'd have to first find an equation for the amount of air at different altitudes. Then calculate the level of air resistance and apply it to the amount of air equation. Then figure out how that applies to gravity at different altitudes.
I hope that helps clear up what I'm getting at.
Staff Emeritus
Homework Helper
It's not clear what you mean by 'the change in gravity due to the change in air resistance against an object'.
You should review Newton's Law of Gravitational attraction: F = Gm1m2/r^2
mephestopheles
I believe he is talking about rate of descent and the effect varying atmospheric densities would have on a falling object of something like the size of a person. So that rate of descent would become variable due to the area resistance would be placed upon.
Staff Emeritus
Homework Helper
Eventually, a terminal velocity should be reached when the weight of the body balances the drag force.
Mentor
The terminal velocity depends on the density of the medium (air). An object falling from a very great height would first reach a rather high terminal velocity corresponding to the low density of air at high altitude. As it falls into denser air, the terminal velocity decreases. To put it another way, the object's drag coefficient increases with air density.
If the drag coefficient is constant, it's possible to solve the differential equation of motion exactly for certain cases. With a variable drag coefficient, you'd have to solve the differential equation numerically.
pitchwest
At extreme heights there's less gravity. A falling person would fall slower, but there's less air resistance at those heights, so a person would fall faster. The guy that did a free fall from 128,100 ft broke the speed of sound. I'd like to calculate the change in speeds he fell and view it on a graph. I'd imagine he broke the speed of sound but slowed down as he got closer to earth.
I'd also like to know if he slowed down to the standard terminal velocity eventually.
To put it another way, the object's drag coefficient increases with air density.
This is not true. Drag coefficient is independent of air density. In fact, that's a large part of the reason why you want a drag coefficient in the first place - it's dimensionless, and largely independent of both air density and airspeed (across a fairly broad range of speeds, though if you start approaching the speed of sound or higher, this can change), and is effectively only dependent on the geometry of the object in the flow.
Drag force increases with air density (all else equal), but not drag coefficient.
Staff Emeritus
You can come fairly close assuming a constant coefficient of drag and gravity that decreases linearly with altitude:
$$F(h) = \frac 1 2 \rho(h) v^2 C_dA - mg(h)$$
Here, h is the person's altitude. I'll use meters rather than feet and force is a signed quantity (not a vector; I'm only looking at the "falling straight down" problem here). A negative value means the person is accelerating downward, positive means upward.
The first part is drag to air resistance. The term ρ(h) is the density as a function of altitude. You'll need some model of air density. I'd suggest using the 76 US Standard Atmosphere model. It's fairly simple, something you can easily implement in a spread sheet or simple computer program. Unfortunately, you might have to wait a few days to get your hands on that model. You might be able to find it with some good internet searching. The US government websites where you would normally find it are down right now The final part of that term, CdA, is the (unit less) coefficient of drag times the cross section to drag. It might be easier to make this one item rather than two in your model. If you do this, try using 1 square meter as a first cut.
The final term is just gravitational force. A simple linear model of gravitational acceleration g(h) will work quite nicely for this problem. The linear term is called the "free air correction". Subtract 3.084×10-6 m/s2 for every meter of altitude from the ground level gravitational acceleration. For New Mexico, I'd suggest using (9.796 -3.084×10-6h) m/s2, where height h is in meters.
voko
At extreme heights there's less gravity.
At 128100 feet, the force of gravity is about 1% percent less than at the sea level. This can be ignored for any practical purpose, because the other errors and uncertainties in the model will most likely be far greater than that.
Gold Member
This is not true. Drag coefficient is independent of air density. In fact, that's a large part of the reason why you want a drag coefficient in the first place - it's dimensionless, and largely independent of both air density and airspeed (across a fairly broad range of speeds, though if you start approaching the speed of sound or higher, this can change), and is effectively only dependent on the geometry of the object in the flow.
Drag force increases with air density (all else equal), but not drag coefficient.
That is not true either.
Drag coefficient may be dimensionless but it does vary with speed, density, and viscosity of the medium. Drag coefficient can be plotted against another dimensionless number called the Reynolds number which incorporates velocity, dynamic viscosity, density and a characteristic length. One can use kinematic viscoscity in favour of density and dynamic viscoscity.
In addition, temperature of the medium will also affect the viscosity, and for gases the kinematic viscosity will increase with temperature.
As the speed of the object increases through the medium particular flow seperation points are reached which will change the coefficient of drag. The range of CD will vary from attached laminar flow at slow speeds, where the coefficient can be quite high ( > 1.0 ), to that at detached turbulent flow where it can be much lower. | 1,380 | 6,483 | {"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": 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} | 3.921875 | 4 | CC-MAIN-2022-33 | latest | en | 0.947112 |
http://mathhelpforum.com/calculus/213403-find-vertical-asymptotes-print.html | 1,505,984,130,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687711.44/warc/CC-MAIN-20170921082205-20170921102205-00351.warc.gz | 220,616,289 | 2,794 | # Find the vertical asymptotes
• Feb 19th 2013, 08:40 AM
dangbeau
Find the vertical asymptotes
Find the vertical asymptotes (if any) of the graph of the function. (Use n as an arbitrary integer if necessary. If an answer does not exist, enter DNE.)
f(x) = 3/x^2
0 is not the answer? why?
edit: so sorry, the answer is x=0, i don't know i have to type the "x="(Worried)
• Feb 19th 2013, 09:16 AM
MINOANMAN
Re: Find the vertical asymptotes
Because limF(x) = infinity when x goes to 0 so the y-axis ( x = 0) is the vertical asymptote of this function.
it is simple.
Minoas
• Feb 19th 2013, 12:45 PM
HallsofIvy
Re: Find the vertical asymptotes
Quote:
Originally Posted by MINOANMAN
Because limF(x) = infinity when x goes to 0 so the y-axis ( x = 0) is the vertical asymptote of this function.
it is simple.
Minoas
Yes, it is simple. But the original question what "why is 0 NOT the answer!" And should have been "why does this machine not accept 0 as the answer". | 301 | 965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-39 | longest | en | 0.892291 |
https://www.diychatroom.com/f19/how-calculate-joist-span-table-3x8s-67036/ | 1,596,806,527,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737178.6/warc/CC-MAIN-20200807113613-20200807143613-00561.warc.gz | 640,386,354 | 26,035 | How To Calculate Joist Span Table For 3x8's - Building & Construction - DIY Chatroom Home Improvement Forum
DIY Chatroom Home Improvement Forum How to calculate Joist span table for 3x8's
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03-17-2010, 03:30 PM #1
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## How to calculate Joist span table for 3x8's
I want to put a loft in a barn. The span is 16 foot. I have a bunch of 16 foot 3x8's I also have a bunch of Simpson HU38's to hang them.
The problem: All span tables I've seen are for 2 bys.
Since a 3x8 is 50% bigger than a 2x8 can I multiply the maximum length a 2x8 can span by 1.5 to get the manimum length a 3x8 can span?
If so; I can span the 16 feet using the 3x8's on a 12 inch center.
This sound like a good idea?
03-17-2010, 05:13 PM #2 Member Join Date: Mar 2010 Location: North Central Kansas Posts: 11,651 Rewards Points: 826 What are you going to be storing in that loft?
03-17-2010, 05:32 PM #3 Civil Engineer Join Date: Mar 2009 Location: Boston Posts: 5,832 Rewards Points: 5,246 No, you cannot multiply the length by 1.5. The allowable span for a beam is governed by the moment of inertia of the beam (depends on the size of the beam), the maximum allowable fiber bending stress (depends on the type of wood and the grade), the modulus of elasticity of the material (depends on the species of wood), and the loading (dead plus live load, often governed by code). Maximum span is governed by the more restrictive of the deflection of the beam at the center, or the bending strength of the beam. Deflection is related to the fourth power of the span length, while maximum fiber bending stress is related to the square of the length, so neither factor scales directly with the beam moment of inertia. If none of this makes any sense, let me make it a little simpler. Any structural engineer worth a nickel can tell you based upon an inspection of your site and a review of the loading what size beam you need, or in your case how much you can span with a specific size beam. There is more to the design than simply sizing the beam, there are connection details and an analysis of the flex of the beam that should also be done at the same time the beam is evaluated for strength.
The Following User Says Thank You to Daniel Holzman For This Useful Post: drtbk4ever (03-19-2010)
03-17-2010, 06:14 PM #4 Member Join Date: Nov 2009 Location: central virginia mountains Posts: 1,857 Rewards Points: 1,000 you the man daniel __________________ The older I get the better I was
03-17-2010, 06:31 PM #5
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Quote:
Originally Posted by kwikfishron What are you going to be storing in that loft?
Just general storage stuff. Nothing too heavy to carry up a ladder. I figured if I could get to a "40 live load" it would do the trick.
03-17-2010, 06:33 PM #6
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Quote:
Originally Posted by Daniel Holzman No, you cannot multiply the length by 1.5. The allowable span for a beam is governed by the moment of inertia of the beam (depends on the size of the beam), the maximum allowable fiber bending stress (depends on the type of wood and the grade), the modulus of elasticity of the material (depends on the species of wood), and the loading (dead plus live load, often governed by code). Maximum span is governed by the more restrictive of the deflection of the beam at the center, or the bending strength of the beam. Deflection is related to the fourth power of the span length, while maximum fiber bending stress is related to the square of the length, so neither factor scales directly with the beam moment of inertia. If none of this makes any sense, let me make it a little simpler. Any structural engineer worth a nickel can tell you based upon an inspection of your site and a review of the loading what size beam you need, or in your case how much you can span with a specific size beam. There is more to the design than simply sizing the beam, there are connection details and an analysis of the flex of the beam that should also be done at the same time the beam is evaluated for strength.
You could have just said "I don't know".
03-17-2010, 06:38 PM #7 Member Join Date: Nov 2009 Location: central virginia mountains Posts: 1,857 Rewards Points: 1,000 daniel gave alot of good info there sorry he did'nt have the reply you needed to do what your going to do anyways __________________ The older I get the better I was
03-17-2010, 06:52 PM #8 Member Join Date: May 2008 Location: Near Philly Posts: 2,902 Rewards Points: 2,716 I know Daniel knows how to do it but he isn't where you are and therefore can't know the issues he outlined, he can't say. There is a difference.
03-17-2010, 07:38 PM #9
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What he is trying to say, in layman terms, is NO, you can not simply multiple by 1.5.
It should be inspected and decided by a professional.
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Short of cutting off a body part, the worst that can happen in woodworking is manufacturing really nice looking kindling. --- Quoted from lenaitch
03-17-2010, 08:35 PM #10
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Quote:
Originally Posted by WillSTX You could have just said "I don't know".
If he was on site & could see what you are doing, existing support, existing structure, WHERE you are located, local codes, etc etc etc then he would be able to run the numbers
What you want to do is beyond the normal span tables
As such you need someone to evaluate YOUR specific installation & base the decision on that
03-17-2010, 10:06 PM #11 I have gas! Join Date: Mar 2007 Location: Massachusetts Posts: 2,303 Rewards Points: 2,004 According to this website calculator, if you double the 3x8s (4.5 x 7.25 actual) 16" OC, you can span 16' and get an L / 417. But I agree with Daniel Holzman, there's more to it than just sizing the beam. __________________ I tear things down and build them up.
03-18-2010, 02:23 AM #12 Member Join Date: Feb 2010 Location: MI's Western UP Posts: 603 Rewards Points: 506 if you can find a table for a 2x9, use that. 3x8 is a little bit stronger than 2x9 (maybe 3/4s of a 2x10, so dont use that chart), but will have to carry more of it own weight too. another option would be to take a 2x8 chart, and space them out further. 6 3x8s should be as strong as 9 2x8s covering the same area. anyone disagree? Last edited by forresth; 03-18-2010 at 02:25 AM.
03-18-2010, 06:59 AM #13
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Quote:
Originally Posted by forresth if you can find a table for a 2x9, use that. 3x8 is a little bit stronger than 2x9 (maybe 3/4s of a 2x10, so dont use that chart), but will have to carry more of it own weight too. another option would be to take a 2x8 chart, and space them out further. 6 3x8s should be as strong as 9 2x8s covering the same area. anyone disagree?
Yes, I disagree with guessing
That's why floors sag
03-18-2010, 11:50 AM #14
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Quote:
Originally Posted by Scuba_Dave Yes, I disagree with guessing That's why floors sag
math and logic is not guessing
03-18-2010, 12:22 PM #15
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Quote:
Originally Posted by forresth if you can find a table for a 2x9, use that. 3x8 is a little bit stronger than 2x9 (maybe 3/4s of a 2x10, so dont use that chart), but will have to carry more of it own weight too. another option would be to take a 2x8 chart, and space them out further. 6 3x8s should be as strong as 9 2x8s covering the same area. anyone disagree?
You are taking one size & trying to guess at what a 3x8 will carry
A 2x9 is not the same as a 3x8
Its not 3/4 of a 2x10
a 2x8 is not the same as a 3x8
"should be" That is guessing
Quote:
Originally Posted by forresth math and logic is not guessing
Without a proper span chart & calculation it is guessing
Show me your math
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Top of Page | View New Posts | 2,509 | 9,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-34 | latest | en | 0.854117 |
https://discourse.julialang.org/t/how-to-convert-symbolics-jl-expressions-to-functions-while-keeping-speed/98111 | 1,685,610,267,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647639.37/warc/CC-MAIN-20230601074606-20230601104606-00350.warc.gz | 245,549,547 | 5,074 | # How to convert Symbolics.jl expressions to functions while keeping speed?
Hello!
I’m using Symbolics.jl to perform some symbolic computations and then convert the resulting expressions to functions for further use. However, when I use `build_function` to generate functions, the speed is only a quarter of that when I directly define the function. How can I optimize this?
Following is an example:
``````using Symbolics
using BenchmarkTools
@variables x y
f(x,y) = x^2 + sin(x+y)
D = Differential(x)
expr = expand_derivatives(D(f(x,y))) # output: 2x + cos(x + y)
# generate the function from the expression
f_expr = build_function(expr, x, y, expression = Val{false})
# define the function directly
f_defn(x, y) = 2x + cos(x + y)
# benchmarks
@btime for x in rand(100), y in rand(100)
f_expr(x, y)
end
# output: 239.600 μs (30101 allocations: 557.12 KiB)
@btime for x in rand(100), y in rand(100)
f_defn(x, y)
end
# output: 60.300 μs (101 allocations: 88.38 KiB)
``````
``````const fexpr = build_function(expr, x, y, expression = Val{false})
``````
`f_defn` is `const`, while `fexpr` is not.
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# AtiyahSolutions - Adam Allan Solutions to Atiyah Macdonald...
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Unformatted text preview: Adam Allan Solutions to Atiyah Macdonald 1 Chapter 1 : Rings and Ideals 1.1. Show that the sum of a nilpotent element and a unit is a unit. If x is nilpotent, then 1- x is a unit with inverse ∑ ∞ i =0 x i . So if u is a unit and x is nilpotent, then v = 1- (- u- 1 x ) is a unit since- u- 1 x is nilpotent. Hence, u + x = uv is a unit as well. 1.2. Let A be a ring with f = a + a 1 x + ··· + a n x n in A [ x ] . a. Show that f is a unit iff a is a unit and a 1 ,...,a n are nilpotent. If a 1 ,...,a n are nilpotent in A , then a 1 x,...,a n x n are nilpotent in A [ x ]. Since the sum of nilpotent elements is nilpotent, a 1 x + ··· + a n x n is nilpotent. So f = a + ( a 1 x + ··· + a n x n ) is a unit when a is a unit by exercise 1.1. Now suppose that f is a unit in A [ x ] and let g = b + b 1 x + ··· + b m x m satisfy fg = 1. Then a b = 1, and so a is a unit in A [ x ]. Notice that a n b m = 0, and suppose that 0 ≤ r ≤ m- 1 satisfies a r +1 n b m- r = a r n b m- r- 1 = ··· = a n b m = 0 Notice that 0 = fg = m + n X i =0 i X j =0 a j b i- j x i = m + n X i =0 c i x i where we define a j = 0 for j > n and b j = 0 for j > m . This means that each c i = 0, and so 0 = a r +1 n c m + n- r- 1 = n X j =0 a j a r +1 n b m + n- r- 1- j = a r +2 n b m- r- 1 since m + n- r- 1- j ≥ m- r for j ≤ n- 1. So by induction a m +1 n b = 0. Since b is a unit, we conclude that a n is nilpotent. This means that f- a n x n is a unit since a n x n is nilpotent and f is a unit. By induction, a 1 ,...,a n are all nilpotent. b. Show that f is nilpotent iff a ,...,a n are nilpotent. Clearly f = a + a 1 x + ... + a n x n is nilpotent if a ,...,a n are nilpotent. Assume f is nilpotent and that f m = 0 for m ∈ N . Then in particular ( a n x n ) m = 0, and so a n x n is nilpotent. Thus, f- a n x n is nilpotent. By induction, a k x k is nilpotent for all k . This means that a ,...,a n are nilpotent. c. Show that f is zero-divisor iff bf = 0 for some b 6 = 0 . If there is b 6 = 0 for which bf = 0, then f is clearly a zero-divisor. So suppose f is a zero-divisor and choose a nonzero g = b + b 1 x + ··· + b m x m of minimal degree for which fg = 0. Then in particular, a n b m = 0. Since a n g · f = 0 and a n g = a n b + ··· + a n b m- 1 x m- 1 , we conclude that a n g = 0 by minimality. Hence, a n b k = 0 for all k . Suppose that a n- r b k = a n- r +1 b k = ··· = a n b k = 0 for all k Then as in part a we obtain the equation 0 = m + n- r- 1 X j =0 a m + n- r- 1- j b j = a n- r- 1 b m 2 Again we conclude that a n- r- 1 g = 0. Hence, by induction a j b k = 0 for all j,k . Choose k so that b = b k 6 = 0. Then bf = 0 with b 6 = 0. d. Prove that f,g are primitive iff fg is primitive....
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This first assessment for lab which covers conversion of units, precision accuracy and measuring length. Please answer this by tuesday afternoon.
• 1.
### Which is the best instrument to measure the volume of a block of wood
• A.
Meterstick
• B.
Vernier caliper
• C.
Micrometer caliper
A. Meterstick
Explanation
A meterstick is the best instrument to measure the volume of a block of wood because it is a long, straight ruler that can provide accurate measurements of length. To find the volume of a block, one would measure the length, width, and height of the block using a meterstick, and then multiply these measurements together. The meterstick's length and simplicity make it a suitable tool for measuring the dimensions of a block of wood accurately. Vernier calipers and micrometer calipers are more suitable for measuring smaller, more precise dimensions.
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• 2.
### Prefix in the SI system which means 1/1000
• A.
Micro
• B.
Mega
• C.
Milli
• D.
Pico
C. Milli
Explanation
The prefix "milli" in the SI system means 1/1000. It is used to denote a unit that is one thousandth of the base unit. For example, millimeter represents one thousandth of a meter, milligram represents one thousandth of a gram, etc. This prefix is commonly used in measurements and calculations to indicate a smaller quantity or scale.
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• 3.
### Prefix in the SI system which means 1/1000000
• A.
Centi
• B.
Micro
• C.
Milli
• D.
Femto
B. Micro
Explanation
The prefix "micro" in the SI system means 1/1000000. It is used to denote a factor of one millionth. This prefix is commonly used in scientific and technical fields to represent very small quantities or measurements. For example, micrograms (µg) and micrometers (µm) are units that indicate one millionth of a gram and one millionth of a meter, respectively.
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• 4.
### Prefix which means 1000
• A.
Hecto
• B.
Kilo
• C.
Mega
• D.
Nano
B. Kilo
Explanation
The prefix "kilo" means 1000. It is commonly used in the metric system to indicate a multiple of 1000. For example, a kilogram is equal to 1000 grams. Therefore, "kilo" is the correct answer as it is the prefix that means 1000.
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• 5.
### The volume of an irregular rigid solid can be best measured through water displacement (compared to geometric measurements).
• A.
True
• B.
False
A. True
Explanation
Water displacement is a method commonly used to measure the volume of irregular rigid solids. This involves placing the solid in a container filled with water and measuring the increase in water level. Since irregular solids do not have regular geometric shapes that can be easily measured using formulas, water displacement provides a more accurate and reliable measurement of their volume. Therefore, the statement that the volume of an irregular rigid solid can be best measured through water displacement is true.
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• 6.
### The volume of a sphere is just . Suppose the diameter of a sphere is 10 inches. What is its volume?
• A.
25*pi/3
• B.
100*pi/2
• C.
500*pi/3
• D.
460*pi/5
C. 500*pi/3
Explanation
The volume of a sphere is given by the formula V = (4/3) * π * r^3, where r is the radius of the sphere. In this question, the diameter of the sphere is given as 10 inches, which means the radius is 5 inches (since radius = diameter/2). Plugging this value into the formula, we get V = (4/3) * π * (5^3) = 500 * π/3. Therefore, the correct answer is 500 * π/3.
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• 7.
### What is the volume of a rectangular solid with length =50.0 cm, width = 2.00 m, and height =1.00 m?
• A.
0.50 cubic meters
• B.
1.00 cubic meters
• C.
2.00 cubic meters
C. 2.00 cubic meters
Explanation
The volume of a rectangular solid is calculated by multiplying its length, width, and height. In this case, the length is given as 50.0 cm, the width is given as 2.00 m, and the height is given as 1.00 m. To calculate the volume, we need to convert the length and width to meters, since the height is already given in meters. Converting 50.0 cm to meters gives us 0.50 m. Therefore, the volume is calculated as 0.50 m * 2.00 m * 1.00 m = 1.00 cubic meters. Therefore, the correct answer is 1.00 cubic meters.
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• 8.
### Which is NOT a part of a vernier caliper
• A.
Main scale
• B.
Air hole
• C.
Locking screw
• D.
Vernier scale
B. Air hole
Explanation
The air hole is not a part of a vernier caliper. Vernier calipers consist of a main scale, a vernier scale, and a locking screw. The main scale is used to measure the larger dimensions, while the vernier scale allows for more precise measurements. The locking screw is used to secure the caliper in place once the measurement has been taken. However, the air hole does not serve any functional purpose in the operation of the vernier caliper and is therefore not a part of it.
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• 9.
### What is the volume of a rectangular solid with length = 2.00 m, width = 0.50 km, and height of about one meter?
• A.
10 cubic meters
• B.
200 cubic meters
• C.
100 cubic meters
• D.
1000 cubic meters
D. 1000 cubic meters
Explanation
The volume of a rectangular solid is calculated by multiplying its length, width, and height. In this case, the length is given as 2.00 m, the width is given as 0.50 km (which can be converted to 500 m), and the height is given as about one meter. Multiplying these values together (2.00 m x 500 m x 1 m) gives a volume of 1000 cubic meters.
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• 10.
### Prefix in the metric system meaning 1/1000000000
• A.
Centi
• B.
Milli
• C.
Micro
• D.
Nano
D. Nano
Explanation
The prefix "nano" in the metric system represents a value of 1/1000000000. It is used to indicate a very small unit of measurement. This prefix is commonly used in scientific and technological contexts where extremely precise measurements are required. For example, nanotechnology deals with materials and devices at the nanoscale level, which is one billionth of a meter. In summary, "nano" is the correct answer because it represents the smallest unit of measurement in the given options.
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# The real number k for which the equation \$2{x^3} + 3x + k = 0\$, has two distinct real roots in [0, 1] (A). Lies between 2 and 3(B). Lies between 1 and 0(C). Does not exist(D). Lies between 1 and 2
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Hint: Start by taking the given equation as function of x or f(x) and differentiate with respect to x . Check whether f’(x) obtained is increasing or decreasing , if it is increasing then it will have at most 1 real root.
Given, \$2{x^3} + 3x + k = 0\$
Let \$f(x) = 2{x^3} + 3x + k\$
Differentiating with respect to x , we get
Here we will use the formula \$\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}\$
\$
f'(x) = 6{x^2} + 3 \\
f'(x) > 0 \\
\$
As for any value of x, f’(x) can never be negative because of the square term involved.
f’(x) is a strictly increasing function and has at most 1 real root .
And we know , if a polynomial of odd degree, in this case it is 3, has exactly 1 real root.
So, f(x) = has exactly one real root.
We see that the results found do not satisfy the conditions.
Therefore, k does not exist.
So , option C is the correct answer.
Note: Students must know the principle of differentiation , nature of function , graph plotting etc in order to solve such similar problems. Questions can also be asked in such a manner which would demand the application of Lagrange’s mean value theorem(LMVT) , Intermediate value theorem(IVT) , Rolle’s theorem, and are recommended to be practised very well as they make the approach to the solution very easy meanwhile giving valuable information about the function too. | 487 | 1,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.908852 |
https://www.learnvest.com/2017/03/can-you-predict-how-much-you-spend-each-week/ | 1,493,493,751,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123560.51/warc/CC-MAIN-20170423031203-00051-ip-10-145-167-34.ec2.internal.warc.gz | 919,819,811 | 25,543 | # Can You Predict How Much You Spend Each Week?
Posted
You can’t predict stock prices or how much you’ll end up with in your IRA when it’s time for you to retire. But could you predict how much money you spend on small day-to-day expenses and household costs each week — those little expenditures like groceries, gas, movie tickets and drugstore runs?
Considering how variable these expenses can be, we decided to launch an experiment: We asked three people to channel their psychic powers and estimate how much they spent on food, entertainment and similar non-fixed costs each week. Then we had our psychics track how much they actually shelled out. Here’s who had decent cash ESP and who failed at reading the tea leaves.
## Ben Rogers, 37 Highland, Utah
As a father of five, much of this traveling technology consultant’s income goes to keeping his kids fed and entertained. “Food is always an uncertain variable,” he jokes. “Some weeks we eat out three times and sometimes not at all.” Something as seemingly low-budget as going to the movies or filling the minivan with gas can constitute a major expenditure that isn’t easy to plan for.
### The Prediction: \$350
Eating out: \$90
Gas: \$160
Miscellaneous/Entertainment: \$100
### The Reality: \$255
Eating out: \$19
Gas: \$107
Miscellaneous/Entertainment: \$97
Groceries: \$32
Ben says: “Gas must still be cheap enough that I actually overestimated how much I would spend on an overnight drive from Madison, Wisconsin, to Atlanta, Georgia,” says Ben, who chalks up his overestimation on his personality. “I’m fundamentally pessimistic about my ability to budget … usually my pessimism is justified.”
Entertaining his large brood proved to be more costly than he thought, but he found a way to cut back. “Taking myself, three kids and some nieces to [a movie] is expensive, even with kids’ priced tickets — I saved my budget by being the mean dad who didn’t buy anyone popcorn.”
## Amanda Griffith, 39 Norton, Massachusetts
“I’m part of a dual-income family with three young children and a dog,” says Amanda. “I have more debt than I would like, and since my children are a little older, I’m working as a media relations and content creation consultant full-time.” Going back to work means daycare and after-school care costs, which are high. “We are also spending more as our children get busier,” she says. “From travel ice hockey costs to two daughters starting the orthodontist process, life is only getting more expensive.”
### The Prediction: \$1,030
Daycare: \$180
Babysitter: \$200
Groceries: \$200
Household goods: \$150
Eating out/Gas: \$300
### The Reality: \$1,573
Daycare: \$180
Babysitter: \$275
Groceries: \$279
Household goods: \$357
Eating out/Gas: \$54
Kids clothes: \$143
Veterinarian: \$143
Drugstore: \$21
Nail care: \$40
Date night: \$56
Amanda says: Having three kids during a particularly stressful week threw her numbers way off. “Add to that an unexpected vet bill because our golden retriever clearly ate some treat he shouldn’t have, and this week got expensive,” says Amanda. “I guess I’m not good at budgeting or predicting.”
Amanda says her not-so-great money psychic abilities showed her that she needs to keep closer tabs on unnecessary spending. “It’s hard, but I’m saying no more to my children when they want random snacks or toys,” she says. “This is helping me get a better handle on my finances and teaching my children to better appreciate things. Ultimately, I’d like to keep finding ways to recoup or save money so that I can get out of debt and then start an emergency fund.”
## Sarah Kennedy, 26 New York, New York
As a recent graduate of NYU’s School of Social Work, Sarah works with adults with disabilities and moonlights as a babysitter once or twice a week to earn extra spending money.
“The majority of my spending goes toward eating out, drinks with friends, thrift-store shopping and the occasional museum, though there are many free events in the city.” One of the most challenging aspects of life in the Big Apple is managing her temptation to eat out all the time or order food every night, which gets pricey.
### The Prediction: \$500
Eating out/Drinks: \$200
Clothing: \$200
Miscellaneous: \$100
### The Reality: \$291
Eating out/Drinks: \$100
Clothing: \$155
Miscellaneous: \$4
Drugstore: \$7
Household goods: \$25
Sarah says: She underestimated her expenses because she was traveling to Oregon to visit family and spent less than she would have in New York City. “I ate meals with my relatives and ate in more than I usually do,” says Sarah. “Also, Oregon has no sales tax. That was a nice change compared to New York state’s tax upwards of 8%.”
Sarah says that overall, playing psychic definitely made her more mindful of where she spends money. “I wouldn’t say I’ve curbed my weekly spending, but it did make me consider cost-of-living expenses and the cost of shopping and eating out in a city other than New York and seeing how my dollar could go further in Portland.”
As our three psychics can attest, it’s not easy to calculate day-to-day spending. But since these costs account for a sizable part of your budget, knowing how much you lay out and where it goes is all part of smart budgeting. One way to get a handle on it is to know your flex spending number, which is a component of the one-number budgeting strategy, a LearnVest financial principle. | 1,252 | 5,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-17 | latest | en | 0.961369 |
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_15&diff=111749&oldid=111748 | 1,610,757,085,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00389.warc.gz | 232,773,406 | 11,059 | # Difference between revisions of "2019 AMC 8 Problems/Problem 15"
## Problem 15
On a beach $50$ people are wearing sunglasses and $35$ people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is $\frac{2}{5}$. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap? $\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}$
## Solution 1
The number of people wearing caps and sunglasses is $$\left(\frac{2}{5}\right)$$*35=14. So then 14 people out of the 50 people wearing sunglasses also have caps. $$\left(\frac{14}{50}\right)$$=$\boxed{\textbf{(B)}\frac{7}{25}}$~heeeeeeheeeeee | 267 | 887 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-04 | longest | en | 0.887118 |
https://uk.mathworks.com/matlabcentral/cody/problems/2075-twins-or-duplicate/solutions/473506 | 1,604,121,793,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107912807.78/warc/CC-MAIN-20201031032847-20201031062847-00073.warc.gz | 527,502,250 | 16,891 | Cody
# Problem 2075. twins or duplicate
Solution 473506
Submitted on 16 Jul 2014 by Martin
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
%% ListA = {'Peter PARKER (27/12/1983)', 'Bob DYLAN (24/08/1983)', 'Bryan DYLAN (24/08/1983)','Victor HUGO (27/12/1902)', 'Barby GIRL (25/12/1856)','Maria EIFEL (06/06/1666)'}; ListB = {'Pat RAYDOUE (21/12/1983)', 'Bob LEPONGE (24/08/1983)', 'Pascal LEGITIM (29/03/1983)','PETER Parquer (27/12/1983)', 'Vector HUGO (27/12/1902)'}; y_correct = {'Victor HUGO (27/12/1902)'}; assert(isequal(TwinOrDuplicate(ListA,ListB),y_correct))
2 Pass
%% ListA = {'Peter PARKER (27/12/1983)', 'Jean MARTINOT (28/04/1983)', 'Bryan DYLAN (24/08/1983)','Victor HUGO (27/12/1902)', 'Barby GIRL (25/12/1856)','Maria EIFEL (06/06/1666)','Jean MARTINOT (28/04/1983)'}; ListB = {'Jean MARTINOD (28/04/1983)','Pat RAYDOUE (21/12/1983)', 'Bob LEPONGE (24/08/1983)','Pascal LEGITIM (29/03/1983)','PETER Parquer (27/12/1983)', 'Vectir HUGO (27/12/1902)','Jena MARTINOT (28/04/1983)','MARIA EIFEL (06/06/1666)'}; y_correct = {'Jean MARTINOT (28/04/1983)','Maria EIFEL (06/06/1666)'}; assert(isequal(TwinOrDuplicate(ListA,ListB),y_correct))
3 Pass
%% ListA = {'Peter PARKER (27/12/1983)', 'Jean MARTINOT (28/04/1983)', 'Bryan DYLAN (24/08/1983)','Victor HUGO (27/12/1902)', 'Barby GIRL (25/12/1856)','Maria EIFEL (06/06/1666)'}; ListB = {'Pat RAYDOUE (21/12/1983)', 'Bob LEPONGE (24/08/1983)', 'Pascal LEGITIM (29/03/1983)','PETER Parquer (27/12/1983)', 'Vectir HUGO (27/12/1902)','Jena MARTINOT (28/04/1983)'}; y_correct = {}; assert(isempty(TwinOrDuplicate(ListA,ListB)))
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Start Hunting! | 716 | 1,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-45 | latest | en | 0.459036 |
https://xo-b.com/which-system-is-represented-by-the-graph/ | 1,674,980,210,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00712.warc.gz | 1,080,252,520 | 20,032 | # Which System is Represented by the Graph
Which System is Represented by the Graph.
### Learning Outcomes
• Graph systems of equations
• Graph a organization of two linear equations
• Graph a system of two linear inequalities
• Evaluate ordered pairs as solutions to systems
• Determine whether an ordered pair is a solution to a organization of linear equations
• Make up one’s mind whether an ordered pair is a solution to a system of linear inequalities
• Classify solutions to systems
• Place what blazon of solution a organisation volition have based on its graph
The way a river flows depends on many variables including how big the river is, how much water it contains, what sorts of things are floating in the river, whether or non it is raining, and and so forth. If you want to best describe its menstruation, you lot must take into business relationship these other variables. A organisation of linear equations can help with that.
A
system of linear equations
consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. Yous will find systems of equations in every awarding of mathematics. They are a useful tool for discovering and describing how behaviors or processes are interrelated. It is rare to discover, for instance, a pattern of traffic period that that is just afflicted past weather. Accidents, fourth dimension of twenty-four hour period, and major sporting events are only a few of the other variables that tin can bear upon the flow of traffic in a city. In this section, nosotros will explore some basic principles for graphing and describing the intersection of ii lines that brand up a arrangement of equations.
## Graph a system of linear equations
In this section, we will look at systems of linear equations and inequalities in two variables. Commencement, nosotros volition practice graphing two equations on the same gear up of axes, and and so nosotros will explore the different considerations y’all need to make when graphing two linear inequalities on the same fix of axes. The same techniques are used to graph a organization of linear equations equally you have used to graph single linear equations. Nosotros tin can utilize tables of values, slope and
y-intercept, or
x– and
y-intercepts to graph both lines on the aforementioned set of axes.
For example, consider the following system of linear equations in two variables.
$\begin{array}{r}2x+y=-eight\\ ten-y=-1\end{array}$
Permit’south graph these using gradient-intercept form on the aforementioned set of axes. Recall that gradient-intercept form looks like $y=mx+b$, and then we will want to solve both equations for $y$.
First, solve for y in $2x+y=-8$ $\brainstorm{array}{c}2x+y=-8\\ y=-2x – viii\cease{array}$
2nd, solve for y in $x-y=-ane$ $\begin{array}{r}x-y=-ane\,\,\,\,\,\\ y=x+ane\finish{array}$
The system is at present written as
$\brainstorm{array}{c}y=-2x – 8\\y=x+1\end{array}$
Now y’all can graph both equations using their slopes and intercepts on the aforementioned set of axes, as seen in the figure below. Note how the graphs share one betoken in common. This is their point of intersection, a point that lies on both of the lines. In the adjacent section we will verify that this point is a solution to the system.
Read: Which Candidate Will Most Likely Get the Job
In the following instance, y’all volition be given a organisation to graph that consists of ii parallel lines.
### Instance
Graph the system $\brainstorm{array}{c}y=2x+1\\y=2x-three\terminate{array}$ using the slopes and y-intercepts of the lines.
In the next example, you lot will be given a arrangement whose equations look different, only later graphing, turn out to be the aforementioned line.
### Example
Graph the organisation $\begin{assortment}{c}y=\frac{1}{2}ten+2\\2y-ten=iv\end{assortment}$ using the x – and y-intercepts.
Graphing a system of linear equations consists of choosing which graphing method you want to utilise and drawing the graphs of both equations on the aforementioned set up of axes. When you lot graph a system of linear inequalities on the same set of axes, in that location are a few more things you volition need to consider.
## Graph a system of 2 inequalities
Call up from the module on graphing that the graph of a unmarried linear inequality splits the
coordinate plane
into ii regions. On ane side lie all the solutions to the inequality. On the other side, there are no solutions. Consider the graph of the inequality $y<2x+five$.
The dashed line is $y=2x+5$. Every ordered pair in the shaded area below the line is a solution to $y<2x+5$, as all of the points below the line will make the inequality true. If you doubt that, effort substituting the
ten
and
y
coordinates of Points A and B into the inequality—yous’ll see that they work. So, the shaded area shows all of the solutions for this inequality.
The boundary line divides the coordinate aeroplane in half. In this example, it is shown as a dashed line every bit the points on the line don’t satisfy the inequality. If the inequality had been $y\leq2x+5$, then the boundary line would accept been solid.
Let’southward graph some other inequality: $y>−x$. You can check a couple of points to determine which side of the boundary line to shade. Checking points M and N yield true statements. So, we shade the area above the line. The line is dashed as points on the line are not true.
To create a system of inequalities, you need to graph ii or more inequalities together. Let’s employ $y<2x+5$ and $y>−x$ since we accept already graphed each of them.
The royal area shows where the solutions of the ii inequalities overlap. This surface area is the solution to the
system of inequalities. Any point within this imperial region will exist true for both $y>−x$ and $y<2x+five$.
In the next example, you lot are given a system of two inequalities whose purlieus lines are parallel to each other.
### Examples
Graph the system $\brainstorm{assortment}{c}y\ge2x+1\\y\lt2x-3\cease{assortment}$
In the adjacent section, we will come across that points can be solutions to systems of equations and inequalities. We will verify algebraically whether a point is a solution to a linear equation or inequality.
## Determine whether an ordered pair is a solution for a arrangement of linear equations
The lines in the graph above are defined as
$\brainstorm{array}{r}2x+y=-eight\\ x-y=-1\end{assortment}$.
They cross at what appears to be $\left(-3,-two\right)$.
Using algebra, we tin can verify that this shared betoken is actually $\left(-3,-two\right)$ and non $\left(-2.999,-ane.999\right)$. By substituting the
x– and
y-values of the ordered pair into the equation of each line, you tin exam whether the betoken is on both lines. If the exchange results in a true statement, then you have plant a solution to the system of equations!
Read: Which Physical Property Can Be Measured Color Density Odor Shape
Since the solution of the system must be a solution to
all
the equations in the arrangement, you volition demand to check the point in each equation. In the following example, we volition substitute -3 for
x
and -ii for
y
in each equation to test whether it is actually the solution.
### Example
Is $\left(-3,-ii\right)$ a solution of the organisation
$\begin{assortment}{r}2x+y=-8\\ x-y=-i\end{array}$
### Instance
Is (iii, nine) a solution of the system
$\begin{array}{r}y=3x\\2x–y=half-dozen\finish{array}$
Is $(−2,4)$ a solution for the system
$\begin{array}{r}y=2x\\3x+2y=1\finish{assortment}$
Before you do any calculations, look at the indicate given and the first equation in the system. Can you predict the answer to the question without doing whatsoever algebra?
Remember that in order to be a solution to the organisation of equations, the values of the point must be a solution for both equations. Once yous discover i equation for which the bespeak is imitation, yous have determined that it is not a solution for the system.
Nosotros can use the same method to determine whether a point is a solution to a system of linear inequalities.
## Make up one’s mind whether an ordered pair is a solution to a system of linear inequalities
On the graph above, you can see that the points B and N are solutions for the system because their coordinates volition make both inequalities true statements.
In contrast, points K and A both prevarication outside the solution region (purple). While betoken M is a solution for the inequality $y>−10$ and bespeak A is a solution for the inequality $y<2x+5$, neither signal is a solution for the
system. The following example shows how to examination a point to see whether it is a solution to a system of inequalities.
### Example
Is the point (2, ane) a solution of the system $x+y>1$ and $2x+y<8$?
Here is a graph of the organization in the example higher up. Discover that (2, 1) lies in the majestic area, which is the overlapping expanse for the two inequalities.
### Example
Is the betoken (2, i) a solution of the organisation $x+y>one$ and $3x+y<4$?
Here is a graph of this system. Notice that (2, 1) is not in the majestic area, which is the overlapping expanse; it is a solution for i inequality (the ruby region), but it is not a solution for the second inequality (the blue region).
As shown above, finding the solutions of a system of inequalities can be done past graphing each inequality and identifying the region they share. Below, you lot are given more than examples that show the entire process of defining the region of solutions on a graph for a arrangement of two linear inequalities. The full general steps are outlined below:
• Graph each inequality every bit a line and determine whether it will be solid or dashed
• Make up one’s mind which side of each boundary line represents solutions to the inequality past testing a betoken on each side
• Shade the region that represents solutions for both inequalities
In this department nosotros have seen that solutions to systems of linear equations and inequalities can be ordered pairs. In the next section, nosotros will work with systems that accept no solutions or infinitely many solutions.
## Utilise a graph to classify solutions to systems
Recall that a linear equation graphs every bit a line, which indicates that all of the points on the line are solutions to that linear equation. There are an infinite number of solutions. As we saw in the last department, if you have a system of linear equations that intersect at one point, this point is a solution to the system. What happens if the lines never cross, as in the instance of parallel lines? How would yous depict the solutions to that kind of system? In this section, we will explore the three possible outcomes for solutions to a system of linear equations.
### Three possible outcomes for solutions to systems of equations
Recall that the solution for a arrangement of equations is the value or values that are true for
all
equations in the organisation. There are three possible outcomes for solutions to systems of linear equations. The graphs of equations within a organization can tell y’all how many solutions exist for that system. Expect at the images below. Each shows ii lines that make up a organisation of equations.
I Solution No Solutions Infinite Solutions
If the graphs of the equations intersect, and so there is one solution that is true for both equations. If the graphs of the equations exercise not intersect (for example, if they are parallel), so there are no solutions that are true for both equations. If the graphs of the equations are the same, then at that place are an infinite number of solutions that are true for both equations.
• One Solution: When a arrangement of equations intersects at an ordered pair, the organization has one solution.
• Infinite Solutions:
Sometimes the two equations will graph as the aforementioned line, in which instance we have an infinite number of solutions.
• No Solution:
When the lines that brand up a system are parallel, there are no solutions because the two lines share no points in common.
### Instance
Using the graph of $\begin{array}{r}y=x\\x+2y=half dozen\end{array}$, shown beneath, determine how many solutions the system has.
Using the graph of $\begin{array}{r}y=3.5x+0.25\\14x–4y=-4.5\end{assortment}$, shown below, determine how many solutions the system has.
### Example
How many solutions does the system $\begin{array}{r}y=2x+1\\−4x+2y=2\end{array}$ accept?
In the side by side section, we volition learn some algebraic methods for finding solutions to systems of equations. Recall that linear equations in one variable can have i solution, no solution, or many solutions and we can verify this algebraically. Nosotros will use the same ideas to allocate solutions to systems in ii variables algebraically.
## Which System is Represented by the Graph
Source: https://courses.lumenlearning.com/beginalgebra/chapter/introduction-to-systems-of-linear-equations/
## Which Book Citations Are Formatted Correctly Check All That Apply
By Vladimir Gjorgiev/Shutterstock Concealer is an essential part of any makeup routine. It’s many women’s … | 3,080 | 13,265 | {"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} | 4.65625 | 5 | CC-MAIN-2023-06 | latest | en | 0.933378 |
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Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
1. Which of the following functions is an even function?
-2x + 3
2x2 – 3x
2x2 – 3
-2x3 + 3
-2x + 3x
2. The bases of a trapezoid are 6m and 8m. The altitude of the trapezoid is 10m. The area of the trapezoid in m2 is?
10
14
50
70
60
3. If f(x) = 6x- 4, then the value of f(-2) – f(2) is?
-24
-16
-8
24
16
4. If f(x) = (x + 3)/ (x2 – x – 30), then the value of ‘x’ can be any real number except?
15, 2
-6, 5
-2, 15
-5, 6
-10, 3
5. If the equation of a circle is given by (x – 4)2 + (y + 3)2 = 49, then the area of the circle is?
49pae
14pae
7pae
12pae
36pae
6. If f(x) = 2x + 1 and g(x) = 3 – x, then the value of fog(1) is?
1
2
3
4
5 | 350 | 792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-23 | latest | en | 0.64595 |
https://howtotrust.com/answers/understand-how-many-water-bottles-make-a-gallon-70907/ | 1,680,046,461,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948900.50/warc/CC-MAIN-20230328232645-20230329022645-00450.warc.gz | 354,394,215 | 11,541 | ## Is 4 bottles of water a gallon?
If the water bottle is 32 ounces, then only 4 of these make up a gallon. A 1-liter bottle of water is approximately 33.8 ounces, so approximately 3.8 of these bottles make up a gallon.
## How many 16 oz water bottles does it take to make a gallon?
8 bottles
Answer: 8 bottles of 16 oz are required to make one gallon.
## How many bottles of water should you drink a day?
To prevent dehydration, you need to get plenty of water from drink and food every day. There are many different opinions on just how much water you should be drinking every day. Health experts commonly recommend eight 8-ounce glasses, which equals about 2 liters, or half a gallon a day.
## How many 16 oz bottles of water should I drink a day?
Because there are 8 fluid ounces in a cup, you should drink eight cups of water per day. Most disposable water bottles are around 16 ounces, so that would mean you should drink three to four bottles of water each day.
## Is 32 oz half a gallon?
One gallon is 128 ounces, so that’s four times 32 ounces. 32 ounces is one quart, meaning one quarter of a gallon. 64 ounces is a half-gallon, also known as two quarts.
## How many 8 oz are in a gallon?
16 glasses
There are 16 glasses of 8 ounce in a gallon. We know that 1 gallon = 128 ounces.
## How many water bottles should a 14 year old drink?
Children between 4 and 8 years old should drink 40 ounces per day, or 5 cups. This amount increases to 56 to 64 ounces, or 7 to 8 cups, by ages 9 to 13. For ages 14 to 18, the recommended water intake is 64 to 88 ounces, or 8 to 11 cups.
## What happens if you drink 3 bottles of water a day?
Keep in mind that excessive water intake can be dangerous. Drinking too much can disrupt your body’s electrolyte balance, leading to hyponatremia, or low levels of sodium in your blood ( 21 ). Symptoms of hyponatremia include weakness, confusion, nausea, vomiting, and — in severe cases — even death ( 22 ).
## Can a person drink to much water?
When you drink too much water, your kidneys can’t get rid of the excess water. The sodium content of your blood becomes diluted. This is called hyponatremia and it can be life-threatening.
## Should a 12 year old drink a gallon of water?
The short answer to that question is: it depends. According to the IOM, toddlers and 4- to 8-year-olds should get 4 to 5 cups of water a day—from drinking water and other sources—while 9- to 13-year olds should aim for 7 to 8 cups, and 14- to 18-year-olds should get 8 to 11 cups. …
## Can a teenager drink a gallon of water?
The Institute of Medicine says children and teenagers should consume about two to three quarts of water a day (1.7 to 3.3 liters, the IOM says), depending on age, size and sex. Adolescent boys generally need to drink more water than girls do, research suggests.
## Can a child drink to much water?
Fortunately, actual overhydration — the kind that creates health problems — is rare. However, it’s possible (though uncommon) for your child to overdo it to the point where they experience so-called water intoxication. This can lead to hyponatremia, a serious imbalance of sodium in your toddler’s system.
## Is 1 gallon of water a day too much?
For most people, there is really no limit for daily water intake and a gallon a day is not harmful. But for those who have congestive heart failure or end stage kidney disease, sometimes water needs to be restricted because the body can’t process it correctly.
## Can you lose weight drinking a gallon of water a day?
Drinking Water Curbs Cravings
A third benefit to drinking a gallon of water each day is that water consumption helps curb hunger cravings, and without as much appetite for snacks or second helpings, you might even see some weight loss.
## Is 2 gallons of water a day too much?
There are many different opinions on just how much water you should be drinking every day. Health experts commonly recommend eight 8-ounce glasses, which equals about 2 liters, or half a gallon a day. This is called the 8×8 rule and is very easy to remember.
If water is continuously taken in too much quantity, it may lead to kidney stones and chronic kidney diseases.” He added that sudden dehydration may lead to acute kidney failure and unconsciousness. People who had kidney or cardiac failures are usually unable to tolerate excessive fluid intake.
## Can I lose weight by drinking more water?
Water can be really helpful for weight loss. It is 100% calorie-free, helps you burn more calories and may even suppress your appetite if consumed before meals. The benefits are even greater when you replace sugary beverages with water. It is a very easy way to cut back on sugar and calories.
## How many times a day should I pee if I drink a lot of water?
How often you have to urinate is a good indicator of your body’s overall state of hydration. It’s considered normal to have to urinate about six to eight times in a 24-hour period.
## What color is urine when your kidneys are failing?
Brown, red, or purple urine
Kidneys make urine, so when the kidneys are failing, the urine may change. How? You may urinate less often, or in smaller amounts than usual, with dark-colored urine. Your urine may contain blood.
## Why do I wake up at night to drink water?
The bottom line
If you wake up during the night because you’re feeling thirsty, the cause could be your sleeping environment, hydration habits, or a medication you’re taking.
## Why does water give me diarrhea?
Nausea or vomiting.
When you have too much water in the body, the kidneys can’t remove the excess liquid. It starts collecting in the body, leading to nausea, vomiting, and diarrhea.
## Why is my pee black?
Dark urine is most commonly due to dehydration. However, it may be an indicator that excess, unusual, or potentially dangerous waste products are circulating in the body. For example, dark brown urine may indicate liver disease due to the presence of bile in the urine. | 1,382 | 5,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-14 | latest | en | 0.950602 |
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# Crystel Llamas Ruacho
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Given the number x, y must be the summation of all integers from 1 to 2^x. For instance if x=2 then y must be 1+2+3+4=10.
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# Problem 51. Find the two most distant points
Solution 290244
Submitted on 26 Jul 2013 by James White
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
%% p = [0 0; 1 0; 2 2; 0 1]; ix_correct = [1 3]; assert(isequal(mostDistant(p),ix_correct))
ans = 0 0 0 0 0 1 2 0 0 2 8 2 0 0 2 1
2 Pass
%% p = [0 0; 1 0; 2 2; 0 10]; ix_correct = [2 4]; assert(isequal(mostDistant(p),ix_correct))
ans = 0 0 0 0 0 1 2 0 0 2 8 20 0 0 20 100
3 Pass
%% p = [0 0; -1 50]; ix_correct = [1 2]; assert(isequal(mostDistant(p),ix_correct))
ans = 0 0 0 2501
4 Pass
%% p = [5 5; 1 0; 2 2; 0 10; -100 20; 1000 400]; ix_correct = [5 6]; assert(isequal(mostDistant(p),ix_correct))
ans = 50 5 20 50 -400 7000 5 1 2 0 -100 1000 20 2 8 20 -160 2800 50 0 20 100 200 4000 -400 -100 -160 200 10400 -92000 7000 1000 2800 4000 -92000 1160000 | 429 | 943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-16 | latest | en | 0.463377 |
https://www.physicsforums.com/threads/euler-lagrange-equation.206046/ | 1,606,992,602,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141727627.70/warc/CC-MAIN-20201203094119-20201203124119-00129.warc.gz | 777,830,807 | 14,827 | # Euler-Lagrange equation
## Homework Statement
Let $$P$$ be a rectangle , $$f_{0} : \partial P \rightarrow R)$$ continuous and Lipschitz, $$C_{0} = \{ f \in C^{2}(P) : f=f_{0} \ on \ \partial P \}$$. and finally $$S : C_{0} \rightarrow R$$ a functional:
$$S(f) = \int^b_a (\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy)\,dx + \int^d_c (\int^a_b (\frac{\partial f}{\partial y})^{2}\,dx)\,dy$$.
Write Euler-Lagrange equation for S.
## The Attempt at a Solution
I tried writing: $$S(f) = \int^b_a (\int^d_c (\frac{\partial f}{\partial x})^{2} + (\frac{\partial f}{\partial y})^{2}\,dy)\,dx$$, so the proper Lagrangian would be $$L(x) = \int^d_c (\frac{\partial f}{\partial x})^{2} + (\frac{\partial f}{\partial y})^{2}\,dy$$.
Then the Euler-Lagrange equation should be $$\frac{d}{dx}\frac{\partial L}{\partial f^{'}_{x}} = 0 \leftrightarrow \frac{d}{dx}\frac{\partial }{\partial f^{'}_{x}}\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy = 0$$, ($$L^{'} = \int^d_c (\frac{\partial f}{\partial x})^{2}\,dy$$) and now since $$\frac{dL^{'}}{dx} = \frac{\partial L^{'}}{\partial f_{x}^{'}}\frac{\partial f_{x}^{'}}{\partial x}$$, we can rewrite that as $$\frac{d}{dx}\frac{\frac{d}{dx}\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy}{f_{xx}^{''}} = 0 \leftrightarrow \frac{d}{dx}\frac{\int^d_c \frac{\partial}{\partial x}(\frac{\partial f}{\partial x})^{2}\,dy}{f_{xx}^{''}} = 0 \leftrightarrow \frac{d}{dx}\frac{\int^d_c 2f_{x}^{'}f_{xx}^{''}\,dy}{f_{xx}^{''}} = 0$$. but what then?
Last edited:
You are making this too hard. Let's call $$\frac{\partial f}{\partial x}=f_x$$ and $$\frac{\partial f}{\partial y}=f_x$$. Then the form of the Euler-Lagrange equations for two independent variables is $$\frac{\partial L}{\partial f}-\frac{\partial}{\partial x} \frac{\partial L}{\partial f_x}-\frac{\partial}{\partial y}\frac{\partial L}{\partial f_y}=0$$ where $$L=(f_x)^2+(f_y)^2$$. | 702 | 1,888 | {"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} | 3.734375 | 4 | CC-MAIN-2020-50 | latest | en | 0.630577 |
https://www.physicsforums.com/threads/how-to-find-the-number-of-oscillations-a-block-goes-through.975238/ | 1,726,719,262,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00671.warc.gz | 860,888,265 | 20,623 | # How to find the number of oscillations a block goes through
• kileigh
In summary: After the collision, how much further will it move from equilibrium before reaching maximum compression?
kileigh
Homework Statement
How would I find the number of oscillations that block 2 would undergo if block 2 rested on a patch of ground with friction.
Relevant Equations
Displacement amplitude=x0-n(4f/k)
This is the image provided with the problem, the values given include:
d= 4.00 m, the mass of block one=0.200 kg, speed of block one=8.00 m/s, the period of oscillations for block two without friction=0.140 s, and the spring constant= 1208.5 N/m.
I know how to solve the oscillations if block two was pulled and released:
For this equation, I have the values for k, but I'm not sure how to change it so rather than being from the distance pulled(x0), instead of for the energy from block one when they collide.
Thanks!
So block one, having mass 2 kg and speed 8 m/s, so kinetic energy 65 Joules, hits block 2, compressing the spring. Are we not told the mass of block 2?
I calculated the mass of block two with the period without friction. I got a mass of 0.6 kg, but I was not sure if that would change since the period of the block is technically changing when friction is considered.
Well, the mass will not change just because there is friction! My first thought was to use the period and mass to calculate k buy I didn't notice that we are given k but not the mass!
kileigh said:
I calculated the mass of block two with the period without friction. I got a mass of 0.6 kg, but I was not sure if that would change since the period of the block is technically changing when friction is considered.
You don't need to know the period with friction.
There must be rather more to the whole question. If you were to provide the whole we would also be able to confirm that you've not overlooked anything, like further interaction with block 1.
haruspex said:
You don't need to know the period with friction.
There must be rather more to the whole question. If you were to provide the whole we would also be able to confirm that you've not overlooked anything, like further interaction with block 1.
This was the whole first question, but my professor asked the question in my orginal post: Figure 15.15 shows block 1 of mass0.200 kg0.200 kg sliding to the right over a frictionless elevated surface at a speed of8.00 m/s8.00 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant1208.5 N/m1208.5 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of0.140 s0.140 s, and block 1 slides off the opposite end of the elevated surface, landing a distancedd from the base of that surface after falling heighth=4.90 mh=4.90 m. What is the value ofdd?
kileigh said:
This was the whole first question, but my professor asked the question in my orginal post: Figure 15.15 shows block 1 of mass0.200 kg0.200 kg sliding to the right over a frictionless elevated surface at a speed of8.00 m/s8.00 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant1208.5 N/m1208.5 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of0.140 s0.140 s, and block 1 slides off the opposite end of the elevated surface, landing a distancedd from the base of that surface after falling heighth=4.90 mh=4.90 m. What is the value ofdd?
Ok, so there will be no further interaction with block 1.
As I posted, you don't need the period with friction. You found the mass from the period without friction, and you can use that in your formula for the number of oscillations.
I assume you found the initial speed of block 2 ok.
But which equation should I use, my original equation asks for the distance the spring is pulled and I'm not sure what to put for the x0.
kileigh said:
But which equation should I use, my original equation asks for the distance the spring is pulled and I'm not sure what to put for the x0.
During one half oscillation the frictional force is constant. This means it is like vertical oscillation under gravity, so it is SHM. Find the SHM equation for that and use the known initial speed. But note that the initial position will not be the equilibrium position.
Would it be similar to my original equation? Sorry I struggle with finding my SHM equations.
kileigh said:
Would it be similar to my original equation? Sorry I struggle with finding my SHM equations.
As I posted, for the purposes of the left to right movement, we can treat friction as a constant force to the left. Where would the equilibrium position be on that basis?
After the collision, how much further will it move from equilibrium before reaching maximum compression?
## 1. How do I measure the number of oscillations a block goes through?
To find the number of oscillations, you can use a stopwatch and count the number of times the block moves back and forth in a set amount of time. Alternatively, you can record the movement using a video camera and use slow-motion playback to count the oscillations.
## 2. What is the formula for calculating the number of oscillations?
The number of oscillations can be calculated using the formula: N = t/T, where N is the number of oscillations, t is the total time elapsed, and T is the period of oscillation.
## 3. Can the number of oscillations change over time?
Yes, the number of oscillations can change over time if the amplitude or frequency of the oscillations changes. For example, if the amplitude decreases, the number of oscillations will decrease as well.
## 4. How does the mass of the block affect the number of oscillations?
The mass of the block does not affect the number of oscillations. The number of oscillations is only affected by the amplitude and frequency of the oscillations.
## 5. Is there a difference between the number of oscillations and the number of cycles?
No, the number of oscillations and the number of cycles are the same. Both terms refer to the number of times the block completes a full back-and-forth motion.
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893 | 1,555 | 6,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-38 | latest | en | 0.963288 |
https://biology.stackexchange.com/questions/27784/what-s-the-reason-for-isovolumic-contraction-and-isovolumic-relaxation/27806 | 1,632,346,751,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057388.12/warc/CC-MAIN-20210922193630-20210922223630-00117.warc.gz | 187,989,014 | 39,105 | # What’s the reason for isovolumic contraction and isovolumic relaxation?
During cardiac cycle, there are two periods in which the heart volume doesn’t change, but there is a change in tension/pressure. It takes about 0.25-0.35 second to achieve this change.
I searched in Google and Wikipedia, but could not find the reason, so I decided to ask this question.
To put it in very simple words the reason for isovolumetric contaction and relaxation is to make the necessary pressure changes that is necessary to allow blood to flow into or flow out of the ventricles.
The basic formula you need to remember to understand this is:-
• Fluids flow from high pressure area to low pressure area along the pressure gradient (Eg: Injecting a drug into blood stream - as the pressure in the syringe is higher than the pressure in blood vessel, the drug flows into the blood vessel)
• Pressure is inversely proportional to volume (when temperature is constant) - this is Boyle's law. In a container with a fixed amount of fluid an increase in the volume of container will cause a proportionate amount of fall in the pressure in the container and vice-versa
Isovolumetric contraction:
The end-diastolic pressure in the aorta is in the range of approximately 90-100 mm of Hg. For blood to flow from left ventricle to aorta, the pressure in the left ventricle should exceed the pressure in the aorta.
As long as the pressure in the left ventricle is lesser than the pressure in the left atrium, the blood flows from Left Atrium to Left Ventricle. When Left Ventricle starts contracting the pressure in the LV rapidly overcomes the pressure in left atrium leading to closure of the mitral valve.
Now the pressure in the LV is greater than the pressure in the LA. So no blood flows in. At the same time the pressure in the LV is not sufficient to overcome the pressure in the aorta. So the semilunar valve remains closed. The ventricle continues contracting causing a time period in which there is increase in pressure without volume change as both inlet and outlet are closed. This period is called period of isovolumic contraction.
The isovolumic contraction ends when the pressure in the left ventricle becomes higher than the pressure in the aorta, which will cause the semilunar valves to open and blood to flow from left ventricle to aorta.
Isovolumic Relaxation:
The reverse happens in isovolumic relaxation.
As the blood volume in the ventricle decreases and the left ventricle starts relaxing the pressure in the left ventricle falls. When the pressure gets below that of the aorta, the semilunar valve closes. But the pressure in the ventricle is still higher than the pressure in the left atrium. So the mitral valve remains closed. So here too both the inlet and outlet are closed but as the ventricle is relaxing the pressure keeps falling. So till the pressure in the left ventricle falls below that of the left atrium, there is a time period volume is constant and the pressure keeps falling - this is called Isovolumic Relaxation.
This period is followed by opening of the mitral valve (as the pressure in the LV falls below that of the LA) and blood flows into the ventricle. | 707 | 3,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.911167 |
http://www.mathworks.co.uk/help/signal/ref/periodogram.html?nocookie=true | 1,412,106,889,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663135.34/warc/CC-MAIN-20140930004103-00237-ip-10-234-18-248.ec2.internal.warc.gz | 740,061,020 | 14,587 | Accelerating the pace of engineering and science
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periodogram
Periodogram power spectral density estimate
Syntax
• [pxx,w] = periodogram(___)
• [pxx,f] = periodogram(___,fs) example
• [pxx,w] = periodogram(x,window,w) example
• [pxx,f] = periodogram(x,window,f,fs) example
• [___] = periodogram(x,window,___,freqrange) example
• [___] = periodogram(x,window,___,spectrumtype) example
• [pxx,f,pxxc] = periodogram(___,'ConfidenceLevel',probability) example
• periodogram(___)
Description
example
pxx = periodogram(x) returns the periodogram power spectral density (PSD) estimate of the input signal, x, using a rectangular window. If x is real-valued, pxx is a one-sided PSD estimate. If x is complex-valued, pxx is a two-sided PSD estimate. The number of points, nfft, in the discrete Fourier transform (DFT) is the maximum of 256 or the next power of two greater than the signal length.
example
pxx = periodogram(x,window) returns the modified periodogram PSD estimate using the window, window. window is a vector the same length as x.
example
pxx = periodogram(x,window,nfft) uses nfft points in the discrete Fourier transform (DFT). If nfft is greater than the signal length, x is zero-padded to length nfft. If nfft is less than the signal length, the signal is wrapped modulo nfft and summed using datawrap. For example, the input signal [1 2 3 4 5 6 7 8] with nfft equal to 4 results in the periodogram of sum([1 5; 2 6; 3 7; 4 8],2).
[pxx,w] = periodogram(___) returns the normalized frequency vector, w. If pxx is a one-sided periodogram, w spans the interval [0,π] if nfft is even and [0,π) if nfft is odd. If pxx is a two-sided periodogram, w spans the interval [0,2π).
example
[pxx,f] = periodogram(___,fs) returns a frequency vector, f, in cycles per unit time. The sampling frequency, fs, is the number of samples per unit time. If the unit of time is seconds, then f is in cycles/sec (Hz). For real–valued signals, f spans the interval [0,fs/2] when nfft is even and [0,fs/2) when nfft is odd. For complex-valued signals, f spans the interval [0,fs).
example
[pxx,w] = periodogram(x,window,w) returns the two-sided periodogram estimates at the normalized frequencies specified in the vector, w. The vector, w, must contain at least 2 elements.
example
[pxx,f] = periodogram(x,window,f,fs) returns the two-sided periodogram estimates at the frequencies specified in the vector, f. The vector, f, must contain at least 2 elements. The frequencies in f are in cycles per unit time. The sampling frequency, fs, is the number of samples per unit time. If the unit of time is seconds, then f is in cycles/sec (Hz).
example
[___] = periodogram(x,window,___,freqrange) returns the periodogram over the frequency range specified by freqrange. Valid options for freqrange are: 'onesided', 'twosided', or 'centered'.
example
[___] = periodogram(x,window,___,spectrumtype) returns the PSD estimate if spectrumtype is specified as 'psd' and returns the power spectrum if spectrumtype is specified as 'power'.
example
[pxx,f,pxxc] = periodogram(___,'ConfidenceLevel',probability) returns the probabilityx100% confidence intervals for the PSD estimate in pxxc.
periodogram(___) with no output arguments plots the periodogram PSD estimate in dB per unit frequency in the current figure window.
Examples
expand all
Periodogram Using Default Inputs
Obtain the periodogram of an input signal consisting of a discrete-time sinusoid with an angular frequency of π/4 radians/sample with additive N(0,1) white noise.
Create a sine wave with an angular frequency of π/4 radians/sample with additive N(0,1) white noise. The signal is 320 samples in length. Obtain the periodogram using the default rectangular window and DFT length. The DFT length is the next power of two greater than the signal length, or 512 points. Because the signal is real-valued and has even length, the periodogram is one-sided and there are 512/2+1 points.
```n = 0:319;
x = cos(pi/4*n)+randn(size(n));
pxx = periodogram(x);
plot(10*log10(pxx))```
Modified Periodogram with Hamming Window
Obtain the modified periodogram of an input signal consisting of a discrete-time sinusoid with an angular frequency of π/4 radians/sample with additive N(0,1) white noise.
Create a sine wave with an angular frequency of π/4 radians/sample with additive N(0,1) white noise. The signal is 320 samples in length. Obtain the modified periodogram using a Hamming window and default DFT length. The DFT length is the next power of two greater than the signal length, or 512 points. Because the signal is real-valued and has even length, the periodogram is one-sided and there are 512/2+1 points.
```n = 0:319;
x = cos(pi/4*n)+randn(size(n));
pxx = periodogram(x,hamming(length(x)));
plot(10*log10(pxx))```
DFT Length Equal to Signal Length
Obtain the periodogram of an input signal consisting of a discrete-time sinusoid with an angular frequency of π/4 radians/sample with additive N(0,1) white noise. Use a DFT length equal to the signal length.
Create a sine wave with an angular frequency of π/4 radians/sample with additive N(0,1) white noise. The signal is 320 samples in length. Obtain the periodogram using the default rectangular window and DFT length equal to the signal length. Because the signal is real-valued, the one-sided periodogram is returned by default with a length equal to 320/2+1.
```n = 0:319;
x = cos(pi/4*n)+randn(size(n));
nfft = length(x);
pxx = periodogram(x,[],nfft);
plot(10*log10(pxx))```
Periodogram of Relative Sunspot Numbers
Obtain the periodogram of the Wolf (relative sunspot) number data sampled yearly between 1700 and 1987.
Load the relative sunspot number data. Obtain the periodogram using the default rectangular window and number of DFT points (512 in this example). The sampling rate for these data is 1 sample/year. Plot the periodogram.
```load sunspot.dat
relNums=sunspot(:,2);
[pxx,f] = periodogram(relNums,[],[],1);
plot(f,10*log10(pxx))
xlabel('Cycles/Year'); ylabel('dB');
title('Periodogram of Relative Sunspot Number Data');```
You see in the preceding figure that there is a peak in the periodogram at approximately 0.1 cycles/year, which indicates a period of approximately 10 years.
Periodogram Using Goertzel's Algorithm — Normalized Frequency
Obtain the periodogram of an input signal consisting of two discrete-time sinusoids with an angular frequencies of π/4 and π/2 radians/sample in additive N(0,1) white noise. Use Goertzel's algorithm to obtain the two-sided periodogram estimates at π/4 and π/2 radians/sample. Compare the result to the one-sided periodogram.
```n = 0:319;
x = cos(pi/4*n)+0.5*sin(pi/2*n)+randn(size(n));
[pxx,w] = periodogram(x,[],[pi/4 pi/2]);
pxx
[pxx1,w1] = periodogram(x);
plot(w1,pxx1)```
You see that the periodogram values obtained using Goertzel's algorithm are 1/2 the values in the one-sided periodogram. This is consistent with the fact that using Goertzel's algorithm returns the two-sided periodogram.
Periodogram Using Goertzel's Algorithm — Frequency in Hz
Create a signal consisting of two sine waves with frequencies of 100 and 200 Hz in N(0,1) white additive noise. The sampling frequency is 1 kHz. Use Goertzel's algorithm to obtain the two-sided periodogram at 100 and 200 Hz.
```fs = 1000;
t = 0:0.001:1-0.001;
x = cos(2*pi*100*t)+sin(2*pi*200*t)+randn(size(t));
freq = [100 200];
[pxx,f] = periodogram(x,[],freq,fs);```
Upper and Lower 95%-Confidence Bounds
The following example illustrates the use of confidence bounds with the periodogram. While not a necessary condition for statistical significance, frequencies in the periodogram where the lower confidence bound exceeds the upper confidence bound for surrounding PSD estimates clearly indicate significant oscillations in the time series.
Create a signal consisting of the superposition of 100-Hz and 150-Hz sine waves in additive white N(0,1) noise. The amplitude of the two sine waves is 1. The sampling frequency is 1 kHz.
```t = 0:0.001:1-0.001;
fs = 1000;
x = cos(2*pi*100*t)+sin(2*pi*150*t)+randn(size(t));```
Obtain the periodogram with 95%-confidence bounds. Plot the periodogram along with the confidence interval and zoom in on the frequency region of interest near 100 and 150 Hz.
```[pxx,f,pxxc] = periodogram(x,rectwin(length(x)),length(x),fs,...
'ConfidenceLevel', 0.95);
plot(f,10*log10(pxx)); hold on;
plot(f,10*log10(pxxc),'r--','linewidth',2);
axis([85 175 min(min(10*log10(pxxc))) max(max(10*log10(pxxc)))]);
xlabel('Hz'); ylabel('dB');
title('Periodogram with 95%-Confidence Bounds');```
At 100 and 150 Hz, the lower confidence bound exceeds the upper confidence bounds for surrounding PSD estimates.
Power Estimate of Sinusoid
Estimate the power of sinusoid at a specific frequency using the 'power' option.
Create a 100-Hz sinusoid one second in duration sampled at 1 kHz. The amplitude of the sine wave is 1.8, which equates to a power of 1.82/2 = 1.62. Estimate the power using the 'power' option.
```fs = 1000;
t = 0:1/fs:1-1/fs;
x = 1.8*cos(2*pi*100*t);
[pxx,f] = periodogram(x,hamming(length(x)),length(x),fs,'power');
[pwrest,idx] = max(pxx);
fprintf('The maximum power occurs at %3.1f Hz\n',f(idx));
fprintf('The power estimate is %2.2f\n',pwrest);```
DC-Centered Periodogram
Obtain the periodogram of a 100-Hz sine wave in additive N(0,1) noise. The data are sampled at 1 kHz. Use the 'centered' option to obtain the DC-centered periodogram and plot the result.
```fs = 1000;
t = 0:0.001:1-0.001;
x = cos(2*pi*100*t)+randn(size(t));
[pxx,f] = periodogram(x,[],length(x),fs,'centered');
plot(f,10*log10(pxx))
xlabel('Hz'); ylabel('dB')```
Input Arguments
expand all
x — Input signalvector
Input signal, specified as a row or column vector.
Data Types: single | double
Complex Number Support: Yes
window — Windowrectwin(length(x)) (default) | [] | vector
Window, specified as a row or column vector the same length as the input signal. If you specify window as empty, the default rectangular window is used.
Data Types: single | double
nfft — Number of DFT pointsmax(256,2^nextpow2(length(x)) (default) | integer | []
Number of DFT points, specified as a positive integer. For a real-valued input signal, x, the PSD estimate, pxx has length (nfft/2+1) if nfft is even, and (nfft+1)/2 if nfft is odd. For a complex-valued input signal,x, the PSD estimate always has length nfft. If nfft is specified as empty, the default nfft is used.
Data Types: single | double
fs — Sampling frequencypositive scalar
Sampling frequency specified as a positive scalar. The sampling frequency is the number of samples per unit time. If the unit of time is seconds, the sampling frequency has units of hertz.
w — Normalized frequencies for Goertzel algorithmvector
Normalized frequencies for Goertzel algorithm, specified as a row or column vector with at least 2 elements. Normalized frequencies are in radians/sample.
Example: w = [pi/4 pi/2]
Data Types: double
f — Cyclical frequencies for Goertzel algorithmvector
Cyclical frequencies for Goertzel algorithm, specified as a row or column vector with at least 2 elements. The frequencies are in cycles per unit time. The unit time is specified by the sampling frequency, fs. If fs has units of samples/second, then f has units of Hz.
Example: fs = 1000; f= [100 200]
Data Types: double
freqrange — Frequency range for PSD estimate'onesided' | 'twosided' | 'centered'
Frequency range for the PSD estimate, specified as a one of 'onesided', 'twosided', or 'centered'. The default is 'onesided' for real-valued signals and 'twosided' for complex-valued signals. The frequency ranges corresponding to each option are
• 'onesided' — returns the one-sided PSD estimate of a real-valued input signal, x. If nfft is even, pxx will have length nfft/2+1 and is computed over the interval [0,π] radians/sample. If nfft is odd, the length of pxx is (nfft+1)/2 and the interval is [0,π) radians/sample. When fs is optionally specified, the corresponding intervals are [0,fs/2] cycles/unit time and [0,fs/2) cycles/unit time for even and odd length nfft respectively.
• 'twosided' — returns the two-sided PSD estimate for either the real-valued or complex-valued input, x. In this case, pxx has length nfft and is computed over the interval [0,2π) radians/sample. When fs is optionally specified, the interval is [0,fs) cycles/unit time.
• 'centered' — returns the centered two-sided PSD estimate for either the real-valued or complex-valued input, x. In this case, pxx has length nfft and is computed over the interval (-π,π] radians/sample for even length nfft and (-π,π) radians/sample for odd length nfft. When fs is optionally specified, the corresponding intervals are (-fs/2, fs/2] cycles/unit time and (-fs/2, fs/2) cycles/unit time for even and odd length nfft respectively.
Data Types: char
spectrumtype — Power spectrum scaling'psd' (default) | 'power'
Power spectrum scaling, specified as one of 'psd' or 'power'. Omitting the spectrumtype, or specifying 'psd', returns the power spectral density. Specifying 'power' scales each estimate of the PSD by the equivalent noise bandwidth of the window. Use the 'power' option to obtain an estimate of the power at each frequency.
Data Types: char
probability — Confidence interval for PSD estimate0.95 (default) | scalar in the range (0,1)
Coverage probability for the true PSD, specified as a scalar in the range (0,1). The output, pxxc, contains the lower and upper bounds of the probabilityx100% interval estimate for the true PSD.
Output Arguments
expand all
pxx — PSD estimatevector
PSD estimate, specified as a real-valued, nonnegative column vector. The units of the PSD estimate are in squared magnitude units of the time series data per unit frequency. For example, if the input data is in volts, the PSD estimate is in units of squared volts per unit frequency. For a time series in volts, if you assume a resistance of 1 ohm and specify the sampling frequency in Hz, the PSD estimate is in watts/Hz.
Data Types: single | double
w — Normalized frequenciesvector
Normalized frequencies, specified as a real-valued column vector. If pxx is a one-sided PSD estimate, w spans the interval [0,π] if nfft is even and [0,π) if nfft is odd. If pxx is a two-sided PSD estimate, w spans the interval [0,2π). For a DC-centered PSD estimate, f spans the interval (-π,π] radians/sample for even length nfft and (-π,π) radians/sample for odd length nfft.
Data Types: double
f — Cyclical frequenciesvector
Cyclical frequencies, specified as a real-valued column vector. For a one-sided PSD estimate, f spans the interval [0,fs/2] when nfft is even and [0,fs/2) when nfft is odd. For a two-sided PSD estimate, f spans the interval [0,fs). For a DC-centered PSD estimate, f spans the interval (-fs/2, fs/2] cycles/unit time for even length nfft and (-fs/2, fs/2) cycles/unit time for odd length nfft .
Data Types: double
pxxc — Confidence boundsmatrix
Confidence bounds, specified as an N-by-2 matrix with real-valued elements. The row dimension of the matrix is equal to the length of the PSD estimate, pxx. The first column contains the lower confidence bound and the second column contains the upper confidence bound for the corresponding PSD estimates in the rows of pxx. The coverage probability of the confidence intervals is determined by the value of the probability input.
Data Types: single | double
expand all
Periodogram
The periodogram is a nonparametric estimate of the power spectral density (PSD) of a wide-sense stationary random process. The periodogram is the Fourier transform of the biased estimate of the autocorrelation sequence. For a signal, xn, sampled at fs samples per unit time, the periodogram is defined as
where Δt is the sampling interval. For a one-sided periodogram, the values at all frequencies except 0 and the Nyquist, 1/2Δt, are multiplied by 2 so that the total power is conserved.
If the frequencies are in radians/sample, the periodogram is defined as
The frequency range in the preceding equations has variations depending on the value of the freqrange argument. See the description of freqrange in Input Arguments.
The integral of the true PSD, P(f), over one period, 1/Δt for cyclical frequency and 2π for normalized frequency, is equal to the variance of the wide-sense stationary random process.
For normalized frequencies, replace the limits of integration appropriately.
Modified Periodogram
The modified periodogram multiplies the input time series by a window function. A suitable window function is nonnegative and decays to zero at the beginning and end points. Multiplying the time series by the window function tapers the data gradually on and off and helps to alleviate the leakage in the periodogram. See Bias and Variability in the Periodogram for an example.
If hn is a window function, the modified periodogram is defined by
where Δt is the sampling interval.
If the frequencies are in radians/sample, the modified periodogram is defined as
The frequency range in the preceding equations has variations depending on the value of the freqrange argument. See the description of freqrange in Input Arguments. | 4,418 | 17,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | latest | en | 0.655722 |
https://kids.classroomsecrets.co.uk/resource/year-2-multiplication-from-pictures-game-2/ | 1,716,478,516,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058642.50/warc/CC-MAIN-20240523142446-20240523172446-00470.warc.gz | 301,544,709 | 47,106 | # Year 2 Multiplication from Images
### Teacher Specific Information
This Year 2 Multiplication from Images Game checks pupils’ ability to work out how many there are altogether from the pictures. Pupils will match the representations to the calculations, identify the correct number sentences that match an image and complete a calculation.
If you would like to access additional resources which link to this interactive game, you can purchase a subscription for only £5.31 per month on our sister site, Classroom Secrets.
### National Curriculum Objectives
Multiplication and Division
Mathematics Year 2: (2C6) Recall and use multiplication and division facts for the 2, 5 and 10 multiplication tables, including recognising odd and even numbers
Mathematics Year 2: (2C9b) Show that multiplication of two numbers can be done in any order (commutative) and division of one number by another cannot
Mathematics Year 2: (2C8) Solve problems involving multiplication and division, using materials, arrays, repeated addition, mental methods, and multiplication and division facts, including problems in contexts | 233 | 1,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.87 |
http://www.marshall.edu/irp/facultysalarycalculation/ | 1,369,143,764,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700074077/warc/CC-MAIN-20130516102754-00047-ip-10-60-113-184.ec2.internal.warc.gz | 582,089,083 | 5,618 | Request Info Future Students Visitors Students Faculty & Staff Alumni
IRP Home > Faculty Salary Calculation
Calculation of Faculty Salary Increases
Step 1 – Determine Pools.
Determine the amounts of the faculty salary increase pool designated for equity and for merit. Of the total pool, 51% is set aside for merit purposes, and 49% is set aside for equity purposes. Five percent of the 49% equity portion is set aside for special equity increases as determined by the Provost. The remainder of the 49% is for formula-driven equity increases.
Step 2 – Target salary.
1. Calculate reference salaries based on most recent AAUP salary survey results.
1. Fall 2008 reference salary for professor rank: \$84,824
2. Fall 2008 reference salary for associate professor rank: \$68,199
3. Fall 2008 reference salary for assistant professor rank: \$57,793
4. Multiply the reference salary times the individual’s FTE (normally 1.00)
2. Calculate the extrapolated market salaries by multiplying the reference salary for the individual’s rank times the salary factor for the individual’s rank and discipline. (Download salary factors table, PDF format, 25 KB)
3. Calculate the longevity-adjusted market salary by multiplying results of step 2c by the longevity factor for the individual’s experience in rank. (Download longevity factors table, PDF format, 9 KB)
Note: The Lewis College of Business, the College of Information Technology and Engineering, and the University Libraries have their own plans for distribution of salaries increases. The calculations below do not describe the method for determining increases in those colleges.
Step 3 – Equity Increase.
1. Note: The Lewis College of Business, the College of Information Technology and Engineering, and the University Libraries have their own plans for distributing salary increases.
2. Calculate the equity salary by multiplying the result of step 2d times the equity level for this year: 82.030%. Note: The equity level is set in such a way that the distribution of equity increases exactly equals the amount of funds for that purpose.
3. Subtract the individual’s current 9-month base salary from the result of step 3b. If the result is 0 or less, then there is no equity increase. The result is the equity increase amount if it is positive. Note: the increase can be no greater than 10% of the market salary for full-professors in the discipline.
Step 4 – Merit Increase.
1. Each college receives a proportional share of the merit pool based on the total number of faculty participating in this salary increase process.
2. The OCR scores of all qualifying faculty (OCR scores in the 2.51-4.0 range) are summed.
3. Calculate the individual’s share of the college merit pool by dividing the OCR score by the results of step 4b.
4. Multiply the result of step 4c by the college’s merit pool. The result is the merit increase.
Step 5 – Total.
1. Add the equity increase and the merit increase to the current 9-month base salary to get the new 9-month base salary. If the individual is on a 10, 11, or 12 month appointment, multiply the new 9-month base salary times 10/9, 11/9, or 12/9 respectively to get the length-of-appointment salary.
2. Round the result from 5a UP to the next whole dollar. This amount is the new base salary.
3. c) If an individual does not have an OCR because the requisite annual report was not submitted, then the individual does not receive a salary increase. | 763 | 3,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2013-20 | latest | en | 0.840794 |
http://www.stumblingrobot.com/2016/01/12/find-a-constant-so-the-given-function-has-a-finite-non-zero-limit/?shared=email&msg=fail | 1,576,395,415,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541307797.77/warc/CC-MAIN-20191215070636-20191215094636-00123.warc.gz | 236,339,294 | 15,094 | Home » Blog » Find a constant so the given function has a finite, non-zero limit
# Find a constant so the given function has a finite, non-zero limit
Find the value for the constant so that
is finite and is not equal to zero. Compute the value of the limit.
We know from the previous exercise that
To apply this result, we first make the substitution , then as and we have
Now, since as we can apply the result of the previous exercise we stated above which tells us
Therefore,
Now, for this limit to be finite and non-zero we need the term in the numerator to have no constant terms (otherwise the in the denominator will make the limit infinite). So, we need which means or . Substituting this value of we then evaluate the limit | 162 | 739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-51 | latest | en | 0.901965 |
https://blog.mie-solutions.com/tag/cycle-time | 1,563,832,094,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528220.95/warc/CC-MAIN-20190722201122-20190722223122-00139.warc.gz | 323,110,303 | 21,707 | • # MIE Solutions Blog
By Dave Ferguson
April 03, 2011
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Now that we have been investigating many of the details of material costs in quotations, we will now take a look at the mathematical calculations required.
By Dave Ferguson
March 29, 2011
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, By Dave Ferguson
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, By Dave Ferguson
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, By Dave Ferguson
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, By Dave Ferguson
March 29, 2011
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March 29, 2011
, By Dave Ferguson
March 29, 2011
Material costs in job estimates can come in a variety of ways. When you are estimating the cost to manufacture a product and the product is made out of either plate material, sheets, tubing, etc there are a few calculation methods that may be...
By Dave Ferguson
February 08, 2011
, By Dave Ferguson
February 08, 2011
, By Dave Ferguson
February 08, 2011
, By Dave Ferguson
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, By Dave Ferguson
February 08, 2011
, By Dave Ferguson
February 08, 2011
, By Dave Ferguson
February 08, 2011
, By Dave Ferguson
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, By Dave Ferguson
February 08, 2011
, By Dave Ferguson
February 08, 2011
, By Dave Ferguson
February 08, 2011
, By Dave Ferguson
February 08, 2011
, By Dave Ferguson
February 08, 2011
, By Dave Ferguson
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, By Dave Ferguson
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, By Dave Ferguson
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, By Dave Ferguson
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February 08, 2011
The basic cost estimating techniques involve analyzing all factors that make up a cost estimate. The first technique I would follow is to go through the following steps, systematically, and costing each one out in detail. Once you have gone through...
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August 10, 2009
, By Dave Ferguson
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Do you remember when it was time to quote and you sat down with your specifications, pencil and paper and started to write down how long each processes would take to manufacture the given specifications. You would add up all the setup time, then...
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July 16, 2009
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, By Dave Ferguson
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July 16, 2009
Calculating the entire operation costs depends on both setup and cycle time. In order accurately estimate your costs to complete a job you must be as accurate as possible during the estimating process. If you spend too much time estimating you...
By Dave Ferguson
July 07, 2009
, By Dave Ferguson
July 07, 2009
, By Dave Ferguson
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, By Dave Ferguson
July 07, 2009
Shop estimating is pretty complex and shop estimating software needs to be handle all types of scenarios. One of those situations is handling time calculations. Handling time is defined as the time it takes during the cycle time phase of...
By Dave Ferguson
June 29, 2009
, By Dave Ferguson
June 29, 2009
, By Dave Ferguson
June 29, 2009
, By Dave Ferguson
June 29, 2009
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June 29, 2009
, By Dave Ferguson
June 29, 2009
, By Dave Ferguson
June 29, 2009
, By Dave Ferguson
June 29, 2009
, By Dave Ferguson
June 29, 2009
, By Dave Ferguson
June 29, 2009
One of the most difficult aspects of estimating is figuring out what the cycle time may be for a process. There are many factors which affect the cycle time which are not fixed including skill level of the person running the machine, reliability... | 1,877 | 5,811 | {"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} | 2.796875 | 3 | CC-MAIN-2019-30 | latest | en | 0.927659 |
https://stats.stackexchange.com/questions/316770/how-to-determine-whether-one-group-reaches-asymptote-significantly-faster-than-a | 1,563,884,476,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529276.65/warc/CC-MAIN-20190723105707-20190723131707-00114.warc.gz | 541,950,834 | 35,128 | # How to determine whether one group reaches asymptote significantly faster than another?
I'm really struggling to select the correct statistical test.
Basically, I have 2 groups and each group contains 10 people. People in group 1 are treated with a drug and people in group 2 are given a placebo. Both groups are trained on a memory test across 10 days. As both groups are trained in the memory test, their scores increase across days. I notice that for both groups, their averaged score increases to 85% and hovers around that level (the asymptote of learning). However, group 1 reaches this asymptote on day 6 , whilst group 2 reaches the asymptote at day 9.
Could any of you point me in the direction of how I could go about testing whether or not there is a significant difference in the time it takes for one group to reach the asymptote compared to another? | 188 | 868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-30 | longest | en | 0.961717 |
https://essaysontime.us/2020/09/17/marketing-class/ | 1,603,454,258,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00447.warc.gz | 307,515,585 | 8,927 | # Marketing class
Please unconnected your apologys by the talents of the scrutiny. NOTE: the 250+ vocable reserve is waived for this assignment. ***For some usage on this assignment, you could investigate your textbook’s online plight (Links to an exterior plight.)Links to an exterior plight. and do one of the Chapter 13 exercises (this is from an precedent edition of the textbook, so the Chapter aggregate do not align)! First, let’s try a townsman of basic shatter level scrutinys. Remember that you can experience the shatter level apex formula in the Week 6 Lecture. ***Please be strong to “show your work” in your apologys to all of the forthcoming scrutinys to interpret how you got your apologys: Emma owns an Internet vocation where she vends weatherproof car foot mats. Her monthly agricultural consumes are \$50,000. The middle figure per ace is \$25 and the middle changeable consume per ace is \$15. What is her shatter-level apex? Remember that your apology gain be in “units,” not “dollars,” as you are experienceing the reckon of items that scarcity to be sold in prescribe to shatter level. Andrew ruptures localitys in his public-house for an middle of \$91 per confusion. The changeable consume per ruptureed locality is \$38. His agricultural consumes are \$59,000 per month. How manifold localitys does he own to rupture per month in prescribe to shatter level? In this uniformt when you do the calculations, your apology gain not be a complete reckon – there gain be a decimal. In shatter level calculations, you must frequently circular your apology up to the next highest complete reckon, accordingly you cannot vend a duty of an item. Now, let’s try to shatter down the multitudinous consumes vocation owners own into Agricultural Costs and into Changeable Costs. You may insufficiency to re-read the Lecture and/or the textbook to restore your retention on this one. Julia owns a sub sandwich garner and has the forthcoming consumes each month: Labor consumes (conduct & workers) = \$8,000 Insurance = \$1,000 Rent = \$1,200 Utilities = \$500 Average consume of ingredients/packaging for each sub = \$1.17 Julia vends subs for \$6 each. How manifold subs gain she scarcity to vend to shatter level each month fixed on the consumes listed aloft? In prescribe to bring-about that shatter level reckon more easy, Julia has set a new wood and vegetable distributor that can inferior the middle consume of ingredients/packaging down to \$0.98 per sub. If all of the other consumes sojourn the selfsame, what would the new shatter-level apex be? Julia decides to reposition her sub garner as “upscale” delay fresher woods and vegetables, parallel delay reward packaging for the subs. Her new figure apex is \$10 per sub, but her changeable consumes own restored to \$4.11 per sub. If all other consumes sojourn the selfsame, what is the shatter-level apex now? | 640 | 2,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.890755 |
https://homework.cpm.org/category/CCI_CT/textbook/pc/chapter/9/lesson/9.2.2/problem/9-83 | 1,723,595,216,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00507.warc.gz | 237,191,035 | 15,129 | ### Home > PC > Chapter 9 > Lesson 9.2.2 > Problem9-83
9-83.
Describe how your procedure would have been different in the previous problem if you wanted to find the slope of the tangent line at $x = 5$.
Use $5$ and $f\left(5\right)$ instead of $2$ and $f\left(2\right)$ . | 88 | 274 | {"found_math": true, "script_math_tex": 5, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-33 | latest | en | 0.844057 |
http://www.wvcityschool.org/mathematics-equations/ | 1,618,606,975,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038089289.45/warc/CC-MAIN-20210416191341-20210416221341-00402.warc.gz | 180,378,879 | 6,031 | # Mathematics Equations
12. March 2021 · Comments Off on Mathematics Equations · Categories: News · Tags:
For the first time to take exams and so hard, and then there was an unknown earlier and not the usual subject of the decision matrix. The decision matrix is not very complicated, there is no need to cram tchatelno object, you only need to think about it a bit and then many hours of practice. Believe me, here not many rules, you should only get the hand in the decision matrix and everything will be fine. The first semester with a decision matrix preodaleli you a good start – it is a guarantee of success. Vladislav Doronin has similar goals. No less pressing problem – a calculation differential equations, there is a lot of techniques and very few places to turn. Finding solutions to differential equations does not imply any unauthorized, just follow the procedure described in textbook way of solving equations, and you will be "happy." Finding solutions to differential equations, it is one hundred percent opposed to the decision matrix is the practice, as such, is not needed, you just need to learn all the techniques solving equations and then apply them with confidence. University of Michigan is full of insight into the issues. The practice then replaces the blind adherence to the method. One of my favorite tasks in mathematics – is the calculation tasks.
In solving the problems do not have any restrictions, you need only only clear that you have to do and a creative approach to solving the problem. In this branch of mathematics you need lots of practice, not blind memorization and problem solving of various orientation, in different ways. Solving problems – is the most creative and growing segment of Mathematics, thoroughly studying this section you will learn the logical and rational thinking. No less interesting part – this is the solution of equations. The solution of equations involves mastery of the material, the creative flow of ideas and a lot of practice. Believe me, there's nothing better – a lot of hours to solve the equation with a bunch of unknowns, and then realize that you have found the shortest and ideal solution to this equation. Qualitative evaluation of equations, that the key to success in solving equations. In this article, I briefly outlined the course of training for your exams. Acquainted with this article, you may not pass the exam, but clearly understand what went wrong, and will not allow such promashek in the future. | 495 | 2,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-17 | latest | en | 0.95498 |
https://brainmass.com/business/capital-budgeting/capital-budgeting-286578 | 1,481,142,321,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542246.21/warc/CC-MAIN-20161202170902-00351-ip-10-31-129-80.ec2.internal.warc.gz | 822,357,156 | 18,722 | Share
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# Capital budgeting
Your company wants to invest in a new product. The marketing department provided you with the following data:
After-tax cash flows for new product
Year 1 2 3 4 5 6
Low Demand 100 100 100 100 100 100
High Demand 300 500 600 700 700 600
All the cash flows will be received at the end of the year (Ignore taxes and depreciation). The project will require an initial investment today of \$1,200.
A. Standard NPV: Assume that there is a 50% chance of high demand and a 50% chance of low demand. Compute NPV if the project is carried through to the end of Year 6 regardless of the demand. The appropriate discount rate is 11%.
B. Real Options NPV: Using the information above, if the demand is low your company can abandon the project at the end of the first year and sell the equipment for \$550 (includes all tax ramifications of the sale). Use the 11% discount rate to compute the NPV including the option to abandon the project if demand is low.
#### Solution Summary
The solution explains the concept of Real options really well using the example in the question. The solution is really easy to follow for anyone who has a basic understanding of finance. All the steps are clearly shown. Overall, an excellent response.
\$2.19 | 299 | 1,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.888649 |
http://examcrazy.com/slender-shaft-supported-on-two-bearing-carries-disc-eccentricity-from-axis-of-rotation | 1,544,786,235,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825512.37/warc/CC-MAIN-20181214092734-20181214114234-00071.warc.gz | 85,305,007 | 28,422 | Question:-
A slender shaft supported on two bearing at its ends carries a disc with a eccentricity āeā from the axis of rotation. The critical speed of the shaft is N. if the disc is replaced by a second one of same weight but mounted with an eccentricity 2e, critical speed of the shaft In the second case is
Option (A)
1/2N
Option (B)
1/ Ć2 N
Option(C)
N
Option(D)
2 N
Correct Option:
(C)
Question Solution:
Critical speed of shaft is independent of the eccentricity | 126 | 473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.892767 |
https://web2.0calc.com/questions/any-help-would-be-appreciated-melody-or-cphill-help-please | 1,604,087,920,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911229.96/warc/CC-MAIN-20201030182757-20201030212757-00213.warc.gz | 586,073,176 | 6,441 | +0
# Any help would be appreciated Melody or CPhill help please.
+8
81
2
+359
The system of equations
$$\frac{xy}{x+y}=1, \frac{xz}{x+z}=2, \frac{yz}{y+z}=3$$
has exactly one solution. What is $$z$$ in this solution?
Jul 15, 2020
#1
-1
Try subsitution
Jul 15, 2020
#2
0
(x,y,z) = (8/5,8/3,-8)
Jul 15, 2020 | 134 | 315 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-45 | latest | en | 0.850079 |
https://www.reddit.com/r/Cubers/comments/873vzw/ksolve_a_new_fast_general_puzzle_solving_program/ | 1,628,132,243,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155268.80/warc/CC-MAIN-20210805000836-20210805030836-00675.warc.gz | 989,138,495 | 24,329 | ×
[–]Sub 1:45 4x4 (2cep) YT: Plus Two 7 points8 points (0 children)
This is awesome, ben. Thanks for sharing.
[–] 4 points5 points (2 children)
any interest in open sourcing?
[–]Sub 43 hours (150x150x150)[S] 3 points4 points (1 child)
maybe, but not immediately. the solvers are already open source though, obviously
[–]Sub-20 (CFOP) 2 points3 points (4 children)
Wait, is the god’s algorithm for 2*2 known?
[–]Sub 43 hours (150x150x150)[S] 7 points8 points (3 children)
of course, there are only 3.6 million positions. it's been known since 1981.
[–]Sub-20 (CFOP) 1 point2 points (2 children)
Wait just to make sure we’re talking about the same god’s algorithm: there’s an algorithm which cycles through every 2*2 permutation (effectively solving every scramble) and we know of it? I just couldn’t find anything about it online
[–]Sub 43 hours (150x150x150)[S] 10 points11 points (1 child)
no, god's algorithm is just the optimal solution to every scramble. you are talking about a hamiltonian cycle which isn't related to this. a hamiltonian cycle is known for both 2x2 and 3x3
[–]Sub-20 (CFOP) 0 points1 point (0 children)
Ah my bad. I remembered wrong.
[–] 2 points3 points (3 children)
Hey Ben, could you run your program to see if you can find a move optimal single dedge flip algorithm for the 5x5x5 in <2R,4u>? I just found a 39 mover for the nxnxn.
(Applied to the inner orbit of the 7x7x7, for example.)
3R 6u2 3R' 6u2 3R2 6u2 3R2 6u2 3R2 6u2 3R 6u2 3R2 6u2 3R 6u2 3R2 6u2 3R 6u2 3R 6u' 3R' 6u' 3R2 6u 3R 6u 3R' 6u' 3R' 6u' 3R2 6u 3R 6u' 3R 6u2 3R'
[–]Sub 43 hours (150x150x150)[S] 4 points5 points (0 children)
The optimal solution is 29 moves (which is also optimal on 4x4)
[–][🍰] 0 points1 point (1 child)
Why is that useful/interesting to you?
[–]Sub 43 hours (150x150x150)[S] 2 points3 points (0 children)
they are not useful, most likely no one will ever use these algorithms for anything. I (and probably cmowla too) like finding them simply because I can. actually the whole reason why ksolve++ exists is because I wanted to find the shortest way to do an R in <Rw,U> on 4x4, just because I knew it was possible to do and I wondered what the shortest alg was. I tried ksolve+ several years ago but it was way too slow, so in june 2017 I made a solver specifically for 4x4 <Rw,U> and used that. while running it and improving that solver, I came up with the idea for ksolve++ and then started making it
[–]Sub-18 (ZZ-WV) PB 10.19 Avg: 14.19 Main: Stickerless GAN 356X 2 points3 points (3 children)
So what is God's number for the 4x4?
[–]Sub 43 hours (150x150x150)[S] 5 points6 points (2 children)
what does that have to do with this post? also no one will ever know
[–]Sub-18 (ZZ-WV) PB 10.19 Avg: 14.19 Main: Stickerless GAN 356X 0 points1 point (1 child)
You said that you used it to figure it out on a super computer. It may be because you mixed up past tense
[–]Sub 43 hours (150x150x150)[S] 2 points3 points (0 children)
oh, I said 4x4 <2R, U>
[–]Ao100 PB: 19.85 (CFOP 2.5LLL) PB 10.50 0 points1 point (1 child)
[–] 0 points1 point (4 children)
Do you have a way to run this on Linux?
[–][deleted] (2 children)
[removed]
[–] 0 points1 point (1 child)
awesome, this is cool! You created the javascript version with Emscripten?
[–]Sub 43 hours (150x150x150)[S] 0 points1 point (0 children)
yes
[–]Sub 43 hours (150x150x150)[S] 0 points1 point (0 children)
no, although if you can run it on windows then you can copy the generated code to linux and compile it there
[–] 0 points1 point (4 children)
So this is like Cube Explorer? Is it better at finding LL algorithms?
[–]Sub 43 hours (150x150x150)[S] 1 point2 points (3 children)
it depends how good you are at using it. you can do things that cube explorer can't, e.g. <r,R,U> algorithms. for standard 3x3 it's much slower than cube explorer because it's not as optimized
[–] 0 points1 point (2 children)
I think I'd prefer transforming CE algs to <r,R,U> if I need it. Thanks anyway! Also big cube algs obviously.
Can I pause alg search and continue after restarting the pc? Because you can't do that in CE and that bothers me a lot...
[–]Sub 43 hours (150x150x150)[S] 1 point2 points (1 child)
um thats significantly more work though, and almost every alg you'd get from cube explorer is not r,R,U, and cube explorer would be way slower for something like that anyway. also there are a lot more things you can do listed in the readme file, including starting the search from a certain depth
[–] 0 points1 point (0 children)
I use another program to transform the CE algs to r, x etc moves. | 1,536 | 4,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-31 | latest | en | 0.859272 |
https://www.physicsforums.com/threads/common-electrical-misconceptions.68428/ | 1,631,828,787,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053759.24/warc/CC-MAIN-20210916204111-20210916234111-00710.warc.gz | 1,006,089,379 | 22,721 | # Common Electrical misconceptions?
Forgive me if I've fallen prey to random google searching but I was wondering if what this guy has to say has any bearing on reality. I know almost nothing about electrical engineering and something steered my surfing into that field so I happened upon the site.
It is supposedly a list of common misconceptions about electricity taught in schools. No need to commentary on his confusing and perhaps unprofessional communication style, I already noticed that and am unconcerned. Just curious if this is totally misleading, has some truth and some crap, or is totally correct presented in an odd fashion.
http://amasci.com/miscon/eleca.html#frkel
Thanks!
re
The article is totally true and brings out important facts. When Benjamin Franklin did his first experiments with electricity he reasoned that there are 2 charges + and - . He went further to say that one of the charges was always stationary while the other had to flow. He couldn't tell which one so he took a wild guess and said the + charges flow and - are stationary. Until 1900's J.J. Thompson did an experiment with a cathode ray tube and discoveter that the charge that flows is infact negative. Another noble peace price for him, but the damage has already been done. All the engineering community used Benjamins convection that + charges flow and use it in ther designs even to this day.
Well I'll tell you one specific area that is kinda confusing to me is the flow of electrons actually being extremely slow.
I understand a wave of kenetic energy traveling across a medium still transmitting energy very quickly. That's fine. If I had a tube full of billiard balls a mile long and I started cramming new ones in at a rate of 1 per second, the flow of 'electrons' would be quite slow but the energy required to pop one out the other end would transmit very quickly.
Here's my problem:
If electrons travel rather slowly through a wire and the way in which electrical energy (pretty much kenetic) is transmitted always through a medium. Isn't a vacuum an absolute removal of medium required to transmit energy electrically? Perhaps it's just because I don't really understand the differences between conductivity, permittivity and permeability, but I thought I understood that vacuum has a greater than 0 permittivity.
Doesn't that mean that it is possible to transmit electrical energy across a vacuum? Are the individual electrons jumping the gap and traveling much faster than they would in a wire?
Averagesupernova
Gold Member
I didn't take alot of time to read through that site. Most of it looks legit. But the author has some misconceptions of his own. Bill Beaty in the past has been a big free energy nut. Yep, you guessed it, perpetual motion.
hehe a bit off the wall eh?
Well can anyone answer my question in my second post for me?
Thanks!
Cliff_J
Start by thinking about the voltage required for lightning to occur, the air is an extremely poor conductor (else we'd never have open circuits!) but the air is still able to conduct at those levels. Not an answer, that's more just changing the medium the charge is sent over, you're asking about removing the medium completely.
Someone will correct if I'm off here but electricity requires matter to flow - the electrons are going from one atom to the next of the conductor. And a vacum would be an absence of matter, so with no matter present there would be no electrons available to flow.
Now convert the electricity to some form of electromagnetic wave and its a different story.
Cliff_J said:
Someone will correct if I'm off here but electricity requires matter to flow - the electrons are going from one atom to the next of the conductor. And a vacum would be an absence of matter, so with no matter present there would be no electrons available to flow.
This seems right to me.
However I think it should be pointed out electrons, and thus electricity, can exist independently of atoms (for instance in an electron beam http://physics.nist.gov/MajResFac/EBIT/intro.html). Notice the beam has both current (basically the number of electrons passing a point per second) and voltage (basically the number of electrons in one place compared to another).
I think one should be able to transmit electrical energy across a vacuum. Give the electron some momentum, then electron enters the vacuum and pops out the other side some time later. Although I guess technically, while the electron is in transit, the space is not a vacuum...
NoTime
Homework Helper
egsmith said:
Although I guess technically, while the electron is in transit, the space is not a vacuum...
There is tunneling as well.
First the electron is at point A then it is at point B.
Transit doesn't seem to apply.
Averagesupernova
Gold Member
Has everyone here forgotten about vacuum tubes? Electrons are 'boiled' off of the cathode by thermionic emission. They form a sort of cloud around it. With the proper voltages on the plate (anode) and cathode we form a current through the vacuum. So technically electrons need NO medium to travel through. Of course we don't build our circuits out of 'nothing' so we end up with WHAT I THOUGHT was the familiar vacuum tube with electrons jumping between 2 conductors through space. Any material that gets in the way, such as air, will impede the flow of electrons and that means no more current.
Averagesupernova said:
Has everyone here forgotten about vacuum tubes?
Actually I had. Great example though.
Now that I think about it, a CRT contains a vacuum as well. Probably not a perfect one though.
Averagesupernova
Gold Member
egsmith said:
Actually I had. Great example though.
Now that I think about it, a CRT contains a vacuum as well. Probably not a perfect one though.
No vacuum is perfect. A CRT is highly evacuated though. That electron has a LONG way to go from the gun to the phosphor when you consider its size. Therefor very little can be allowed to get in the way.
Ouabache
Homework Helper
waht said:
When Benjamin Franklin did his first experiments with electricity he reasoned that there are 2 charges + and - . He went further to say that one of the charges was always stationary while the other had to flow. He couldn't tell which one so he took a wild guess and said the + charges flow and - are stationary. Until 1900's J.J. Thompson did an experiment with a cathode ray tube and discoveter that the charge that flows is infact negative. Another noble peace price for him, but the damage has already been done. All the engineering community used Benjamins convection that + charges flow and use it in ther designs even to this day.
That always bugged me.. I learned early on in chemistry and physics that electrons are the particles that move from orbital to orbital within an atom, and highly energetic ones can leave an atom, leaving a positive ion in its wake. If that free electron comes upon another positive ion that is missing an electron, the electron would be attracted to it and fill that orbit balancing the charge and creating a neutral atom. Then I get to electrical engineering, and find current defined in terms of the flow of holes (atoms missing an electron). Seemed abit silly to carry on that idea. Though it is often rationalized that a net movement of electrons in one direction can be equated to a movement of holes in the opposite direction.
Regarding his other thoughts, it looks to me that he is just being picky about choice of wording.
Cliff_J
Ok, I'll stand as corrected as technically that is true.
But if we remove the heat and the edison effect is gone then no electron flow occurs as would in any other matter, correct? As in regardless of buildup stored on the plate, electrons wouldn't jump the gap without the thermonic emission, right? Or does that even occur at room temperatures?
(sorry, tubes were a little before my time and now it seems they are only used as fodder in some sort of sonic attribute debates between solid-state and tube affectionados)
Averagesupernova
Gold Member
Not sure cliff but I think there are some tubes that have no filament. I believe they were used as high voltage rectifiers. They may have been gas filled. Have to research this.
Averagesupernova said:
Not sure cliff but I think there are some tubes that have no filament. I believe they were used as high voltage rectifiers. They may have been gas filled. Have to research this.
Yes, these do exist and are called cold cathode tubes. They are filled with a low pressure gas (Nobel gasses) and operate by avalanche. CC Tubes operate in a manner very similar to semiconductors.In fact, I believe the old Nixie tubes where cold cathode as well. Need to look that up myself.
NoTime
Homework Helper
Yep, Nixie tubes are cold cathode. No different than a neon lamp except for the multiple electrode structure works as a display.
SGT
Cliff_J said:
Ok, I'll stand as corrected as technically that is true.
But if we remove the heat and the edison effect is gone then no electron flow occurs as would in any other matter, correct? As in regardless of buildup stored on the plate, electrons wouldn't jump the gap without the thermonic emission, right? Or does that even occur at room temperatures?
(sorry, tubes were a little before my time and now it seems they are only used as fodder in some sort of sonic attribute debates between solid-state and tube affectionados)
The cathode, being made of a metal, has plenty of free electrons on it. Those electrons are bound in the material of the cathode by the attraction of the charged nuclei, but if you give them enough energy, they can leave the material.
The easiest way to provide energy to the material is by heating it. That is why we heat the cathode. But you can provide energy by keeping the cathode in a strong enough electric field.
So, if the anode and the cathode are close enough and have a great potential difference between them (anode positive with respect to cathode), electrons can leave the cathode and jump to the anode, even in absolute vacuum.
The originial question here was actually whether electrical "energy" could be transmitted through a vacuum. And you don't need electrons or cathodes for that. Electromagnetic induction works fine, and it DOES travel at the speed of light.
jdavel said:
The originial question here was actually whether electrical "energy" could be transmitted through a vacuum. And you don't need electrons or cathodes for that. Electromagnetic induction works fine, and it DOES travel at the speed of light.
I guess I might have stated the question wrong. I didn't mean EM induction.
Basically I was meaning that even aside from electron clouds produced in a variety of items, it is my understanding that electrical permittivity of a vacuum is 1 and other mediums are given values related to that. The problem I have with that is that it is not 0. This seems confusing because it seems as though this is saying that electrons will jump across a vacuum gap if there is enough potential or whatever. (plz excuse my ignorance)
Can any of you answer what is meant by electrical permittivity of 1 instead of 0? I'm pretty sure that's a known and accepted constant but it seems I'm missing something.
SGT
TheAntiRelative said:
I guess I might have stated the question wrong. I didn't mean EM induction.
Basically I was meaning that even aside from electron clouds produced in a variety of items, it is my understanding that electrical permittivity of a vacuum is 1 and other mediums are given values related to that. The problem I have with that is that it is not 0. This seems confusing because it seems as though this is saying that electrons will jump across a vacuum gap if there is enough potential or whatever. (plz excuse my ignorance)
Can any of you answer what is meant by electrical permittivity of 1 instead of 0? I'm pretty sure that's a known and accepted constant but it seems I'm missing something.
Electric permittivity has nothing to do with electric current. Electric permittivity is introduced in Coulomb's law in order relate force to charges and distance. The permittivity of vacuum ε0 is 8.85x10-22 F/m and not 1. The dielectric constant is the quotient of the permittivity of a medium and the permittivity of the vacuum. Of course, the dielectric constant of the vacuum is 1.
SGT said:
Electric permittivity has nothing to do with electric current. Electric permittivity is introduced in Coulomb's law in order relate force to charges and distance. The permittivity of vacuum ε0 is 8.85x10-22 F/m and not 1. The dielectric constant is the quotient of the permittivity of a medium and the permittivity of the vacuum. Of course, the dielectric constant of the vacuum is 1.
What is the purpose/use of relating force to charges and distance then. What is the permittivity of a medium used for. What does it tell us?
Is the permittivity of the vacuum kind of an imaginary concept or something?
SGT
TheAntiRelative said:
What is the purpose/use of relating force to charges and distance then. What is the permittivity of a medium used for. What does it tell us?
Is the permittivity of the vacuum kind of an imaginary concept or something?
From Coulomb's law, the electrostatic force between two charges q1 and q2 is:
F = k.q1 .q2 /d2
Where k is Coulomb's constant k = 1/4.π.ε0
The 3 constants concerning electric and magnetic fields propagation are the speed of light in a vacuum: 3.108 m/s, the magnetic permeability of free space: μ0 = 4π.10-7 and the electric permittivity of free space ε0.
They are related such that c = 1/sqrt(μ0ε0). This sets the value for ε0.
Last edited by a moderator:
chroot
Staff Emeritus
Gold Member
SGT said:
k is chosen such as if both charges are 1 coulomb and the distance between them 1 meter, the force must be 1 newton.
This is quite false! The force between two one-coulomb charges one meter apart is almost ten billion newtons.
- Warren
SGT
chroot said:
This is quite false! The force between two one-coulomb charges one meter apart is almost ten billion newtons.
- Warren | 3,096 | 14,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.977235 |
http://oeis.org/A037590 | 1,571,043,346,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649841.6/warc/CC-MAIN-20191014074313-20191014101313-00049.warc.gz | 195,830,040 | 3,600 | This site is supported by donations to The OEIS Foundation.
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A037590 Base-4 digits are, in order, the first n terms of the periodic sequence with initial period 1,0,3. 0
1, 4, 19, 77, 308, 1235, 4941, 19764, 79059, 316237, 1264948, 5059795, 20239181, 80956724, 323826899, 1295307597, 5181230388, 20724921555, 82899686221, 331598744884, 1326394979539, 5305579918157, 21222319672628, 84889278690515 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Index entries for linear recurrences with constant coefficients, signature (4,0,1,-4). FORMULA G.f.: x*(1+3*x^2) / ( (x-1)*(4*x-1)*(1+x+x^2) ). - R. J. Mathar, Apr 27 2015 MATHEMATICA Table[FromDigits[PadRight[{}, n, {1, 0, 3}], 4], {n, 30}] (* Harvey P. Dale, Apr 28 2018 *) CROSSREFS Sequence in context: A094734 A094578 A130132 * A037681 A156760 A320088 Adjacent sequences: A037587 A037588 A037589 * A037591 A037592 A037593 KEYWORD nonn,base,easy AUTHOR STATUS approved
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Last modified October 14 04:44 EDT 2019. Contains 327995 sequences. (Running on oeis4.) | 464 | 1,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-43 | latest | en | 0.557244 |
http://dynoinsight.com/Help/Interface/OCC/GeomAPI/IKO_GeomAPI_PointsToBSpline.aspx | 1,585,647,443,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500426.22/warc/CC-MAIN-20200331084941-20200331114941-00128.warc.gz | 60,536,521 | 41,023 | DInsight Home
IKO_GeomAPI_PointsToBSpline
# IKO_GeomAPI_PointsToBSpline Interface
Used to approximate a BsplineCurve passing through an array of points, with a given Continuity. Describes functions for building a 3D BSpline curve which approximates a set of points. A PointsToBSpline object provides a framework for: defining the data of the BSpline curve to be built, implementing the approximation algorithm, and consulting the results.
Init
Init2
Init3
Init4
Curve
HRESULT Init(IKO_TColgp_Array1OfPnt* Points, int DegMin, int DegMax, int GeomAbs_Shape_continuity, double Tol3D)
Approximate a BSpline Curve passing through an array of Point. The resulting BSpline will have the following properties: 1- his degree will be in the range [Degmin,Degmax] 2- his continuity will be at least 3- the distance from the point to the BSpline will be lower to Tol3D
HRESULT Init2(IKO_TColgp_Array1OfPnt* Points, int Approx_ParametrizationType_ParType, int DegMin, int DegMax, int GeomAbs_Shape_continuity, double Tol3D)
Approximate a BSpline Curve passing through an array of Point. The resulting BSpline will have the following properties: 1- his degree will be in the range [Degmin,Degmax] 2- his continuity will be at least 3- the distance from the point to the BSpline will be lower to Tol3D
HRESULT Init3(IKO_TColgp_Array1OfPnt* Points, IKO_TColStd_Array1OfReal* Parameters, int DegMin, int DegMax, int GeomAbs_Shape_continuity, double Tol3D)
Approximate a BSpline Curve passing through an array of Point, which parameters are given by the array . The resulting BSpline will have the following properties: 1- his degree will be in the range [Degmin,Degmax] 2- his continuity will be at least 3- the distance from the point to the BSpline will be lower to Tol3D
HRESULT Init4(IKO_TColgp_Array1OfPnt* Points, double Weight1, double Weight2, double Weight3, int DegMax, int GeomAbs_Shape_continuity, double Tol3D)
Approximate a BSpline Curve passing through an array of Point using variational smoothing algorithm, which tries to minimize additional criterium: Weight1*CurveLength + Weight2*Curvature + Weight3*Torsion
HRESULT Curve(IKO_Geom_BSplineCurve** curve)
Returns the computed BSpline curve. Raises StdFail_NotDone if the curve is not built | 589 | 2,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.898807 |
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