url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://converter.ninja/volume/us-customary-teaspoons-to-centiliters/41-usteaspoon-to-cl/ | 1,686,274,232,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655244.74/warc/CC-MAIN-20230609000217-20230609030217-00341.warc.gz | 202,414,073 | 5,372 | # 41 US customary teaspoons in centiliters
## Conversion
41 US customary teaspoons is equivalent to 20.208578534375 centiliters.[1]
## Conversion formula How to convert 41 US customary teaspoons to centiliters?
We know (by definition) that: $1\mathrm{usteaspoon}\approx 0.492892159375\mathrm{centiliter}$
We can set up a proportion to solve for the number of centiliters.
$1 usteaspoon 41 usteaspoon ≈ 0.492892159375 centiliter x centiliter$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{centiliter}\approx \frac{41\mathrm{usteaspoon}}{1\mathrm{usteaspoon}}*0.492892159375\mathrm{centiliter}\to x\mathrm{centiliter}\approx 20.208578534375\mathrm{centiliter}$
Conclusion: $41 usteaspoon ≈ 20.208578534375 centiliter$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 centiliter is equal to 0.0494839356612337 times 41 US customary teaspoons.
It can also be expressed as: 41 US customary teaspoons is equal to $\frac{1}{\mathrm{0.0494839356612337}}$ centiliters.
## Approximation
An approximate numerical result would be: forty-one US customary teaspoons is about twenty point two one centiliters, or alternatively, a centiliter is about zero point zero five times forty-one US customary teaspoons.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 404 | 1,466 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2023-23 | latest | en | 0.679324 |
https://nl.mathworks.com/matlabcentral/cody/solutions/1819712 | 1,566,468,195,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317037.24/warc/CC-MAIN-20190822084513-20190822110513-00447.warc.gz | 563,155,304 | 15,415 | Cody
# Problem 52. What is the next step in Conway's Life?
Solution 1819712
Submitted on 19 May 2019 by SUPRABHA MUKHOPADHYAY
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 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0]; B = [ ... 1 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0]; assert(isequal(gameOfLife(A),B))
2 Pass
A = [ ... 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0]; B = [ ... 1 0 1 1 1 0 0 0 0 0 1 1 0 1 1 0]; assert(isequal(gameOfLife(A),B)) | 258 | 548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.700851 |
https://study.com/academy/answer/sketch-the-graph-of-the-given-function-by-demanding-the-appropriate-information-and-points-from-the-first-and-second-derivatives-y-2x-3-24x-7-a-what-are-the-coordinates-of-the-relative-maxima-b-what-are-the-coordinates-of-the-relative-minima-c.html | 1,585,968,781,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370519111.47/warc/CC-MAIN-20200404011558-20200404041558-00476.warc.gz | 717,728,244 | 23,439 | # Sketch the graph of the given function by demanding the appropriate information and points from...
## Question:
Sketch the graph of the given function by demanding the appropriate information and points from the first and second derivatives:
{eq}y = 2x^3 - 24x - 7 {/eq}
a. What are the coordinates of the relative maxima?
b. What are the coordinates of the relative minima?
c. What are the coordinates of the points of inflection?
## Relative Extremes:
The relative extremes of a function are critical points, we can find them by matching the first derivative to zero or when the first derivative does not exist. Inflection points are points where concavity changes.
We have the function:
{eq}f(x)= 2x^3 - 24x - 7 \\ {/eq}
Differentiating the function
{eq}f'(x)=6\,{x}^{2}-24 \\ {/eq}
{eq}f'(x)=0 {/eq} when {eq}x=2 \\ x=-2 \\ {/eq}
{eq}\begin{array}{r|D{.}{,}{5}} Interval & {-\infty<x<-2} & {-2<x<2} & {2<x<\infty } \\ \hline Test \space{} value & \ x=-5 & \ x=1 & \ x=3 \\ Sign \space{} of \ f'(x) & \ f'(-5)>0 & \ f'(1)<0 & \ f'(3)>0 \\ Conclusion & increasing & decreasing & increasing \\ \end{array} \\ {/eq}
Relative maximum at:
{eq}(-2, 25) \\ {/eq}
Relative minimum at: {eq}(2, -39) \\ {/eq}
Differentiating the function
{eq}f'(x)=6\,{x}^{2}-24 \\ f''(x)=12\,x \\ {/eq}
{eq}f''(x)=0 {/eq} when {eq}x=0 {/eq}
{eq}\begin{array}{r|D{.}{,}{5}} Interval & {-\infty<x<0} & {0<x<\infty} \\ \hline Test \space{} value & \ x=-1 & \ x=1 \\ Sign \space{} of \ f'' (x) & \ f'' (-1.5)<0 & \ f'' (1)>0 \\ Conclusion & concave \space down & concave \space up \\ \end{array} \\ {/eq}
Inflection point:
{eq}(0, -7) \\ {/eq} | 569 | 1,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-16 | latest | en | 0.551428 |
statisticsassignmenthelp43837.fitnell.com | 1,544,784,531,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825512.37/warc/CC-MAIN-20181214092734-20181214114234-00141.warc.gz | 277,389,951 | 5,678 | # Everything about statistics homework help
Color quantities in Pascal's Triangle by rolling a amount and afterwards clicking on all entries that have exactly the same remainder when divided with the variety rolled, thereby practising division and remainders, investigating variety designs, and investigating fractal patterns. Coloring Remainders in Pascal's Triangle is one of the Interactivate assessment explorers.
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Give enter to The full Variety Cruncher and try to guess what it did in the output it generates. This exercise only generates multiplication and addition capabilities in order to avoid outputting any damaging numbers. Complete Quantity Cruncher is one of the Interactivate assessment explorers. | 620 | 3,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.913974 |
http://www.dummies.com/how-to/content/how-torque-causes-angular-acceleration.html | 1,469,272,167,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257822172.7/warc/CC-MAIN-20160723071022-00181-ip-10-185-27-174.ec2.internal.warc.gz | 431,583,964 | 14,967 | In physics, when you rotate an extended object, such as a rod, disk, or cube, which has its mass distributed through space, you have to take into account where the force is applied. Enter torque. Torque is a measure of the ability of a force to cause rotation. In physics terms, the torque exerted on an object depends on the force itself (its magnitude and direction) and where you exert the force. You go from the strictly linear idea of force as something that acts in a straight line (such as when you push a refrigerator up a ramp) to its angular counterpart, torque.
Just as a net force causes acceleration, a net torque causes angular acceleration, so you can think of torque as the angular equivalent of force.
Torque brings forces into the rotational world. Most objects aren’t just points or rigid masses, so if you push them, they not only move but also turn. For example, if you apply a force tangentially to a merry-go-round, you don’t move the merry-go-round away from its current location — you cause it to start spinning.
A seesaw demonstrates torque in action.
The figure shows a seesaw with a mass m on it. If you want to balance the seesaw, you can’t have a larger mass, M, placed on a similar spot on the other side of the seesaw. Where you put the larger mass M determines whether the seesaw balances. As you can see in diagram A, if you put the mass M on the pivot point — also called the fulcrum — of the seesaw, you don’t have balance. The larger mass exerts a force on the seesaw, but the force doesn’t balance it.
As you can see in diagram B, as you increase the distance you put the mass M away from the fulcrum, the balance improves. In fact, if M = 2m, you need to put the mass M exactly half as far from the fulcrum as the mass m is.
The torque is a vector. The magnitude of the torque tells you the ability of the torque to generate rotation; more specifically, the magnitude of the torque is proportional to the angular acceleration it generates. The direction of the torque is along the axis of this angular acceleration. | 461 | 2,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2016-30 | latest | en | 0.927031 |
https://gateoverflow.in/101486/ram-chips-questions | 1,675,147,306,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00770.warc.gz | 293,706,467 | 21,848 | 2,318 views
A RAM chip has a capacity of 1024 words of 8 bits each (1K × 8). The number of 2 × 4 decoders with enable line needed to construct a 16K × 16 RAM from 1K × 8 RAM is
Note:it is previous yr gate ques , i asked it again bcz that answer i did not get. After getting answer i hide it. So don't close it.
RAM chip size is 1K × 8 =213
To construct 16K × 16 RAM =218
total output lines 218/213 =25
number of 2 × 4 decoders with enable line hence
25/23=22
hence 4 decoder required
1 comment
No we need 5 decoders as number of address lines are 16K/1K = 16, so we need to realise 4X16 decoder to 2X4 decoder. We can do the same by 16/4 = 4 decoders and for these 4 decoders we also need one more decoder. So total 5 decoders are needed.
Capacity of RAM Chip = 1k*8 bit
MEMORY CAPACITY = NUMBER OF CHIP * ONE RAM CHIP CAPACITY
Number of chip = memory capacity / one chip capacity
# chip =16k * 16bit / 1k * 8bit
# chip =(16 * 2 ) = 32 chip ➔➔➔ (16 * 2 ): Meaning is there are 16 Rows , where 2 chips in each rows .
so,we could required (4 * 16) Decoder , Now get ( 4* 16) Decoder using ( 2* 4) it would be like below…
so the number of decoder of size(2*4) is 5
1 vote
1
24,633 views | 413 | 1,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-06 | latest | en | 0.831692 |
http://talkchess.com/forum3/viewtopic.php?f=2&t=74550&start=80 | 1,610,940,799,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00233.warc.gz | 98,818,638 | 6,570 | ## What are Stockfish and Komodo estimated rating for these 5 depths ?
Discussion of anything and everything relating to chess playing software and machines.
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lkaufman
Posts: 4476
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### Re: What are Stockfish and Komodo estimated rating for these 5 depths ?
Chessqueen wrote:
Mon Aug 10, 2020 3:39 am
Chessqueen wrote:
Fri Jul 24, 2020 4:06 am
1st Stockfish at depth 1 Please provide estimated rating = Komodo =
1st Stockfish at depth 2 Please provide estimated rating = Komodo =
1st Stockfish at depth 3 Please provide estimated rating = Komodo =
1st Stockfish at depth 4 Please provide estimated rating = Komodo =
1st Stockfish at depth 5 Please provide estimated rating = Komodo =
We all know why Stockfish is soo strong, because it is the Program that reaches the highest depth withing a certain time control, but has anybody done a 100 match games between Komodo vs Stockfish with depth set to 22 or 24 ?
I have a curious question to Larry Kaufman, Why is it that Komodo is limited to a maximum Skill 25 and NOT 30? I am pretty sure that on very fast computer like your or the one used by TCEC it could reach a Depth of 17 or Skill = 30.
The depth = skill - 13 rule is only valid for skill levels 19 thru 23. Skill 24 is depth 12 (not 11), and Skill 25 is just normal Komodo using all its available time. We didn't go beyond depth 12 with the skill levels because I estimated that depth 12 was strong enough to give Magnus Carlsen a good fight in rapid (15' + 10") chess, so who would need a crippled level beyond that? I may have been a bit too optimistic in my estimate, I now think that we might need depth 13 to equal MC in rapid, but we need more rapid games with top humans to really tell.
Komodo rules! | 489 | 1,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-04 | latest | en | 0.939155 |
http://www.indiabix.com/placement-papers/tcs/4219 | 1,558,717,808,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257699.34/warc/CC-MAIN-20190524164533-20190524190533-00023.warc.gz | 278,954,324 | 9,191 | # Placement Papers - TCS
## Why TCS Placement Papers?
Learn and practice the placement papers of TCS and find out how much you score before you appear for your next interview and written test.
## Where can I get TCS Placement Papers with Answers?
IndiaBIX provides you lots of fully solved TCS Placement Papers with answers. You can easily solve all kind of placement test papers by practicing the exercises given below.
## How to solve TCS Placement Papers?
You can easily solve all kind of questions by practicing the following exercises.
### TCS Placement Paper Pattern (English, Aptitude)
Posted By : D.Devender Rating : +147, -22
Paper : TCS Placement Paper Pattern (English, Aptitude)
Hi, Its Vineet Mishra. TCS Test Consists of 3 Sections..
1. Test on Synonyms and Antonyms (40 questions, 20 minutes: Also Q's on Sentence Completion).
2. Aptitude Test (32 questions(may vary), 40 minutes: General Questions on Mathematics).
3. Critical Reasoning (30 Minutes: 3 Analysis Paragraphs and 12 questions Relating them
Written Test:
Written test is on line test paper. There were three sections. plz prepare last papers too & BARRONS Properly. Some of Antonyms are
Adhesive = tenacious, sticky, glue, gum, bonding agent
Alienate = estrange
Bileaf = big screen, big shot, big success
Belief = conviction
Baffle = puzzle
Brim = edge
Covet = to desire
Caprice = whim
Concur = similar, acquiesce
Confiscate = appropriate, to take charge, to annex
Dispel = scatter
Divulge = reveal, make known, disclose
Discretion = prudence
Emancipate = liberate
Efface = obliterate
Embrace = hug, hold, cuddle
Furtive = stealthy
Heap = to gather
Hamper = obstruct
Heap = to pile
Hover = linger
Incentive = spur
Instigate = incite
Inert = passive
Latitude = scope
Lethargy = stupor
Lamont = lakes, lamentable
Lament = wail
Latent = potential
Merry = Enjoy
Meager = small, little, scanty
Misery = distress
Momentary = for small time
Merit = to deserve
Obstinate = stubborn
Overt = obvious, clear, explicit, evident
Pretentious = ostentatious
Potential = ability
Rinaile = rigorous
Renounce= reject
Solicit = Humble, urge
Subside = wane
Stifle = snits
Tranquil = calm, silent, serene
To merit- to deserve
Volume = quantity
Veer = diverge
Wethargy = well wisher
2. General Aptitude:
For this section it is more than enough to solve all the previous papers available in this site. Then for some concepts refer R.S Agarwal aptitude book. Most of the problems will be repeated only with the numbers changed. But u will have the same model. I could not able to recollect all the questions in my test, but I'm just giving you the models for few questions, what I have in my mind.
http://www.TowardsJob.com
1.If g (0)=1,g (1)=1 and g (n)= g (n-1) + g (n-2). Find g (6).
a) 9 b) 13 c) 21 d) 7
3. Select the odd one out.
a) Java b) Lisp c) Smalltalk d) Eiffel.
a) Lisp. b) Java. c) Eiffel. d) Smalltalk.
4. Select the odd one out.
a) Linux b) Oracle c) DB2 d) Ingress.
5. Number of faces, vertices and edges of a cube
a) 6,8,12 b) 8,12,8 c) 8,6,12 d) 12,8,6
6) in a two dimensional array X(9,7) which each element occupying 8 bytes of memory with address of the first element X(1,1) is 1072, find the address of X(8,5).
7) Which of the following highest Standard deviation
a) 5,-5,5,-5, 5,-5 b) 5,5,5,5,5,5 c) -5,-5,-5,-5,-5,-5 d) -5,5,-5,5,-5,5
3. Critical Reasoning:
questions were on one passage "politicians attending a party. They are seated in some way". 4 questions were on the passage "Red and Brown group in a community and conditions for their marriage". .
Technical & HR Round:
In Technical questions bassically on JAVA & C++
In HR They asked me abt my family background. They asked me to tell abt it. They asked whether I am ready to sign a bond for 2yrs. I said OK. They asked my strengths, why shud I select u?
ALL ARE FROM 12 TH EDITION BARRONS. SO SEE ALL THE MODEL TEST PAPERS FOR THIS PART. | 1,147 | 4,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.876812 |
http://mathhelpforum.com/advanced-statistics/78140-normal-distribution-problem-print.html | 1,529,859,523,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866984.71/warc/CC-MAIN-20180624160817-20180624180817-00308.warc.gz | 207,793,569 | 3,320 | # Normal distribution problem
• Mar 11th 2009, 08:16 AM
Fnus
Normal distribution problem
Hey, I've always been bad at these, and I don't even know what to plot into the calculator, so any help is great appreciated, thanks!
The weights of adult males of a type og dog may be assumed to be normally distributed with mean 25 kg and a standard deviation of 3 kg.
Given that 30% of the weights lie between 25 kg and x kg, where x > 25, find the value of x.
• Mar 11th 2009, 03:09 PM
mr fantastic
Quote:
Originally Posted by Fnus
Hey, I've always been bad at these, and I don't even know what to plot into the calculator, so any help is great appreciated, thanks!
The weights of adult males of a type og dog may be assumed to be normally distributed with mean 25 kg and a standard deviation of 3 kg.
Given that 30% of the weights lie between 25 kg and x kg, where x > 25, find the value of x.
\$\displaystyle \Pr(25 < X < x) = 0.3\$
\$\displaystyle \Rightarrow \Pr(X < x) - \Pr(X < 25) = 0.3\$
\$\displaystyle \Rightarrow \Pr(X < x) - 0.5 = 0.3\$
\$\displaystyle \Rightarrow \Pr(X < x) = 0.8\$
which is now just a routine inverse normal problem.
• Mar 11th 2009, 03:20 PM
josh_amsterdam
First thing you do it simply picture a normal distribution curve. Right in the middle is the average, which is 25 kg in this case. You are looking for an area under the curve in between x and the mean, and since x > 25 it will be the right side of the mean.
Since both on the left an the right side of the mean you have exactly 50% you know that left of the mean you have 50%, then you have the area mean to x which is 30% and then you have a remaining 20%.
I assume you have a TI calculator, if not, what I'll say is useless. You press [2ND] [Vars] [2] [ENTER] and you'll have the program 'normalcdf'. If you enter a left boundy, a right boundry, the mean, the standard deviation and it will give you the area contained withing the boundaries. (for instance: normalcdf(25,10^90,25,3) will give exactly .5 because that's half of the whole area.)
You are looking for not halve, but 30% = .3 of the area. Go to [tex] solver and enter eqn: 0= normalcdf(25,x,25,3) - 0,3 and then press down [APLHA] [ENTER] and it will give you the answer.
I hope you understand. And also, you mentioned plotting, which is what most schools teach, but using the solver is a LOT easier and faster. | 675 | 2,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 4 | CC-MAIN-2018-26 | latest | en | 0.927626 |
http://www.mrbartonmaths.com/blog/angles-and-kandinsky-tes-maths-resource-of-the-week/ | 1,529,542,670,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863980.55/warc/CC-MAIN-20180621001211-20180621021211-00362.warc.gz | 455,006,733 | 10,956 | # Angles and Kandinsky: TES Maths Resource of the Week
To see all of the work I do for TES Maths, including Resource of the Week, Inspect the Spec, Maths Newsletters and Topic Collections, please visit the TES Maths Blog here
What is it?
Time for a bit of culture on the Resource of the Week! Measuring angles is one of those skills that students need to develop, but can often be a little dull. Well, why not spice it up by having students measure angles and check relationships on the paintings of everyone’s favourite 20th century Russian painter, Wassily Kandinsky! The artist made beautiful use of straight lines, triangles and quadrilaterals, all of which can be measured and checked by students, whilst at the same time exposing them to some lovely art work.Better still, this activity form part of a larger lesson which recaps key angle facts and then goes on to look at constructions, all presented in clear step-by-step fashion.
How can it be used?
The lesson that contains the Kandinsky measuring activity is really well structured. As ever, it is important to adapt it to the needs of your class and your teaching style, inserting, removing and modifying slides accordingly. This task could also lead to students finding other mathematically inspired artwork for homework and bringing in their findings. Or, why not see if you have your own budding Kandinsky in the class by setting them the challenge of creating an piece of art that contains at least three geometrical facts? If you do, and given that Kandinsky’s paintings regularly sell for multi millions of pounds, make sure to negotiate 20% of any future profits.
Thanks so much for sharing
Craig Barton
View the author’s other resources | 360 | 1,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.954501 |
http://xn----7sbarcbawbtqc2eidh2f0d9av.xn--p1ai/en/etudes/monge-problem/ | 1,560,704,701,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998288.34/warc/CC-MAIN-20190616162745-20190616184745-00327.warc.gz | 333,298,960 | 7,946 | # Leaving the plane
in collaboration with Sergei Markelov
## VideoFull screenZoomDownloadAVI video (768×576, 7.2 MB) MP4 video (768×576, 1.8 MB) Zip archive (AVI, 5.8 MB)
00:00|00:00
Living on the Earth surface people thought for a long time that it was flat. We had to build some theories to guess that the Earth looks more like a ball. And only in the second part of the XX century we could look at our planet from the space and certify it visually.
The same happens in mathematics: considering the ambient space we can often learn more about some object.
Consider three arbitrary circles and draw common tangents to every pair. What can we say about the single line. Well, a picture is not a proof, it's just a source for our conjecture. Let's try to prove it.
Both the problem and the image consider plane objects. But what if we look at them from outside, from the ambient three-dimensional space.
Consider three spheres whose equators are the circles. The cones enveloping the pairs of spheres have the common tangents as generatrices. The points we expect to lie on the same line are the apices of the cones.
Put a plane on the cones. The upper generatrices intersect pairwise and define this plane unambiguously. The points we a interested in, the cones apices, lie both on this plane and the «equatorial» one. But this two nonparallel planes have intersect in a line! So, as we conjectured, the three points of intersection of common tangents to three arbitrary circles lie on a single line.
Nowadays this theorem we've just proved is named after a french mathematician Gasrpard Monge. | 374 | 1,604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-26 | longest | en | 0.933437 |
https://ancientgreecereloaded.com/files/ancient_greece_reloaded_website/great_persons/dinostratus.php | 1,721,357,909,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00208.warc.gz | 93,805,302 | 12,024 | TheDome
## ANCIENT GREECE RELOADED
ENTER THE WORLD OF MYTHS AND LEGENDS
## Dinostratus
Dinostratus (Greek: Δεινόστρατος, c. 390 BCE – c. 320 BCE) was a Greek mathematician and geometer, and the brother of Menaechmus. He is known for using the quadratrix to solve the problem of squaring the circle.
According to Proclus (Commentary on Euclid, Book I; Friedlein, ed., 67.8–12),“Amyclas of Heraclea, one of the associates of Plato, and Menaechmus, a pupil of Eudoxus who had also studied with Plato, and his brother Dinostratus made the whole of geometry still more perfect.” Dinostratus therefore lived in the middle of the fourth century b.c., and although there is no direct evidence his Platonic associations point to Athens as the scene of his activities. He must have ranged over the whole field of geometry, although only one of his achievements is recorded and the record bristles with difficulties. This is the application of the curve known as the quadratrix to the squaring of the circle.
The evidence rests solely on Pappus (Collection, IV. 30; Hultsch ed., 250.33–252.3), whose account is probably derived from Sporus (third century). Pappus says: “For the squaring of the circle there was used by Dinostratus, Nicomedes and certain other later persons a certain curve which took its name from this property; for it is called by them square-forming” (τєτραγωνíζоνσα sc. γραµµή quadratrix). The curve was not discovered by Dinostratus but by Hippias, for Proclus, whose account is derived from Eudemus, says: “Nicomedes trisected any rectilineal angle by means of the conchoidal curves, of which he had handed down the origin, order and properties, being himself the discoverer of their special characteristic.
Others have done the same thing by means of the quadratrices of Hippias and Nicomedes” (Friedlein, ed., 272.3–10). It has been usual, following Bretschneider, to deduce that Hippias first discovered the curve and that Dinostratus first applied it to finding a square equal in area to a circle, whence it came to be called quadratrix. It is no objection that Proclus writes of the “quadratrix of Hippias,” for we regularly speak of Dinostratus’ brother Menaechmus as discovering the parabola and hyperbola, although these terms were not employed until Apollonius; nor is there any significance in the plural “quadratrices.” It is a more serious objection that Proclus (Friedlein, ed., 356.11) says that different mathematicians have been accustomed to discourse about curves, showing the special property of each kind, as “Hippias with the quadratrices,” for this suggests that Hippias may have written a whole treatise on such curves, and he could hardly have failed to omit the circle-squaring aspect; against this may be set the fact that the angle-dividing property of the curve is more fundamental than its circle-squaring property. It is also odd that Proclus does not mention the name of Dinostratus in connection with the quadratrix; nor does Iamblichus, as quoted by Simplicius (On the Categories of Aristotle, 7; Kalbfleisch, ed., 192.15–25), who writes of the quadrature of the circle as having been effected by the spiral of Archimedes, the quadratrix of Nicomedes, the “sister of the cochloid” invented by Apollonius, and a curve arising from double motion found by Carpus.
Despite all these difficulties, posterity has firmly associated the name of Dinostratus with the quadrature of the circle by means of the quadratrix.
Pappus, IV.30 (Hultsch, ed., 252.5–25), describes how the curve is formed. Let ABCD be a square and BED a quadrant of a circle with center A. If the radius of the circle moves uniformly from AB to AD and in the same time the line BC moves, parallel to its original
position, from BC to AD, then at any given time the intersection of the moving radius and the moving straight line will determine a point F. The path traced by F is the quadratrix. If G is the point where it meets AD, it can be shown by reductio per impossibile (Pappus, IV.31–32; Hultsch, ed., 256.4-258.11) that
arc BED:AB = AB:AG.
This gives the circumference of the circle, the area of which may be deduced by using the proposition, later proved by Archimedes, that the area of a circle is equal to a right triangle in which the base is equal to the circumference and the perpendicular to the radius. If Dinostratus rectified the circle in the manner of Pappus’ proof, it is one of the earliest examples in Greek mathematics of the indirect proof per impossible so widely employed by Euclid. (Pythagoras before him is said to have used the method to prove the irrationality of and Eudoxus must have used it for his proofs by exhaustion.) It is not out of the question that a mathematician of the Platonic school could have proved Archimedes, Measurement of a Circle, proposition 1, which is also proved per impossible, but he may only have suspected its truth without a rigorous proof.
According to Pappus, IV.31 (Hultsch, ed., 252.26–256.3), Sporus was displeased with the quadrature because the very thing that the construction was designed to achieve was assumed in the hypothesis. If G is known, the circle can indeed be rectified and thence squared, but Sporus asks two questions: How is it possible to make the two points moving from B reach their destinations at the same time unless we first know the ratio of the straight line AB to the circumference BED? Since in the limit the radius and the moving line do not intersect but coincide, how can G be found without knowing the ratio of the circumference to the straight line? Pappus endorsed these criticisms. Most modern mathematicians have agreed that the second is valid, for G can be found only by closer and closer approximation, but some, such as Hultsch, have thought that modern instrument makers would have no difficulty in making the moving radius and the moving straight line reach AD together. It is difficult, however, as Heath argues, to see how this could be done without, at some point, a conversion of circular into rectilinear motion, which assumes a knowledge of the thing sought. Both objections would therefore seem to be valid.
BIBLIOGRAPHY
For further reading see the following, listed chronologically: C. A. Bretschneider, Die Geometric und dieGeometer von Euklides (Leipzig, 1870), pp. 95–96, 153–155; Paul Tannery, “Pour l’histoire des lignes et des surfaces courbes dans l’antiquityé,” in Bulletin des sciences mathématiques et astronomiques, 2nd ser., 7, pt. 1 (1883), 278–284; G. J. Allman, Greek Geometry From Thales to Euclid (Dublin, 1889), pp. 180–193; Gino Loria, Le scienze esatte nell’antica Grecia, 2nd ed., (Milan, 1914), pp.160–164; T. L. Heath, A History of Greek Mathematics, I (Oxford, 1921), 225–230; Ivor Thomas, Selections Illustrating the History of Greek Mathematics, I (London-Cambridge, Mass., 1939), 334–347; B. L. van der Waerden, Science Awakening (Groningen, 1954), pp. 191–193; and Robert Böker, in Der kleine Pauly, I (Stuttgart, 1964), cols. 1429–1431.
[1]
### Sources
[1] "Encyclopedia.com" by Ivor Bulmer-Thomas
### Our Mobile Application
Check out Our Mobile Application "Ancient Greece Reloaded" | 1,813 | 7,159 | {"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} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.965262 |
http://physics.stackexchange.com/questions/91628/really-nothing-special-when-falling-into-a-black-hole | 1,469,500,784,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824570.25/warc/CC-MAIN-20160723071024-00020-ip-10-185-27-174.ec2.internal.warc.gz | 193,930,411 | 19,611 | # Really nothing special when falling into a black hole?
It has been said time and again, that an observer who falls into a black hole will not notice anything special. Is this really true?
There is of course the problem with the tidal forces, but I assume that these can be lowered beyond any limit by making the black hole big enough, right?
But what if I take a cigar-shaped spaceship and let it free-fall into a black hole? There will be a moment, where the head is already inside, but the tail is still outside. This of course requires some increase in gravity along the axis of the spaceship, but this increase can be so small, that an observer won't have to worry about it, but still the event horizon will be inside the spaceship.
Now the obeserver could send a light beam from the head to the tail of his spaceship and have it reflected there. Since the observer is in free fall and tidal forces are minimal, I would expect that the beam would just bounce back. This would be in line with the statement, that he does not observe anything special.
However that cannot be true, because he has just sent a beam of light across the event horizon.
-
If I remember correctly the bigger the black hole is, the tidal forces that the body feels are weaker, and the smaller it is they are stronger, but in all reality if you are falling in the black hole you are doomed one way or another :D – dingo_d Dec 28 '13 at 22:43
– Rob Jeffries Jul 8 at 5:35
The "nothing special" is a local property and applies only to those who are small enough. In extreme situations like black holes, the usual allowed area of locality becomes much smaller due to the extreme curvature. A cigar-shaped spaceship is not local as it spans some radial distance.
In addition, from an outside observer's point of view, there is never a point when the spaceship is straddling the horizon. That is because from an outsider's POV, infalling material gets "stuck" to the horizon in a black hole.
-
Are you saying, that nothing straddling the even horizon can be small enough to be considered "local"? If so, then there is something special for an observer free-falling into a black hole! – Martin Drautzburg Dec 28 '13 at 22:21
@MartinDrautzburg Well, the other thing is that from the observer's point of view, the true event horizon always lies below him. When he passes the externally visible event horizon, nothing special happens. – Manishearth Dec 28 '13 at 22:26
I didn't know that for the falling observer the event horizon is always below him. That shomewhat shoots this thought experiment, doesn't it? Mind elaborating a bit and posting this as an answer? – Martin Drautzburg Dec 28 '13 at 23:17
@MartinDrautzburg from here "In terms of visual appearance, observers who fall into the hole perceive the black region constituting the horizon as lying at some apparent distance below them, and never experience crossing this visual horizon". The falling observer does cross a "point of no return", but that point is nothing special otherwise. But from their point of view (moving frame), light particles ahead of them have a different point of no return. – Manishearth Dec 28 '13 at 23:20
I'm a bit fuzzy on this though, it would be best if someone more experienced answered :) – Manishearth Dec 28 '13 at 23:21
To observe "nothing special" near a black hole, you would have to be staring very intently away from it. If you could see it as more than a "missing" point source, you could see the entire universe wrapped around it in perfect Einstein rings. The closer you approached the black hole, the more it would intrude into your remaining field of vision. Around 1.5 Schwarzschild radii, the only stable orbit available to light is a perfect circle, meaning the rest of the universe would appear to be piling up onto a perfectly flat infinite nothing. Below the photon sphere, there are no stable orbits. The black hole would be clearly visible as a gigantic missing thing curling around you — only light from a shrinking circle above could reach you.
Robert Nemiroff has simulated the effects of some high-gravity environments:
Virtual Trips to Black Holes and Neutron Stars
-
Just a hunch, but I am guessing that the time it would take for the light beam to reach the tail of the ship would be longer than it takes for the tail to pass through the event horizon.
Keep in mind that "nothing special" is a massive understatement. It's not like nothing is happening when you are that close to a black hole :) But, relative to what goes on outside of the horizon, nothing else happens (at first).
- | 1,029 | 4,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.959523 |
http://acscihotseat.org/index.php?qa=1160&qa_1=economic-interpretation-of-the-fair-forward-zcb-price&show=1161 | 1,553,417,446,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203409.36/warc/CC-MAIN-20190324083551-20190324105551-00050.warc.gz | 5,682,604 | 7,184 | # Economic Interpretation of the Fair Forward ZCB Price
66 views
The fair forward ZCB price is given as $$P_{tTS}=\frac{P_{tS}}{P_{tT}}$$ and there's a bullet point in the slides saying "economically, the bond forward price allows one (at $$t$$) to lock into a price for $$P_{TS}$$ (at $$T$$)".
The ZCB forward is a contract which allows one to lock into a price at time $$t$$ for the bond at time $$T$$. We can see this in the formula for the fair forward price. The fair forward price is the price one will pay for $$P_{TS}$$ at $$T$$, however it can be calculated at time $$t$$ since both $$P_{tT}$$ and $$P_{tS}$$ are measurable at time $$t$$. One therefore 'locks in' to this future price that one will pay for the bond by agreeing at time $$t$$ to pay $$P_{tTS}$$ for the bond at time $$T$$. | 234 | 799 | {"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} | 2.75 | 3 | CC-MAIN-2019-13 | longest | en | 0.92851 |
https://www.unitsconverters.com/en/Acre-To-Squaremetre/Unittounit-308-302 | 1,627,260,303,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00613.warc.gz | 1,108,553,747 | 36,373 | Formula Used
1 Acre = 4046.85611888692 Square Meter
1 Square Meter = 1 Square Meter
1 Acre = 4046.85611888692 Square Meter
## Acres to Square Meters Conversion
ac stands for acres and m² stands for square meters. The formula used in acres to square meters conversion is 1 Acre = 4046.85611888692 Square Meter. In other words, 1 acre is 4047 times bigger than a square meter. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
## Convert Acre to Square Meter
How to convert acre to square meter? In the area measurement, first choose acre from the left dropdown and square meter from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from square meter to acre? You can check our square meter to acre converter.
How many Square Meter is 1 Acre?
1 Acre is equal to 4046.85611888692 Square Meter. 1 Acre is 4046.85611888692 times Bigger than 1 Square Meter.
How many Square Kilometer is 1 Acre?
1 Acre is equal to 0.00404685611888692 Square Kilometer. 1 Acre is 247.1054 times Smaller than 1 Square Kilometer.
How many Square Centimeter is 1 Acre?
1 Acre is equal to 40468561.1888692 Square Centimeter. 1 Acre is 40468561.1888692 times Bigger than 1 Square Centimeter.
How many Square Millimeter is 1 Acre?
1 Acre is equal to 4046856118.88692 Square Millimeter. 1 Acre is 4046856118.88692 times Bigger than 1 Square Millimeter.
## Acres to Square Meters Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like area finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like ac to m² through multiplicative conversion factors. When you are converting area, you need a Acres to Square Meters converter that is elaborate and still easy to use. Converting Acre to Square Meter is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Acre to Square Meter conversion along with a table representing the entire conversion.
Let Others Know | 602 | 2,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-31 | latest | en | 0.870785 |
http://www.expertsmind.com/questions/write-a-matlab-program-30162664.aspx | 1,670,170,768,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710974.36/warc/CC-MAIN-20221204140455-20221204170455-00656.warc.gz | 61,810,232 | 13,227 | ## Write a matlab program, MATLAB Programming
Assignment Help:
Write a MATLAB program that calculates the arithmetic mean, the geometric mean, and the root-mean-square average for a given set of values.
Your program must use 3 functions to calculate the 3 required means. The output should be formatted as follows:
Statistical Package
arithmetic mean = x.xxxxx
geometric mean = x.xxxxx
RMS average = x.xxxxx
Background and Discussions:
N number of values
Test your program on the following values: 1.1, 3.3, 3.00, 2.22, 2.00, 2.72, 4.00, 4.62 and 5.37. Your main program calls 3 functions.
The data should be read by the main program from a text file, stored in an array and then passed to the functions.
The results and any other output should be printed by the main program. Notice how the output results are aligned. Also notice that the results are printed accurate to 5 decimal places.
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#### Rank order clustering, I want the code for rank order clustering
I want the code for rank order clustering | 655 | 2,924 | {"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} | 2.828125 | 3 | CC-MAIN-2022-49 | latest | en | 0.851127 |
http://astronomy.stackexchange.com/questions/tagged/orbit?page=4&sort=newest&pagesize=15 | 1,469,521,308,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824757.62/warc/CC-MAIN-20160723071024-00200-ip-10-185-27-174.ec2.internal.warc.gz | 15,791,203 | 25,806 | # Tagged Questions
Questions regarding an object 'falling around' another object, due to a combination of gravity and momentum.
252 views
### What are the azimuths of the planets' orbits?
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Researching a sci-fi story involving a "twin-planet" system. Is such an arrangement (however unlikely) physically possible? And if so, must they be tidally locked with each other, similar to the moon, ...
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Why Earth's perihelion occurs 3rd January rather than 1st January? Is there any effort to correct this discrepancy?
100 views
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659 views
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While browsing through Physics SE, I noticed a question about satellites in geostationary orbit (unrelated to the one I'm asking here), and for a moment I interpreted it as referring to natural ...
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211 views
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### Asteroids turning out to be Meteorites [closed]
What causes an asteroid to loose control over it's orbit and enter in to another objects atmosphere? | 2,830 | 12,991 | {"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} | 2.578125 | 3 | CC-MAIN-2016-30 | latest | en | 0.927933 |
https://socratic.org/questions/the-sum-of-two-consecutive-integers-is-5-what-are-the-integers | 1,575,869,221,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517557.43/warc/CC-MAIN-20191209041847-20191209065847-00327.warc.gz | 544,237,650 | 6,083 | The sum of two consecutive integers is 5, what are the integers?
$n + \left(n + 1\right) = 5$ simplified to $2 n = 4$ divide by 2 gives $n = 2 \mathmr{and} \left(n + 1\right) = 3$ | 69 | 180 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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} | 3.671875 | 4 | CC-MAIN-2019-51 | longest | en | 0.902173 |
https://oeis.org/A002561 | 1,696,477,006,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00386.warc.gz | 472,785,908 | 4,367 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A002561 a(n) = n^5 + 1. 6
0, 1, 2, 33, 244, 1025, 3126, 7777, 16808, 32769, 59050, 100001, 161052, 248833, 371294, 537825, 759376, 1048577, 1419858, 1889569, 2476100, 3200001, 4084102, 5153633, 6436344, 7962625, 9765626, 11881377 (list; graph; refs; listen; history; text; internal format)
OFFSET -1,3 LINKS Vincenzo Librandi, Table of n, a(n) for n = -1..1000 Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1). FORMULA a(n) = A000584(n) + 1, for n >= 0. a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), n > 4. - Harvey P. Dale, Aug 13 2013 G.f.: ( 1 - 4*x + 36*x^2 + 56*x^3 + 31*x^4 ) / (x-1)^6. - R. J. Mathar, Jul 28 2014 E.g.f.: (1 + x + 15*x^2 + 25*x^3 + 10*x^4 + x^5)*exp(x). - G. C. Greubel, Oct 24 2018 Sum_{n>=1} 1/a(n) = 1/2 + Sum_{n>=1} (-1)^(n+1) * (zeta(5*n) - 1) = 0.5359628431... - Amiram Eldar, Nov 06 2020 MATHEMATICA Table[n^5+1, {n, -1, 40}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 2, 33, 244, 1025}, 40] (* Harvey P. Dale, Aug 13 2013 *) PROG (Magma) [n^5+1: n in [-1..30]]; // Vincenzo Librandi, Jun 07 2013 (PARI) a(n)=n^5+1 \\ Charles R Greathouse IV, Oct 07 2015 (GAP) List([1..30], n->n^5+1); # Muniru A Asiru, Oct 23 2018 (Sage) [n^5 + 1 for n in (-1..30)] # G. C. Greubel, Nov 20 2018 CROSSREFS Cf. A000584. Sequence in context: A362538 A006558 A228542 * A181547 A030448 A093992 Adjacent sequences: A002558 A002559 A002560 * A002562 A002563 A002564 KEYWORD nonn,easy AUTHOR N. J. A. Sloane STATUS approved
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Last modified October 4 23:18 EDT 2023. Contains 365888 sequences. (Running on oeis4.) | 868 | 2,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2023-40 | latest | en | 0.528564 |
https://issuu.com/wiley_publishing/docs/r_fd_-_sample | 1,501,048,928,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425766.58/warc/CC-MAIN-20170726042247-20170726062247-00068.warc.gz | 648,763,835 | 26,988 | g Easier! Making Everythin
R Learn to: • Use R for data analysis and processing • Write functions and scripts for repeatable analysis • Create high-quality charts and graphics • Perform statistical analysis and build models
Andrie de Vries Joris Meys
Contents at a Glance Introduction ................................................................ 1
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Part I: R You Ready? ................................................... 7
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Chapter 1: Introducing R: The Big Picture ...................................................................... 9 Chapter 2: Exploring R .................................................................................................... 15 Chapter 3: The Fundamentals of R ................................................................................ 31
Part II: Getting Down to Work in R ............................. 43
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Chapter 4: Getting Started with Arithmetic .................................................................. 45 Chapter 5: Getting Started with Reading and Writing ................................................. 71 Chapter 6: Going on a Date with R ................................................................................. 93 Chapter 7: Working in More Dimensions .................................................................... 103
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Part III: Coding in R ................................................ 137
GH
Chapter 8: Putting the Fun in Functions ..................................................................... 139 Chapter 9: Controlling the Logical Flow ..................................................................... 159 Chapter 10: Debugging Your Code .............................................................................. 179 Chapter 11: Getting Help ............................................................................................... 193
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Part IV: Making the Data Talk .................................. 203
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Chapter 12: Getting Data into and out of R ................................................................. 205 Chapter 13: Manipulating and Processing Data ......................................................... 219 Chapter 14: Summarizing Data ..................................................................................... 253 Chapter 15: Testing Differences and Relations .......................................................... 275
Part V: Working with Graphics ................................. 299 Chapter 16: Using Base Graphics ................................................................................. 301 Chapter 17: Creating Faceted Graphics with Lattice................................................. 317 Chapter 18: Looking At ggplot2 Graphics ................................................................... 333
Part VI: The Part of Tens .......................................... 347 Chapter 19: Ten Things You Can Do in R That You Would’ve Done in Microsoft Excel.............................................................................................. 349 Chapter 20: Ten Tips on Working with Packages ...................................................... 359
Appendix: Installing R and RStudio ........................... 365 Index ...................................................................... 371
Table of Contents Introduction ................................................................. 1 About This Book .............................................................................................. 1 Conventions Used in This Book ..................................................................... 2 What You’re Not to Read ................................................................................ 3 Foolish Assumptions ....................................................................................... 4 How This Book Is Organized .......................................................................... 4 Part I: R You Ready? .............................................................................. 4 Part II: Getting Down to Work in R ....................................................... 5 Part III: Coding in R ................................................................................ 5 Part IV: Making the Data Talk ............................................................... 5 Part V: Working with Graphics ............................................................. 5 Part VI: The Part of Tens ....................................................................... 6 Icons Used in This Book ................................................................................. 6 Where to Go from Here ................................................................................... 6
Part I: R You Ready? .................................................... 7 Chapter 1: Introducing R: The Big Picture . . . . . . . . . . . . . . . . . . . . . . . . .9 Recognizing the Benefits of Using R ............................................................ 10 It comes as free, open-source code ................................................... 10 It runs anywhere .................................................................................. 11 It supports extensions......................................................................... 11 It provides an engaged community ................................................... 11 It connects with other languages ....................................................... 12 Looking At Some of the Unique Features of R............................................ 12 Performing multiple calculations with vectors ................................ 12 Processing more than just statistics ................................................. 13 Running code without a compiler...................................................... 14
Chapter 2: Exploring R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15 Working with a Code Editor ......................................................................... 16 Exploring RGui...................................................................................... 16 Dressing up with RStudio.................................................................... 19 Starting Your First R Session ....................................................................... 22 Saying hello to the world .................................................................... 22 Doing simple math ............................................................................... 22 Using vectors ........................................................................................ 23 Storing and calculating values ........................................................... 23 Talking back to the user...................................................................... 25 Sourcing a Script............................................................................................ 25
x
R For Dummies Navigating the Workspace............................................................................ 28 Manipulating the content of the workspace..................................... 28 Saving your work ................................................................................. 28 Retrieving your work ........................................................................... 29
Chapter 3: The Fundamentals of R. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31 Using the Full Power of Functions ............................................................... 31 Vectorizing your functions ................................................................. 32 Putting the argument in a function .................................................... 33 Making history...................................................................................... 34 Keeping Your Code Readable ...................................................................... 35 Following naming conventions .......................................................... 35 Structuring your code ......................................................................... 38 Adding comments ................................................................................ 40 Getting from Base R to More ........................................................................ 40 Finding packages.................................................................................. 40 Installing packages............................................................................... 41 Loading and unloading packages....................................................... 41
Part II: Getting Down to Work in R .............................. 43 Chapter 4: Getting Started with Arithmetic . . . . . . . . . . . . . . . . . . . . . . .45 Working with Numbers, Infinity, and Missing Values ............................... 45 Doing basic arithmetic ........................................................................ 46 Using mathematical functions ............................................................ 48 Calculating whole vectors .................................................................. 51 To infinity and beyond ........................................................................ 52 Organizing Data in Vectors........................................................................... 54 Discovering the properties of vectors .............................................. 54 Creating vectors ................................................................................... 56 Combining vectors............................................................................... 57 Repeating vectors ................................................................................ 58 Getting Values in and out of Vectors .......................................................... 58 Understanding indexing in R .............................................................. 59 Extracting values from a vector ......................................................... 59 Changing values in a vector................................................................ 60 Working with Logical Vectors ...................................................................... 61 Comparing values ................................................................................ 62 Using logical vectors as indices ......................................................... 63 Combining logical statements ............................................................ 64 Summarizing logical vectors .............................................................. 65 Powering Up Your Math with Vector Functions ........................................ 66 Using arithmetic vector operations................................................... 66 Recycling arguments ........................................................................... 69
Table of Contents Chapter 5: Getting Started with Reading and Writing. . . . . . . . . . . . . .71 Using Character Vectors for Text Data ....................................................... 71 Assigning a value to a character vector............................................ 72 Creating a character vector with more than one element ............. 72 Extracting a subset of a vector .......................................................... 73 Naming the values in your vectors .................................................... 74 Manipulating Text.......................................................................................... 76 String theory: Combining and splitting strings ................................ 76 Sorting text ........................................................................................... 79 Finding text inside text ........................................................................ 80 Substituting text ................................................................................... 83 Revving up with regular expressions ................................................ 84 Factoring in Factors ...................................................................................... 86 Creating a factor................................................................................... 86 Converting a factor .............................................................................. 87 Looking at levels .................................................................................. 89 Distinguishing data types ................................................................... 90 Working with ordered factors ............................................................ 91
Chapter 6: Going on a Date with R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .93 Working with Dates ....................................................................................... 93 Presenting Dates in Different Formats ........................................................ 95 Adding Time Information to Dates .............................................................. 97 Formatting Dates and Times ........................................................................ 98 Performing Operations on Dates and Times .............................................. 99 Addition and subtraction .................................................................... 99 Comparison of dates ......................................................................... 100 Extraction............................................................................................ 101
Chapter 7: Working in More Dimensions. . . . . . . . . . . . . . . . . . . . . . . .103 Adding a Second Dimension....................................................................... 103 Discovering a new dimension .......................................................... 103 Combining vectors into a matrix ..................................................... 106 Using the Indices ......................................................................................... 107 Extracting values from a matrix ....................................................... 108 Replacing values in a matrix............................................................. 110 Naming Matrix Rows and Columns ........................................................... 111 Changing the row and column names ............................................. 111 Using names as indices ..................................................................... 112 Calculating with Matrices ........................................................................... 113 Using standard operations with matrices ...................................... 113 Calculating row and column summaries......................................... 114 Doing matrix arithmetic .................................................................... 115 Adding More Dimensions ........................................................................... 117 Creating an array ............................................................................... 117 Using dimensions to extract values................................................. 118
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R For Dummies Combining Different Types of Values in a Data Frame ........................... 119 Creating a data frame from a matrix ............................................... 119 Creating a data frame from scratch ................................................. 121 Naming variables and observations ................................................ 122 Manipulating Values in a Data Frame........................................................ 123 Extracting variables, observations, and values ............................. 124 Adding observations to a data frame .............................................. 125 Adding variables to a data frame ..................................................... 127 Combining Different Objects in a List ....................................................... 129 Creating a list...................................................................................... 129 Extracting elements from lists ......................................................... 131 Changing the elements in lists ......................................................... 132 Reading the output of str() for lists................................................. 134 Seeing the forest through the trees ................................................. 135
Part III: Coding in R ................................................. 137 Chapter 8: Putting the Fun in Functions . . . . . . . . . . . . . . . . . . . . . . . . .139 Moving from Scripts to Functions ............................................................. 139 Making the script ............................................................................... 140 Transforming the script .................................................................... 140 Using the function .............................................................................. 141 Reducing the number of lines .......................................................... 143 Using Arguments the Smart Way ............................................................... 145 Adding more arguments ................................................................... 145 Conjuring tricks with dots ................................................................ 147 Using functions as arguments .......................................................... 148 Coping with Scoping.................................................................................... 150 Crossing the borders ......................................................................... 150 Using internal functions .................................................................... 152 Dispatching to a Method ............................................................................ 154 Finding the methods behind the function ...................................... 154 Doing it yourself ................................................................................. 156
Chapter 9: Controlling the Logical Flow. . . . . . . . . . . . . . . . . . . . . . . . .159 Making Choices with if Statements ........................................................... 160 Doing Something Else with an if...else Statement .................................... 162 Vectorizing Choices .................................................................................... 163 Looking at the problem ..................................................................... 164 Choosing based on a logical vector................................................. 164 Making Multiple Choices ............................................................................ 166 Chaining if...else statements ............................................................. 166 Switching between possibilities ....................................................... 167 Looping Through Values ............................................................................ 168 Constructing a for loop ..................................................................... 169 Calculating values in a for loop ........................................................ 169
Table of Contents Looping without Loops: Meeting the Apply Family ................................ 171 Looking at the family features .......................................................... 172 Meeting three of the members ......................................................... 173 Applying functions on rows and columns ...................................... 173 Applying functions to listlike objects .............................................. 175
Chapter 10: Debugging Your Code. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .179 Knowing What to Look For ......................................................................... 179 Reading Errors and Warnings .................................................................... 180 Reading error messages .................................................................... 180 Caring about warnings (or not) ....................................................... 181 Going Bug Hunting ....................................................................................... 182 Calculating the logit ........................................................................... 182 Knowing where an error comes from.............................................. 183 Looking inside a function .................................................................. 184 Generating Your Own Messages ................................................................ 187 Creating errors ................................................................................... 187 Creating warnings .............................................................................. 188 Recognizing the Mistakes You’re Sure to Make ....................................... 189 Starting with the wrong data ............................................................ 189 Having your data in the wrong format ............................................ 190
Chapter 11: Getting Help . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .193 Finding Information in the R Help Files .................................................... 193 When you know exactly what you’re looking for........................... 193 When you don’t know exactly what you’re looking for ................ 194 Searching the Web for Help with R ........................................................... 196 Getting Involved in the R Community ....................................................... 197 Using the R mailing lists .................................................................... 197 Discussing R on Stack Overflow and Stack Exchange ................... 198 Tweeting about R ............................................................................... 199 Making a Minimal Reproducible Example ................................................ 199 Creating sample data with random values ..................................... 199 Producing minimal code ................................................................... 201 Providing the necessary information .............................................. 201
Part IV: Making the Data Talk................................... 203 Chapter 12: Getting Data into and out of R. . . . . . . . . . . . . . . . . . . . . . .205 Getting Data into R ...................................................................................... 205 Entering data in the R text editor .................................................... 205 Using the Clipboard to copy and paste........................................... 207 Reading data in CSV files................................................................... 209 Reading data from Excel ................................................................... 211 Working with other data types ........................................................ 212 Getting Your Data out of R ......................................................................... 214
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R For Dummies Working with Files and Folders ................................................................. 215 Understanding the working directory ............................................. 215 Manipulating files ............................................................................... 216
Chapter 13: Manipulating and Processing Data. . . . . . . . . . . . . . . . . .219 Deciding on the Most Appropriate Data Structure ................................. 219 Creating Subsets of Your Data ................................................................... 221 Understanding the three subset operators .................................... 221 Understanding the five ways of specifying the subset.................. 221 Subsetting data frames...................................................................... 222 Adding Calculated Fields to Data .............................................................. 227 Doing arithmetic on columns of a data frame ................................ 227 Using with and within to improve code readability ...................... 227 Creating subgroups or bins of data ................................................. 228 Combining and Merging Data Sets ............................................................ 230 Creating sample data to illustrate merging .................................... 231 Using the merge() function ............................................................... 232 Working with lookup tables.............................................................. 234 Sorting and Ordering Data.......................................................................... 236 Sorting vectors ................................................................................... 237 Sorting data frames............................................................................ 237 Traversing Your Data with the Apply Functions ..................................... 240 Using the apply() function to summarize arrays ........................... 241 Using lapply() and sapply() to traverse a list or data frame........ 242 Using tapply() to create tabular summaries .................................. 243 Getting to Know the Formula Interface..................................................... 245 Whipping Your Data into Shape ................................................................ 246 Understanding data in long and wide format ................................. 247 Getting started with the reshape2 package .................................... 248 Melting data to long format .............................................................. 249 Casting data to wide format ............................................................. 250
Chapter 14: Summarizing Data. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .253 Starting with the Right Data ....................................................................... 253 Using factors or numeric data .......................................................... 254 Counting unique values..................................................................... 254 Preparing the data ............................................................................. 255 Describing Continuous Variables .............................................................. 256 Talking about the center of your data............................................. 256 Describing the variation.................................................................... 257 Checking the quantiles ...................................................................... 257 Describing Categories ................................................................................. 258 Counting appearances....................................................................... 259 Calculating proportions .................................................................... 259 Finding the center .............................................................................. 260 Describing Distributions ............................................................................. 261 Plotting histograms ........................................................................... 261 Using frequencies or densities ......................................................... 262
Table of Contents Describing Multiple Variables .................................................................... 264 Summarizing a complete dataset ..................................................... 265 Plotting quantiles for subgroups ..................................................... 266 Tracking correlations ........................................................................ 268 Working with Tables ................................................................................... 270 Creating a two-way table................................................................... 271 Converting tables to a data frame ................................................... 272 Looking at margins and proportions ............................................... 273
Chapter 15: Testing Differences and Relations . . . . . . . . . . . . . . . . . .275 Taking a Closer Look at Distributions ...................................................... 276 Observing beavers ............................................................................. 276 Testing normality graphically .......................................................... 276 Using quantile plots ........................................................................... 277 Testing normality in a formal way ................................................... 280 Comparing Two Samples ............................................................................ 281 Testing differences ............................................................................ 281 Comparing paired data ..................................................................... 283 Testing Counts and Proportions ............................................................... 284 Checking out proportions ................................................................. 284 Analyzing tables ................................................................................. 286 Extracting test results ....................................................................... 287 Working with Models .................................................................................. 288 Analyzing variances ........................................................................... 288 Evaluating the differences ................................................................ 290 Modeling linear relations .................................................................. 292 Evaluating linear models................................................................... 295 Predicting new values ....................................................................... 297
Part V: Working with Graphics .................................. 299 Chapter 16: Using Base Graphics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .301 Creating Different Types of Plots .............................................................. 301 Getting an overview of plot .............................................................. 301 Adding points and lines to a plot ..................................................... 302 Different plot types ............................................................................ 306 Controlling Plot Options and Arguments ................................................. 308 Adding titles and axis labels ............................................................. 308 Changing plot options ....................................................................... 309 Putting multiple plots on a single page ........................................... 312 Saving Graphics to Image Files .................................................................. 314
Chapter 17: Creating Faceted Graphics with Lattice. . . . . . . . . . . . . .317 Creating a Lattice Plot................................................................................. 318 Loading the lattice package .............................................................. 319 Making a lattice scatterplot .............................................................. 319 Adding trend lines ............................................................................. 320
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R For Dummies Changing Plot Options ................................................................................ 321 Adding titles and labels..................................................................... 322 Changing the font size of titles and labels ...................................... 322 Using themes to modify plot options .............................................. 324 Plotting Different Types .............................................................................. 325 Making a bar chart ............................................................................. 325 Making a box-and-whisker plot ........................................................ 326 Plotting Data in Groups............................................................................... 327 Using data in tall format .................................................................... 327 Creating a chart with groups ............................................................ 329 Adding a key ....................................................................................... 330 Printing and Saving a Lattice Plot.............................................................. 331 Assigning a lattice plot to an object ................................................ 331 Printing a lattice plot in a script ...................................................... 331 Saving a lattice plot to file................................................................. 332
Chapter 18: Looking At ggplot2 Graphics. . . . . . . . . . . . . . . . . . . . . . . .333 Installing and Loading ggplot2 ................................................................... 333 Looking At Layers ........................................................................................ 334 Using Geoms and Stats ............................................................................... 335 Defining what data to use ................................................................. 336 Mapping data to plot aesthetics ...................................................... 336 Getting geoms..................................................................................... 337 Sussing Stats................................................................................................. 340 Adding Facets, Scales, and Options .......................................................... 343 Adding facets ...................................................................................... 344 Changing options ............................................................................... 345 Getting More Information ........................................................................... 346
Part VI: The Part of Tens ........................................... 347 Chapter 19: Ten Things You Can Do in R That You Would’ve Done in Microsoft Excel . . . . . . . . . . . . . . . . . . . . . . . . . . . . .349 Adding Row and Column Totals ................................................................ 349 Formatting Numbers ................................................................................... 350 Sorting Data .................................................................................................. 352 Making Choices with If ................................................................................ 352 Calculating Conditional Totals................................................................... 353 Transposing Columns or Rows .................................................................. 353 Finding Unique or Duplicated Values ....................................................... 354 Working with Lookup Tables ..................................................................... 355 Working with Pivot Tables ......................................................................... 355 Using the Goal Seek and Solver ................................................................. 356
Chapter 20: Ten Tips on Working with Packages . . . . . . . . . . . . . . . .359 Poking Around the Nooks and Crannies of CRAN ................................... 359 Finding Interesting Packages ..................................................................... 360
Table of Contents Installing Packages ...................................................................................... 360 Loading Packages ........................................................................................ 361 Reading the Package Manual and Vignette .............................................. 361 Updating Packages ...................................................................................... 362 Unloading Packages .................................................................................... 362 Forging Ahead with R-Forge ....................................................................... 363 Conducting Installations from BioConductor .......................................... 364 Reading the R Manual ................................................................................. 364
Appendix: Installing R and RStudio ........................... 365 Installing and Configuring R ....................................................................... 365 Installing R .......................................................................................... 365 Configuring R ...................................................................................... 366 Installing and Configuring RStudio ............................................................ 368 Installing RStudio ............................................................................... 368 Configuring RStudio ........................................................................... 368
Index ....................................................................... 371
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R For Dummies
Chapter 1
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Introducing R: The Big Picture In This Chapter
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▶ Discovering the benefits of R
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▶ Identifying some programming concepts that make R special
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ith an estimated worldwide user base of more than 2 million people, the R language has rapidly grown and extended since its origin as an academic demonstration language in the 1990s.
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Some people would argue — and we think they’re right — that R is much more than a statistical programming language. It’s also: ✓ A very powerful tool for all kinds of data processing and manipulation
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✓ A community of programmers, users, academics, and practitioners
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✓ A tool that makes all kinds of publication-quality graphics and data visualizations ✓ A collection of freely distributed add-on packages
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✓ A toolbox with tremendous versatility
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In this chapter, we fill you in on the benefits of R, as well as its unique features and quirks. You can download R at www.r-project.org. This website also provides more information on R and links to the online manuals, mailing lists, conferences and publications.
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Part I: R You Ready?
Tracing the history of R Ross Ihaka and Robert Gentleman developed R as a free software environment for their teaching classes when they were colleagues at the University of Auckland in New Zealand. Because they were both familiar with S, a commercial programming language for statistics, it seemed natural to use similar syntax in their own work. After Ihaka and Gentleman announced their software on the S-news mailing list, several people became interested and started to collaborate with them, notably Martin Mächler. Currently, a group of 18 people has rights to modify the central archive of source code. This group is referred to as the R Development Core Team. In addition, many other people have contributed new code and bug fixes to the project. Here are some milestone dates in the development of R: ✓ Early 1990s: The development of R began. ✓ August 1993: The software was announced on the S-news mailing list. Since then, a set of active R mailing lists has been created.
The web page at www.r-project. org/mail.html provides descriptions of these lists and instructions for subscribing. (For more information, turn to “It provides an engaged community,” later in this chapter.) ✓ June 1995: After some persuasive arguments by Martin Mächler (among others) to make the code available as “free software,” the code was made available under the Free Software Foundation’s GNU General Public License (GPL), Version 2. ✓ Mid-1997: The initial R Development Core Team was formed (although, at the time, it was simply known as the core group). ✓ February 2000: The first version of R, version 1.0.0, was released. Ross Ihaka wrote a comprehensive overview of the development of R. The web page http:// cran.r-project.org/doc/html/ interface98-paper/paper.html provides a fascinating history.
Recognizing the Benefits of Using R Of the many attractive benefits of R, a few stand out: It’s actively maintained, it has good connectivity to various types of data and other systems, and it’s versatile enough to solve problems in many domains. Possibly best of all, it’s available for free, in more than one sense of the word.
It comes as free, open-source code R is available under an open-source license, which means that anyone can download and modify the code. This freedom is often referred to as “free as in speech.” R is also available free of charge — a second kind of freedom, sometimes referred to as “free as in beer.” In practical terms, this means that you can download and use R free of charge.
Chapter 1: Introducing R: The Big Picture Another benefit, albeit slightly more indirect, is that anybody can access the source code, modify it, and improve it. As a result, many excellent programmers have contributed improvements and fixes to the R code. For this reason, R is very stable and reliable. Any freedom also has associated obligations. In the case of R, these obligations are described in the conditions of the license under which it is released: GNU General Public License (GPL), Version 2. The full text of the license is available at www.r-project.org/COPYING. It’s important to stress that the GPL does not pertain to your usage of R. There are no obligations for using the software — the obligations just apply to redistribution. In short, if you change or redistribute the R source code, you have to make those changes available for anybody else to use.
It runs anywhere The R Development Core Team has put a lot of effort into making R available for different types of hardware and software. This means that R is available for Windows, Unix systems (such as Linux), and the Mac.
It supports extensions R itself is a powerful language that performs a wide variety of functions, such as data manipulation, statistical modeling, and graphics. One really big advantage of R, however, is its extensibility. Developers can easily write their own software and distribute it in the form of add-on packages. Because of the relative ease of creating these packages, literally thousands of them exist. In fact, many new (and not-so-new) statistical methods are published with an R package attached.
It provides an engaged community The R user base keeps growing. Many people who use R eventually start helping new users and advocating the use of R in their workplaces and professional circles. Sometimes they also become active on the R mailing lists (www.r-project.org/mail.html) or question-and-answer (Q&A) websites such as Stack Overflow, a programming Q&A website (www. stackoverflow.com/questions/tagged/r) and CrossValidated, a statistics Q&A website (http://stats.stackexchange.com/questions/ tagged/r). In addition to these mailing lists and Q&A websites, R users participate in social networks such as Twitter (www.twitter.com/search/ rstats) and regional R conferences. (See Chapter 11 for more information on R communities.)
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Part I: R You Ready?
It connects with other languages As more and more people moved to R for their analyses, they started trying to combine R with their previous workflows, which led to a whole set of packages for linking R to file systems, databases, and other applications. Many of these packages have since been incorporated into the base installation of R. For example, the R package foreign (http://cran.r-project.org/ web/packages/foreign/index.html) is part of the standard R distribution and enables you to read data from the statistical packages SPSS, SAS, Stata, and others (see Chapter 12). Several add-on packages exist to connect R to database systems, such as the RODBC package, to read from databases using the Open Database Connectivity protocol (ODBC) (http://cran.r-project.org/web/packages/RODBC/ index.html), and the ROracle package, to read Oracle data bases (http:// cran.r-project.org/web/packages/ROracle/index.html). Initially, most of R was based on Fortran and C. Code from these two languages easily could be called from within R. As the community grew, C++, Java, Python, and other popular programming languages got more and more connected with R. Because many statisticians also worked with commercial programs, the R Development Core Team (and others) wrote tools to read data from those programs, including SAS Institute’s SAS and IBM’s SPSS. By now, many of the big commercial packages have add-ons to connect with R. Notably, SPSS has incorporated a link to R for its users, and SAS has numerous protocols that show you how to move data and graphics between the two packages.
Looking At Some of the Unique Features of R R is more than just a domain-specific programming language aimed at statisticians. It has some unique features that make it very powerful, including the notion of vectors, which means that you can make calculations on many values at the same time.
Performing multiple calculations with vectors R is a vector-based language. You can think of a vector as a row or column of numbers or text. The list of numbers {1,2,3,4,5}, for example, could be
Chapter 1: Introducing R: The Big Picture a vector. Unlike most other programming languages, R allows you to apply functions to the whole vector in a single operation without the need for an explicit loop. We’ll illustrate with some real R code. First, we’ll assign the values 1:5 to a vector that we’ll call x: > x <- 1:5 > x [1] 1 2 3 4 5
Next, we’ll add the value 2 to each element in the vector x and print the result: > x + 2 [1] 3 4 5 6 7
You can also add one vector to another. To add the values 6:10 elementwise to x, you do the following: > x + 6:10 [1] 7 9 11 13 15
To do this in most other programming language would require an explicit loop to run through each value of x. This feature of R is extremely powerful because it lets you perform many operations in a single step. In programming languages that aren’t vectorized, you’d have to program a loop to achieve the same result. We introduce the concept of vectors in Chapter 2 and expand on vectors and vectorization in much more depth in Chapter 4.
Processing more than just statistics R was developed by statisticians to make statistical processing easier. This heritage continues, making R a very powerful tool for performing virtually any statistical computation. As R started to expand away from its origins in statistics, many people who would describe themselves as programmers rather than statisticians have become involved with R. The result is that R is now eminently suitable for a wide variety of nonstatistical tasks, including data processing, graphic visualization, and analysis of all sorts. R is being used in the fields of finance, natural language processing, genetics, biology, and market research, to name just a few.
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Part I: R You Ready? R is Turing complete, which means that you can use R alone to program anything you want. (Not every task is easy to program in R, though.) In this book, we assume that you want to find out about R programming, not statistics, although we provide an introduction to statistics in R in Part IV.
Running code without a compiler R is an interpreted language, which means that — contrary to compiled languages like C and Java — you don’t need a compiler to first create a program from your code before you can use it. R interprets the code you provide directly and converts it into lower-level calls to pre-compiled code/functions. In practice, it means that you simply write your code and send it to R, and the code runs, which makes the development cycle easy. This ease of development comes at the cost of speed of code execution, however. The downside of an interpreted language is that the code usually runs slower than compiled code runs. If you have experience in other languages, be aware that R is not C or Java. Although you can use R as a procedural language such as C or an object-oriented language such as Java, R is mostly based on the paradigm of functional programming. As we discuss later in this book, especially in Part III, this characteristic requires a bit of a different mindset. Forget what you know about other languages, and prepare for something completely different.
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In this book, Velleman does three things: Still, the part that I got through almost to proofs section was very insightful and fun. Introduction to the Representation Theory of Algebras. The book begins with the basic concepts of logic and set theory, to familiarize students with the language of mathematics and how it is interpreted. Another chapter on functions.
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Author does not expect much from the reader and begins with very basic concepts and slowly progresses towards quantifiers, then set theory, relation and functions, mathematical induction and finally, infinite sets.
Inside introduction, author gives proof of few theorems in an intuitive way. Later when armed with all the proofing techniques all of those proofs were revisited and reader can clearly Highly recommended for beginners as it helps tremendously in understanding the mathematical rigour. Later when armed with all the proofing techniques all of those proofs were revisited and reader can clearly see the difference in his understanding for reading and writing proofs. All the techniques of proofs except induction are covered in chapter Post that, book introduced other topics like relations and functions and employs proof techniques for proving theorem in these topics.
It was a great way to demonstrate that techniques learned for writing proofs are independent of any area and can be applied anywhere in mathematics. I loved the treatment of proof by contradiction and mathematical induction. Cracking the corresponding exercises was a very rewarding experience.
In many proofs when no approach seems to be working, proof by contradiction comes to the rescue. Similarly power of proof by induction was on display in solving many humongous problems. All exercises were ordered from easy to moderate preparing the reader along the way to learn writing proofs for easier to challenging ones.
Many exercises are built on top of the theorems from earlier exercises. This is a good thing as it helped me in two ways: revising the older chapters and discovering errors in my proofs. There were many exercises asking the reader if the given proof is correct. Many times proof looked correct but turned out wrong because of a conceptual mistake. This helped tremendously in clearing many misconceptions.
In most of the sections, author also explains about how he arrived at a solution which helped in understanding how to approach a problem. Finally in the last chapter author picked up a relatively advanced topic and employs all the proof techniques learned.
In this chapter author does not go into explaining the proof structure but writes in a mathematical rigour so that reader should be able to read those proofs and gets an overall idea about reading and writing proofs by giving more focus to the topic than the proof technique.
One small thing that could have been better is the treatment of empty sets. I got confused while solving many exercises and felt like missing on some concepts regarding empty sets specially while dealing with family of sets. To summarise,.
DESCARGAR LIBRO UN PUENTE HACIA TERABITHIA PDF
## How to Prove It: A Structured Approach
Jun 28, 0vai5 rated it really liked it Highly recommended for beginners as it helps tremendously in understanding the mathematical rigour. Author does not expect much from the reader and begins with very basic concepts and slowly progresses towards quantifiers, then set theory, relation and functions, mathematical induction and finally, infinite sets. Inside introduction, author gives proof of few theorems in an intuitive way. Later when armed with all the proofing techniques all of those proofs were revisited and reader can clearly Highly recommended for beginners as it helps tremendously in understanding the mathematical rigour. Later when armed with all the proofing techniques all of those proofs were revisited and reader can clearly see the difference in his understanding for reading and writing proofs.
EEN PRAKTIJKGERICHTE BENADERING VAN ORGANISATIE EN MANAGEMENT DRUK 6 PDF
. | 865 | 4,367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.952327 |
https://www.studypool.com/discuss/1235641/Draw-a-diagram-and-write-an-equation-to-solve?free | 1,496,136,406,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463614620.98/warc/CC-MAIN-20170530085905-20170530105905-00118.warc.gz | 1,193,690,783 | 14,130 | ##### Draw a diagram and write an equation to solve
Mathematics Tutor: None Selected Time limit: 1 Day
i need to draw a diagram and write a equation to solve:
Julia SAW 5 TIMES AS MANY CARS AS TRUCKS IN A PARKING LOT IF SHE SAW 30 CARS AND TRUCKS TOTAL HOW MANY WERE TRUCKS
Oct 26th, 2015
Let the number of trucks be x, number of cars will be 5x, since total is 30
we have 5x+x = 30
6x = 30, dividing through by 6, we have, x = 30/6 = 5, so number of trucks is 5 trucks
whereas the number of cars is 5 by 5 = 25 cars
Thanks
Oct 26th, 2015
I understand that but it saw to draw a diagram how do u draw this into a diagram
Oct 26th, 2015
...
Oct 26th, 2015
...
Oct 26th, 2015
May 30th, 2017
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https://www.celestialessays.com/papers/?paper_id=11483107 | 1,596,952,516,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738425.43/warc/CC-MAIN-20200809043422-20200809073422-00572.warc.gz | 623,085,954 | 7,439 | ### The Rocky Mountain district sales manager of Rath Publishing Inc.
#### Description
The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales representatives make an average of 37 sales calls per week on professors. Several reps say that this estimate is too low. To investigate, a random sample of 43 sales representatives reveals that the mean number of calls made last week was 41. The standard deviation of the sample is 2.6 calls. Using the 0.010 significance level, can we conclude that the mean number of calls per salesperson per week is more than 37? H0 : μ ≤ 37H1 : μ > 37(1)Compute the value of the test statistic. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.) Value of the test statistic (2)What is your decision regarding H0? (Click to select)Do not rejectReject H0. The mean number of calls is (Click to select)lessgreater than 37 per week.
#### Question
The Rocky Mountain district sales manager of Rath Publishing Inc.
• Written in: 17-Oct-2019
• Paper ID: 11483107
Price: \$ 10
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# Michelson Morley And Gravitational Lensing Question
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What is a straight line locally can be curved in a wider context, it's not a paradox, it's just an arbitrary choice of how you want to define your coordinate system.
Then there is equivalence of straight and curved lines. The question is then, why is there acceleration in local frames? If everything follows curvelinear lines of motions, there should be no acceleration.
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SR deals specifically with local coodinate systems and inertial observers. That's why it's special, not general.
Dave, and Mod, and Alex, and even that freak FreeThinker are missed. Haven't seen hide nor hair of most of the old crew in a long time. On your question: One intuitive way to grasp it is that Gravit
Yes gravity slows the speed of light, it would have to through time dilation and length contraction. The speed of light is always c locally, and then only to interial observers.
Then there is equivalence of straight and curved lines. The question is then, why is there acceleration in local frames? If everything follows curvelinear lines of motions, there should be no acceleration.
Why? Inertial objects follow straight lines in four dimensions, accelerating objects follow curved paths.
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Why? Inertial objects follow straight lines in four dimensions, accelerating objects follow curved paths.
You don't know that. Object follow whatever paths and you do not know whether they are accelerating or moving at constant speed along those lines.
Is the Sun an accelerating object, does it rotate at constant speed, or is it stationary? You do not know. Whatever you say, I can say something different, and my claim is just as good as yours, it's a guess.
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You don't know that. Object follow whatever paths and you do not know whether they are accelerating or moving at constant speed along those lines.
Is the Sun an accelerating object, does it rotate at constant speed, or is it stationary? You do not know. Whatever you say, I can say something different, and my claim is just as good as yours, it's a guess.
You need to define a coordinate system and then you can talk about the paths that objects follow. If that coordinate system includes the gravitational effects of a massive object then it will cause objects to follow curved paths, but each object will be following what is locally a straight path. It's not a contradiction or a paradox, it's just a why of describing gravity that works.
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Yes, in many ways it is about 'space density' I have seen the term 'RSD' related to this subject that stands for 'Relative Space Density', but what it is, is more like 'relative matter density' in that point in space.
Thanks for the answer, I am looking at various space density theories now. The both concepts are interesting.https://www.researchgate.net/publication/228736243_Proving_that_Space_Density_Theory_is_Different_and_More_Complete_than_Spacetime
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Thanks for the answer, I am looking at various space density theories now. The both concepts are interesting.https://www.researchgate.net/publication/228736243_Proving_that_Space_Density_Theory_is_Different_and_More_Complete_than_Spacetime
Thankyou, the reference to RSD was from a paper I read awhile ago where they were talking redshift/distance attributed to the big bang and they were exploring how it is more reasonable to accept the redshift is from gravitational shift (or Einstein shift), of light as a function of the relative masses of the objects observed (galaxies).
So they were referring to 'relative space density' total 'density' of the matter in that region of space due to the amount of matter in that region, so if the RSD was higher to galaxy would look more red.
This goes against the Hubble constant where the redshift is a function of distance, not relative mass (and with Doppler relative velocity). Which puts into question if redshift is a function of distance, and if it is not, then no big bang! (which I expect is the case).
I read that abstract, I'm not sure what 'spacetime theory' is, in my model of relativity time (the length of time) is an emergent property of space length, as is the length of space is the length of time. (otherwise c would not be constant).
If my space length model, matter makes space longer (and everything in that space bigger), that means more matter present in the system the lower the space density (it's longer so lower density).
Here is something to think about in regards to density or space and matter, think about the various states of matter can only exist in that state in a narrow range of space length.
Low mass objects has shorter length and can have binding forces able to makes solids, earth exists in this region.
Higher mass and you may only be able to have weakly bound liquids and gases, these would be gas giants, they have enough mass that the local space density is too high for the length of the binding forces for solids.
Step up to lots more mass, and you get the Sun, very low density (1/3 the density of the earth), so much mass that now not even electrons can bind and you have the state of a plasma.
BLACK HOLES: Do they follow this trend? If the RSD so low that neutrons and protons can no longer stay bound? Does that make active galactic nucelli mini matter to energy to matter big bangs?
The big shock from the observation of M-87 was how freaking HUGE it is, far larger than expected, not the compact massive object but the HUGE massive object, because all that mass makes for very long space!
This is all the e=mc^2 matter (energy into space length), to energy (space length to energy)… This is a balance, black holes when large enough convert matter to energy by making space length shorter (extracting the energy from the length of space itself, because that is what energy as matter is). then creating high energy matter (jets), that is the creation of new matter.
It makes a lot of sense to me at least !!
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Thankyou, the reference to RSD was from a paper I read awhile ago where they were talking redshift/distance attributed to the big bang and they were exploring how it is more reasonable to accept the redshift is from gravitational shift (or Einstein shift), of light as a function of the relative masses of the objects observed (galaxies).
So they were referring to 'relative space density' total 'density' of the matter in that region of space due to the amount of matter in that region, so if the RSD was higher to galaxy would look more red.
This goes against the Hubble constant where the redshift is a function of distance, not relative mass (and with Doppler relative velocity). Which puts into question if redshift is a function of distance, and if it is not, then no big bang! (which I expect is the case).
I read that abstract, I'm not sure what 'spacetime theory' is, in my model of relativity time (the length of time) is an emergent property of space length, as is the length of space is the length of time. (otherwise c would not be constant).
If my space length model, matter makes space longer (and everything in that space bigger), that means more matter present in the system the lower the space density (it's longer so lower density).
Here is something to think about in regards to density or space and matter, think about the various states of matter can only exist in that state in a narrow range of space length.
Low mass objects has shorter length and can have binding forces able to makes solids, earth exists in this region.
Higher mass and you may only be able to have weakly bound liquids and gases, these would be gas giants, they have enough mass that the local space density is too high for the length of the binding forces for solids.
Step up to lots more mass, and you get the Sun, very low density (1/3 the density of the earth), so much mass that now not even electrons can bind and you have the state of a plasma.
BLACK HOLES: Do they follow this trend? If the RSD so low that neutrons and protons can no longer stay bound? Does that make active galactic nucelli mini matter to energy to matter big bangs?
The big shock from the observation of M-87 was how freaking HUGE it is, far larger than expected, not the compact massive object but the HUGE massive object, because all that mass makes for very long space!
This is all the e=mc^2 matter (energy into space length), to energy (space length to energy)… This is a balance, black holes when large enough convert matter to energy by making space length shorter (extracting the energy from the length of space itself, because that is what energy as matter is). then creating high energy matter (jets), that is the creation of new matter.
It makes a lot of sense to me at least !!
I still only have the abstract from research gate it is messing me about for some reason.
I am interested in any theory not involving big bangs as origins of the universe, although I suspect inflation could produce matter and cause a big bang.
The accepted view is redshift is a function of distance, the more red shifted the further away they are due to space expanding between objects. Explaining the Hubble not so constant constant
If light is leaving a gravitational field it will lose energy and so be red shifted.
Are you suggesting galaxies at the limit of the visible horizon have much more mass than previously thought?
How do you balance red shifted galaxies with huge blue shifted ones like Andromeda. Would you agree its still coming this way?
I am guessing you think Hoyle was on the ball, how does your view of space deal with creation of matter? Why isnt new matter being created today.?
Does space have a definable shape/limit in your model?
The length of space (density) varying around objects could perhaps be equated to quantum inflow theory, by Reg Cahill.
and a whole heap of other questions ???????????????????????
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I still only have the abstract from research gate it is messing me about for some reason.
I am interested in any theory not involving big bangs as origins of the universe, although I suspect inflation could produce matter and cause a big bang.
The accepted view is redshift is a function of distance, the more red shifted the further away they are due to space expanding between objects. Explaining the Hubble not so constant constant
If light is leaving a gravitational field it will lose energy and so be red shifted.
Are you suggesting galaxies at the limit of the visible horizon have much more mass than previously thought?
How do you balance red shifted galaxies with huge blue shifted ones like Andromeda. Would you agree its still coming this way?
I am guessing you think Hoyle was on the ball, how does your view of space deal with creation of matter? Why isnt new matter being created today.?
Does space have a definable shape/limit in your model?
The length of space (density) varying around objects could perhaps be equated to quantum inflow theory, by Reg Cahill.
and a whole heap of other questions ???????????????????????
The abstract messed with me too, some idea's and terms used there that don't sit too well yet.. Anyway
I liked your post because you are asking the same questions I have been asking (mostly), and I have some good news for you :)
I am interested in any theory not involving big bangs as origins of the universe, although I suspect inflation could produce matter and cause a big bang.
First, I'm not really interested in what did happen or how the universe got here, if it is not a big bang 13.8 BY ago, then I think the question is beyond an answer, I think it IS here, it has probably always been here, and it has always been EVERYWHERE, it's not contained, there is no 'beyond' there is NOTHING BUT Universe. (and it is history and not science). How or why it exists I will leave up to philosophers, religions and wild speculation. We don't know, and I don't think we can know that..
Inflation is one of those things to 'fix' some other problem with the BB models, and of course it created many more problems.
The accepted view is redshift is a function of distance, the more red shifted the further away they are due to space expanding between objects. Explaining the Hubble not so constant constant
The accepted view IS correct, there IS a redshift with distance relationship that Hubble observed, that is a fact.... BUT, what is not a fact at all is that this redshift is the consequence of 'space expanding between objects'. That's the problem.
Space expanding between objects would not produce a redshift! There is no relative velocity so it is not Doppler shift, so if space expands between when the photon leaves the source and arrives at the destination that will only produce a longer path so a greater transit time, but no shift in light frequency.
Light is 'uncoupled' to space, as far as light is concerned space (and time) do not exist, as such with no coupling or interaction with space the light cannot 'give up' it's energy to space or have that energy taken away from it by gravity.
If light is leaving a gravitational field it will lose energy and so be red shifted.
This is for me what is so beautiful about understanding that matter (gravity) gives space a fundamental length property, and we live in a universe where we are that is the SUM of all the space length contributions of all the masses as a function of the mass and the distance from that mass.
(you could also say that about 'gravity', we exist in all the gravity influences of all the mass as a function of the mass and distance to that mass).
SO instead of thinking about 'light losing energy' to be redshifted, think about it as relative size or relative length or space between the object and the observer.
So what we want to achieve here is a redshift with distance relationship using a known, tested and somewhat understood mechanism.
We only have two to choose from Doppler shift and Gravitational shift, (BB inflation/expansion shift is unknown and untested and no understood).
WE can rule out Doppler shift it is a real shift but it is not a redshift/distance function, it is relative velocity. So not Doppler shift (although it is often stated as such, but people who should know better IMO).
So that leave the only other KNOWN physics to account for a redshift/distance relationship. (I am discounting for now diffusion and particles that give up a red sunset because that is a filtering process not a shift process).
That know effect is Gravitational shift or Einstein shift, it's the same thing just called two different things.
Einstein Shift produces a redshift with distance effect in a homogeneous and isotropic universe that is BIDIRECTIONAL. That is in BOTH directions the further away you are the redder the object will appear.
Before I describe a simple model to explain this, IF the universe was perfectly homogeneous and isotropic the Hubble constant would be absolutely constant and the plot of distance to redshift would be perfect, as you stated, it is a complete mess, there data is all over the place, and the more precisely we look the worse it gets, this is a part of the 'crisis in cosmology', the big bang model is falling apart. (I'm starting to rant !!!)..
OK, lets have a very simple universe and look at Einstein Shift... JUST THE EARTH, that is the universe right now and here we are on it. No other mass except the earth and somehow it is just like this earth in terms of space and time length.
The earth is matter and a ball, the center of the earth is the points where there is the most amount of mass as a function of distance, that point is the point of longest space length (highest gravity). That point is ALL the way down, if you are there every direction is up, every direction puts you into shorter space.
The length of space and time on the surface of the earth is a little bit shorter, because you are further away from the points of most mass to distance.
The amount of length change is very small something like 2.5 and 4.5billion, so it is a slight slop straight line.
Now we put a spaceship some distance away from the earth (everything is stationary here, so disregard orbits of velocity or doppler shift).
What the ship is at the surface of the earth it sees not redshift the ship is in the same length space.
Now the ship is 10,000Km away from the earth, the ship is in shorter space, you are in the ship and you see that the earth is REDSHIFTED...
You go 20,000km away, and the earth is even more redshifted, but this is NOT light losing energy because of it leaving a gravitational field.
You are observing the earth getting redder because you are in shorter space away from the earth and you are in a higher energy state, relative to you the earth is at a lower energy, it is in a higher 'gravity well', or for me, you are shorter (smaller) looking at something relatively longer (bigger).
You are in shorter space looking into longer space:
OK, this only gives us a single direction redshift to distance relationship, the ship will see the earth has redshifted, but the earth will see the ship as more blue when it is further away.
This is because the only significant source of matter to give the length (gravity) property to space is the earth, the ship is low mass and contributes far less.
I really have to learn how to post diagrams here, because it is so much simpler to diagram out.
Now replace the ship with a second identical earth, then you will so no shift at all to an observer on the NEAR side of earth, the contribution of space length (gravity) will be equal, you will both be at the same 'level' in the gravity 'well'. No shift at all regardless of the distance.
But if you go to the FAR side of one of the earths and look at the other earth, it will be redder, for either observer on either earth bidirectionally as a function of the distance between the two earths.
This is because there is an asymmetry of matter between each observer, if you are on earth 2 on the far side the mass of your earth is contributing more to earths 1 length and at the point you at. You are in shorter space relative to the distant object, because your local objects contribute more to making the distant object larger..
It's hard to explain, but it is trivial to show with a simple diagram example. So that is how you get redshift/distance in both directions without a big bang..
Are you suggesting galaxies at the limit of the visible horizon have much more mass than previously thought?
No, not at all, in a perfectly smooth and uniform universe with evenly distributed equal masses evenly spread out then the redshift distance Einstein shift would be perfect and useful for measuring distance. BUT IT'S NOT. It's pretty blotchy, and things are different sizes and masses, and on top of that things are moving about, frankly it's a mess!
That's why if you look at Hubble's paper and his Hubble constant plots they are all over the place, it's a VERY poor data fit, Even Hubble did not think it was from expansion.
So a more distant galaxy will be redder, but so will a more massive galaxy.
How do you balance red shifted galaxies with huge blue shifted ones like Andromeda. Would you agree its still coming this way?
Exactly, This is the same effect as I explained above, there is a redshift with distance because of the 'shared' masses both locally and to the observed object.
The local galaxies contribute more to our length so we are relative to them 'bigger' (in longer space) and locally they are relative to use 'smaller', Light from shorter space is blue shifted.
So Andromeda might be moving towards us and it's doppler shift, or it might not be and it's just local Einstein shift..
Light from longer space is redshifted.
I am guessing you think Hoyle was on the ball, how does your view of space deal with creation of matter? Why isnt new matter being created today.?
Not sure that Hoyle says LOL...
I think active Galactic nucelli and probably quasars are matter/energy/matter recyclers, breaking up matter into some fundaments energy/matter state (with very long space length), and exchanging that energy for space length (distributing energy over space and time). I think new matter is being created all the time.
Galaxies are universes, they create matter and space (for that matter), that gives the universe it's length property.
Does space have a definable shape/limit in your model?
No, it has no shape, it cannot have a shape and it has no limit, it also does not have a size!, there cannot be a multiverse, IF you created a universe that has space in it, it WILL become THIS universe, that's why I see galaxies are Universes, self contained except they share 'space'.
Even for our universe that is only the earth, the further you go away from the earth the shorter the length of space, eventually all you will really have is your own length from your own mass, but the length contribution from the earth is never be zero. 10,000Km on the surface of earth might be 1mm where you are very far away from earth, your speedo might say you are going 10,000km/h but I am on earth seeing you go 1mm per hour.
Single earth universe, the earth is the only mass contributing to space length, now start removing mass from that earth, as you do the length of space to become shorter it will approach zero length as you approach zero mass.
You take out that last unit of lass, you have a universe in your lab that is infinite in extent if you were in it you could go in any direction at the speed of light,, forever.
But the length of space is zero so the distance between every points is zero, an event will occur everywhere all at once. Any energy is infinite (dissipated over zero area in zero time). This is zero length infinite energy no mass universe.
Put a unit of mass in that zero length universe, and you have NON-ZERO length universe, finally you get away from infinite energy, you have some place some space to put things like matter and energy.. Now we live in a universe that is a balance between matter and energy moderated by space length. (that gives us space based time or 'spacetime).
Now we are all happy with places to go, and things to do and time to do it.. A happy universe...
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deleted - posted to wrong thread
Edited by AnssiH
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Dave, and Mod, and Alex, and even that freak FreeThinker are missed. Haven't seen hide nor hair of most of the old crew in a long time.
On your question: One intuitive way to grasp it is that Gravity can pack more "space per area."
Light still travels at C, but there is more distance-per-area for it to travel when there's more stress on space itself so it takes more time for it to get past that area.
That's drastically over-simplified, but it's an intuitive way to grasp at the invisible forces at play.
GAHD is correct and anyone that disagrees with that statement is wrong.
Edited by VictorMedvil
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Actually I suppose that would be gravity also packs less time per volume, time moves slower the more gravity there is. More space, less time.
After revisiting this, I thought it would be useful to point out that "time" is measured in frequency. If there is more "space" in a give area/volume, that frequency is "slowed" because it takes longer for it to move across that "more-space area/volume."
A lot of time dilation can be explained with frequency changes in that physical way. I'm still waiting on a recovery of a long-term atomic clock from orbital-frame to see if the decay rates also change along with the frequency-counter adjustments we use. First principles really matter in this kind of interaction. The nature of dilation being from "momentum-over-area frequency change" or from actual "entropy acts differently in these conditions" is an important FP that changes a lot of the resulting (in/de)ductions.
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After revisiting this, I thought it would be useful to point out that "time" is measured in frequency. If there is more "space" in a give area/volume, that frequency is "slowed" because it takes longer for it to move across that "more-space area/volume."
A lot of time dilation can be explained with frequency changes in that physical way. I'm still waiting on a recovery of a long-term atomic clock from orbital-frame to see if the decay rates also change along with the frequency-counter adjustments we use. First principles really matter in this kind of interaction. The nature of dilation being from "momentum-over-area frequency change" or from actual "entropy acts differently in these conditions" is an important FP that changes a lot of the resulting (in/de)ductions.
This is really interesting, it's a subject I have had stuck in my mind for the past 4 or 5 years. (specifically this subject or what space and time is from first principles)
And how space and time work in terms of gravity and relativity. (I want to do a full thread on this but I would like to post some diagrams with it first, and it's hard to word..)
I thought it would be useful to point out that "time" is measured in frequency. If there is more "space" in a give area/volume, that frequency is "slowed" because it takes longer for it to move across that "more-space area/volume."
What I had a lot of trouble with, and others do as well, is the idea that time is measured in frequency. Someone asked on reddit or somewhere if we could take advantage of the 'higher frequency of altitude' and would a computer produce output faster in 'higher frequency space'.
So would a computer on top of Everest produce results faster than one in the center of the earth, where time is a 'lower frequency' ?
You put the same question with nuclear decay, would a particle decay faster in higher frequency time?
This is a testing of the 'twins paradox', you have two people of identical age, send on orbiting a black hole for some time and when they come back they have aged much less than the person on earth.
This is where I think the mainstream treatment of relativity at best paints a confusing picture/model.
The 4 Dimensional geometric model of relativity treats time as a frequency.
You run an experiment with two clocks, you end up with two different numbers on each clock, one clock might say 5 and the other 10 (seconds or whatever).
You started and stopped the clocks at the same time in the same location, so how do you calculate and justify the different readings?
In the geometric treatment, you draw a curve (a worldline, or geodesic or a 'like path'). In this case you curve the 10 so that the start and finish match that of the 5.
Curved space and time.
Frequency is not fundamental! it's a function of two different terms, the number of somethings over the number of something else. (cycles over seconds or whatever).
This for me was very confusing, they talk about time slowing down and even stopping with mass or in a black hole. So if time slows or stops does that imply infinite frequency?
So, what if what if time is not a frequency or measured in frequency, but is measured as a period 1/f ?
In this case in the experiment where two clocks show 5 and 10 you do not curve the 10 to match the start and end points in a curved geometrical model.
Instead, you understand that time is a period, it has a fundamental length property (or if you like a duration), it is as such a variable and not a constant.
It's locally constant (in your frame of reference) but it is relatively variable.
This is a non-geometrical treatment, you have two flat lines starting and stopping at the same time (and reference frame), but the clock that displays 5 has simply been in time where time is twice as long as the clock that displays 10.
They both could the same (local) time, one counts 5 longer seconds and the other counts 10 shorter seconds.
Time is a variable its length varies from place to place (we observe that from relativity tests).
The speed of light is constant in every reference frame (so regardless of the length of time).
SPEED is the length of some unit of space over the length of some unit of time. So, if the speed of light is constant and the length of time is a variable, then the length of space also needs to be a variable fixed to the length of time.
You could consider this a volume or an area, but what you need to get your head around is that it is not more miles or kilometers or seconds, it's LONGER (or shorter) miles or kilometers or seconds.
AM I CORRECT???
IF my model is correct, I would make a testable prediction that we will not measure any spacetime curvature (we'll I win that one), but also this: The twins paradox is wrong, the twin will come back and will have aged the same amount as the one left behind. AND, the nuclear decay of the atomic clock will be no different that one on earth.
WHY?
Because the same amount of work is being done. The time is longer but the length of space is equally longer, you have more time but you have to cover a greater distance.
Or in shorter time, you cover less distance but do it faster, you use the same amount of energy either way.
Light is a waveLENGTH, it is not a frequency, so a photons LENGTH is a function of how much time it takes to create the photon, this is the length of space and time measurement we see with Gravitational/Einstein shift of light. Bigger photons from bigger space, or smaller photons from shorter space.
IF time was a frequency we would not see blue shifted light from the ISS, we would see a brighter signal from the ISS (a higher frequency of photos, as opposed a frequency of higher photons).
(WOW, WHAT!!)
You would see a higher number of UNSHIFTED photons, not the same number of shifted photons.. (I hope that makes sense)..
(I will stop now).. Thanks for any comments.
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Hey everyone! Where are Dave and Modest and all the dudes from the old forum?
I cannot wrap my head around something. If light path "bends" in presence of gravity, gravitational lensing. This is observed. Then, it follows, the speed of light will change according to coordinate system.
On the other hand, there is Michelson Morley experiment that says the light does not bend, and consequently Einstein said the speed of light is constant.
How do we explain this paradox? Am I wrong or what? Does the speed depend on gravity (coordinate system) or does it not depend on gravity?
Funnily I tend to agree so long as gravity is the aether. Notice that their experiment ruled out a very narrow class.
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Einstein also revised his statement saying that light was only a constant in a vacuum devoid of gravitational fields, hence he came to realize it did indeed spatially variate.
Edited by Dubbelosix
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Einstein also revised his statement saying that light was only a constant in a vacuum devoid of gravitational fields, hence he came to realize it did indeed spatially variate.
And that's the point. So, if the the light accelerates in gravitational field, what is the point of Lorentz transformations and special relativity? Do those have any real application?
I mean, we normalized everything to constant speed of light because it is invariant based on MM experiment, and that is the universal coordinate system in which we use Lorentz transformations, but in real life where light indeed bends, special relativity and its math gymnastics are meaningless. We could've just stayed with our own frame of reference and use Galilean transformations. It's the same thing as tying everything to speed of light which is arbitrary.
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And that's the point. So, if the the light accelerates in gravitational field, what is the point of Lorentz transformations and special relativity? Do those have any real application?
I mean, we normalized everything to constant speed of light because it is invariant based on MM experiment, and that is the universal coordinate system in which we use Lorentz transformations, but in real life where light indeed bends, special relativity and its math gymnastics are meaningless. We could've just stayed with our own frame of reference and use Galilean transformations. It's the same thing as tying everything to speed of light which is arbitrary.
Yes, the constancy of the speed of light is the very foundation of relativity, if that is wrong, you have nothing to work with. Also, there is no such thing as anything 'devoid of a gravitational field' in our universe.
The reason why there is an apparent change in the speed of light in a non-vacuum is due to the longer path the light travels (at c), not the value of c changing.
It's sort of like the Hubble constant, if it is not constant and changes from place to place then you might as well throw out the whole model, because everything stops making sense.
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And that's the point. So, if the the light accelerates in gravitational field, what is the point of Lorentz transformations and special relativity? Do those have any real application?
I mean, we normalized everything to constant speed of light because it is invariant based on MM experiment, and that is the universal coordinate system in which we use Lorentz transformations, but in real life where light indeed bends, special relativity and its math gymnastics are meaningless. We could've just stayed with our own frame of reference and use Galilean transformations. It's the same thing as tying everything to speed of light which is arbitrary.
Well sure, in the sense that it can take light longer to reach a point in spacetime depending on the bend in space and time, but this will still require non Euclidean transformations. We also have to remain aware that no matter what inertial reference frame we take, the peculiarities of relativity states that we should always measure the speed of light as a constant.. But yes, given the right set up, light does vary in the gravitational field.
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https://www.physicsforums.com/threads/coefficient-of-friction-from-a-graph.503538/ | 1,623,895,088,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626465.55/warc/CC-MAIN-20210617011001-20210617041001-00323.warc.gz | 853,664,740 | 15,391 | # Coefficient of friction from a graph
Is it possible to find the coefficient of friction just by looking at a graph?
A velocity-time or acceleration-time graph which models a particle going up a ramp and down a ramp.
You would need at least some kind of additional information.
Even if there are no external forces but gravity on the block, and your acceleration-time graph is just a constant you would still be unable to find the coefficient of friction without at least knowing the angle the ramp makes with the horizontal.
You would need at least some kind of additional information.
No that's not right
Even if there are no external forces but gravity on the block, and your acceleration-time graph is just a constant you would still be unable to find the coefficient of friction without at least knowing the angle the ramp makes with the horizontal.
You do not know inclination as well. So you have got two equations and two variables. What else do you need?
I could generate two identical velocity-time and acceleration-time graphs, for two different situations with different ramp angles and different coefficients of friction.
We are talking about one graph of moving up, and the other moving down. The two situations are well connected. You cannot change the coefficient of friction or angle in between the motion. If you feel you can generate the graphs then please do and post.
There should only be a single acceleration graph.. unless there are other forces acting on the block other than gravity. The gravitational acceleration on the block does not care if the block is going up or down, the acceleration is constant. Therefore the acceleration-time graph is just a straight line, and fully describes the block moving up and down the ramp together with the initial velocity and position, no second graph is required.
Well the acceleration experienced by the block regardless of whether it is going up or down the ramp and regardless of whatever its velocity is will be found by:
in general
$$F_{friction} = \mu F_{normal}$$
$$F_{friction} = \mu mgcos\theta$$
the component of gravity down the ramp will be:
$$F_{g_{x}} = mgsin\theta$$
So the total force is:
$$F = F_{g_{x}} - F_{friction}$$
$$F = mgsin\theta - \mu mgcos\theta$$
So the acceleration is in general:
$$a = g(sin\theta - \mu cos\theta)$$
Then from this the velocity is simply the integral, we find:
$$v = g(sin\theta - \mu cos\theta)t + v_{0}$$
So for example, with a ramp angle of pi/4 and a coefficient of friction of 0.2, the acceleration is directed down the ramp with a magnitude of 5.544m/s^2
However, it is easy to verify that a ramp angle of pi/3 and a coefficient of friction of 0.601 also has an acceleration down the ramp of 5.544m/s^2
There should actually be an infinite amount of combinations of ramp angle and coefficients of friction that will give the same acceleration.
If the acceleration is the same, then the velocity and position graphs are also the same. (provided that all situations have the same starting point and initial velocity).
Last edited: | 676 | 3,063 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-25 | latest | en | 0.950148 |
https://www.techwhiff.com/issue/how-did-john-locke-define-liberty--504957 | 1,660,828,215,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573197.34/warc/CC-MAIN-20220818124424-20220818154424-00378.warc.gz | 863,676,615 | 12,194 | # How did John Locke define liberty?
###### Question:
How did John Locke define liberty?
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### Who is correct between the two of them?
Who is correct between the two of them?... | 1,095 | 4,894 | {"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} | 3.015625 | 3 | CC-MAIN-2022-33 | latest | en | 0.935996 |
https://www.enotes.com/topics/math/questions/x-3t-2-y-t-3-t-determine-open-t-intervals-which-810612 | 1,709,493,320,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00590.warc.gz | 763,940,024 | 16,579 | # `x=3t^2 , y=t^3-t` Determine the open t-intervals on which the curve is concave downward or concave upward.
Given parametric equations are:
`x=3t^2,y=t^3-t`
We need to find the second derivative, to determine the concavity of the curve.
`dy/dx=(dy/dt)/(dx/dt)`
Let's take the derivative of x and y with respect to t,
`dx/dt=3*2t=6t`
`dy/dt=3t^2-1`
`dy/dx=(3t^2-1)/(6t)`
`dy/dx=(3t^2)/(6t)-1/(6t)`
`dy/dx=t/2-1/(6t)`
`(d^2y)/dx^2=d/dx[dy/dx]`
`=(d/dt[dy/dx])/(dx/dt)`
`=(d/dt(t/2-1/(6t)))/(6t)`
`=(1/2-1/6(-1)t^(-2))/(6t)`
`=(1/2+1/(6t^2))/(6t)`
`=((3t^2+1)/(6t^2))/(6t)`
`=(3t^2+1)/(6t^2(6t))`
`=(3t^2+1)/(36t^3)`
Curve is concave upwards if second derivative is positive and concave downwards if it is negative,
So, the curve is concave upward forĀ `t>0`
Curve is concave downward forĀ `t<0` | 362 | 815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-10 | latest | en | 0.755328 |
https://forum.audacityteam.org/t/frequency-analysis-on-white-noise-depending-on-window-size/53252 | 1,709,598,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476592.66/warc/CC-MAIN-20240304232829-20240305022829-00360.warc.gz | 260,354,855 | 5,606 | # Frequency analysis on White Noise depending on window size?
Hello,
I’m working on a white noise, generated with Audacity:
Then I need to analyse the spectrum, and I notice that the result is depending on the window size.
At size 2048, level is around -30 dB.
At size 4096, level is around -33 dB.
Basically, when I double the window size, the level decreases by 3 dB.
However, with another signal (e.g. sine) the level stays consistent whatever the window size I chose. A full scale sine [-1;1] will return a spectrum level of 0 dB no matter what window size.
That is correct (and an interesting question).
“Plot Spectrum” uses FFT analysis, which basically splits the sound into multiple “frequency buckets”.
For a pure sine tone, all of the sound will be in one “bucket” (or in the special case, on the border between two buckets). The graph is normalized (scaled) such that a 0 dB sine tone will measure (approximately) 0 dB in Plot Spectrum.
For white noise, the audio is distributed approximately equally across all of the buckets. The number of buckets depends on the “window size”. If the window size is doubled, then the number of buckets also doubles. Thus, each bucket contains only half the “amount” of sound because the sound is being distributed across twice as many buckets.
“Plot Spectrum” shows the “sound intensity” (power). Doubling the sound intensity is an increase of +3 dB. Halving the power intensity is a decrease of -3 dB. | 329 | 1,460 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-10 | latest | en | 0.9147 |
https://www.edumple.com/cbse-class-6/science/let-us-recapitulate/notes/rc@gm_1130 | 1,718,979,932,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00886.warc.gz | 671,151,541 | 12,417 | 1. In ancient time man used to move only on foot and carry goods either on his back or on the back of some animals.
2. A great change in the modes of transport was made:
(i) by the invention of wheel. (ii) by the invention of steam engine.
3. Non-standard measures: The lengths of steps, arms, hands, or fingers of different people are different, therefore the distance measured with their help is not always reliable. These methods are, therefore, called non-standard measures.
4. Standard measures: Measures that are the same allover the world are known as standard measures.
5. In October 1960, the 12th general conference on weight and measures adopted the International system of units. "The System International Units" is the set of units to maintain uniformity all over the world.
6. Metre: It is the standard unit of length. The symbol of metre is m.
Each metre (m) is divided into 100 equal divisions, called centimetre (cm). Each centimetre has ten equal divisions, called millimetre (mm). Thus
1 m = 100 cm
1 cm = 10 mm
For measuring large distances, metre is not a convenient unit. We define a larger unit of length. It is called kilometre (km).
1 km = 1000 m.
7. Simple multiples of units: Units that are used for the measurement of larger distances are the multiples of SI unit. For example: deca, hecto, kilo.
1 decametre = 10 m
1 hectometre = 100 m
1 kilometre = 1000 m
8. Sub-multiples of units: Units used for measuring smaller distances are the submultiples of SI units. For example, milli, centi, deci.
1 m = 10 decimetre
1 m = 100 centimetre
1 m = 1000 millimetre.
9. Making measurement of a length: In making measurement of length of an object, we should follow the following procedure:
(a) Place the scale in contact with the object along its length as shown in Fig.
(b) Measurement with a scale with broken ends
(i) Avoid taking measurements from zero mark.
(ii) Use any other full mark of the scale, say 1.0 cm. [Fig. (a)]
(iii) Subtract the reading of this mark from the reading at the other end. For example, in Fig. (b), the reading at starting mark 1.0 cm and at the other end it is 6.5 cm. Therefore, the length of the object (6.5 – 1.0) cm = 5.5 cm.
(c) Correct position of the eye is also important for making measurement. Your eye must be exactly above the point where the measurement is to be taken.
10. Least count: A scale is marked in centimetres and millimetres. With the scales of this kind we can measure correctly up to one millimetre, that is one-tenth of a centimetre. This is called the least count of a (15 cm) scale.
11. Measuring the length of a curved line: We cannot measure the length of a curved line directly by using a metre scale. We can use a thread or divider to measure the length of a curved line.
12. Motion: It is a state of objects in which they are moving, that is, they are changing their place with the changing time.
13. Rest: All the stationary objects which are not in motion, that is, do not change their place with time are said to be at rest.
14. Rectilinear motion: When the objects change their position with time along a straight line, this type of motion is called rectilinear motion.
15. Circular motion
(i) When a body moves in a circular path, its motion is known as circular motion.
Examples: (a) motion of stone tied in a thread and whirled.
(b) motion of a blade of an electric fan.
(c) motion of second's hand in a clock.
(ii) In circular motion, the object remains at the same distance from a fixed point.
16. Rotational motion: Motion in which a whole body moves about an axis is called a rotational motion. Example: motion of a top.
17. Periodic motion: Motion in which an object repeats its motion after a fixed interval of time is called periodic motion.
Examples:
(i) Oscillations of a pendulum
(ii) Motion of a swing.
18. Combination of two or more types of motions: In some situations, the motion of an object may be a combination of two or more of the blow mentioned types of motion.
Examples:
(i) Motion of a ball on the ground. Here, the ball is rotating about an axis but the axis itself is moving along a straight line. Thus, the ball executes a rectilinear motion as well as rotational motion.
(ii) Motion of earth-earth executes rotations on its axis and also revolves around the sun.
19. Unit of measurements
(i) It involves the comparison of an unknown quantity with some known quantity of the same kind.
(ii) This known fixed quantity is called unit.
(iii) The result of measurement is expressed in two parts. One part is a number the other part is the unit of measurement.
KEY WORDS
1. Circular motion: When a body moves in a circular path, its motion is known as circular motion.
2. Distance: Measurement of gap between two points in certain units is called distance.
3. Measurement: Measurement means the comparison of an unknown quantity with some known quantity.
4. Motion : It is .a state of objects in which they are moving that is, they are changing their place with time.
5. Periodic motion: Motion in which an object repeats its motion after a fixed interval of time is called periodic motion.
6. Rectilinear motion: When the objects change their position with time along a straight line, this type of motion is called rectilinear motion.
7. SI units: In October, 1960 the 12th general conference on weight and measures adopted in International system of units to maintain uniformly all over the world. This system of units is called SI units.
8. Units of measurement: Measurement means the comparison of an unknown quantity with some known quantity. This known fixed quantity is called a unit of measurement. | 1,401 | 5,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-26 | latest | en | 0.892911 |
targetcalculator.cloud | 1,720,895,914,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00661.warc.gz | 517,716,080 | 3,299 | # A
## Annualised Revenue
Two terms you might come across are monthly recurring revenue and annualised recurring revenue, sometimes shortened to MRR and ARR respectively. MRR is the revenue that happens repeatedly every month, and ARR is the 12-month view of that recurring revenue. For example, if you add a new customer who's consuming \$1,000 of services half way through a year, you'd get to bill them six times. This means their MRR is \$1,000, and their total revenue for that first year is \$6,000 (6 months * \$1,000). Assuming you keep the customer, going into the next year you'll get to bill them 12 times instead of 6, so the ARR is 12x the MRR, or 12 * \$1,000 = \$12,000.
# B
## Baseline
Your baseline is the existing recurring revenue that you're building your future plans upon. For example, if you finish the last month of your last fiscal period on \$10,000 of recurring revenue, then that's your baseline. This matters because you probably aren't starting from \$0 and it's important to factor in your existing baseline when trying to calculate your targets. A \$10,000 per month baseline equates to \$120,000 of total revenue for the next year (12 months * \$10,000 of recurring revenue) - if your target is \$240,000, then you've already got half of that 'in the bank'!
# C
Adding customers is one method of generating more revenue. A customer 'add' is simply establishing a new relationship with a customer where they consume services that you bill them for each month. These services might include several things, such as the cloud costs (for example the Microsoft Azure services), your support and management costs, margin, etc. These costs might be bespoke per customer, or more 'off the shelf' as a standard managed service, IP you deliver in a SaaS model, etc. However you think about it, a customer is worth an amount of money to your business each month, and you should have a goal in mind for the minimum amount a customer should consume per month in order to calculate how many customers you'd need in order to achieve your targets.
The other main method of generating revenue is by growing the spend of your existing customers by selling them more services. Deciding the minimum amount you want a customer to spend per month will help you understand how much of your target should come from adding more new customers versus growing existing ones. There'll be many factors in this decision, including your capacity to add and manage more customers and your customer lifecycle management process to be able to grow existing customers.
# R
## Recurring Revenue
Recurring revenue is the amount of money you're receving each month from the sale and consumption of your services. This is different to traditional purchasing. For example, if I purchase a physical server costing \$5,000 you'll get to bill me once for \$5,000. You might not sell me anything else for weeks, months, or even years. If, however, you're selling me a cloud service at \$1,000 per month, you get to bill me every month that I use the service - this means you'd have \$1,000 of recurring revenue. One of the biggest differences in sales between traditional 'up front' billing, and recurring billing, is how you approach hitting your targets.
Let's say you've got 12 months and a target of \$100,000 of revenue to generate. In a traditional model, where you're selling \$5,000 physical servers, you'd need to sell 20 servers to hit your target. If you build your pipeline through the year, you could wait until the last day of the year and close all those deals and see all of that revenue immediately.
In the recurring revenue world, where you're selling \$1,000 per month cloud services, you need to think differently. If you sell one 'server' in the first month of your year, it's worth \$12,000 (12 months * \$1,000 per month), if you sell one in the second month, it's worth \$11,000 (11 months * \$1,000 per month). If you sell your 'server' in the last month then it's worth only \$1,000 (1 month * \$1,000 per month). If you leave closing deals at the end of the year, you'd need to add 100 'servers' in the last month!
Instead, if you add one 'server' per month (1.3 to be exact in this example!), you'd only need to add 15.4 to generate \$100,000 of revenue in the same 12 month period. | 983 | 4,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.956654 |
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# Are the integers z and f to the right of 0 on the number
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Are the integers z and f to the right of 0 on the number [#permalink] 21 Mar 2012, 12:33
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Are the integers z and f to the right of 0 on the number line?
(1) The product of z and f is positive.
(2) The sum of z and f is positive.
[Reveal] Spoiler: OA
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Re: Number Line [#permalink] 21 Mar 2012, 12:43
Here is my take:
ZF>0 tells you that z and f have same sign - insuficient because both can be positive or negative
z + f > 0 means either z or f is positive or both are positive so its insufficient
Together sufficient because Z & F have same sign and one is positive so both must be positive
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Re: Are the integers z and f to the right of 0 on the number [#permalink] 21 Mar 2012, 13:11
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Are the integers z and f to the right of 0 on the number line?
Question basically asks whether z and f are both positive.
(1) The product of z and f is positive --> zf>0 --> z and f have the same sign, they are either both positive or both negative. Not sufficient.
(2) The sum of z and f is positive --> z+f>0 --> it's certainly possible that both are positive but it's also possible one to be positive and another negative (for example z=2 and f=-1). Not sufficient.
(1)+(2) Since from (2) z+f>0 then from (1) it's not possible that z and f to be both negative. Hence, z and f are both positive. Sufficient.
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Re: Are the integers z and f to the right of 0 on the number [#permalink] 21 Mar 2012, 13:11
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Display posts from previous: Sort by | 914 | 2,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2013-20 | latest | en | 0.841056 |
https://hub.jmonkeyengine.org/t/rendering-superellipsoids/24017 | 1,709,075,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00873.warc.gz | 305,149,899 | 8,796 | # Rendering superellipsoids
Hi. I want to render superellipsoids with jme3. http://paulbourke.net/geometry/superellipse/
I use the following code for calculating one vertex:
[java]public Vector3f EvalSuperEllipse(float t1,float t2,float p1,float p2)
{
float tmp;
float ct1,ct2,st1,st2;
ct1 = (float) Math.cos(t1);
ct2 = (float) Math.cos(t2);
st1 = (float) Math.sin(t1);
st2 = (float) Math.sin(t2);
tmp = (float) Math.pow(fabs(ct1),p1)sign(ct1);
Vector3f pp = new Vector3f(sign(ct2)15(float) (tmp
Math.pow(fabs(ct2),p2)),
sign(st1)15(float) Math.pow(fabs(st1),p1),
sign(st2)15(float) (tmp * Math.pow(fabs(st2),p2))
);
return pp;
}[/java]
The following code calculates all the vertices and sets the arrays for them, the indexes array and the textures array:
[java]for (int j=0;j<n/2;j++) {
theta1 = j * TWOPI / n - PID2;
theta2 = (j + 1) * TWOPI / n - PID2;
for (int i=0;i<=n;i++) {
if (i == 0 || i == n)
theta3 = 0;
else
theta3 = i * TWOPI / n;
p = EvalSuperEllipse(theta2,theta3,e1,e2);
p1 = EvalSuperEllipse(theta1+delta,theta3,e1,e2);
p2 = EvalSuperEllipse(theta1,theta3+delta,e1,e2);
en = calcNormal(p1,p,p2);
tex = new Vector2f(i/(float)n,2j/(float)n);
p = EvalSuperEllipse(theta1,theta3,e1,e2);
p1 = EvalSuperEllipse(theta2+delta,theta3,e1,e2);
p2 = EvalSuperEllipse(theta2,theta3+delta,e1,e2);
en = calcNormal(p1,p,p2);
tex = new Vector2f(i/(float)n,2
(j+1)/(float)n);
}
}
vertices = new Vector3f[vertexArray.size()];
vertices = vertexArray.toArray(vertices);
texCoord = new Vector2f[textureArray.size()];
texCoord = textureArray.toArray(texCoord);
int k = 0;
while(k < vertices.length) {
int d = k;
for(int j=d; j < d+3; j++) {
if(j < vertices.length) {
k++;
}
}
}
indexes = new int[indexList.size()];
for(int i=0; i<indexes.length; i++) {
indexes = indexList.get(i);
}
setBuffer(Type.Position, 3, BufferUtils.createFloatBuffer(vertices));
setBuffer(Type.TexCoord, 2, BufferUtils.createFloatBuffer(texCoord));
setBuffer(Type.Index, 3, BufferUtils.createIntBuffer(indexes));
updateBound();
}[/java]
I used these codes from the link given above. In the end i render the mesh a standard way:
[java]SuperEllipsoid se = new SuperEllipsoid(1.0f, 1.0f, 100);
Geometry seGeo = new Geometry(“SuperEllipsoid”, se);
Material seMat = new Material(assetManager, “Common/MatDefs/Misc/Unshaded.j3md”);
seMat.setColor(“Color”, ColorRGBA.Blue);
seGeo.setMaterial(seMat);
rootNode.attachChild(seGeo);[/java]
Somehow i get this:
http://i.imgur.com/Fq7li.png
What is the problem? I’ve read the Custom Mesh tutorial. I think the problem is the indexes, but i dont really know how to solve it…
That should be a sphere, for the parameters 1.0, 1.0
When each triangle is projected onto screenspace, the vertices should be in counterclockwise order (on screen), otherwise they are culled. You may need to orient your indices to generate triangles in that order.
1 Like
The main problem is that these superellipsoids can be very complicated shapes, so its a little hard to imagine an order of the vertices.
Well, you could cheat and turn off back face culling but later when you go to have normals and stuff, you’ll wish you hadn’t.
Edit: you “might” wish you hadn’t. Could be that it works ok for you and your use-case.
1 Like
Without backface culling it looks good, but as you said: when i switch to Lightning.j3m material, its a mess again. I am atill pretty sure, that my indexes are wrong, but I just couldnt find out how to arrange them in the correct order. I am attaching a txt file containing the coordinates of the sphere, if someone could just take a look at it, maybe somebody else can figure it out, i would appriciate that Currently the indexes look like this: 0,1,2, 1,2,3, 2,3,4, 3,4,5, 4,5,6, etc, you see the logic.
Btw dont open it in notepad, because you wont see sh*t, wordpad works fine.
http://www.fileswap.com/dl/vPiZmEOZeq/
1----2
| |
| |
0----3
Assuming you have indexed your angles this way the correct order is:
0,3,2
0,2,1
1 Like
Just to see how does it look like now (with backface culling ON, and the indexes mentioned in my previous post):
Seems he’s making a triangle strip mesh, why not do that as well? Then you dont need to specify the index order. Otherwise why not look at the indexing of the jME sphere? Maybe even start from that class and just change the equations, and maybe do some extra alternatives for degenerate shapes etc.
1 Like
I set it to TriangleStrip, and removed the indexes. Its better now thx for the tip, but still something is wrong. Maybe not the indexes are the problem? The objects are look like this now, all have these stripes:
Just got something in my mind…maybe something wrong about the normals…I’ll check it.
YES! It was the normals. Problem ALMOST solved Now only one issue left, every object has a “hole” (like a butthole :D) on the top and bottom side, here how it looks like on the sphere:
Thx for the help everyone btw, I try to solve this last little problem (i hope its a little problem), but any ideas are welcomed
I dunno if anyone reading this or i’m talking with myself, but heres a hint, if someone is into solving this last problem: it looks like this isnt a hole (which is btw much bigger at objects like a cylinder for example), but it looks like the top and bottom of the object are mixed up. This black circle u see on the picture above is gray on the bottom. TLDR: the top is on the bottom, and the bottom is on the top.
PROBLEM SOLVED! It was the normals again. I wrote my own method to normalize the vectors and something was wrong in it, so i just picked the built in normalize() method and it works like a charm Thanks everyone again! | 1,623 | 5,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.255447 |
https://matsci.org/t/how-to-setup-velocity-inlet-boundary-condition-in-lammps/19389 | 1,721,074,833,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00757.warc.gz | 335,073,898 | 5,583 | # how to setup velocity inlet boundary condition in lammps
10.10.2013, 00:30, "Fubing BAO" <[email protected]...>:
Hi Oleg, Thank you for your suggestion. But in my opinion, the fix append/atoms command is used to enlarge the box in one direction by appending the atoms, e.g., zhi
Yes. And you would have to "melt" this added part.
``````In my case, the target sphere is fixed in the box\. I just want to generate an uniform flow with fixed convection velocity as inlet boundary\.
``````
You can move sphere instead of a liquid (isn't it the same?), repeatedly append atoms in front of the sphere and delete them and shrink box behind the sphere. So, I think, technically it can be done.
That being said, I like Axel's suggestion much more.
Oleg.
Clearly, the most straightforward approach is to apply a constant
force to the fluid particles. This will achieve something that is
close to a velocity inlet boundary condition. There are several
problems that I can think of:
1) the steady state flow rate can not be specified, only measured
2) the body force or pressure gradient will be proportional to local
fluid density which does not correspond precisely to a pressure-driven
flow
3) the flow at the boundary will not be perfectly uniform due to
periodic images of the wake
Solutions:
1) This is not really a problem, just a matter of convenience, as Axel
pointed out
2) This can be solved by applying the force to the sphere instead of
the fluid. Note that if the cell contains no momentum sink such as a
fixed wall, then at steady state the fluid will simply reach
equilibrium with the sphere with both having the same streaming
velocity.
3) Until you do the simulation, you will not know to what extent this
is a problem, so you should first assume it is not. You can test this
by running simulations with different box lengths and comparing. My
guess is that for a cell that is 10x longer than the sphere, there
will be no effect, and you can probably get away with much less than
that.
I think adding and deleting chunks of atoms at both boundaries is a
terrible idea. It is complicated to do and will introduce additional
problems that will have to be managed. We use this method in shock
simulations of crystals, but that is a non-steady state problem in
which we would normally have to simulate a length of material
proportional to the duration of the simulation, which quickly becomes
prohibitively expensive.
Maybe an easier way to handle this (also suggest by Axel I think) is
to break up the flow by applying a strong Langevin thermostat biased
to a specified streaming velocity to the fluid that is farthest from
the sphere. This would require adding some extract logic to
fix_langevin.cpp. | 626 | 2,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-30 | latest | en | 0.944738 |
https://webapps.stackexchange.com/questions/82789/how-to-count-rows-in-google-sheets-that-do-not-contain-a-specific-text | 1,656,157,055,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00221.warc.gz | 654,582,379 | 65,759 | # How to count rows in Google Sheets that do not contain a specific text
I need a formula that will tell me how many projects are going on at any given time in my Google spreadsheet. My table header looks like this:
``````Project | Jan 1-5 | Jan 6-10 | Jan 11-15...
``````
And each row will either be empty, or have text in the different cells (to show what was happening on a given date). For instance:
``````Project 1 | U | U | L
Project 2 | P | M | M
Project 3 | M | O | K
``````
I found this formula which works very well, however, it counts each row with a value. I need to add a criteria that says that says not to count "M" or "P" as a value.
``````=ArrayFormula(
SUM(SIGN(MMULT(LEN(Sheet1!B3:E),TRANSPOSE(SIGN(COLUMN(Sheet1!B3:E))))))
)
``````
For the example above, this would give me the answer of 3, but I am looking for it to say 2.
Here is an approach similar to yours; the filtering is done by `regexmatch` in the middle.
``````=countif(mmult(
arrayformula(if(regexmatch(B3:D,"^[^MP]\$"),1,0)),
transpose(arrayformula(column(B3:D)))
),">0")
``````
Explanation:
1. `regexmatch(B3:D,"^[^MP]\$")` matches the cells that contain one letter which is neither M nor P.
2. `if(...,1,0)` converts boolean match result to 1 or 0.
3. `arrayformula` applies the above to the whole array, obtaining a matrix of 1s and 0s. It remains to count the number of rows that contain at least one 1.
4. As in your formula, this is done by multiplying the matrix by a vector with positive entries and counting the number of positive results. The vector I use is `transpose(arrayformula(column(B3:D)))`
5. After the multiplication, `countif(..., ">0")` counts the positive entries.
• Thank you SO much for your response and explanation! We are really close but though this is working in theory, it is not counting the rows that have any M or P anywhere. So in my above example, it would count "project 1" but not "project 2" or "3" and I would want it to count both 1 and 3. Basically I want it to pretend like the M and P are not even there. Is that possible? Or should I think of a way to make this work? Aug 27, 2015 at 16:53
• I tested this command, and it returned 2 for your example.
– user79865
Aug 27, 2015 at 17:07
• So strange, I put my example into a sheet to test and I am getting 3 now - it's counting every row. I'll keep playing with it but I really appreciate your help, this is closer than I've been in days! :) Aug 27, 2015 at 17:21
• Ah, I see what it's doing. It is working for "MP" but not "M" or "P" in separate cells. Aug 27, 2015 at 17:23
• It's meant to work for M or P individually. Try testing the regex part of the formula, `=arrayformula(if(regexmatch(B3:D,"^[^MP]\$"),1,0))`
– user79865
Aug 27, 2015 at 17:32 | 797 | 2,738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-27 | latest | en | 0.929159 |
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Topic: Matheology § 222 Back to the roots
Replies: 606 Last Post: Apr 18, 2013 2:04 AM
Messages: [ Previous | Next ]
mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: Matheology § 222 Back to the roots
Posted: Feb 10, 2013 3:55 PM
On 10 Feb., 18:40, William Hughes <wpihug...@gmail.com> wrote:
> On Feb 10, 10:51 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
>
>
> > On 9 Feb., 17:36, William Hughes <wpihug...@gmail.com> wrote:
>
> > > > > the arguments are yours
> > > > > and the statements are yours-
>
> > > > Of course. But the wrong interpretation is yours.
>
> > > How does one interpret
> > > we have shown m does not exist
>
> > > to mean that
>
> > > m might still exist
>
> > > ?
>
> > TND is invalid in the infinite.
>
> > Regards, WM
>
> In Wolkenmeukenheim, we can have
> for a potentially infinite set
>
> we know that x does not exist
> we don't know that x does not exist
>
> true at the same time.
Is it so hard to conclude from facts without believing in matheology?
The diagonal of the list
1
11
111
...
is provably not in a particular line.
But the diagonal is in the list, since it is defined in the list only.
Nothing of the diagonal can be proven to surpass the lines and rows of
the list.
It is so easy. But it is contrary to the mess taught by matheologians.
Regards, WM
Date Subject Author
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2/7/13 Virgil | 7,185 | 15,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-17 | latest | en | 0.728799 |
http://www.efunda.com/glossary/units/units--surface_density--microgram_per_square_millimeter.cfm | 1,586,503,195,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371886991.92/warc/CC-MAIN-20200410043735-20200410074235-00249.warc.gz | 231,754,993 | 7,287 | Surface Density Units: Microgram Per Square Millimeter
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Glossary » Units » Surface Density » Microgram Per Square Millimeter
Microgram Per Square Millimeter (µg/mm2) is a unit in the category of Surface density. It is also known as microgram/square millimeter. Microgram Per Square Millimeter (µg/mm2) has a dimension of ML-2 where M is mass, and L is length. It essentially the same as the corresponding standard SI unit kg/m2.
Note that the seven base dimensions are M (Mass), L (Length), T (Time), Q (Temperature), N (Aamount of Substance), I (Electric Current), and J (Luminous Intensity).
Other units in the category of Surface density include Gram Per Square Centimeter (g/cm2), Kilogram Per Square Centimeter (kg/cm2), Kilogram Per Square Meter (kg/m2), Ounce Mass Per Square Inch (ozm/in2), Pound Mass Per Square Foot (lbm/ft2), and Pound Mass Per Square Inch (lbm/in2). | 269 | 1,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.759836 |
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Amin
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Cuch, very sorry for delay in my reply. Below is the Mathematica code to solve any 1st order or nth order ODE. I am writing the code in text format below and explain it again in a next code window.
.
Clear[t,y,Zi,Wi,Yi,Y,y0,i,i1,p,n,ZAns];
n=1;
Array[y,n+1,0];
y[1]:= y[0]^2;
Array[y0,n,0];
y0[0]=1;
p=8;
Y:=y0[0];
For[k=1,k<n,k++,Y=Y+y0[k] *t^k/k!];
Array[Zi,n+p-1,0];
Array[Wi,n+p-1,0];
Array[Yi,n+p-1,0];
Array[ti,n+p-1,0];
Zi[n]=(y[n]/.t-> ti[n]);
For [i=n,i<p+n-1,i++,(Zi[i+1]=0;For [k=0,k<n,k++,(Zi[i+1]=Zi[i+1]+D[Zi[i],y[k]]*(y[k+1]/.t-> ti[i+1]));]);];
For[i=n,i<= p+n-1,i++,(Wi[i]=Zi[i];For[k=0,k<n,k++,(Wi[i]=(Wi[i]/.y[k]-> y0[k]))];)];
For[i=n,i<= p+n-1,i++,(Yi[i]=Wi[i];For[i1=i,i1>=1,i1--,(Yi[i1-1]=Integrate[Yi[i1],{ti[i1],0,ti[i1-1]}])];Y=Y+(Yi[0]/.ti[0]-> t);)]
ZAns=Collect[Y,t,Simplify]//PolynomialForm[#,TraditionalOrder -> False]&
.
In Below I explain the above code with some comments.
Clear[t,y,Zi,Wi,Yi,Y,y0,i,i1,p,n,ZAns]; This line is handy to clear variables especially when you want to solve for more than one ODE.
n=1; n is order of the ODE. For 1st order ODE n=1;
Array[y,n+1,0]; Define a workhorse variable.
y[1]:= y[0]^2; Specify the ODE. y[1]= dy/dt. y[0]=y[t] and similarly for higher order ODE's y[2]=d2y/dt2
Array[y0,n,0];
y0[0]=1; Specify initial conditions. y0[0]=y(0) similalry for 2nd order ODE, you will also have to specify y0[1]=dy(0)/dt and so on.
p=8; Number of powers in expansion of series.
Y:=y0[0]; Allocate first initial condition.
For[k=1,k<n,k++,Y=Y+y0[k] *t^k/k!]; Allocation of higher order initial conditions.
Array[Zi,n+p-1,0];
Array[Wi,n+p-1,0];
Array[Yi,n+p-1,0];
Array[ti,n+p-1,0];
Zi[n]=(y[n]/.t-> ti[n]);
For [i=n,i<p+n-1,i++,(Zi[i+1]=0;For [k=0,k<n,k++,(Zi[i+1]=Zi[i+1]+D[Zi[i],y[k]]*(y[k+1]/.t-> ti[i+1]));]);];
For[i=n,i<= p+n-1,i++,(Wi[i]=Zi[i];For[k=0,k<n,k++,(Wi[i]=(Wi[i]/.y[k]-> y0[k]))];)];
For[i=n,i<= p+n-1,i++,(Yi[i]=Wi[i];For[i1=i,i1>=1,i1--,(Yi[i1-1]=Integrate[Yi[i1],{ti[i1],0,ti[i1-1]}])];Y=Y+(Yi[0]/.ti[0]-> t);)]; This is integration loop which iteratively integrates starting from zero. replace zero with initial time when it is different from zero.
ZAns=Collect[Y,t,Simplify]//PolynomialForm[#,TraditionalOrder -> False]&
.
I think your ODE(2) can be very easily solved. ODE(1) seems singular at t=0 but can possibly be solved by changing starting value to some positive time and specifying initial conditions there.
Sorry, I have been pretty busy with work on my trading algorithms and want to apply them on NASDAQ stocks so little time left for other things. Will come back soon to tackle the discussion.
In my paper, I have given a mathematica routine that can solve arbitrary systems of nth order ODEs (at least in theory). I have also given an example explaining the algorithm where I used the algorithm to solve a system of three 2nd order non-linear ODEs.
You think life is a secret, Life is only love of flying, It has seen many ups and downs, But it likes travel more than the destination. Allama Iqbal
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Thanks, Amin
Code is important but it is not the driver. It just produces a number.
We are not on the same wavelength, unfortunately. My two examples were concrete insights into the bespoke qualitative behaviour of ODEs.
„A good technical writer, trying not to be obvious about it, but says everything twice: formally and informally. Or maybe three times.“ — Donald Ervin Knuth
They sought it with thimbles, they sought it with care; They pursued it with forks and hope; They threatened its life with a railway-share; They charmed it with smiles and soap.
Cuchulainn
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation
For completeness
$dx/dt = x^2, x(0) = 1$ general solution $x = 1/(1-t)$ (B)
. The solution can be found by inspection or Picard iteration serie
. The solution does blow up afer t > 1
. Solving (B) by ODE solvers, including yours ... a prioiri quantitatve results on stability and accurscy
These are the compelling issues in numerical analysis, like gravity in physics.
They sought it with thimbles, they sought it with care; They pursued it with forks and hope; They threatened its life with a railway-share; They charmed it with smiles and soap.
Amin
Topic Author
Posts: 2079
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Sorry Cuch, a lot of other things to do. I will come back to your examples soon when I will have some more time. I am sure they are very simple.
Friends, I have decided to write 3-5 formal papers about the material in this thread that would ideally be submitted to Wilmott magazine and I will soon be approaching Wilmott people for their requirements and guidelines.
The research subject of first paper would be finding the dynamics of variance/volatility SDE so that volatility surface is closely fitted across time given that asset SDE in two dimensional correlated SDE system is given either by lognormal or CEV dynamics.
All through this paper, we will remain mainly in a single dimensional SDE framework other than just a few small excursions into two dimensions.
For my first paper, I am cleaning up my work on one dimensional SDEs with addition of cumulants. And also on time integral of the mean-reverting and other SDEs. I want to solve for distribution of integral of volatility/variance SDE by backing it out from option prices using lognormal or CEV formula. And then I will find the dynamics of volatility SDE so that it perfectly fits the implied distribution of integral of variance. While backing out integral of volatility, I will also have to do accounting for correlation and contribution of correlated part of SV SDE but this should be simple given the insights we have learnt about variances and their addition within Z-series framework. All of this will be done in a one dimensional SDE framework. We have already done 70% of the work in this thread. I will be changing a few things like I do not want to go into cumulants at all (even though they should be used a lot for inference) and will be using new alternative methods. I will also be making some changes to calculation of time integrals of SDEs. Only major part that remains to be done is bootstrapping of integral of variance SDE and correlated part of the SDE from the cross-section of option prices. Since we want to remain in the one dimensional domain, I really hope that computational time for option chains out to 4-5 years will be in seconds and not in minutes.
And my time frame to complete the unfinished part is a few weeks. I will be posting the code on Wilmott as usual.
You think life is a secret, Life is only love of flying, It has seen many ups and downs, But it likes travel more than the destination. Allama Iqbal
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Sorry Cuch, a lot of other things to do. I will come back to your examples soon when I will have some more time. I am sure they are very simple.
Take your time. Don't want be a party pooper, but the mathematical and numerical issues are far from simple.
They sought it with thimbles, they sought it with care; They pursued it with forks and hope; They threatened its life with a railway-share; They charmed it with smiles and soap.
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Cuch, very sorry for delay in my reply. Below is the Mathematica code to solve any 1st order or nth order ODE. I am writing the code in text format below and explain it again in a next code window.
.
Clear[t,y,Zi,Wi,Yi,Y,y0,i,i1,p,n,ZAns];
n=1;
Array[y,n+1,0];
y[1]:= y[0]^2;
Array[y0,n,0];
y0[0]=1;
p=8;
Y:=y0[0];
For[k=1,k<n,k++,Y=Y+y0[k] *t^k/k!];
Array[Zi,n+p-1,0];
Array[Wi,n+p-1,0];
Array[Yi,n+p-1,0];
Array[ti,n+p-1,0];
Zi[n]=(y[n]/.t-> ti[n]);
For [i=n,i<p+n-1,i++,(Zi[i+1]=0;For [k=0,k<n,k++,(Zi[i+1]=Zi[i+1]+D[Zi[i],y[k]]*(y[k+1]/.t-> ti[i+1]));]);];
For[i=n,i<= p+n-1,i++,(Wi[i]=Zi[i];For[k=0,k<n,k++,(Wi[i]=(Wi[i]/.y[k]-> y0[k]))];)];
For[i=n,i<= p+n-1,i++,(Yi[i]=Wi[i];For[i1=i,i1>=1,i1--,(Yi[i1-1]=Integrate[Yi[i1],{ti[i1],0,ti[i1-1]}])];Y=Y+(Yi[0]/.ti[0]-> t);)]
ZAns=Collect[Y,t,Simplify]//PolynomialForm[#,TraditionalOrder -> False]&
.
In Below I explain the above code with some comments.
Clear[t,y,Zi,Wi,Yi,Y,y0,i,i1,p,n,ZAns]; This line is handy to clear variables especially when you want to solve for more than one ODE.
n=1; n is order of the ODE. For 1st order ODE n=1;
Array[y,n+1,0]; Define a workhorse variable.
y[1]:= y[0]^2; Specify the ODE. y[1]= dy/dt. y[0]=y[t] and similarly for higher order ODE's y[2]=d2y/dt2
Array[y0,n,0];
y0[0]=1; Specify initial conditions. y0[0]=y(0) similalry for 2nd order ODE, you will also have to specify y0[1]=dy(0)/dt and so on.
p=8; Number of powers in expansion of series.
Y:=y0[0]; Allocate first initial condition.
For[k=1,k<n,k++,Y=Y+y0[k] *t^k/k!]; Allocation of higher order initial conditions.
Array[Zi,n+p-1,0];
Array[Wi,n+p-1,0];
Array[Yi,n+p-1,0];
Array[ti,n+p-1,0];
Zi[n]=(y[n]/.t-> ti[n]);
For [i=n,i<p+n-1,i++,(Zi[i+1]=0;For [k=0,k<n,k++,(Zi[i+1]=Zi[i+1]+D[Zi[i],y[k]]*(y[k+1]/.t-> ti[i+1]));]);];
For[i=n,i<= p+n-1,i++,(Wi[i]=Zi[i];For[k=0,k<n,k++,(Wi[i]=(Wi[i]/.y[k]-> y0[k]))];)];
For[i=n,i<= p+n-1,i++,(Yi[i]=Wi[i];For[i1=i,i1>=1,i1--,(Yi[i1-1]=Integrate[Yi[i1],{ti[i1],0,ti[i1-1]}])];Y=Y+(Yi[0]/.ti[0]-> t);)]; This is integration loop which iteratively integrates starting from zero. replace zero with initial time when it is different from zero.
ZAns=Collect[Y,t,Simplify]//PolynomialForm[#,TraditionalOrder -> False]&
.
I think your ODE(2) can be very easily solved. ODE(1) seems singular at t=0 but can possibly be solved by changing starting value to some positive time and specifying initial conditions there.
Sorry, I have been pretty busy with work on my trading algorithms and want to apply them on NASDAQ stocks so little time left for other things. Will come back soon to tackle the discussion.
In my paper, I have given a mathematica routine that can solve arbitrary systems of nth order ODEs (at least in theory). I have also given an example explaining the algorithm where I used the algorithm to solve a system of three 2nd order non-linear ODEs.
I have non-hands on experience of Mathematica from a previous article with Alan and Paul.
AFAIR the routine "Integrate" is a magic black box ODE solver and has hidden know-how on "tricky" ODEs. It does noty bring you closer to an understanding of the essential complexity. It is also slow when used as a method of lines (MOL) , and relatively few quants use it (?). For speed and insights, I reckon hand-crafted C++ code is needed. Quants want to know what is happening under the radar.
They sought it with thimbles, they sought it with care; They pursued it with forks and hope; They threatened its life with a railway-share; They charmed it with smiles and soap.
Amin
Topic Author
Posts: 2079
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Cuch, Integrate command in my code is nothing more than symbolic integration. It gives an analytic answer which is very helpful. It is not at all what you suspect and nothing more than symbolic computing. The above code solves nth order ODEs. So in order to automate everything, there are loops that solve iterated integrals but only symbolic integration is performed in the final line with loops.
Another thing when we specify ODE
y[1]:= y[0]^2; in the code means dy/dt=y(t)^2;
you could have specified an ODE with t in it as
y[1]:=t y[0]^2 -t y[0]; which would mean the ODE is: dy/dt=t y(t)^2 - t y(t);
It is this specified ODE that is symbolically integrated in iterated integrals using loops. There are loops both to tackle change of variables and also iterated symbolic integrations. The final answer of mathematica code is a symbolic series expansion of the true solution of the ODE. It is not anything like numerical integration or MOL or automatic ODE solutions etc. as you fear.
Mathematic code is a wizard. Paste it in mathematica with your own ODE and initial conditions and you get the series expansion solution right away.
Friends, I went back to my work on solution of SDEs with addition of cumulants. I had earlier solved one dimensional SDEs with this method but Z-series I used was only up till 5th power of Z. The previous solution also had relatively poor convergence because root-search was done over cumulants. With only five powers in the Z-series solution done in Bessel coordinates meant when we converted the solution into Z-series coefficients in original coordinates, the approximation in tails was poor at best and it could be easily seen that Z-series coefficients in original coordinates diverged from true solution to quite large extent in both tails.
Now, I have re-done the above solution of SDEs with a seven power Z-series. I have also updated the root-search and changed that from cumulants to moments which is very helpful. Z-series root-search solution almost always converges(though there are relatively rare exceptions when cumulants/moments change too fast and root-search solution to moments does not converge but I know that can be tackled by adaptively decreasing the step size when moments/cumulants change too fast. I have not introduced this adaptive step size setting in present program but want to do it sometime in the future soon.)
But I am absolutely in love with the new program when I noticed that converting Z-series SDE coefficients in Bessel coordinates to Z-series coefficients in original coordinates very faithfully mimics the true solution. When solution is plotted from original coordinates Z-series, right tail is almost always indistinguishable from the true right tail till the end. Left tail close to zero sometimes sneaks into negative territory but that is mostly very little and very insignificant and very close to true solution.
This is very important since all the stochastic volatility analytics are done on Z-series representation of SDE in original coordinates. And if conversion of Z-series coefficients from Bessel coordinates to original coordinates is not faithful, all analytics will be very approximate which is luckily not the case when Z-series coefficients are taken to seventh power of Z in Bessel coordinates to solve the SDE.
I will post my new programs for one dimensional SDEs that solve the SDE in Bessel coordinates using 7th order Z-series and also find analytic Z-series in original coordinates in another day or two latest by Tuesday. Again I am very happy that we have been able to get a faithful analytical representation of SDEs in original coordinates.
And we will need this analytical representation of stochastic volatility SDE in original coordinates again and again when we would solve for the SV dependent asset SDE using new methods.
You think life is a secret, Life is only love of flying, It has seen many ups and downs, But it likes travel more than the destination. Allama Iqbal
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation
It is not anything like numerical integration or MOL or automatic ODE solutions etc. as you fear.
Me fear?
Actually, I was confused with NSolve.
It is possible ODE solvers
https://library.wolfram.com/infocenter/ ... aPart1.pdf
And probably better performance than Shlow Symbolix.
They sought it with thimbles, they sought it with care; They pursued it with forks and hope; They threatened its life with a railway-share; They charmed it with smiles and soap.
Amin
Topic Author
Posts: 2079
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, I am posting my code for solution for densities of one dimensional SDEs from addition of cumulants and calculation of their variable representation as a Z-series which is a series in powers of standard normal with variable coefficients (which we calculate for every specific SDE).
I want to submit this work for publication in Wilmott magazine. I am still unsure whether to submit it as one large article that covers all the work up to solution of system of stochastic volatility SDEs or as in various smaller articles. But here is the code.
.
.
function [] = FPESeriesCoeffsFromMomentsAndVarianceAdditionAdaptStep()
%Copyright Ahsan Amin. Infiniti derivatives Technologies.
%or skype ahsan.amin2999
%In this program, I am simulating the SDE given as
%dy(t)=mu1 x(t)^beta1 dt + mu2 x(t)^beta2 dt +sigma x(t)^gamma dz(t)
%I have simulated the transformed
%Besse1l process version of the SDE and then changed coordinates to retreive
%the SDE in original coordinates.
dt=.125/2/2*1; % Simulation time interval.%Fodiffusions close to zero
%decrease dt for accuracy.
Tt=128/2/2*2;%*4*4*1;%*4;%128*2*2*2; % Number of simulation levels. Terminal time= Tt*dt; //.125/32*32*16=2 year;
T=Tt*dt;
%Below I have done calculations for smaller step size at start.
ds(1:3)=dt/4;
%Below order for monte carlo
OrderA=4; %
OrderM=4; %
if(T>=1)
dtM=1/32;%.0625/1;
TtM=T/dtM;
else
dtM=T/16;%.0625/1;
TtM=T/dtM;
end
dNn=.1/2; % Normal density subdivisions width. would change with number of subdivisions
Nn=45*4; % No of normal density subdivisions
x0=1; % starting value of SDE
beta1=0.0;
beta2=1.0; % Second drift term power.
gamma=.95;%50; % volatility power.
kappa=3.2;%.950; %mean reversion parameter.
theta=.25;%mean reversion target
sigma0=1.0;%Volatility value
%StepSize Management. decrease the step size if MomentsIncreaseRatio is larger
%than MaxMomentRatio and increase the step size if
%MomentsIncreaseRatio is smaller than MinMomentRatio
%Look at the code in the body and feel free to play with stepsize
%variables
MaxMomentRatio=1.01;
MinMomentRatio=1.002;
MaxStepSize=1/128;
MinStepSize=1/512/16;
%you can specify any general mu1 and mu2 and beta1 and beta2.
mu1=1*theta*kappa; %first drift coefficient.
mu2=-1*kappa; % Second drift coefficient.
%mu1=0;
%mu2=0;
alpha=1;% x^alpha is being expanded. This is currently for monte carlo only.
alpha1=1-gamma;%This is for expansion of integrals for calculation of drift
%and volatility coefficients
%yy(1:Nn)=x0;
w0=x0^(1-gamma)/(1-gamma);
NnMid=((1+Nn)/2)*dNn;
Z(1:Nn)=(((1:Nn)*dNn)-NnMid);
%Z(1:Nn)=(((1:Nn)-20.5)*dNn-NnMid);
Z
str=input('Look at Z');
ZProb(1)=normcdf(.5*Z(1)+.5*Z(2),0,1)-normcdf(.5*Z(1)+.5*Z(2)-dNn,0,1);
ZProb(Nn)=normcdf(.5*Z(Nn)+.5*Z(Nn-1)+dNn,0,1)-normcdf(.5*Z(Nn)+.5*Z(Nn-1),0,1);
ZProb(2:Nn-1)=normcdf(.5*Z(2:Nn-1)+.5*Z(3:Nn),0,1)-normcdf(.5*Z(2:Nn-1)+.5*Z(1:Nn-2),0,1);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%This block is analytics for higher order monte carlo. This is explained in
sigma11(1:OrderA+1)=0;
mu11(1:OrderA+1)=0;
mu22(1:OrderA+1)=0;
sigma22(1:OrderA+1)=0;
% index 1 correponds to zero level since matlab indexing starts at one.
sigma11(1)=1;
mu11(1)=1;
mu22(1)=1;
sigma22(1)=1;
for k=1:(OrderA+1)
if sigma0~=0
sigma11(k)=sigma0^(k-1);
end
if mu1 ~= 0
mu11(k)=mu1^(k-1);
end
if mu2 ~= 0
mu22(k)=mu2^(k-1);
end
if sigma0~=0
sigma22(k)=sigma0^(2*(k-1));
end
end
%Ft(1:TtM+1,1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0; %General time powers on hermite polynomials
Fp(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;%General x powers on coefficients of hermite polynomials.
Fp1(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;%General x powers for bessel transformed coordinates.
%YCoeff0 and YCoeff are coefficents for original coordinates monte carlo.
%YqCoeff0 and YqCoeff are bessel/lamperti version monte carlo.
YCoeff0(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;
YqCoeff0(1:(OrderA+1),1:(OrderA+1),1:(OrderA+1),1:(OrderA+1))=0;
%Pre-compute the time and power exponent values in small multi-dimensional arrays
YCoeff = ItoTaylorCoeffsNew(alpha,beta1,beta2,gamma); %expand y^alpha where alpha=1;
YqCoeff = ItoTaylorCoeffsNew(alpha1,beta1,beta2,gamma);%expand y^alpha1 where alpha1=(1-gamma)
YqCoeff=YqCoeff/(1-gamma); %Transformed coordinates coefficients have to be
%further divided by (1-gamma)
for k = 0 : (OrderA)
for m = 0:k
l4 = k - m + 1;
for n = 0 : m
l3 = m - n + 1;
for j = 0:n
l2 = n - j + 1;
l1 = j + 1;
%Ft(l1,l2,l3,l4) = dtM^((l1-1) + (l2-1) + (l3-1) + .5* (l4-1));
Fp(l1,l2,l3,l4) = (alpha + (l1-1) * beta1 + (l2-1) * beta2 + (l3-1) * 2* gamma + (l4-1) * gamma ...
- (l1-1) - (l2-1) - 2* (l3-1) - (l4-1));
Fp1(l1,l2,l3,l4) = (alpha1 + (l1-1) * beta1 + (l2-1) * beta2 + (l3-1) * 2* gamma + (l4-1) * gamma ...
- (l1-1) - (l2-1) - 2* (l3-1) - (l4-1));
YCoeff0(l1,l2,l3,l4) =YCoeff(l1,l2,l3,l4).*mu11(l1).*mu22(l2).*sigma22(l3).*sigma11(l4);
YqCoeff0(l1,l2,l3,l4) =YqCoeff(l1,l2,l3,l4).*mu11(l1).*mu22(l2).*sigma22(l3).*sigma11(l4);
end
end
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Below analytics to calculate the Z-series of SDE at different time steps.
%We use the principle of addition of cumulants to find the resulting
%density of SDE at each step.
ExpnOrder=7;
SeriesOrder=7;
NMoments=8;
wnStart=1;
tic
%Tx is running time.
%T is terminal time.
%tt is time index.
Tx=0;
tt=0;
while(Tx<T)
tt=tt+1;
% t1=(tt-1)*dt;
% t2=(tt)*dt;
if(tt==1)
[wMu0dt,dwMu0dtdw] = BesselDriftAndDerivativesH0(w0,YqCoeff0,Fp1,gamma,ds(tt),8);
c0=wMu0dt+w0;
c(1)=sigma0*sqrt(ds(tt))
c(2:10)=0.0;
w0=c0;
Tx=Tx+ds(tt);
elseif(tt==2)
%dt=ds(tt);
%[wMu0dt,dwMu0dtdw] = BesselDriftAndDerivatives02A(w0,mu1,mu2,beta1,beta2,sigma0,gamma,ds(tt),8);
[wMu0dt,dwMu0dtdw] = BesselDriftAndDerivativesH0(w0,YqCoeff0,Fp1,gamma,ds(tt),8);
[b0,b] = CalculateDriftbCoeffs08(wMu0dt,dwMu0dtdw,c,SeriesOrder);
c0=c0+b0;
c(1)=b(1)+sqrt(c(1)^2+sigma0^2*ds(tt));
c(2)=b(2);
c(3)=b(3);
c(4)=b(4);
c(5)=b(5);
c(6)=b(6);
c(7)=b(7);
Tx=Tx+ds(tt);
else
%below wMu0dta is drift and dwMu0dtdwa are its derivatives.
%wMu1dt is its first derivative and dwMu1dtdw are derivatives of first
%derivative.
%[wMu0dta,dwMu0dtdwa,wMu1dt,dwMu1dtdw] = BesselDriftAndDerivatives08(w0,mu1,mu2,beta1,beta2,sigma0,gamma,dt,10);
%[wMu0dt,dwMu0dtdw] = BesselDriftAndDerivatives04A(w0,mu1,mu2,beta1,beta2,sigma0,gamma,ds(tt),8);
[wMu0dt,dwMu0dtdw] = BesselDriftAndDerivativesH0(w0,YqCoeff0,Fp1,gamma,ds(tt),8);
[b0,b] = CalculateDriftbCoeffs08(wMu0dt,dwMu0dtdw,c,SeriesOrder);
%Below Calculate original terms(associated with first hermite and second
%hermite polynomial in volatility)and its eight derivatives
[wVol0dt,dwVol0dtdw] = BesselVolAndDerivativesH1(w0,YqCoeff0,Fp1,gamma,ds(tt),8);
[wVol2dt,dwVol2dtdw] = BesselVolAndDerivativesH2(w0,YqCoeff0,Fp1,gamma,ds(tt),8);
%Below Calculate Z-Series expansions of volatility terms associated with
%first hermite polynomial and 2nd hermite polynomial.
[g0,g] = CalculateDriftbCoeffs08(wVol0dt,dwVol0dtdw,c,SeriesOrder);
[h0,h] = CalculateDriftbCoeffs08(wVol2dt,dwVol2dtdw,c,SeriesOrder);
%Principal term in first hermite was not included in expansion. It is
g0=g0+sigma0*sqrt(ds(tt));
%Below function calculates Moments and Cumulants of diffusion term of
%the SDE
[Moments,kk] = CalculateVolTermMoments(g0,g,h0,h,SeriesOrder,NMoments);
%Below drift series is linearly added to SDE series to get new intermediate
%SDE series.
c0=c0+b0;
c(1:7)=c(1:7)+b(1:7);
%Below calculate cumulants and moments of intermediate SDE series
[CC] = CalculateCumulants(c0,c,SeriesOrder,NMoments);
[MM1] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
%Below Calculate Cumulants of target density by arithmetic addition of
%first eight cumulants of diffusion term and intermediate SDE term.
k1=CC+kk;
%Below Calculate Standardized Cumulants of the target density.
k(1)=0;
k(2)=1;
k(3)=k1(3)/k1(2).^1.5;
k(4)=k1(4)/k1(2).^2.0;
k(5)=k1(5)/k1(2).^2.5;
k(6)=k1(6)/k1(2).^3.0;
k(7)=k1(7)/k1(2).^3.5;
k(8)=k1(8)/k1(2).^4.0;
%Below Calculate Standardized Moments of target density.
rmu(1)=k(1);
rmu(2)=k(1)*rmu(1)+k(2);
rmu(3)=k(1)*rmu(2)+2*k(2)*rmu(1)+k(3);
rmu(4)=k(1)*rmu(3)+3*k(2)*rmu(2)+3*k(3)*rmu(1)+k(4);
rmu(5)=k(1)*rmu(4)+4*k(2)*rmu(3)+6*k(3)*rmu(2)+4*k(4)*rmu(1)+k(5);
rmu(6)=k(1)*rmu(5)+5*k(2)*rmu(4)+10*k(3)*rmu(3)+10*k(4)*rmu(2)+5*k(5)*rmu(1)+k(6);
rmu(7)=k(1)*rmu(6)+6*k(2)*rmu(5)+15*k(3)*rmu(4)+20*k(4)*rmu(3)+15*k(5)*rmu(2)+6*k(6)*rmu(1)+k(7);
rmu(8)=k(1)*rmu(7)+7*k(2)*rmu(6)+21*k(3)*rmu(5)+35*k(4)*rmu(4)+35*k(5)*rmu(3)+21*k(6)*rmu(2)+7*k(7)*rmu(1)+k(8);
%We provide initial guess for calibration procedure
if(tt==3)
a0Guess=(c0-k1(1))/sqrt(CC(2));
aGuess(1)=c(1)/sqrt(CC(2));
aGuess(2:7)=c(2:7)/sqrt(CC(2));
else
a0Guess=(c0-k1(1))/sqrt(CC(2));
aGuess(1:7)=(c(1:7))/sqrt(CC(2));
end
%We calculate coefficients of Standardized target density
[c0,c] = CalculateDensityFromStandardizedMoments(rmu,a0Guess,aGuess);
%From standardized Z-seriesCoefficients of target density, we change to
%actual non-standardized Z-series coefficients.
c0=c0*sqrt(k1(2))+k1(1);
c=c*sqrt(k1(2));
%Actual Raw Moments after the Calculation of target density.
[MM2] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
%Calculate the Ratio of Moments between intermediate density and the
%new target density
MomentsIncreaseRatio=max(MM2(2:8)./MM1(2:8));
%Below decrease the step size if MomentsIncreaseRatio is larger
%than MaxMomentRatio and increase the step size if
%MomentsIncreaseRatio is smaller than MinMomentRatio
IncreaseRatioFlag=0;
DecreaseRatioFlag=0;
if(MomentsIncreaseRatio>MaxMomentRatio)
if (ds(tt)>MinStepSize)
DecreaseRatioFlag=1;
end
elseif (MomentsIncreaseRatio<MinMomentRatio)
if (ds(tt)<MaxStepSize)
IncreaseRatioFlag=1;
end
end
if(DecreaseRatioFlag==1)
ds(tt+1)=ds(tt)/2;
elseif(IncreaseRatioFlag==1)
ds(tt+1)=ds(tt)*2;
else
ds(tt+1)=ds(tt);
end
%Tx is running time.
Tx=Tx+ds(tt);
%At last step make sure that step size is appropriate.
if(Tx+ds(tt+1)>T)
ds(tt+1)=T-Tx;
end
%Assign the median of density to w0
w0=c0;
end
end
ttFirst=tt;
wnStart=1;
%Below w which is the SDE in bessel coordinates and it is calulated from
%Z-series in Bessel coordinates.
w(1:Nn)=c0;
for nn=1:ExpnOrder
w(1:Nn)=w(1:Nn)+c(nn)*Z(1:Nn).^nn;
end
Flag=0;
for nn=Nn-1:-1:1
if((w(nn)<0)&&(Flag==0))
wnStart=nn+1;
Flag=1;
end
end
%Below we directly transform the density of SDE in bessel coordinates into
%density of SDE in original coordinates.
yy(wnStart:Nn)=((1-gamma).*w(wnStart:Nn)).^(1/(1-gamma));
%yy0=((1-gamma).*w0).^(1/(1-gamma));
%Below we make calculations of density in original coordinates by doing
%change of probability derivative with respect to gaussian density.
Dfyy(wnStart:Nn)=0;
for nn=wnStart+1:Nn-1
Dfyy(nn) = (yy(nn + 1) - yy(nn - 1))/(Z(nn + 1) - Z(nn - 1));
%Change of variable derivative for densities
end
Dfyy(Nn)=Dfyy(Nn-1);
Dfyy(1)=Dfyy(2);
pyy(1:Nn)=0;
for nn = wnStart:Nn
pyy(nn) = (normpdf(Z(nn),0, 1))/abs(Dfyy(nn));
end
toc
%Below we do calculations for change of Z-series coefficients from
%coordinates in Bessel to Z-series coefficients in original coordinates of
%the SDE.
%The new series is expanded around c0 which is median of the density in
%Bessel coordinates.
Bmu=c0;%+c(2)+3*c(4)+15*c(6)
Bmu
%Calculate the first seven derivatives of transformation from Bessel
%coordinates to original coordinates of SDE at the median value Bmu=c0.
yB0=((1-gamma).*Bmu).^(1/(1-gamma));
dyB0(1)=((1-gamma).*Bmu).^(1/(1-gamma)-1);
dyB0(2)=(1/(1-gamma)-1).*((1-gamma).*Bmu).^(1/(1-gamma)-2)*(1-gamma);
dyB0(3)=(1/(1-gamma)-1).*(1/(1-gamma)-2).*((1-gamma).*Bmu).^(1/(1-gamma)-3)*(1-gamma)^2;
dyB0(4)=(1/(1-gamma)-1).*(1/(1-gamma)-2)*(1/(1-gamma)-3)*((1-gamma).*Bmu).^(1/(1-gamma)-4)*(1-gamma)^3;
dyB0(5)=(1/(1-gamma)-1).*(1/(1-gamma)-2)*(1/(1-gamma)-3)*(1/(1-gamma)-4)*((1-gamma).*Bmu).^(1/(1-gamma)-5)*(1-gamma)^4;
dyB0(6)=(1/(1-gamma)-1).*(1/(1-gamma)-2)*(1/(1-gamma)-3)*(1/(1-gamma)-4)*(1/(1-gamma)-5)*((1-gamma).*Bmu).^(1/(1-gamma)-6)*(1-gamma)^5;
dyB0(7)=(1/(1-gamma)-1).*(1/(1-gamma)-2)*(1/(1-gamma)-3)*(1/(1-gamma)-4)*(1/(1-gamma)-5)*(1/(1-gamma)-6)*((1-gamma).*Bmu).^(1/(1-gamma)-7)*(1-gamma)^6;
% dyB0(8)=(1/(1-gamma)-1).*(1/(1-gamma)-2)*(1/(1-gamma)-3)*(1/(1-gamma)-4)*(1/(1-gamma)-5)*(1/(1-gamma)-6)*(1/(1-gamma)-7)*((1-gamma).*Bmu).^(1/(1-gamma)-8)*(1-gamma)^7;
% dyB0(9)=(1/(1-gamma)-1).*(1/(1-gamma)-2)*(1/(1-gamma)-3)*(1/(1-gamma)-4)*(1/(1-gamma)-5)*(1/(1-gamma)-6)*(1/(1-gamma)-7)*(1/(1-gamma)-8)*((1-gamma).*Bmu).^(1/(1-gamma)-9)*(1-gamma)^8;
% dyB0(10)=(1/(1-gamma)-1).*(1/(1-gamma)-2)*(1/(1-gamma)-3)*(1/(1-gamma)-4)*(1/(1-gamma)-5)*(1/(1-gamma)-6)*(1/(1-gamma)-7)*(1/(1-gamma)-8)*(1/(1-gamma)-9)*((1-gamma).*Bmu).^(1/(1-gamma)-10)*(1-gamma)^9;
c(8:10)=0;
%Below finally calculate the Z-series coefficients from Bessel coordinates
%to original coordinates using derivatives calculated at previous step.
[d0,d] = CalculateDriftbCoeffs08(yB0,dyB0,c,7);
%Below calculate SDE variable in original coordinates after expanding the
%Z-series coefficients in original coordinates.
Yw(1:Nn)=d0;
for nn=1:7
Yw(1:Nn)=Yw(1:Nn)+d(nn)*Z(1:Nn).^nn;
end
%Below calculate the density from Z-series Coefficients in original
%coordinates. We want to see how exact they are since we would need to use
%them in analytics later when calculating integrals of variance for
%stochastic volatility models.
DfYw(wnStart:Nn)=0;
for nn=wnStart+1:Nn-1
DfYw(nn) = (Yw(nn + 1) - Yw(nn - 1))/(Z(nn + 1) - Z(nn - 1));
%Change of variable derivative for densities
end
DfYw(Nn)=DfYw(Nn-1);
DfYw(1)=DfYw(2);
pYw(1:Nn)=0;
for nn = wnStart:Nn
pYw(nn) = (normpdf(Z(nn),0, 1))/abs(DfYw(nn));
end
%AnalyticMean1 is calculated from direct transformation of density in Bessel coordinates
AnalyticMean1=sum(yy(wnStart:Nn).*ZProb(wnStart:Nn)) %Original process average from transformed density
%AnalyticMean2 is calculated from Z-series coordinates of density in Original coordinates
AnalyticMean2=d0+d(2)+3*d(4)+15*d(6) %Original process average from coordinates
disp('true Mean only applicable to standard SV mean reverting type models otherwise disregard');
TrueMean=theta+(x0-theta)*exp(-kappa*T)%Mean reverting SDE original variable true average
%yy0
%yB0
%theta1=1;
%Monte Carlo starts here. For details see my thread on wilmott.com
%where I gave details of higher order monte carlo.
rng(29079137, 'twister')
paths=50000;
YY(1:paths)=x0; %Original process monte carlo.
Random1(1:paths)=0;
for tt=1:TtM
Random1=randn(size(Random1));
HermiteP1(1,1:paths)=1;
HermiteP1(2,1:paths)=Random1(1:paths);
HermiteP1(3,1:paths)=Random1(1:paths).^2-1;
HermiteP1(4,1:paths)=Random1(1:paths).^3-3*Random1(1:paths);
HermiteP1(5,1:paths)=Random1(1:paths).^4-6*Random1(1:paths).^2+3;
YY(1:paths)=YY(1:paths) + ...
(YCoeff0(1,1,2,1).*YY(1:paths).^Fp(1,1,2,1)+ ...
YCoeff0(1,2,1,1).*YY(1:paths).^Fp(1,2,1,1)+ ...
YCoeff0(2,1,1,1).*YY(1:paths).^Fp(2,1,1,1))*dtM + ...
(YCoeff0(1,1,3,1).*YY(1:paths).^Fp(1,1,3,1)+ ...
YCoeff0(1,2,2,1).*YY(1:paths).^Fp(1,2,2,1)+ ...
YCoeff0(2,1,2,1).*YY(1:paths).^Fp(2,1,2,1)+ ...
YCoeff0(1,3,1,1).*YY(1:paths).^Fp(1,3,1,1)+ ...
YCoeff0(2,2,1,1).*YY(1:paths).^Fp(2,2,1,1)+ ...
YCoeff0(3,1,1,1).*YY(1:paths).^Fp(3,1,1,1))*dtM^2 + ...
((YCoeff0(1,1,1,2).*YY(1:paths).^Fp(1,1,1,2).*sqrt(dtM))+ ...
(YCoeff0(1,1,2,2).*YY(1:paths).^Fp(1,1,2,2)+ ...
YCoeff0(1,2,1,2).*YY(1:paths).^Fp(1,2,1,2)+ ...
YCoeff0(2,1,1,2).*YY(1:paths).^Fp(2,1,1,2)).*dtM^1.5) .*HermiteP1(2,1:paths) + ...
((YCoeff0(1,1,1,3).*YY(1:paths).^Fp(1,1,1,3) *dtM) + ...
(YCoeff0(1,1,2,3).*YY(1:paths).^Fp(1,1,2,3)+ ...
YCoeff0(1,2,1,3).*YY(1:paths).^Fp(1,2,1,3)+ ...
YCoeff0(2,1,1,3).*YY(1:paths).^Fp(2,1,1,3)).*dtM^2).*HermiteP1(3,1:paths) + ...
((YCoeff0(1,1,1,4).*YY(1:paths).^Fp(1,1,1,4)*dtM^1.5 )).*HermiteP1(4,1:paths) + ...
(YCoeff0(1,1,1,5).*YY(1:paths).^Fp(1,1,1,5)*dtM^2.0).*HermiteP1(5,1:paths);
end
YY(YY<0)=0;
disp('Original process average from monte carlo');
MCMean=sum(YY(:))/paths %origianl coordinates monte carlo average.
disp('Original process average from our simulation');
ItoHermiteMean=sum(yy(wnStart:Nn).*ZProb(wnStart:Nn)) %Original process average from coordinates
disp('true Mean only applicble to standard SV mean reverting type models otherwise disregard');
TrueMean=theta+(x0-theta)*exp(-kappa*T)%Mean reverting SDE original variable true average
MaxCutOff=100;
NoOfBins=round(1*500*gamma^2*4*sigma0/sqrt(MCMean)/(1+kappa));%Decrease the number of bins if the graph is too
[YDensity,IndexOutY,IndexMaxY] = MakeDensityFromSimulation_Infiniti_NEW(YY,paths,NoOfBins,MaxCutOff );
plot(Yw(wnStart+1:Nn-1),pYw(wnStart+1:Nn-1),'b',yy(wnStart+1:Nn-1),pyy(wnStart+1:Nn-1),'r',IndexOutY(1:IndexMaxY),YDensity(1:IndexMaxY),'g');
%plot(y_w(wnStart+1:Nn-1),fy_w(wnStart+1:Nn-1),'r',IndexOutY(1:IndexMaxY),YDensity(1:IndexMaxY),'g',Z(wnStart+1:Nn-1),fy_w(wnStart+1:Nn-1),'b');
dtAvg=T/ttFirst;
title(sprintf('x0 = %.4f,theta=%.3f,kappa=%.2f,gamma=%.3f,sigma=%.2f,T=%.2f,dtAvg=%.5f,M=%.4f,TM=%.4f', x0,theta,kappa,gamma,sigma0,T,dtAvg,ItoHermiteMean,TrueMean));%,sprintf('theta= %f', theta), sprintf('kappa = %f', kappa),sprintf('sigma = %f', sigma0),sprintf('T = %f', T));
legend({'SDE density From Z-series in Original Coordinates','SDE density Transformed Directly From Bessel Coordinates','Monte Carlo Density'},'Location','northeast')
str=input('red line is density of SDE from Addition of Cumulants, green is monte carlo.');
plot((wnStart+1:Nn-1)*dNn-NnMid,Yw(wnStart+1:Nn-1),'b',(wnStart+1:Nn-1)*dNn-NnMid,yy(wnStart+1:Nn-1),'r');
title('SDE Stochastic Variable as a Function of Standard Normal');%,sprintf('theta= %f', theta), sprintf('kappa = %f', kappa),sprintf('sigma = %f', sigma0),sprintf('T = %f', T));
legend({'SDE Variable From Z-series in Original Coordinates','SDE Variable Transformed Directly From Bessel Coordinates'},'Location','northwest')
str=input('red line is density of SDE from addition of Cumulants method, green is monte carlo.');
end
.
.
.
function [Y] = ItoTaylorCoeffsNew(alpha,beta1,beta2,gamma)
%In the coefficient calculation program which calculates Y(l1,l2,l3,l4),
%I have used four levels of looping each for relevant expansion order.
%The first loop takes four values and second loop takes 16 values and
%third loop takes 64 values and so on. And then each coefficient
%term can be individually calculated while carefully accounting
%for path dependence.
%So for example in a nested loop structure
%m1= 1:mDim
% m2=1:mDim
% m3=1:mDim
% l(m1)=l(m1)+1;
% l(m2)=l(m2)+1;
% l(m3)=l(m3)+1;
%in the above looping loop takes values from one to four with one
%indicating the first drift term, two indicating the second drift term
%and three indicating quadratic variation term and
%four indicating the volatility term. And with this looping structure
%we can So in the above looping m1=1 would mean that all terms are
%descendent of first drift term and m2=4 would mean that all terms are
%descendent of first drift term on first expansion order and descendent
%of volatility term on the second order and so we can keep track of path
%dependence perfectly.
%And on each level, we individually calculate the descendent terms. While
%keeping track of path dependence and calculating the coefficients with
%careful path dependence consideration, we update the appropriate element
%in our polynomial like expansion coefficient array
%explaining the part of code
%m1= 1:mDim
% m2=1:mDim
% m3=1:mDim
% l(m1)=l(m1)+1;
% l(m2)=l(m2)+1;
% l(m3)=l(m3)+1;
%Y(l(1)+1,l(2),l(3),l(4))=Y(l(1)+1,l(2),l(3),l(4))+Coeff1st*IntegralCoeff(1,1,1,2);
%Here l(1) denotes l1 but written as l(1) so it can be conveniently
%updated with the loop variable when the loop variable takes value one
%indicating first drift term . And l(2) could be conveniently updated when
%the loop variable takes value equal to two indicating second
%drift term and so on.
%Here is the part of code snippet for that
%for m1=1:mDim
% l(1)=1;
% l(2)=1;
% l(3)=1;
% l(4)=1;
% l(m1)=l(m1)+1;
%CoeffDX1 = alpha + (l(1)-1) *beta1 + (l(2)-1) *beta2 + (l(3)-1) *2*gamma + (l(4)-1)*gamma - (l(1)-1) - (l(2)-1) - 2*(l(3)-1) - (l(4)-1);
% CoeffDX2 = CoeffDX1 - 1;
% ArrIndex0=m1;
% ArrIndex=(m1-1)*mDim;
% Coeff1st=Y1(ArrIndex0)*CoeffDX1;
% Coeff2nd=Y1(ArrIndex0)*.5*CoeffDX1*CoeffDX2;
% Y2(1+ArrIndex)=Coeff1st;
% Y(l(1)+1,l(2),l(3),l(4))=Y(l(1)+1,l(2),l(3),l(4))+Coeff1st*IntegralCoeff(1,1,2,n1(m1));
% Y2(2+ArrIndex)=Coeff1st;
% Y(l(1),l(2)+1,l(3),l(4))=Y(l(1),l(2)+1,l(3),l(4))+Coeff1st*IntegralCoeff(1,1,2,n1(m1));
% Y2(3+ArrIndex)=Coeff2nd;
% Y(l(1),l(2),l(3)+1,l(4))=Y(l(1),l(2),l(3)+1,l(4))+Coeff2nd*IntegralCoeff(1,1,2,n1(m1));
% Y2(4+ArrIndex)=Coeff1st;
% Y(l(1),l(2),l(3),l(4)+1)=Y(l(1),l(2),l(3),l(4)+1)+Coeff1st*IntegralCoeff(1,1,3,n1(m1));
%The first four lines update the array indices according to the parent term.
%And then CoeffDX1 and CoeffDX2 are calculated according to algebraic exponents on parent terms.
%ArrIndex0=m1; calculates the array index of the parent term
%And ArrIndex=(m1-1)*mDim; calculates the array index of the descendent terms
%And coefficient of the drift and volatility descendent terms is
%calculated by multiplying the coefficient of the parent term by
%Coeff1st=Y1(ArrIndex0)*CoeffDX1;
%And coefficient of the quadratic variation descendent terms is
%calculated by multiplying the coefficient of the parent term by
%Coeff2nd=Y1(ArrIndex0)*.5*CoeffDX1*CoeffDX2;
%And then each of the four descendent terms are updated with Coeff1st
%if they are drift or volatility descendent terms or Coeff2nd if
%they are quadratic variation descendent terms.
%Here Y1 indicates the temporary coefficient array with parent terms on
%first level. And Y2 denotes temporary coefficient array with parent terms
%on second level and Y3 indicates temporary coefficient array with parent terms on third level.
[IntegralCoeff,IntegralCoeffdt,IntegralCoeffdz] = ComputeIntegralCoeffs();
n1(1)=2;
n1(2)=2;
n1(3)=2;
n1(4)=3;
%n1(1), n1(2), n1(3) are drift and quadratic variation term variables
%and take a value equal to 2 indicating a dt integral.
%n1(4) is volatility term variable and indicates a dz-integral by taking a
%value of 3.
mDim=4; % four descendent terms in each expansion
Y(1:5,1:5,1:5,1:5)=0;
Y1(1:mDim)=0;
Y2(1:mDim*mDim)=0;
Y3(1:mDim*mDim*mDim)=0;
%First Ito-hermite expansion level starts here. No loops but four
%descendent terms.
l(1)=1;
l(2)=1;
l(3)=1;
l(4)=1;
CoeffDX1 = alpha;
CoeffDX2 = CoeffDX1 - 1;
Coeff1st=CoeffDX1;
Coeff2nd=.5*CoeffDX1*CoeffDX2;
Y1(1)=Coeff1st;
Y(l(1)+1,l(2),l(3),l(4))=Y(l(1)+1,l(2),l(3),l(4))+Coeff1st*IntegralCoeff(1,1,1,2);
Y1(2)=Coeff1st;
Y(l(1),l(2)+1,l(3),l(4))=Y(l(1),l(2)+1,l(3),l(4))+Coeff1st*IntegralCoeff(1,1,1,2);
Y1(3)=Coeff2nd;
Y(l(1),l(2),l(3)+1,l(4))=Y(l(1),l(2),l(3)+1,l(4))+Coeff2nd*IntegralCoeff(1,1,1,2);
Y1(4)=Coeff1st;
Y(l(1),l(2),l(3),l(4)+1)=Y(l(1),l(2),l(3),l(4)+1)+Coeff1st*IntegralCoeff(1,1,1,3);
%Second Ito-hermite expansion level starts. It has a loop over four parent
%terms and there are four descendent terms for each parent term.
%The coefficient terms are then lumped in a polynomial-like expansion
%array of coefficients.
for m1=1:mDim
l(1)=1;
l(2)=1;
l(3)=1;
l(4)=1;
l(m1)=l(m1)+1;
CoeffDX1 = alpha + (l(1)-1) *beta1 + (l(2)-1) *beta2 + (l(3)-1) *2*gamma + (l(4)-1)*gamma ...
- (l(1)-1) - (l(2)-1) - 2*(l(3)-1) - (l(4)-1);
CoeffDX2 = CoeffDX1 - 1;
ArrIndex0=m1;
ArrIndex=(m1-1)*mDim;
Coeff1st=Y1(ArrIndex0)*CoeffDX1;
Coeff2nd=Y1(ArrIndex0)*.5*CoeffDX1*CoeffDX2;
Y2(1+ArrIndex)=Coeff1st;
Y(l(1)+1,l(2),l(3),l(4))=Y(l(1)+1,l(2),l(3),l(4))+Coeff1st*IntegralCoeff(1,1,2,n1(m1));
Y2(2+ArrIndex)=Coeff1st;
Y(l(1),l(2)+1,l(3),l(4))=Y(l(1),l(2)+1,l(3),l(4))+Coeff1st*IntegralCoeff(1,1,2,n1(m1));
Y2(3+ArrIndex)=Coeff2nd;
Y(l(1),l(2),l(3)+1,l(4))=Y(l(1),l(2),l(3)+1,l(4))+Coeff2nd*IntegralCoeff(1,1,2,n1(m1));
Y2(4+ArrIndex)=Coeff1st;
Y(l(1),l(2),l(3),l(4)+1)=Y(l(1),l(2),l(3),l(4)+1)+Coeff1st*IntegralCoeff(1,1,3,n1(m1));
%Third Ito-hermite expansion level starts and it is a nested loop with
%a total of sixteen parents and each parent takes four descendent
%terms.
for m2=1:mDim
l(1)=1;
l(2)=1;
l(3)=1;
l(4)=1;
l(m1)=l(m1)+1;
l(m2)=l(m2)+1;
CoeffDX1 = alpha + (l(1)-1) *beta1 + (l(2)-1) *beta2 + (l(3)-1) *2*gamma + (l(4)-1)*gamma ...
- (l(1)-1) - (l(2)-1) - 2*(l(3)-1) - (l(4)-1);
CoeffDX2=CoeffDX1-1;
ArrIndex0=(m1-1)*mDim+m2;
ArrIndex=((m1-1)*mDim+(m2-1))*mDim;
Coeff1st=Y2(ArrIndex0)*CoeffDX1;
Coeff2nd=Y2(ArrIndex0)*.5*CoeffDX1*CoeffDX2;
Y3(1+ArrIndex)=Coeff1st;
Y(l(1)+1,l(2),l(3),l(4))=Y(l(1)+1,l(2),l(3),l(4))+Coeff1st*IntegralCoeff(1,2,n1(m2),n1(m1));
Y3(2+ArrIndex)=Coeff1st;
Y(l(1),l(2)+1,l(3),l(4))=Y(l(1),l(2)+1,l(3),l(4))+Coeff1st*IntegralCoeff(1,2,n1(m2),n1(m1));
Y3(3+ArrIndex)=Coeff2nd;
Y(l(1),l(2),l(3)+1,l(4))=Y(l(1),l(2),l(3)+1,l(4))+Coeff2nd*IntegralCoeff(1,2,n1(m2),n1(m1));
Y3(4+ArrIndex)=Coeff1st;
Y(l(1),l(2),l(3),l(4)+1)=Y(l(1),l(2),l(3),l(4)+1)+Coeff1st*IntegralCoeff(1,3,n1(m2),n1(m1));
%fourht Ito-hermite expansion level starts and it is a triply-nested loop with
%a total of sixteen parents and each parent takes four descendent
%terms. We then lump the terms in a relatively sparse polynomial
%like expansion coefficient array that has smaller number of
%non-zero terms.
for m3=1:mDim
l(1)=1;
l(2)=1;
l(3)=1;
l(4)=1;
l(m1)=l(m1)+1;
l(m2)=l(m2)+1;
l(m3)=l(m3)+1;
CoeffDX1 = alpha + (l(1)-1) *beta1 + (l(2)-1) *beta2 + (l(3)-1) *2*gamma + (l(4)-1)*gamma ...
- (l(1)-1) - (l(2)-1) - 2*(l(3)-1) - (l(4)-1);
CoeffDX2=CoeffDX1-1;
ArrIndex0=((m1-1)*mDim+(m2-1))*mDim+m3;
%ArrIndex=(((m1-1)*mDim+(m2-1))*mDim+(m3-1))*mDim;
Coeff1st=Y3(ArrIndex0)*CoeffDX1;
Coeff2nd=Y3(ArrIndex0)*.5*CoeffDX1*CoeffDX2;
%Y4(1+ArrIndex)=Coeff1st;
Y(l(1)+1,l(2),l(3),l(4))=Y(l(1)+1,l(2),l(3),l(4))+Coeff1st*IntegralCoeff(2,n1(m3),n1(m2),n1(m1));
%Y4(2+ArrIndex)=Coeff1st;
Y(l(1),l(2)+1,l(3),l(4))=Y(l(1),l(2)+1,l(3),l(4))+Coeff1st*IntegralCoeff(2,n1(m3),n1(m2),n1(m1));
%Y4(3+ArrIndex)=Coeff2nd;
Y(l(1),l(2),l(3)+1,l(4))=Y(l(1),l(2),l(3)+1,l(4))+Coeff2nd*IntegralCoeff(2,n1(m3),n1(m2),n1(m1));
%Y4(4+ArrIndex)=Coeff1st;
Y(l(1),l(2),l(3),l(4)+1)=Y(l(1),l(2),l(3),l(4)+1)+Coeff1st*IntegralCoeff(3,n1(m3),n1(m2),n1(m1));
end
end
end
end
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.
.
function [IntegralCoeff0,IntegralCoeffdt,IntegralCoeffdz] = ComputeIntegralCoeffs()
IntegralCoeff(1:3,1:3,1:3,1:3)=0;
IntegralCoeffdt(1:3,1:3,1:3,1:3)=0;
IntegralCoeffdz(1:3,1:3,1:3,1:3)=0;
IntegralCoeff0(1:3,1:3,1:3,1:3)=0;
IntegralCoeff0(1,1,1,2)=1;
IntegralCoeff0(1,1,1,3)=1;
IntegralCoeff0(1,1,2,2)=1/2;
IntegralCoeff0(1,1,3,2)=1-1/sqrt(3);
IntegralCoeff0(1,1,2,3)=1/sqrt(3);
IntegralCoeff0(1,1,3,3)=1/2;
IntegralCoeff0(1,2,2,2)=1/6;
IntegralCoeff0(1,3,2,2)=(1-1/sqrt(3))*1/2*(1-1/sqrt(5));
IntegralCoeff0(1,2,3,2)=1/sqrt(3)/2*(1-1/sqrt(5));
IntegralCoeff0(1,3,3,2)=1/2*(1-sqrt(2)/2);
IntegralCoeff0(1,2,2,3)=1/2*1/sqrt(5);
IntegralCoeff0(1,3,2,3)=(1-1/sqrt(3))*1/sqrt(2)*1/2;
IntegralCoeff0(1,2,3,3)=1/sqrt(3)/sqrt(2)*1/(2);
IntegralCoeff0(1,3,3,3)=1/6;
IntegralCoeff0(2,2,2,2)=1/24;
IntegralCoeff0(2,2,3,2)=1/2*1/sqrt(5)*1/3*(1-1/sqrt(7));
IntegralCoeff0(2,3,2,2)=1/sqrt(3)*1/2*(1-1/sqrt(5))*1/3*(1-1/sqrt(7));
IntegralCoeff0(3,2,2,2)=(1-1/sqrt(3))*1/2*(1-1/sqrt(5))*1/3*(1-1/sqrt(7));
IntegralCoeff0(3,3,2,2)=1/2*(1-sqrt(2)/2)*1/2*(1-sqrt(2)/sqrt(6));
IntegralCoeff0(2,3,3,2)=1/sqrt(3)*1/sqrt(2)*1/2*1/2*(1-sqrt(2)/sqrt(6));
IntegralCoeff0(3,2,3,2)=(1-1/sqrt(3))*1/sqrt(2)*1/2*1/2*(1-sqrt(2)/sqrt(6));
IntegralCoeff0(3,3,3,2)=1/6*(1-sqrt(3)/sqrt(5));
IntegralCoeff0(2,2,2,3)=1/6*1/sqrt(7);
IntegralCoeff0(2,2,3,3)=1/2*1/sqrt(5)*1/sqrt(2)*1/sqrt(6);
IntegralCoeff0(2,3,2,3)=1/sqrt(3)*1/2*(1-1/sqrt(5))*1/sqrt(2)*1/sqrt(6);
IntegralCoeff0(3,2,2,3)=(1-1/sqrt(3))*1/2*(1-1/sqrt(5))*1/sqrt(2)*1/sqrt(6);
IntegralCoeff0(3,3,2,3)=1/2*(1-sqrt(2)/2)*1/sqrt(3)*1/sqrt(5);
IntegralCoeff0(2,3,3,3)=1/sqrt(3)*1/sqrt(2)*1/sqrt(4)*1/sqrt(3)*(1/sqrt(5));
IntegralCoeff0(3,2,3,3)=(1-1/sqrt(3))*1/sqrt(2)*1/2*1/sqrt(3)*1/sqrt(5);
IntegralCoeff0(3,3,3,3)=1/24;
%Can also be calculated with algorithm below.
%here IntegralCoeffdt indicates the coefficients of a dt-integral.
%This dt-integral means that you are calculating the Ito-hermite expansion
%of Integral X(t)^alpha dt with X(t) dynamics given by the SDE
%here IntegralCoeffdz indicates the coefficients of a dz-integral.
%This dz-integral means that you are calculating the Ito-hermite expansion
%of Integral X(t)^alpha dz(t) with X(t) dynamics given by the SDE
%IntegralCoeff is is associated with expansion of X(t)^alpha.
%IntegralCoeff below is not returned and IntegralCoeff0 manually calculated
%above is returned but both are the same.
l0(1:2)=1;
for m4=1:2
l0(1)=1;
l0(2)=1;
%IntegralCoeff4(m4,1,1,1)=1;
%IntegralCoeff4(m4,1,1,1)=1;
%1 is added to m4 since index 1 stands for zero, 2 for one and three
%for two.
IntegralCoeff(1,1,1,m4+1)=1;
l0(m4)=l0(m4)+1;
IntegralCoeffdt(1,1,1,m4+1)=IntegralCoeff(1,1,1,m4+1)* ...
1/(l0(1)-1+1)*(1-sqrt(l0(2)-1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+2));
IntegralCoeffdz(1,1,1,m4+1)= IntegralCoeff(1,1,1,m4+1)* ...
1/sqrt(l0(2)-1+1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+1);
for m3=1:2
l0(1)=1;
l0(2)=1;
l0(m4)=l0(m4)+1;
if(m3==1)
IntegralCoeff(1,1,m4+1,m3+1)=1/(l0(1)-1+1)*(1-sqrt(l0(2)-1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+2));
end
if(m3==2)
IntegralCoeff(1,1,m4+1,m3+1)= 1/sqrt(l0(2)-1+1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+1);
end
l0(m3)=l0(m3)+1;
%IntegralCoeff(1,1,m4+1,m3+1)=IntegralCoeff4(m4,m3,1,1);
IntegralCoeffdt(1,1,m4+1,m3+1)=IntegralCoeff(1,1,m4+1,m3+1)* ...
1/(l0(1)-1+1)*(1-sqrt(l0(2)-1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+2));
IntegralCoeffdz(1,1,m4+1,m3+1)= IntegralCoeff(1,1,m4+1,m3+1)* ...
1/sqrt(l0(2)-1+1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+1);
for m2=1:2
l0(1)=1;
l0(2)=1;
l0(m4)=l0(m4)+1;
l0(m3)=l0(m3)+1;
if(m2==1)
IntegralCoeff(1,m4+1,m3+1,m2+1)=IntegralCoeff(1,1,m4+1,m3+1)*1/(l0(1)-1+1)*(1-sqrt(l0(2)-1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+2));
end
if(m2==2)
IntegralCoeff(1,m4+1,m3+1,m2+1)= IntegralCoeff(1,1,m4+1,m3+1)*1/sqrt(l0(2)-1+1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+1);
end
l0(m2)=l0(m2)+1;
%IntegralCoeff(1,m4+1,m3+1,m2+1)=IntegralCoeff4(m4,m3,m2,1);
IntegralCoeffdt(1,m4+1,m3+1,m2+1)=IntegralCoeff(1,m4+1,m3+1,m2+1)* ...
1/(l0(1)-1+1)*(1-sqrt(l0(2)-1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+2));
IntegralCoeffdz(1,m4+1,m3+1,m2+1)= IntegralCoeff(1,m4+1,m3+1,m2+1)* ...
1/sqrt(l0(2)-1+1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+1);
for m1=1:2
l0(1)=1;
l0(2)=1;
l0(m4)=l0(m4)+1;
l0(m3)=l0(m3)+1;
l0(m2)=l0(m2)+1;
if(m1==1)
IntegralCoeff(m4+1,m3+1,m2+1,m1+1)=IntegralCoeff(1,m4+1,m3+1,m2+1)*1/(l0(1)-1+1)*(1-sqrt(l0(2)-1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+2));
end
if(m1==2)
IntegralCoeff(m4+1,m3+1,m2+1,m1+1)= IntegralCoeff(1,m4+1,m3+1,m2+1)*1/sqrt(l0(2)-1+1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+1);
end
l0(m1)=l0(m1)+1;
%IntegralCoeff(m4+1,m3+1,m2+1,m1+1)=IntegralCoeff4(m4,m3,m2,m1);
IntegralCoeffdt(m4+1,m3+1,m2+1,m1+1)=IntegralCoeff(m4+1,m3+1,m2+1,m1+1)* ...
1/(l0(1)-1+1)*(1-sqrt(l0(2)-1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+2));
IntegralCoeffdz(m4+1,m3+1,m2+1,m1+1)= IntegralCoeff(m4+1,m3+1,m2+1,m1+1)* ...
1/sqrt(l0(2)-1+1)/sqrt(2*(l0(1)-1)+(l0(2)-1)+1);
end
end
end
end
end
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.
.
function [wMu0dt,dwMu0dtdw] = BesselDriftAndDerivativesH0(w0,YqCoeff0,Fp1,gamma,dt,ExpnOrder)
NoOfTerms=19;
%NoOfTerms=9;
YqCoeffa(1:NoOfTerms)=0.0;
%Fp1
Fp1=Fp1/(1-gamma);
%Fp1
%gamma
YqCoeffa(1)=YqCoeff0(1,1,2,1)*dt;%*(1-gamma)^Fp1(1,1,2,1)*dt;
YqCoeffa(2)=YqCoeff0(1,2,1,1)*dt;%*(1-gamma)^Fp1(1,2,1,1)*dt;
YqCoeffa(3)=YqCoeff0(2,1,1,1)*dt;%*(1-gamma)^Fp1(2,1,1,1)*dt;
YqCoeffa(4)=YqCoeff0(1,1,3,1)*dt^2;%*(1-gamma)^Fp1(1,1,3,1)*dt^2;
YqCoeffa(5)=YqCoeff0(1,2,2,1)*dt^2;%*(1-gamma)^Fp1(1,2,2,1)*dt^2;
YqCoeffa(6)=YqCoeff0(2,1,2,1)*dt^2;%*(1-gamma)^Fp1(2,1,2,1)*dt^2;
YqCoeffa(7)=YqCoeff0(1,3,1,1)*dt^2;%*(1-gamma)^Fp1(1,3,1,1)*dt^2;
YqCoeffa(8)=YqCoeff0(2,2,1,1)*dt^2;%*(1-gamma)^Fp1(2,2,1,1)*dt^2;
YqCoeffa(9)=YqCoeff0(3,1,1,1)*dt^2;%*(1-gamma)^Fp1(3,1,1,1)*dt^2;
YqCoeffa(10)=YqCoeff0(1,1,4,1)*dt^3;%;*(1-gamma)^Fp1(1,1,4,1)*dt^3;
YqCoeffa(11)=YqCoeff0(1,2,3,1)*dt^3;%;*(1-gamma)^Fp1(1,2,3,1)*dt^3;
YqCoeffa(12)=YqCoeff0(2,1,3,1)*dt^3;%*(1-gamma)^Fp1(2,1,3,1)*dt^3;
YqCoeffa(13)=YqCoeff0(1,3,2,1)*dt^3;%*(1-gamma)^Fp1(1,3,2,1)*dt^3;
YqCoeffa(14)=YqCoeff0(2,2,2,1)*dt^3;%*(1-gamma)^Fp1(2,2,2,1)*dt^3;
YqCoeffa(15)=YqCoeff0(3,1,2,1)*dt^3;%*(1-gamma)^Fp1(3,1,2,1)*dt^3;
YqCoeffa(16)=YqCoeff0(1,4,1,1)*dt^3;%*(1-gamma)^Fp1(1,4,1,1)*dt^3;
YqCoeffa(17)=YqCoeff0(2,3,1,1)*dt^3;%*(1-gamma)^Fp1(2,3,1,1)*dt^3;
YqCoeffa(18)=YqCoeff0(3,2,1,1)*dt^3;%*(1-gamma)^Fp1(3,2,1,1)*dt^3;
YqCoeffa(19)=YqCoeff0(4,1,1,1)*dt^3;%*(1-gamma)^Fp1(4,1,1,1)*dt^3;
Fp2(1)=Fp1(1,1,2,1);
Fp2(2)=Fp1(1,2,1,1);
Fp2(3)=Fp1(2,1,1,1);
Fp2(4)=Fp1(1,1,3,1);
Fp2(5)=Fp1(1,2,2,1);
Fp2(6)=Fp1(2,1,2,1);
Fp2(7)=Fp1(1,3,1,1);
Fp2(8)=Fp1(2,2,1,1);
Fp2(9)=Fp1(3,1,1,1);
Fp2(10)=Fp1(1,1,4,1);
Fp2(11)=Fp1(1,2,3,1);
Fp2(12)=Fp1(2,1,3,1);
Fp2(13)=Fp1(1,3,2,1);
Fp2(14)=Fp1(2,2,2,1);
Fp2(15)=Fp1(3,1,2,1);
Fp2(16)=Fp1(1,4,1,1);
Fp2(17)=Fp1(2,3,1,1);
Fp2(18)=Fp1(3,2,1,1);
Fp2(19)=Fp1(4,1,1,1);
%YqCoeffa
%Fp2
%str=input('Look at numbers');
wMu0dt0=0;
dwMu0dt(1:ExpnOrder)=0.0;
wMu0dt=0;
dwMu0dtdw(1:ExpnOrder)=0.0;
for mm=1:NoOfTerms
wMu0dt0=YqCoeffa(mm).*((1-gamma)*w0).^Fp2(mm);
for nn=1:ExpnOrder
if(nn==1)
dwMu0dt(nn)=wMu0dt0*Fp2(mm)/w0;
else
dwMu0dt(nn)=dwMu0dt(nn-1)*(Fp2(mm)-(nn-1))/w0;
end
end
wMu0dt=wMu0dt+wMu0dt0;
for nn=1:ExpnOrder
dwMu0dtdw(nn)=dwMu0dtdw(nn)+dwMu0dt(nn);
end
end
end
.
.
.
function [b0,b] = CalculateDriftbCoeffs08(wMu0dt,dwMu0dtdw,a,SeriesOrder)
b0=wMu0dt;
b(1)=a(1) *dwMu0dtdw(1);
b(2)=1/2*(2* a(2) *dwMu0dtdw(1)+a(1)^2 *dwMu0dtdw(2));
b(3)=1/6*(6 *a(3) *dwMu0dtdw(1)+6 *a(1)* a(2) *dwMu0dtdw(2)+a(1)^3 *dwMu0dtdw(3));
b(4)=1/24*(24* a(4) *dwMu0dtdw(1)+12* a(2)^2 *dwMu0dtdw(2)+24 *a(1)* a(3) *dwMu0dtdw(2)+ ...
12* a(1)^2* a(2) *dwMu0dtdw(3)+a(1)^4 *dwMu0dtdw(4));
if(SeriesOrder>=5)
b(5)=1/120*(120* a(5) *dwMu0dtdw(1)+120* a(2) *a(3) *dwMu0dtdw(2)+120* a(1)* a(4) *dwMu0dtdw(2)+ ...
60 *a(1)* a(2)^2 *dwMu0dtdw(3)+60* a(1)^2* a(3) *dwMu0dtdw(3)+ ...
20 *a(1)^3* a(2) *dwMu0dtdw(4)+a(1)^5 *dwMu0dtdw(5));
end
if(SeriesOrder>=6)
b(6)=1/720*(720* a(6) *dwMu0dtdw(1)+360* a(3)^2 *dwMu0dtdw(2)+720* a(2)* a(4) *dwMu0dtdw(2)+720* a(1)* a(5) *dwMu0dtdw(2)+ ...
120* a(2)^3 *dwMu0dtdw(3)+720* a(1)* a(2)* a(3) *dwMu0dtdw(3)+360* a(1)^2* a(4) *dwMu0dtdw(3)+ ...
180* a(1)^2* a(2)^2 *dwMu0dtdw(4)+120* a(1)^3* a(3) *dwMu0dtdw(4)+ ...
30* a(1)^4 *a(2) *dwMu0dtdw(5)+a(1)^6 *dwMu0dtdw(6));
end
if(SeriesOrder>=7)
b(7)=1/(720*7)*(5040* a(7)*dwMu0dtdw(1)+5040* a(3)* a(4)*dwMu0dtdw(2)+5040* a(2)* a(5) *dwMu0dtdw(2)+5040 *a(1)* a(6) *dwMu0dtdw(2)+ ...
2520* a(2)^2* a(3) *dwMu0dtdw(3)+2520* a(1)* a(3)^2 *dwMu0dtdw(3)+5040 *a(1)* a(2)* a(4) *dwMu0dtdw(3)+2520* a(1)^2 *a(5) *dwMu0dtdw(3)+ ...
840* a(1)* a(2)^3 *dwMu0dtdw(4)+2520* a(1)^2* a(2)* a(3) *dwMu0dtdw(4)+840* a(1)^3* a(4) *dwMu0dtdw(4)+ ...
420* a(1)^3* a(2)^2 *dwMu0dtdw(5)+210* a(1)^4* a(3) *dwMu0dtdw(5)+ ...
42* a(1)^5* a(2) *dwMu0dtdw(6)+a(1)^7 *dwMu0dtdw(7));
end
if(SeriesOrder>=8)
b(8)=1/(720*7*8)*(40320* a(8) *dwMu0dtdw(1)+20160* a(4)^2 *dwMu0dtdw(2)+40320* a(3)* a(5) *dwMu0dtdw(2)+40320* a(2) *a(6) *dwMu0dtdw(2)+ ...
40320* a(1) *a(7) *dwMu0dtdw(2)+ ...
20160* a(2)* a(3)^2 *dwMu0dtdw(3)+20160* a(2)^2* a(4) *dwMu0dtdw(3)+40320* a(1)* a(3)* a(4) *dwMu0dtdw(3)+40320* a(1)* a(2)* a(5) *dwMu0dtdw(3)+ ...
20160* a(1)^2* a(6) *dwMu0dtdw(3)+ ...
1680* a(2)^4 *dwMu0dtdw(4)+20160* a(1)* a(2)^2* a(3) *dwMu0dtdw(4)+10080* a(1)^2* a(3)^2 *dwMu0dtdw(4)+20160* a(1)^2* a(2) *a(4) *dwMu0dtdw(4)+ ...
6720* a(1)^3* a(5) *dwMu0dtdw(4)+ ...
3360* a(1)^2* a(2)^3 *dwMu0dtdw(5)+6720* a(1)^3 *a(2)* a(3) *dwMu0dtdw(5)+1680* a(1)^4* a(4) *dwMu0dtdw(5)+ ...
840* a(1)^4* a(2)^2 *dwMu0dtdw(6)+336* a(1)^5* a(3) *dwMu0dtdw(6)+ ...
56* a(1)^6* a(2) *dwMu0dtdw(7)+a(1)^8 *dwMu0dtdw(8));
end
if(SeriesOrder>=9)
b(9)=1/(720*7*8*9)*(362880* a(9) *dwMu0dtdw(1)+ ...
362880* a(4)* a(5) *dwMu0dtdw(2)+ 362880* a(3)* a(6) *dwMu0dtdw(2) +362880* a(2)* a(7) *dwMu0dtdw(2) +362880* a(1) *a(8) *dwMu0dtdw(2) + ...
60480* a(3)^3 *dwMu0dtdw(3)+362880* a(2)* a(3)* a(4) *dwMu0dtdw(3)+181440* a(1)* a(4)^2 *dwMu0dtdw(3)+181440* a(2)^2* a(5) *dwMu0dtdw(3) + ...
362880* a(1)* a(3)* a(5) *dwMu0dtdw(3)+362880 *a(1)* a(2)* a(6) *dwMu0dtdw(3) +181440 *a(1)^2* a(7) *dwMu0dtdw(3)+ ...
60480* a(2)^3* a(3) *dwMu0dtdw(4)+181440* a(1)* a(2)* a(3)^2 *dwMu0dtdw(4)+181440* a(1)* a(2)^2 *a(4) *dwMu0dtdw(4)+ ...
181440* a(1)^2* a(3)* a(4) *dwMu0dtdw(4)+181440* a(1)^2* a(2)* a(5) *dwMu0dtdw(4)+60480* a(1)^3* a(6) *dwMu0dtdw(4)+ ...
15120* a(1)* a(2)^4 *dwMu0dtdw(5)+90720* a(1)^2* a(2)^2* a(3) *dwMu0dtdw(5)+30240* a(1)^3* a(3)^2 *dwMu0dtdw(5)+ ...
60480 *a(1)^3* a(2)* a(4) *dwMu0dtdw(5)+15120* a(1)^4* a(5) *dwMu0dtdw(5)+ ...
10080* a(1)^3* a(2)^3 *dwMu0dtdw(6)+15120* a(1)^4 *a(2)* a(3) *dwMu0dtdw(6)+3024* a(1)^5* a(4) *dwMu0dtdw(6)+ ...
1512* a(1)^5* a(2)^2 *dwMu0dtdw(7)+504* a(1)^6* a(3) *dwMu0dtdw(7)+ ...
72* a(1)^7 *a(2) *dwMu0dtdw(8)+a(1)^9 *dwMu0dtdw(9));
end
if(SeriesOrder>=10)
b(10)=1/(720*7*8*9*10)*(3628800* a(10)* dwMu0dtdw(1)+ ...
1814400 * a(5)^2 *dwMu0dtdw(2)+3628800 *a(4)* a(6)* dwMu0dtdw(2)+3628800* a(3)* a(7)* dwMu0dtdw(2)+3628800 *a(2)* a(8)* dwMu0dtdw(2)+ ...
3628800* a(1)* a(9)* dwMu0dtdw(2)+ ...
1814400* a(3)^2* a(4)* dwMu0dtdw(3)+1814400* a(2)* a(4)^2* dwMu0dtdw(3)+3628800* a(2)* a(3)* a(5)* dwMu0dtdw(3)+3628800* a(1)* a(4)* a(5) *dwMu0dtdw(3)+ ...
1814400* a(2)^2* a(6)* dwMu0dtdw(3)+3628800* a(1)* a(3)* a(6)* dwMu0dtdw(3)+3628800 *a(1)* a(2)* a(7)* dwMu0dtdw(3)+1814400* a(1)^2* a(8)* dwMu0dtdw(3)+ ...
907200* a(2)^2 *a(3)^2* dwMu0dtdw(4)+604800* a(1)* a(3)^3* dwMu0dtdw(4)+604800* a(2)^3* a(4)* dwMu0dtdw(4)+3628800* a(1)* a(2)* a(3)* a(4)* dwMu0dtdw(4)+ ...
907200* a(1)^2* a(4)^2* dwMu0dtdw(4)+1814400* a(1)* a(2)^2* a(5)* dwMu0dtdw(4)+1814400* a(1)^2* a(3)* a(5)* dwMu0dtdw(4)+ ...
1814400* a(1)^2* a(2)* a(6)* dwMu0dtdw(4)+ 604800* a(1)^3* a(7) *dwMu0dtdw(4)+ ...
30240 *a(2)^5* dwMu0dtdw(5)+604800* a(1)* a(2)^3* a(3)* dwMu0dtdw(5)+907200 *a(1)^2* a(2)* a(3)^2* dwMu0dtdw(5)+907200* a(1)^2* a(2)^2* a(4) *dwMu0dtdw(5)+ ...
604800* a(1)^3* a(3)* a(4)* dwMu0dtdw(5)+604800* a(1)^3* a(2)* a(5)* dwMu0dtdw(5)+151200* a(1)^4* a(6)* dwMu0dtdw(5)+ ...
75600* a(1)^2* a(2)^4* dwMu0dtdw(6)+302400 *a(1)^3* a(2)^2* a(3)* dwMu0dtdw(6)+75600* a(1)^4 *a(3)^2* dwMu0dtdw(6)+151200* a(1)^4* a(2)* a(4) *dwMu0dtdw(6)+ ...
30240* a(1)^5* a(5)* dwMu0dtdw(6)+ ...
25200* a(1)^4* a(2)^3* dwMu0dtdw(7)+30240* a(1)^5* a(2)* a(3)* dwMu0dtdw(7)+5040* a(1)^6* a(4)* dwMu0dtdw(7)+ ...
2520* a(1)^6* a(2)^2* dwMu0dtdw(8)+720* a(1)^7* a(3)* dwMu0dtdw(8)+ ...
90 *a(1)^8* a(2)* dwMu0dtdw(9)+ ...
a(1)^10* dwMu0dtdw(10));
end
end
.
.
.
function [wVol0dt,dwVol0dtdw] = BesselVolAndDerivativesH1(w0,YqCoeff0,Fp1,gamma,dt,ExpnOrder)
% c1(wnStart:Nn)=YqCoeff0(1,1,1,2).*yy(wnStart:Nn).^Fp1(1,1,1,2).*sqrt(dt)+ ...
% (YqCoeff0(1,1,2,2).*yy(wnStart:Nn).^Fp1(1,1,2,2)+YqCoeff0(1,2,1,2).*yy(wnStart:Nn).^Fp1(1,2,1,2)+ ...
% YqCoeff0(2,1,1,2).*yy(wnStart:Nn).^Fp1(2,1,1,2)).*dt^1.5+ ...
% (YqCoeff0(1,1,3,2).*yy(wnStart:Nn).^Fp1(1,1,3,2)+YqCoeff0(1,2,2,2).*yy(wnStart:Nn).^Fp1(1,2,2,2)+ ...
% YqCoeff0(2,1,2,2).*yy(wnStart:Nn).^Fp1(2,1,2,2)+YqCoeff0(1,3,1,2).*yy(wnStart:Nn).^Fp1(1,3,1,2)+ ...
% YqCoeff0(2,2,1,2).*yy(wnStart:Nn).^Fp1(2,2,1,2)+YqCoeff0(3,1,1,2).*yy(wnStart:Nn).^Fp1(3,1,1,2)).*dt^2.5;
NoOfTerms=9;%excluding dt^3 terms
YqCoeffa(1:NoOfTerms)=0.0;
Fp1=Fp1/(1-gamma);
YqCoeffa(1)=YqCoeff0(1,1,2,2)*(1-gamma)^Fp1(1,1,2,2)*dt^1.5;
YqCoeffa(2)=YqCoeff0(1,2,1,2)*(1-gamma)^Fp1(1,2,1,2)*dt^1.5;
YqCoeffa(3)=YqCoeff0(2,1,1,2)*(1-gamma)^Fp1(2,1,1,2)*dt^1.5;
YqCoeffa(4)=YqCoeff0(1,1,3,2)*(1-gamma)^Fp1(1,1,3,2)*dt^2.5;
YqCoeffa(5)=YqCoeff0(1,2,2,2)*(1-gamma)^Fp1(1,2,2,2)*dt^2.5;
YqCoeffa(6)=YqCoeff0(2,1,2,2)*(1-gamma)^Fp1(2,1,2,2)*dt^2.5;
YqCoeffa(7)=YqCoeff0(1,3,1,2)*(1-gamma)^Fp1(1,3,1,2)*dt^2.5;
YqCoeffa(8)=YqCoeff0(2,2,1,2)*(1-gamma)^Fp1(2,2,1,2)*dt^2.5;
YqCoeffa(9)=YqCoeff0(3,1,1,2)*(1-gamma)^Fp1(3,1,1,2)*dt^2.5;
Fp2(1)=Fp1(1,1,2,2);
Fp2(2)=Fp1(1,2,1,2);
Fp2(3)=Fp1(2,1,1,2);
Fp2(4)=Fp1(1,1,3,2);
Fp2(5)=Fp1(1,2,2,2);
Fp2(6)=Fp1(2,1,2,2);
Fp2(7)=Fp1(1,3,1,2);
Fp2(8)=Fp1(2,2,1,2);
Fp2(9)=Fp1(3,1,1,2);
wVol0dt0=0;
dwVol0dt(1:ExpnOrder)=0.0;
wVol0dt=0;
dwVol0dtdw(1:ExpnOrder)=0.0;
for mm=1:NoOfTerms
wVol0dt0=YqCoeffa(mm).*w0.^Fp2(mm);
for nn=1:ExpnOrder
if(nn==1)
dwVol0dt(nn)=wVol0dt0*Fp2(mm)*1/w0;
else
dwVol0dt(nn)=dwVol0dt(nn-1)*(Fp2(mm)-(nn-1))/w0;
end
end
wVol0dt=wVol0dt+wVol0dt0;
for nn=1:ExpnOrder
dwVol0dtdw(nn)=dwVol0dtdw(nn)+dwVol0dt(nn);
end
end
end
.
.
.
function [wVol2dt,dwVol2dtdw] = BesselVolAndDerivativesH2(w0,YqCoeff0,Fp1,gamma,dt,ExpnOrder)
% c2(wnStart:Nn)=YqCoeff0(1,1,1,3).*yy(wnStart:Nn).^Fp1(1,1,1,3) *dt + ...
% (YqCoeff0(1,1,2,3).*yy(wnStart:Nn).^Fp1(1,1,2,3)+YqCoeff0(1,2,1,3).*yy(wnStart:Nn).^Fp1(1,2,1,3)+ ...
% YqCoeff0(2,1,1,3).*yy(wnStart:Nn).^Fp1(2,1,1,3)).*dt^2+ ...
% (YqCoeff0(1,1,3,3).*yy(wnStart:Nn).^Fp1(1,1,3,3)+YqCoeff0(1,2,2,3).*yy(wnStart:Nn).^Fp1(1,2,2,3)+ ...
% YqCoeff0(2,1,2,3).*yy(wnStart:Nn).^Fp1(2,1,2,3) + YqCoeff0(1,3,1,3).*yy(wnStart:Nn).^Fp1(1,3,1,3)+ ...
% YqCoeff0(2,2,1,3).*yy(wnStart:Nn).^Fp1(2,2,1,3)+YqCoeff0(3,1,1,3).*yy(wnStart:Nn).^Fp1(3,1,1,3)).*dt^3;
Fp1=Fp1/(1-gamma);
NoOfTerms=9;%excluding dt^3 terms
YqCoeffa(1:NoOfTerms)=0.0;
YqCoeffa(1)=YqCoeff0(1,1,2,3)*(1-gamma)^Fp1(1,1,2,3)*dt^2;
YqCoeffa(2)=YqCoeff0(1,2,1,3)*(1-gamma)^Fp1(1,2,1,3)*dt^2;
YqCoeffa(3)=YqCoeff0(2,1,1,3)*(1-gamma)^Fp1(2,1,1,3)*dt^2;
YqCoeffa(4)=YqCoeff0(1,1,3,3)*(1-gamma)^Fp1(1,1,3,3)*dt^3;
YqCoeffa(5)=YqCoeff0(1,2,2,3)*(1-gamma)^Fp1(1,2,2,3)*dt^3;
YqCoeffa(6)=YqCoeff0(2,1,2,3)*(1-gamma)^Fp1(2,1,2,3)*dt^3;
YqCoeffa(7)=YqCoeff0(1,3,1,3)*(1-gamma)^Fp1(1,3,1,3)*dt^3;
YqCoeffa(8)=YqCoeff0(2,2,1,3)*(1-gamma)^Fp1(2,2,1,3)*dt^3;
YqCoeffa(9)=YqCoeff0(3,1,1,3)*(1-gamma)^Fp1(3,1,1,3)*dt^3;
Fp2(1)=Fp1(1,1,2,3);
Fp2(2)=Fp1(1,2,1,3);
Fp2(3)=Fp1(2,1,1,3);
Fp2(4)=Fp1(1,1,3,3);
Fp2(5)=Fp1(1,2,2,3);
Fp2(6)=Fp1(2,1,2,3);
Fp2(7)=Fp1(1,3,1,3);
Fp2(8)=Fp1(2,2,1,3);
Fp2(9)=Fp1(3,1,1,3);
wVol2dt0=0;
dwVol2dt(1:ExpnOrder)=0.0;
wVol2dt=0;
dwVol2dtdw(1:ExpnOrder)=0.0;
for mm=1:NoOfTerms
wVol2dt0=YqCoeffa(mm).*w0.^Fp2(mm);
for nn=1:ExpnOrder
if(nn==1)
dwVol2dt(nn)=wVol2dt0*Fp2(mm)*1/w0;
else
dwVol2dt(nn)=dwVol2dt(nn-1)*(Fp2(mm)-(nn-1))/w0;
end
end
wVol2dt=wVol2dt+wVol2dt0;
for nn=1:ExpnOrder
dwVol2dtdw(nn)=dwVol2dtdw(nn)+dwVol2dt(nn);
end
end
end
.
.
.
function [Moments,k] = CalculateVolTermMoments(a0,a,b0,b,SeriesOrder,NMoments)
[EA,EB,EAB] = CalculateSeriesProducts(a0,a,b0,b,SeriesOrder,NMoments);
c0=0;
c(1)=1;
c(2)=0;
d0=-1;
d(1)=0;
d(2)=1;
[EZ1,EZ2,EZ1Z2] = CalculateSeriesProducts(c0,c,d0,d,2,8);
Moments(1)=EA(1)*EZ1(1)+EB(1)*EZ2(1);
Moments(2)=EA(2)*EZ1(2)+2*EAB(1,1)*EZ1Z2(1,1)+EB(2)*EZ2(2);
Moments(3)=EA(3)*EZ1(3)+3*EAB(2,1)*EZ1Z2(2,1)+3*EAB(1,2)*EZ1Z2(1,2)+EB(3)*EZ2(3);
Moments(4)=EA(4)*EZ1(4)+4*EAB(3,1)*EZ1Z2(3,1)+6*EAB(2,2)*EZ1Z2(2,2)+4*EAB(1,3)*EZ1Z2(1,3)+EB(4)*EZ2(4);
Moments(5)=EA(5)*EZ1(5)+5*EAB(4,1)*EZ1Z2(4,1)+10*EAB(3,2)*EZ1Z2(3,2)+10*EAB(2,3)*EZ1Z2(2,3)+5*EAB(1,4)*EZ1Z2(1,4)+EB(5)*EZ2(5);
Moments(6)=EA(6)*EZ1(6)+6*EAB(5,1)*EZ1Z2(5,1)+15*EAB(4,2)*EZ1Z2(4,2)+20*EAB(3,3)*EZ1Z2(3,3)+15*EAB(2,4)*EZ1Z2(2,4)+6*EAB(1,5)*EZ1Z2(1,5)+EB(6)*EZ2(6);
Moments(7)=EA(7)*EZ1(7)+7*EAB(6,1)*EZ1Z2(6,1)+21*EAB(5,2)*EZ1Z2(5,2)+35*EAB(4,3)*EZ1Z2(4,3)+35*EAB(3,4)*EZ1Z2(3,4)+21*EAB(2,5)*EZ1Z2(2,5)+7*EAB(1,6)*EZ1Z2(1,6)+EB(7)*EZ2(7);
Moments(8)=EA(8)*EZ1(8)+8*EAB(7,1)*EZ1Z2(7,1)+28*EAB(6,2)*EZ1Z2(6,2)+56*EAB(5,3)*EZ1Z2(5,3)+70*EAB(4,4)*EZ1Z2(4,4)+56*EAB(3,5)*EZ1Z2(3,5)+28*EAB(2,6)*EZ1Z2(2,6)+8*EAB(1,7)*EZ1Z2(1,7)+EB(8)*EZ2(8);
k(1)=0;
k(2)=Moments(2);
k(3)=Moments(3);
k(4)=Moments(4)-3*Moments(2).^2;
k(5)=Moments(5)-10*Moments(3)*Moments(2);
k(6)=Moments(6)-15*Moments(4)*Moments(2)-10*Moments(3).^2+30*Moments(2).^3;
k(7)=Moments(7)- 21* Moments(2) *Moments(5)- 35* Moments(3) *Moments(4) + 210* Moments(2).^2* Moments(3);
k(8)=Moments(8)- 7 *k(2)* Moments(6) - 21* k(3)* Moments(5) - 35* k(4)* Moments(4) - ...
35* k(5)* Moments(3) - 21* k(6)* Moments(2);
end
.
.
.
%function [u1,u2,u3,u4,u5,u6] = CalculateCumulants(a0,a,SeriesOrder,NMoments)
function [CC] = CalculateCumulants(a0,a,SeriesOrder,NMoments)
%[CC] = CalculateCumulants(a0,a,SeriesOrder,NoOfCumulants);
%a(1)=a1;
%a(2)=a2;
%a(3)=a3;
%a(4)=a4;
%a(5)=a5;
%a(6)=a6;
%a(7)=a7;
aa0=a0;
a0=0;
EZ(1)=0;
EZ(2)=1;
for nn=3:NMoments*SeriesOrder+2
if rem(nn,2)==1
EZ(nn)=0;
else
EZ(nn)=EZ(nn-2)*(nn-1);
EZ(nn);
end
end
EZ;
EXZ(1,1)=1;
%for pp1=1:NZterms
% EXZ(1,pp1+1)=EZ(pp1);
%end
a(SeriesOrder+1:60)=0;
b0=a0;
b=a;
for mm=1:NMoments
if(mm>1)
[b0,b] =SeriesProduct(a0,a,b0,b,SeriesOrder*mm);
b(SeriesOrder*mm+1:60)=0;
end
% b0
% b
%str=input('Look at numbers')
EXZ(mm+1,1)=b0;
for pp2=1:SeriesOrder*mm
EXZ(mm+1,1)=EXZ(mm+1,1)+b(pp2).*EZ(pp2);
end
end
%u1=EXZ(2,1);
if(SeriesOrder>=4)
u1=a(2)+3*a(4);
end
if(SeriesOrder>=6)
u1=u1+15*a(6);
end
u2=EXZ(3,1);
u3=EXZ(4,1);
u4=EXZ(5,1);
if(SeriesOrder>=4)
k1=aa0+a(2)+3*a(4);
end
if(SeriesOrder>=6)
k1=k1+15*a(6);
end
k2=u2-u1^2;
k3=u3-3*u2*u1+2*u1^3;
k4=u4-4*u3*u1-3*u2^2+12*u2*u1^2-6*u1^4;
CC(1)=k1;
CC(2)=k2;
CC(3)=k3;
CC(4)=k4;
if(NMoments>=5)
u5=EXZ(6,1);
k5=u5-5*u4*u1-10*u3*u2+20*u3*u1^2+30*u2^2*u1-60*u2*u1^3+24*u1^5;
CC(5)=k5;
end
if(NMoments>=6)
u6=EXZ(7,1);
k6=u6-6*u5*u1-15*u4*u2+30*u4*u1^2-10*u3^2+120*u3*u2*u1-120*u3*u1^3+30*u2^3 - ...
270*u2^2*u1^2+360*u2*u1^4-120*u1^6;
CC(6)=k6;
end
if(NMoments>=7)
u7=EXZ(8,1);
k7=u7-u1*u6-6*k2*u5-15*k3*u4-20*k4*u3-15*k5*u2-6*k6*u1;
CC(7)=k7;
end
if(NMoments>=8)
u8=EXZ(9,1);
k8 = u8 - u1* u7 - 7 *k2* u6 - 21* k3* u5 - 35* k4* u4 - ...
35* k5* u3 - 21* k6* u2 - 7 *k7* u1;
CC(8)=k8;
end
end
.
.
.
function [Moments] = CalculateMomentsOfZSeries(a0,a,SeriesOrder,NMoments)
%aa0=a0;
%a0=0;% ---1
EZ(1)=0;
EZ(2)=1;
%LogEZ(2)=0;
for nn=3:NMoments*SeriesOrder
if rem(nn,2)==1
EZ(nn)=0;
else
EZ(nn)=EZ(nn-2)*(nn-1);
%LogEZ(nn)=log(EZ(nn-2))+log(nn-1);
%EZ(nn);
end
end
%EZ
%EZ(1:30)
%str=input('Look at numbers')
a(SeriesOrder+1:SeriesOrder*NMoments+1)=0;
b0=a0;
b=a;
for mm=1:NMoments
if(mm>1)
%[b0,b] =SeriesProductLogarithmic(a0,a,b0,b,SeriesOrder*mm);
[b0,b] =SeriesProduct(a0,a,b0,b,SeriesOrder*mm);
%b0-bb0
%b-bb
%str=input('Look at two products')
b(SeriesOrder*mm+1:SeriesOrder*NMoments+1)=0;
end
% Logb=log(abs(b));
% Signb=sign(b);
EXZ(mm,1)=b0;
for pp2=1:SeriesOrder*mm
if rem(pp2,2)==0
%EXZ(mm,1)=EXZ(mm,1)+exp(Logb(pp2)+LogEZ(pp2)).*Signb(pp2);
EXZ(mm,1)=EXZ(mm,1)+b(pp2).*EZ(pp2);
% b(pp2).*EZ(pp2)
% exp(Logb(pp2)+LogEZ(pp2)).*Signb(pp2)
% mm
% pp2
%str=input('Look at moment values')
end
end
end
for nn=1:NMoments
Moments(nn)=EXZ(nn,1);
end
end
.
.
.
function [b0,b] = CalculateDensityFromStandardizedMoments(Moments0,b0Guess,bGuess)
% cmu(1)=0;
% cmu(2)=1;
% cmu(3)=Moments0(3)/Moments0(2).^1.5;
% cmu(4)=Moments0(4)/Moments0(2).^2.0;
% cmu(5)=Moments0(5)/Moments0(2).^2.50;
% cmu(6)=Moments0(6)/Moments0(2).^3.0;
% cmu(7)=Moments0(7)/Moments0(2).^3.50;
% cmu(8)=Moments0(8)/Moments0(2).^4.0;
cmu=Moments0;
%[c0,c] =PreSmoothingGuess02(cmu);
%cc=c;
%cc0=-(cc(2)+cc(4)*3+cc(6)*15);
%cc=bGuess;
%cc0=b0Guess;
%[cc0,cc] =PreSmoothingGuess02(cmu);
[cc0,cc] =PreSmoothingGuess02NewFromGuess(cmu,bGuess);
SeriesOrder=7;
NMoments=8;
NZterms=8;
[Moments] = CalculateMomentsOfZSeries(cc0,cc,SeriesOrder,NMoments)
cmu
%str=input('Look at moments--before')
%Below calculate F, the defect vector from target cumulants. And calculate
%dF which is the matrix consisting of derivatives of defect vector from
%target cumulants.
%[F,dF] = CalculateCumulantsAndDerivativesFromMoments_0(C,cc0,cc,SeriesOrder,NMoments,8)
[F,dF] = CalculateMomentsAndDerivatives_0(cmu,cc0,cc,SeriesOrder,NZterms,NMoments);
da(1,1)=cc0;
da(2:SeriesOrder+1,1)=cc(1:SeriesOrder);
[Moments] = CalculateMomentsOfZSeries(cc0,cc,SeriesOrder,NMoments);
ObjBest=0;
ObjBest=100000*(abs(cmu(1)-Moments(1)))+abs(cmu(2)-Moments(2))+abs((cmu(3)-Moments(3))^(1/1.5))+abs((cmu(4)-Moments(4))^(1/2.0)) + ...
abs((cmu(5)-Moments(5))^(1/2.0))+abs((cmu(6)-Moments(6))^(1/2.0))+abs((cmu(7)-Moments(7))^(1/2.0))+ ...
abs((cmu(8)-Moments(8))^(1/2.0));
b0Best=cc0;
bBest(1:SeriesOrder)=cc(1:SeriesOrder);
nn=0;
while((nn<100)&&((abs(F(1,1))>.000000000001) || (abs(F(2,1))>.000000000001) || (abs(F(3,1))>.000000000001) || (abs(F(4,1))>.00000000001)|| (abs(F(5,1))>.00000000001)|| (abs(F(6,1))>.00000000001)|| (abs(F(7,1))>.00000000001)|| (abs(F(8,1))>.00000000001) ))
nn=nn+1;
%Below Newton matrix equation to improve the Z-series coefficients guess at previous step.
da=da-dF\F;
%b0=median
b0=da(1,1);
b(1:SeriesOrder)=da(2:SeriesOrder+1,1);
%[F,dF] = CalculateCumulantsAndDerivativesFromMoments_0(C,b0,b,SeriesOrder,SeriesOrder,NoOfCumulants);
[F,dF] = CalculateMomentsAndDerivatives_0(cmu,b0,b,SeriesOrder,NZterms,NMoments);
[IsValidFlag] = CheckIsValidDensity(b0,b);
[Moments] = CalculateMomentsOfZSeries(b0,b,SeriesOrder,NMoments);
ObjNew=0;
ObjNew=100000*(abs(cmu(1)-Moments(1)))+abs(cmu(2)-Moments(2))+abs((cmu(3)-Moments(3))^(1/1.5))+abs((cmu(4)-Moments(4))^(1/2.0)) + ...
abs((cmu(5)-Moments(5))^(1/2.0))+abs((cmu(6)-Moments(6))^(1/2.0))+abs((cmu(7)-Moments(7))^(1/2.0))+ ...
abs((cmu(8)-Moments(8))^(1/2.0));
if((ObjBest>ObjNew) &&( IsValidFlag))
ObjBest=ObjNew;
b0Best=b0;
bBest(1:SeriesOrder)=b(1:SeriesOrder);
end
da(1,1)=b0;
da(2:SeriesOrder+1,1)=b(1:SeriesOrder);
end
b0=b0Best;%Best;
b(1:SeriesOrder)=bBest;%Best(1:SeriesOrder);
%Above are the best chosen Z-series coefficients based on the objective function.
%But my objective function can be improved.
[Moments] = CalculateMomentsOfZSeries(b0,b,SeriesOrder,NMoments)
%Above are raw moments calculated from the resulting Z-series from Newton
%Method
%Below are raw moments that were input to calibration.
cmu
%str=input('Look at comparison of moments calibration--00');
%b0
%b
nn
%str=input('Look at comparison of moments calibration--0');
b0=b0*sqrt(Moments0(2));
b=b*sqrt(Moments0(2));
b0;
b;
%str=input('Look at comparison of moments calibration--1');
end
.
.
.
function [XDensity,IndexOut,IndexMax] = MakeDensityFromSimulation_Infiniti_NEW(X,Paths,NoOfBins,MaxCutOff )
%Processes monte carlo paths to return a series Xdensity as a function of IndexOut. IndexMax is the maximum value of index.
%
Xmin=0;
Xmax=0;
for p=1:Paths
if(X(p)>MaxCutOff)
X(p)=MaxCutOff;
end
%if(X(p)<0)
%X(p)=0;
%end
if(Xmin>real(X(p)))
Xmin=real(X(p));
end
if(Xmax<real(X(p)))
Xmax=real(X(p));
end
end
%IndexMax=NoOfBins+1;
BinSize=(Xmax-Xmin)/NoOfBins;
%IndexMax=floor((Xmax-Xmin)/BinSize+.5)+1
IndexMax=floor((Xmax-Xmin)/BinSize+.5)+1
XDensity(1:IndexMax)=0.0;
for p=1:Paths
index=real(floor(real(X(p)-Xmin)/BinSize+.5)+1);
if(real(index)<1)
index=1;
end
if(real(index)>IndexMax)
index=IndexMax;
end
XDensity(index)=XDensity(index)+1.0/Paths/BinSize;
end
IndexOut(1:IndexMax)=Xmin+(0:(IndexMax-1))*BinSize;
end
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You think life is a secret, Life is only love of flying, It has seen many ups and downs, But it likes travel more than the destination. Allama Iqbal
Amin
Topic Author
Posts: 2079
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Here is the graph output of the program.
The first graph shows density of the SDE from analytic methods VS from monte carlo.
Second graph shows how good reconstruction of the SDE variable is as a function of Standard normal when calculated from Z-series in original coordinates and it is compared with SDE variable from direct transformation of density in Bessel coordinates.
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You think life is a secret, Life is only love of flying, It has seen many ups and downs, But it likes travel more than the destination. Allama Iqbal
Amin
Topic Author
Posts: 2079
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Sorry Missed a few matlab functions needed to run the program. Here they are:
If there is still any function missing, please mention and I will post it. Most of the functions (except a few new ones) have already been posted in the thread earlier and you can try finding them in the thread.
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function [c0,c] = PreSmoothingGuess02NewFromGuess(cmu,cin)
Mul=1.0;
SeriesOrder=7;
NMoments=8;
c0=0;
c(1)=1.0;
%At this point, we have set first two coefficients of Z-series. Rest of the coefficients are zero.
if(cmu(3)==0)
c(2)=cin(2);
else
if(cmu(3)>0)
c(2)=.001;
elseif(cmu(3)<0)
c(2)=-.001;
end
end
for nn=1:5*Mul
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
c(2)=c(2)*(cmu(3))/Moments(3);
c0=-c(2);
end
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
c=c/sqrt(Moments(2));
c0=c0/sqrt(Moments(2));
%Above we have standardized the coefficients so that 2nd central moment is
%one.
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
Moments2=Moments;
%At this point first three coefficients have been set and the rest are
%zero.
%Moments2 above has been calculated with only c0, c1, c2 non-zero. All
%higher coefficients upto c7 are zero. Particularly in its calculation c3
%is zero.
%if(cmu(4)-Moments2(4)>0)
% c(3)=.0001;
%elseif(cmu(4)-Moments2(4)<0)
% c(3)=-.0001;
%end
c(3)=cin(3);
for nn=1:5*Mul
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
%please recall that in calculation of Moments2, c3 is zero. We are taking
%slope for fourth moment from where the value of c3 is zero.
Ratio4=(cmu(4)-Moments2(4))/(Moments(4)-Moments2(4));
c(3)=c(3)*Ratio4;
if(cmu(4)<Moments2(4))
c(3)=abs(c(3))*-1;
end
%cmu(4)
%Moments2(4)
%Moments(4)
%Ratio4
%c(3)
% str=input('Look at fourth moment');
%
end
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
%Moments
%str=input('Look at moments--inside my function---M4');
c=c/sqrt(Moments(2));
c0=c0/sqrt(Moments(2));
%Above we have standardized the coefficients so that 2nd central moment is
%one.
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
Moments2=Moments;
%Moments2 above has been calculated with only c0, c1, c2,c3 non-zero. All
%higher coefficients upto c7 are zero. Particularly in its calculation c4
%is zero from where we will calculate slope in next loop..
%if(cmu(5)-Moments2(5)>0)
% c(4)=.0001;
%elseif(cmu(5)-Moments2(5)<0)
% c(4)=-.0001;
%end
c(4)=cin(4);
for nn=1:5*Mul
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
%please recall that in calculation of Moments2, c4 is zero. We are taking
%slope for fifth moment from where the value of c4 is zero.
Ratio5=(cmu(5)-Moments2(5))/(Moments(5)-Moments2(5));
c(4)=c(4)*Ratio5;
c0=-c(2)-3*c(4);
end
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
c=c/sqrt(Moments(2));
c0=c0/sqrt(Moments(2));
%Above we have standardized the coefficients so that 2nd central moment is
%one.
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
Moments2=Moments;
%Moments2 above has been calculated with only c0, c1, c2,c3,c4 non-zero. All
%higher coefficients upto c7 are zero. Particularly in its calculation c5
%is zero from where we will calculate slope in next loop..
%if(cmu(6)-Moments2(6)>0)
% c(5)=.0001;
%elseif(cmu(6)-Moments2(6)<0)
% c(5)=-.0001;
%end
c(5)=cin(5);
for nn=1:5*Mul
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
%please recall that in calculation of Moments2, c5 is zero. We are taking
%slope for sixth moment from where the value of c5 is zero.
Ratio6=(cmu(6)-Moments2(6))/(Moments(6)-Moments2(6));
c(5)=c(5)*Ratio6;
end
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
c=c/sqrt(Moments(2));
c0=c0/sqrt(Moments(2));
%Above we have standardized the coefficients so that 2nd central moment is
%one.
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
Moments2=Moments;
%Moments2 above has been calculated with only c0, c1, c2,c3,c4,c5 non-zero. All
%higher coefficients upto c7 are zero. Particularly in its calculation c6
%is zero from where we will calculate slope in next loop..
% c(6)=0;
% if(cmu(7)-Moments2(7)>0)
% c(6)=.0001;
%
% elseif(cmu(7)-Moments2(7)<0)
% c(6)=-.0001;
% end
c(6)=cin(6);
for nn=1:5*Mul
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
%please recall that in calculation of Moments2, c6 is zero. We are taking
%slope for seventh moment from where the value of c6 is zero.
Ratio7=(cmu(7)-Moments2(7))/(Moments(7)-Moments2(7));
c(6)=c(6)*Ratio7;
c0=-c(2)-3*c(4)-15*c(6);
if(cmu(6)<Moments2(6))
c(5)=abs(c(5))*-1;
end
% cmu(6)
% Moments2(6)
% Moments(6)
% Ratio6
% c(5)
% str=input('Look at sixth moment');
end
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
c=c/sqrt(Moments(2));
c0=c0/sqrt(Moments(2));
%Above we have standardized the coefficients so that 2nd central moment is
%one.
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
Moments2=Moments;
%Moments2 above has been calculated with only c0, c1, c2,c3,c4,c5,c6 non-zero. All
%higher coefficients upto c7 are zero. Particularly in its calculation c7
%is zero from where we will calculate slope in next loop..
% c(7)=0;
% if(cmu(8)-Moments2(8)>0)
% c(7)=.00005;
%
% elseif(cmu(8)-Moments2(8)<0)
% c(7)=-.00005;
% end
c(7)=cin(7);
for nn=1:5*Mul
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
%please recall that in calculation of Moments2, c7 is zero. We are taking
%slope for eighth moment from where the value of c7 is zero.
Ratio8=(cmu(8)-Moments2(8))/(Moments(8)-Moments2(8));
c(7)=c(7)*(abs(Ratio8))*sign(Ratio8);
if(cmu(8)<Moments2(8))
c(7)=abs(c(7))*-1;
% cmu(8)
% Moments2(8)
% Moments(8)
% Ratio8
% c(7)
% str=input('Look at eighth moment');
end
[Moments] = CalculateMomentsOfZSeries(c0,c,SeriesOrder,NMoments);
c=c/sqrt(Moments(2));
c0=c0/sqrt(Moments(2));
%Above we have standardized the coefficients so that 2nd central moment is
%one.
end
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function [F,dF] = CalculateMomentsAndDerivatives_0(Ms,a0,a,SeriesOrder,NZterms,NMoments)
%[F,dF] = CalculateCumulantsAndDerivativesFromMoments(C,a0,a,SeriesOrder,SeriesOrder,NoOfCumulants);
%a(1)=a1;
%a(2)=a2;
%a(3)=a3;
%a(4)=a4;
%a(5)=a5;
%a(6)=a6;
%a(7)=a7;
%aa0=a0;
if(NMoments>8)
a0=-(a(2)+3*a(4)+15*a(6)+105*a(8));% ---1
end
if(NMoments<=8)
a0=-(a(2)+3*a(4)+15*a(6));% ---1
end
EZ(1)=0;
EZ(2)=1;
for nn=3:NMoments*SeriesOrder+NZterms+2
if rem(nn,2)==1
EZ(nn)=0;
else
EZ(nn)=EZ(nn-2)*(nn-1);
EZ(nn);
end
end
EZ;
EXZ(1,1)=1;
for pp1=1:NZterms
EXZ(1,pp1+1)=EZ(pp1);
end
a(SeriesOrder+1:NMoments*SeriesOrder+1)=0;
b0=a0;
b=a;
for mm=1:NMoments
if(mm>1)
[b0,b] =SeriesProduct(a0,a,b0,b,SeriesOrder*mm);
b(SeriesOrder*mm+1:NMoments*SeriesOrder+1)=0;
end
% b0
% b
%str=input('Look at numbers')
EXZ(mm+1,1)=b0;
for pp2=1:SeriesOrder*mm
EXZ(mm+1,1)=EXZ(mm+1,1)+b(pp2).*EZ(pp2);
end
for pp1=1:NZterms
EXZ(mm+1,pp1+1)=b0.*EZ(pp1);
for pp2=1:SeriesOrder*mm
EXZ(mm+1,pp1+1)=EXZ(mm+1,pp1+1)+b(pp2).*EZ(pp2+pp1);
end
end
end
%u1=EXZ(2,1);
u2=EXZ(3,1);
u3=EXZ(4,1);
u4=EXZ(5,1);
u1=a0+a(2);
if(SeriesOrder>=4)
u1=a0+a(2)+3*a(4);
end
if(SeriesOrder>=6)
u1=u1+15*a(6);
end
if(SeriesOrder>=8)
u1=u1+105*a(8);
end
%k2=u2-u1^2;
%k3=u3-3*u2*u1+2*u1^3;
%k4=u4-4*u3*u1-3*u2^2+12*u2*u1^2-6*u1^4;
du1(1)=1;%----2
% du1(2)=0;%----2
% du1(3)=1;%----2
% du1(4)=0;%----2
% du1(5)=1;%----2
% du1(6)=0;%----2
% du1(7)=15;%----2
% du1(8)=0;%----2
% du1(9)=105;%----2
% du1(10)=0;%----2
du2(1)=2*EXZ(2,1);
du3(1)=3*EXZ(3,1);
du4(1)=4*EXZ(4,1);
%du1(1)=1;%----2
%du2(1)=0;
%du3(1)=0;
%du4(1)=0;
for mm=2:SeriesOrder+1
du1(mm)=EXZ(1,mm);
du2(mm)=2*EXZ(2,mm);
du3(mm)=3*EXZ(3,mm);
du4(mm)=4*EXZ(4,mm);
end
if(NMoments>=5)
u5=EXZ(6,1);
%du5(1)=5*EXZ(5,1);
for mm=1:SeriesOrder+1
du5(mm)=5*EXZ(5,mm);
end
end
if(NMoments>=6)
u6=EXZ(7,1);
%du6(1)=0;
for mm=1:SeriesOrder+1
du6(mm)=6*EXZ(6,mm);
end
end
if(NMoments>=7)
u7=EXZ(8,1);
%str=input('Look at k7 and k71')
%du7(1)=0;
for mm=1:SeriesOrder+1
du7(mm)=7*EXZ(7,mm);
end
end
if(NMoments>=8)
u8=EXZ(9,1);
for mm=1:SeriesOrder+1
du8(mm)=8*EXZ(8,mm);
end
end
if(NMoments>=9)
u9=EXZ(10,1);
for mm=1:SeriesOrder+1
du9(mm)=9*EXZ(9,mm);
end
end
if(NMoments>=10)
u10=EXZ(11,1);
for mm=1:SeriesOrder+1
du10(mm)=10*EXZ(10,mm);
end
end
F(1,1)=u1-Ms(1);
F(2,1)=u2-Ms(2);
F(3,1)=u3-Ms(3);
F(4,1)=u4-Ms(4);
if(NMoments>=5)
F(5,1)=u5-Ms(5);
end
if(NMoments>=6)
F(6,1)=u6-Ms(6);
end
if(NMoments>=7)
F(7,1)=u7-Ms(7);
end
if(NMoments>=8)
F(8,1)=u8-Ms(8);
end
if(NMoments>=9)
F(9,1)=u9-Ms(9);
end
if(NMoments>=10)
F(10,1)=u10-Ms(10);
end
for mm=1:SeriesOrder+1
dF(1,mm)=du1(mm);
dF(2,mm)=du2(mm);
dF(3,mm)=du3(mm);
dF(4,mm)=du4(mm);
if(NMoments>=5)
dF(5,mm)=du5(mm);
end
if(NMoments>=6)
dF(6,mm)=du6(mm);
end
if(NMoments>=7)
dF(7,mm)=du7(mm);
end
if(NMoments>=8)
dF(8,mm)=du8(mm);
end
if(NMoments>=9)
dF(9,mm)=du9(mm);
end
if(NMoments>=10)
dF(10,mm)=du10(mm);
end
end
end
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function [EA,EB,EAB] = CalculateSeriesProducts(a0,a,b0,b,SeriesOrder,NMoments)
EA(1:NMoments)=0;
EB(1:NMoments)=0;
EAB(1:NMoments,1:NMoments)=0;
EA(1)=a0+a(2);
EB(1)=b0+b(2);
if(SeriesOrder>=4)
EA(1)=EA(1)+3*a(4);
EB(1)=EB(1)+3*b(4);
end
if(SeriesOrder>=6)
EA(1)=EA(1)+15*a(6);
EB(1)=EB(1)+15*b(6);
end
EZ(1)=0;
EZ(2)=1;
for nn=3:NMoments*SeriesOrder
if rem(nn,2)==1
EZ(nn)=0;
else
EZ(nn)=EZ(nn-2)*(nn-1);
EZ(nn);
end
end
EZ;
aa(1:NMoments,1:SeriesOrder*NMoments)=0;
aa0(1)=a0;
aa(1,1:SeriesOrder)=a(:);
bb(1:NMoments,1:SeriesOrder*NMoments)=0;
bb0(1)=b0;
bb(1,1:SeriesOrder)=b(:);
for mm=1:NMoments-1
[ab0,ab] =SeriesProductHigher(a0,a,aa0(mm),aa(mm,:),SeriesOrder*(mm+1));
aa0(mm+1)=ab0;
aa(mm+1,1:(mm+1)*SeriesOrder)=ab(1:(mm+1)*SeriesOrder);
[ac0,ac] =SeriesProductHigher(b0,b,bb0(mm),bb(mm,:),SeriesOrder*(mm+1));
bb0(mm+1)=ac0;
bb(mm+1,1:(mm+1)*SeriesOrder)=ac(1:(mm+1)*SeriesOrder);
EA(mm+1)=aa0(mm+1);
EB(mm+1)=bb0(mm+1);
for pp=1:SeriesOrder*(mm+1)
EA(mm+1)=EA(mm+1)+aa(mm+1,pp).*EZ(pp);
EB(mm+1)=EB(mm+1)+bb(mm+1,pp).*EZ(pp);
end
end
%b(SeriesOrder*mm+1:NMoments*SeriesOrder+1)=0;
aa0;
aa;
for mm=1:NMoments
for nn=1:NMoments-mm
dd0=aa0(mm);
dd=aa(mm,1:SeriesOrder*NMoments);
ee0=bb0(nn);
ee=bb(nn,1:SeriesOrder*NMoments);
[ab0,ab] =SeriesProductHigher(dd0,dd,ee0,ee,SeriesOrder*(mm+nn));
AB0(mm,nn)=ab0;
AB(mm,nn,1:SeriesOrder*(mm+nn))=ab(1:SeriesOrder*(mm+nn));
EAB(mm,nn)=AB0(mm,nn);
for pp=1:SeriesOrder*(mm+nn)
EAB(mm,nn)=EAB(mm,nn)+AB(mm,nn,pp).*EZ(pp);
end
end
end
end
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function [c0,c] =SeriesProductHigher(a0,a,b0,b,SeriesOrder)
a(size(a,2)+1:SeriesOrder)=0;
b(size(b,2)+1:SeriesOrder)=0;
c0=a0*b0;
c(1:SeriesOrder)=0.0;
%a
%b
%str=input('Look at numbers');
for nn=1:SeriesOrder
c(nn)=c(nn)+a0*b(nn);
c(nn)=c(nn)+b0*a(nn);
for kk=1:nn-1
c(nn)=c(nn)+a(kk)*b(nn-kk);
end
end
end
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function [IsValidFlag] = CheckIsValidDensity(cc0,cc)
%Roughly checks if the density is valid. It is not a definitive check but
%just a rough one but mostly works.
Z=2.0;
Xp1=cc0 + cc(1) * Z+ cc(2) * Z^2+ cc(3) * Z^3+ cc(4) * Z^4+ cc(5) * Z^5+ cc(6) * Z^6+ cc(7) * Z^7;
Z=3.0;
Xp2=cc0 + cc(1) * Z+ cc(2) * Z^2+ cc(3) * Z^3+ cc(4) * Z^4+ cc(5) * Z^5+ cc(6) * Z^6+ cc(7) * Z^7;
Z=4.65;
Xp3=cc0 + cc(1) * Z+ cc(2) * Z^2+ cc(3) * Z^3+ cc(4) * Z^4+ cc(5) * Z^5+ cc(6) * Z^6+ cc(7) * Z^7;
Z=-2.0;
Xn1=cc0 + cc(1) * Z+ cc(2) * Z^2+ cc(3) * Z^3+ cc(4) * Z^4+ cc(5) * Z^5+ cc(6) * Z^6+ cc(7) * Z^7;
Z=-3.0;
Xn2=cc0 + cc(1) * Z+ cc(2) * Z^2+ cc(3) * Z^3+ cc(4) * Z^4+ cc(5) * Z^5+ cc(6) * Z^6+ cc(7) * Z^7;
Z=-4.65;
Xn3=cc0 + cc(1) * Z+ cc(2) * Z^2+ cc(3) * Z^3+ cc(4) * Z^4+ cc(5) * Z^5+ cc(6) * Z^6+ cc(7) * Z^7;
IsValidFlag=0;
if( ((Xp2>Xp1) && (Xp3>Xp2)) &&(Xp1>Xn1) && ((Xn2<Xn1) && (Xn3<Xn2)))
IsValidFlag=1;
end
end
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function [c0,c] =SeriesProduct(a0,a,b0,b,SeriesOrder)
c0=a0*b0;
c(1:SeriesOrder)=0.0;
for nn=1:SeriesOrder
c(nn)=c(nn)+a0*b(nn);
c(nn)=c(nn)+b0*a(nn);
for kk=1:nn-1
c(nn)=c(nn)+a(kk)*b(nn-kk);
end
end
end
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You think life is a secret, Life is only love of flying, It has seen many ups and downs, But it likes travel more than the destination. Allama Iqbal
Amin
Topic Author
Posts: 2079
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, almost everyday when I try to go to sleep, mind control agents leave a pungent like gas in my room. I always keep window of my room open to get fresh air, and as soon as the atmosphere in my room goes from fresh to pungent that feels like slightly acidic as I breathe it, I can right away tell they have released some gas in my room. The gas is so acidic that taste in my mouth changes as I breathe it and I can feel acidity in my mouth. I have been busy for past 4-5 days working on my programs and could not find time to mention it on the forum. But now when mind control agents released gas in my room, I decided to write about it immediately. Please force mind control crooks to stop using ionized, pungent and acidic gases on innocent civilians.
You think life is a secret, Life is only love of flying, It has seen many ups and downs, But it likes travel more than the destination. Allama Iqbal
Amin
Topic Author
Posts: 2079
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, I am travelling tomorrow. I am going from Lahore to Kot Addu where I might stay for a month or slightly more. I want to complete the work for my paper in the meantime. I want to put serious effort into trying to make it a good paper. Kot Addu is a very small city and they can easily drug food at just a few places where I can eat outside. I want to request friends to please ask mind control agencies to not drug any food in the small city. Also an antipsychotic injection is due next week and there is probably just one pharmacy in Kot Addu where I can get the particular injection and it would be very easy for them to give me an injection that is filled with mind control chemicals and my life after that would become extremely difficult. I want to request friends to please keep a vigil and force mind control agencies to not drug the food or injections. I hope that most friends would like if I can present some good research while working without mind control chemicals.
Another interesting thing I want to do is to try to discover if we can find out Z-series representation of a stochastic variable once its closed form density is given. As opposed to finding Z-series from moments of the random variable. I have some ideas and I am very sure it is possible with a bit of hack. This should be very easy to find out if formulas for CDF of the random variable are given. I will come back in a two to three days with examples of lognormal diffusions, CEV diffusions and other cases where closed form formulas exist and will work out how to calculate their Z-series expansions.
You think life is a secret, Life is only love of flying, It has seen many ups and downs, But it likes travel more than the destination. Allama Iqbal
Amin
Topic Author
Posts: 2079
Joined: July 14th, 2002, 3:00 am
### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, I travelled from Lahore to Kot Addu yesterday. It is around six hours of travel. It is not very long but long enough to be stimulating. Another possible reason for stimulation for a mind control victim might be that it is difficult to control the victim and project waves on his mind with precision during long travel despite that most major highways and roads have mind control devices installed in Pakistan and Pak army has ensured that mind control waves could be projected when CIA mind control victims are travelling on those roads and highways. Especially when some victim connects his brain when mind control is weak and brain connects to parts unknown to mind control science, it can continue to create problems for mind control agents for several weeks in properly controlling the victim after the victim has travelled on a long journey. So when I have to travel, I like to make it stimulating while mind control agencies like to make it otherwise and sometimes drug the particular beverages like energy and caffeine drinks along the entire highway.
So something very interesting happened during night before travel. I bought three red bulls (red bull has recently become good in Lahore) and a chicken grilled sandwich the night before travel and kept them in my room so that I would be able to eat and drink good food while travelling next day. I know they can open door bolts that are used to lock the doors from outside and before I sleep, I usually put compressed tissue in the bolt's path so that it could not be opened from outside using magnets or anything. Mostly I put enough tissue that it would be extremely difficult to open the door from outside. But no external intrusion/unlocking had happened for more than five months so I was not being extra careful. I had put compressed tissue beneath the bolt but it was not so super compressed and if enough force could be applied from outside, it would open. I was just not being careful enough since nothing of the sort had happened for months. When I woke up in the morning, compressed tissue was lying on the dressing table on one side(as nobody could place it back from outside using magnets or any devices after locking the door from outside). I knew they had entered my room while I was sleeping to drug the red bulls and the sandwich.
We (my parents and I) were supposed to start the travel well into the day since due to high fog motorways are mostly closed in the morning at this time of the year. I had enough time so I went out and bought a few red bulls and bottled water again quickly and came back to the house and I had good food while travelling.
The travel was very stimulating and I thought of several new ideas and gained clarity about many new things related to my research which I will continue to share with wilmott friends in coming weeks.
Today, after coming to Kot addy yesterday evening, I cleaned my room and removed huge amount of dust and then changed sheets and now I really like my clean and big airy room that has two large windows and lots of sunlight. I will start working on my research tonight and hope that it would be a lot of fun working here in our Kot Addu home. I really enjoy coming to Kot Addu in winters. And drive around on the motorcycle in the morning around the countryside with lush green fields on both sides of the roads or streams. I was able to do very good research when I came here last winters.
I bought bottled water and some beverages from Kot Addu yesterday night and this morning and bottled water and many other drinks in the market are perfectly good. I am very afraid that mind control agencies would start drugging the food, drinks and bottled water in coming days to make life difficult for me. Please ask them to not drug and food or beverages in Kot Addu.
I will start working on my research in a few hours. Working at night is fun in mild winters. And first thing I Want to do is to see if we can easily find Z-series of a random variable when its closed form analytic density is given. And then present some simple examples from lognormal density and other well known densities. I hope some results will be ready in another day or two. So I hope to come back quickly with some new work in another day.
You think life is a secret, Life is only love of flying, It has seen many ups and downs, But it likes travel more than the destination. Allama Iqbal
Amin
Topic Author
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation
Friends, mind control agents are getting desperate to drug my food. And they are coming to my room even when I am present inside the house somewhere else. We have a large old house that was built in late eighties. Upper floor also has three bedrooms and a large living room and was constructed a decade later in late nineties. Stairs to upper floor are right next to main entrance. Main entrance (with stairs) leads to a gallery and other rooms and kitchen have approach from the gallery. There is no living room on ground floor. My room is on first floor and only I am living on first floor. It is a large 3.5 kanals house but since nobody lives here (except for my father) everything including lawns remains dark unless there is somebody living in a room for a few days or weeks.
Somewhat after writing previous post, I went out and brought some food. I quickly had some bit of food in my room on first floor and then went down to kitchen to make green tea in microwave before I start working on my research. OK, I very clearly remember this when I was leaving my room, I turned off lights of my room and then also turned off lights outside my room. And there was dark on first floor and the staircase and no light was coming out of my room even though its door was open. I clearly remember that I thought I can easily go down the stairs in the dark but it would be difficult to come up when I would have hot green tea in my hands and then I turned on the light outside my room. I recall this very clearly that there was dark all over the first floor and then I turned on outside light so I would not face difficulty going up the stairs when I come back. I remained in the kitchen for more than fifteen minutes. Anybody could come inside the house from the dark lawns and go to upper floor without my knowing about it at all since kitchen has approach from gallery and you could not view the entrance or the stairs from there. When I came upstairs after making green tea, not only outside light was on, the light of my room was also on which I knew very well that I had turned off. I was very surprised and came down and explicitly asked my brother, father and mother but they all refused that they went upstairs in my absence. My only guess is that somebody sneaked quickly from the entrance and took stairs to my room possibly to drug the food when I was in the kitchen making green tea. I am going to be careful with the food that was left in my room but I want friends to complain to mind control agencies to end their nasty tactics. I really do not understand what urgency they have that they have to resort to such desperate tactics.
You think life is a secret, Life is only love of flying, It has seen many ups and downs, But it likes travel more than the destination. Allama Iqbal | 36,771 | 91,300 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-14 | latest | en | 0.531554 |
https://documen.tv/question/a-1-375-g-sample-of-mannitol-a-sugar-found-in-seaweed-is-burned-completely-in-oygen-to-give-1-99-24021410-15/ | 1,627,745,201,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00431.warc.gz | 222,469,333 | 18,498 | ## A 1.375 g sample of mannitol, a sugar found in seaweed, is burned completely in oxygen to give 1.993 g of carbon dioxide and 0.9519 g of wat
Question
A 1.375 g sample of mannitol, a sugar found in seaweed, is burned completely in oxygen to give 1.993 g of carbon dioxide and 0.9519 g of water. The empirical formula of mannitol is
in progress 0
1 week 2021-07-21T21:18:33+00:00 1 Answers 3 views 0
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the empirical formula of mannitol contains carbon, hydrogen and oxygen, so that the first step is to calculate the moles of C and H contained in the CO2 and H2O, respectively, as the only sources of these two elements in the formula:
Next, we calculate the grams and moles of O by subtracting the mass of C and H from the mass of the sample:
Finally, we divide the moles of C, H and O by 0.0453 as the fewest moles of both C and O to find the mole ratios in the formula:
To get:
Which must be multiplied by 3 to get whole numbers for all the subscripts, and therefore obtain:
Regards! | 299 | 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} | 3.609375 | 4 | CC-MAIN-2021-31 | latest | en | 0.925411 |
https://www.archivemore.com/why-is-geometry-important-in-the-real-world/ | 1,696,100,242,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00112.warc.gz | 691,407,896 | 10,162 | Why is geometry important in the real world?
Geometry helps us in deciding what materials to use, what design to make and also plays a vital role in the construction process itself. Geometrical tools like the protractor, ruler, measuring tape, and much more are used in construction work, astronomy, for measurements, drawing etc.
How is geometry used in the real world?
Applications of geometry in the real world include computer-aided design for construction blueprints, the design of assembly systems in manufacturing, nanotechnology, computer graphics, visual graphs, video game programming and virtual reality creation.
Is geometry really useful for every student in life?
Geometry has an effect on us that not only teaches us the core fundamentals of the course itself, but also allows us to take those skills and use them throughout our daily life. More specifically, it teaches you how to measure lengths, areas, and volumes.
How is geometry used in real life for kids?
Kid Geometry Have the kids pose as shapes or other geometric concepts. For example, a child sitting on the floor with his back upright is a right angle; two kids standing next to each other represent parallel lines, six students can form a hexagon on the ground.
How do I know if I really apply the concept of geometry in my life?
Geometry has many practical uses in everyday life, such as measuring circumference, area and volume, when you need to build or create something. Geometric shapes also play an important role in common recreational activities, such as video games, sports, quilting and food design.
Why do we need to learn geometry?
Why Geometry Matters At a basic level, geometry is important to learn because it creates a foundation for more advanced mathematical learning. It introduces important formulas, such as the Pythagorean theorem, used across science and math classes. It is also foundational knowledge for certain careers in STEM fields.
Euclid
What math do I need to be a surgeon?
Students who want to become doctors should take all available math courses in high school. In chronological order, most high schools and many colleges offer students courses in pre-algebra, algebra 1, algebra 2 and trigonometry or pre-calculus.
What college should I go to to become a surgeon?
School Overviews Three of the best places to pursue surgical training in the U.S. are Johns Hopkins University, Columbia University, and Emory University. Each of these schools offers four years of medical education and clerkship as well as postgraduate residencies and fellowships.
What type of surgeon makes the most money?
Neurosurgery and thoracic surgery were the highest paying medical specialties in 2019, with average salaries north of \$550,000, according to a new physician employment report from Doximity. Orthopedic surgery, radiation oncology and vascular surgery rounded out the top five.
What are the highest paid doctors?
Top 19 highest-paying doctor jobs
• Surgeon.
• Dermatologist.
• Orthopedist.
• Urologist.
• Neurologist. National average salary: \$237,309 per year.
• Orthodontist. National average salary: \$259,163 per year.
• Anesthesiologist. National average salary: \$328,526 per year.
• Cardiology physician. National average salary: \$345,754 per year.
How much money does a surgeon make a year 2020?
How Much Do Surgeon Jobs Pay per Hour?
Annual Salary Monthly Pay
Top Earners \$400,000 \$33,333
75th Percentile \$390,000 \$32,500
Average \$248,104 \$20,675
25th Percentile \$98,000 \$8,166
What is the highest paid doctor 2020?
According to Medscape, the highest-paying specialty in 2020 is orthopedics, with an average compensation of \$511,000, while public health and preventive medicine and pediatrics are tied for the lowest-paying specialty, with an average compensation of \$232,000. | 810 | 3,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-40 | latest | en | 0.947525 |
https://sec.oercommons.org/EN/courses/multiple-representations-of-numeric-patterning | 1,618,391,071,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077336.28/warc/CC-MAIN-20210414064832-20210414094832-00353.warc.gz | 601,453,419 | 14,136 | Teacher
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# Multiple Representations of Numeric Patterning
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## Description
Overview:
This lesson is a re-engagement lesson designed for learners to revisit a problem-solving task they have already experienced. Students will activate prior knowledge of graphical representations through the 'what's my rule' number talk; compare and contrast two different learners' interpretations of the growing pattern; use multiple representations to demonstrate how one of these learners would represent the numeric pattern; make connections between the different representations to more critically compare the two interpretations. (5th/6th Grade Math)
Subject:
Numbers and Operations
Level:
Upper Primary, Middle School
Material Type:
Activity/Lab, Lecture, Lesson Plan, Teaching/Learning Strategy
Author:
Provider:
Noyce Foundation
Provider Set:
Inside Mathematics
11/30/2011
http://www.insidemathematics.org/
Language:
English
Media Format:
SY HIDOE on Aug 15, 08:21pm
This resource provides a nice visual format for teachers to see how a lesson might run in the classroom. It addresses a portion of 5.OA.3 and 4.OA.5. It shows a nice connection between how you might take a lesson from working with a visual pattern, to the more symbolic level of graphing an equation on a coordinate plane. It is a nice example of helping students develop Math Practice #2, among others.
EK HIDOE on Jul 17, 05:26pm
This video resource aligns with CCSS-M 5.OA.3. It does align with 4.OA.5 as well in how it pertains to generating number patterns but since it refers to ordered pairs and graphing them in a coordinate plane, this resource aligns more closely with the 5th grade standard. The video is broken down into parts which takes the viewer through the process of developing the lesson, execution of the lesson, and a debrief on the learning. Each part offers some insight into how we must attend to both content and the learning we want to take place. That being said, this resource also aligns to several Standards for Mathematical Practice (SMP), specifically SMPs #1, 2, 3, and 7. During various video segments, it displays how we want our students to be engaged in the learning through discussions and hands-on work. The reference to the 'Buttons' task in relation to the use of an input/output table assist in making connections for the learners. | 535 | 2,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2021-17 | longest | en | 0.909816 |
https://www.coursehero.com/file/6028492/Normal-Distribution-slides-update3/ | 1,519,227,458,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813626.7/warc/CC-MAIN-20180221143216-20180221163216-00216.warc.gz | 843,829,677 | 121,320 | {[ promptMessage ]}
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Normal-Distribution-slides-update3
# Normal-Distribution-slides-update3 - The Normal Model Stat...
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The Normal Model Stat C180/C236 Intro Bayesian Statistics Juana Sanchez [email protected] UCLA Department of Statistics Stat C180/C236 Intro Bayesian Statistics/J. Sanchez The Normal Model
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The Normal Model (1) The Normal density function. Review. Properties of the Normal Density. Joint Distribution of n iid Normal random variables (2) Likelihood Interpretation of joint distribution (3) Case 1. Normal model, mean unknown, variance known. Conjugate prior distribution for θ . n=1. Interpretation of the posterior mean of θ . Likelihood, prior and posterior for θ , several examples. R code for figure. (4) Case 2. Normal model n > 1, θ unknown, σ 2 known. Example-midge wing length. Interpretation of the posterior distribution parameters. (5) Case 3. Normal model, variance unknown, mean known (6) Multiparameter models. Averaging over nuisance parameters. Two parameter case example. Interpretation of the marginal posterior distribution for θ . Posterior inference by simulation of the joint posterior distribution. (7) Case 4 Two parameter Normal model with non informative prior. Obtaining marginal posterior distribution for θ . Example: experiment on magnetic fields. (8) Case 5. Joint Inference for the mean and variance with informative priors Stat C180/C236 Intro Bayesian Statistics/J. Sanchez The Normal Model
The Normal density function. Review A random variable Y is said to be normally distributed with mean θ and variance σ 2 > 0 if the density of Y is given by p ( y | θ, σ 2 ) = 1 2 πσ 2 e - 1 2 ( y - θ σ ) 2 - ∞ < y < 0 2 4 6 8 10 0.0 0.2 0.4 0.6 0.8 y p(y | θ , σ 2 ) θ = 2, σ 2 = 0.25 θ = 5, σ 2 = 4 θ = 7, σ 2 = 1 Figure: Some Normal Densities for a few values of θ and σ Stat C180/C236 Intro Bayesian Statistics/J. Sanchez The Normal Model
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Properties of the Normal Density 1 The distribution is symmetric about θ , and the mode, median and mean are all equal to θ 2 95% of the population lies within two standard deviations of the mean (68-95-99.7 rule). 3 If X normal ( μ, τ 2 ), Y normal ( θ, σ 2 ) and X and Y are independent, then aX + bY normal ( a μ + b θ, a 2 τ 2 + b 2 σ 2 ); 4 The dnorm, rnorm, pnorm and qnorm commands in R take the standard deviation σ as their argument, not the variance σ 2 . Be careful about this. Stat C180/C236 Intro Bayesian Statistics/J. Sanchez The Normal Model
Joint Distribution of n iid Normal Random Variables Suppose we have a random sample of i.i.d. y 1 , .... , y n , where y i N ( θ, σ 2 ) Then the joint distribution is p ( y 1 , .... , y n | θ, σ 2 ) = n Y i =1 p ( y i | θ, σ 2 ) = n Y i =1 1 2 πσ 2 e - 1 2 ( y - θ σ ) 2 = (2 2 ) - n 2 exp ( - 1 2 X ± y - θ σ ² 2 ) where ( - 1 2 X ± y - θ σ ² 2 = 1 σ 2 X y 2 i - 2 θ σ 2 X y i + n θ 2 σ 2 ) Stat C180/C236 Intro Bayesian Statistics/J. Sanchez The Normal Model
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Likelihood interpretation of joint Distribution and Nice properties 1 As a function of the parameters θ and σ 2 , the joint distribution is the likelihood of the data for different values of the parameters. Recall in Mathematical Statistics we maximized the likelihood with respect to θ and σ 2 to obtain the maximum Likelihood estimators of θ and σ 2 .
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Ask a homework question - tutors are online | 1,093 | 3,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-09 | latest | en | 0.655435 |
https://www.convertunits.com/from/chaldron/to/ounce-force | 1,653,208,289,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545090.44/warc/CC-MAIN-20220522063657-20220522093657-00410.warc.gz | 822,173,829 | 17,070 | ## ››Convert chaldron to ounce-force
chaldron ounce-force
How many chaldron in 1 ounce-force? The answer is 1.0528992588727E-5.
We assume you are converting between chaldron and ounce-force.
You can view more details on each measurement unit:
chaldron or ounce-force
The SI base unit for mass is the kilogram.
1 kilogram is equal to 0.00037139928394218 chaldron, or 35.27396194958 ounce-force.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between chaldrons and ounce-force.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of chaldron to ounce-force
1 chaldron to ounce-force = 94975.84803 ounce-force
2 chaldron to ounce-force = 189951.69606 ounce-force
3 chaldron to ounce-force = 284927.54409 ounce-force
4 chaldron to ounce-force = 379903.39211 ounce-force
5 chaldron to ounce-force = 474879.24014 ounce-force
6 chaldron to ounce-force = 569855.08817 ounce-force
7 chaldron to ounce-force = 664830.9362 ounce-force
8 chaldron to ounce-force = 759806.78423 ounce-force
9 chaldron to ounce-force = 854782.63226 ounce-force
10 chaldron to ounce-force = 949758.48028 ounce-force
## ››Want other units?
You can do the reverse unit conversion from ounce-force to chaldron, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 496 | 1,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-21 | longest | en | 0.69074 |
www.kraj.in | 1,582,670,943,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146160.21/warc/CC-MAIN-20200225202625-20200225232625-00063.warc.gz | 215,643,763 | 16,559 | Select Page
# Theory
5A Module 20A Module 30A Module
Supply Voltage (VCC) 5Vdc Nominal 5Vdc Nominal 5Vdc Nominal
Measurement Range -5 to +5 Amps -20 to +20 Amps -30 to +30 Amps
Voltage at 0A VCC/2
(nominally 2.5Vdc)
VCC/2
(nominally 2.5Vdc)
VCC/2
(nominally 2.5VDC)
Scale Factor 185 mV per Amp 100 mV per Amp 66 mV per Amp
Chip ACS712ELC-05A ACS712ELC-10A ACS712ELC-30A
## ACS712 Module Pin Outs:
• Always connect load in mentioned direction for positive output.
• If you will connect as illustrated below, the output will be positive ans above 2.5 volt .
• If you will connect it in opposite direction as illustrated in below picture, the output will decrease from the 2.5 volt.
• It will give 2.5 volt when there is no current flowing through it.
# Coding
/*
Arduino Acs712 DC Current Sensing
Function: Sense the DC Current and Display on the Serial Monitor
Website: http://www.Kraj.in
*/
double Voltage = 0;
double Current = 0;
void setup(){
Serial.begin(9600);
}
void loop(){
// Voltage is Sensed 1000 Times for precision
for(int i = 0; i < 1000; i++) {
Voltage = (Voltage + (.0049 * analogRead(A0))); // (5 V / 1024 (Analog) = 0.0049) which converter Measured analog input voltage to 5 V Range
delay(1);
}
Voltage = Voltage /1000;
Current = (Voltage -2.5)/ 0.185; // Sensed voltage is converter to current
Serial.print(“\n Voltage Sensed (V) = “); // shows the measured voltage
Serial.print(Voltage,2); // the ‘2’ after voltage allows you to display 2 digits after decimal point
Serial.print(“\t Current (A) = “); // shows the voltage measured
Serial.print(Current,2); // the ‘2’ after voltage allows you to display 2 digits after decimal point
delay(1000);
} | 514 | 1,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.636173 |
https://oeis.org/A335109 | 1,620,401,781,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988796.88/warc/CC-MAIN-20210507150814-20210507180814-00573.warc.gz | 457,544,842 | 4,473 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A335109 Triangle read by rows: T(n,k) is the number of permutations of length n with each cycle of the permutation containing only elements that are identical (mod k), where 1 <= k <= n. 0
1, 2, 1, 6, 2, 1, 24, 4, 2, 1, 120, 12, 4, 2, 1, 720, 36, 8, 4, 2, 1, 5040, 144, 24, 8, 4, 2, 1, 40320, 576, 72, 16, 8, 4, 2, 1, 362880, 2880, 216, 48, 16, 8, 4, 2, 1, 3628800, 14400, 864, 144, 32, 16, 8, 4, 2, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Let [n] denote {1,2,...,n} and let [n](j,k) denote the subset of [n] consisting of all elements of [n] that equal j mod k. The cardinality of [n](j,k) equals ceiling(n/k) for j = 1..(n mod k) and equals floor(n/k) for j > (n mod k). Therefore, upon permuting the elements of each [n](j,k) subset, we obtain T(n,k) = ceiling(n/k)!)^(n mod k)*(floor(n/k)!)^(k-n mod k). LINKS FORMULA T(n,k) = (ceiling(n/k)!)^(n mod k)*(floor(n/k)!)^(k-n mod k) for 1 <= k <= n. T(n,1) = A000142(n). T(n,2) = A010551(n) for n > 1. T(n,3) = A264557(n) for n > 2. T(n,4) = A264635(n) for n > 3. T(n,5) = A264656(n) for n > 4. T(n,k) = Product_{i=0..k-1} floor((n+i)/k)!. - Alois P. Heinz, May 23 2020 EXAMPLE Triangle begins: 1; 2 1; 6 2 1; 24 4 2 1; 120 12 4 2 1; ... T(6,3) counts the 8 permutations of [6] where all cycle-mates are identical mod 3, namely, (1 4)(2 5)(3 6), (1 4)(2 5)(3)(6), (1 4)(2)(5)(3 6), (1)(4)(2 5)(3 6), (1 4)(2)(5)(3)(6), (1)(4)(2 5)(3)(6), (1)(4)(2)(5)(3 6) and (1)(2)(3)(4)(5)(6). MAPLE seq(seq((ceil(n/k)!)^(n mod k)*(floor(n/k)!)^(k-(n mod k)), k=1..n), n=1..10); MATHEMATICA Table[(Ceiling[n/k]!)^Mod[n, k]*(Floor[n/k]!)^(k - Mod[n, k]), {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Jun 28 2020 *) CROSSREFS Cf. A000142, A010551, A264557, A264635, A264656. Cf. A275062. Sequence in context: A106187 A110135 A114423 * A179863 A069123 A134133 Adjacent sequences: A335106 A335107 A335108 * A335110 A335111 A335112 KEYWORD nonn,tabl AUTHOR Dennis P. Walsh, May 23 2020 STATUS approved
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https://kupdf.net/download/joint-costing_5af6e81be2b6f5f65b15363a_pdf | 1,656,399,336,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103355949.26/warc/CC-MAIN-20220628050721-20220628080721-00609.warc.gz | 389,472,983 | 18,806 | # Joint Costing
November 4, 2017 | Author: Hassan Mohiuddin | Category: Chocolate, Gross Margin, Debits And Credits, Cost Of Goods Sold, Inventory Valuation
#### Short Description
Joint Costing Questions...
#### Description
Problem 16-16: Joint-cost allocation, insurance settlement. Given: Quality Chicken grows and processes chickens. Each chicken is disassembled into five main parts. Information pertaining to production in July 2012 is: Parts Breasts Wings Thighs Bones Feathers
Pounds of Product 100 20 40 80 10
Final Wholesale SP/Pound \$0.55 0.20 0.35 0.10 0.05
Joint cost of production in July 2012 was \$50.
\$50
A special shipment of 40 pounds of breasts and 15 pounds of wings has been destroyed in a fire. Quality Chicken's insurance policy provides reimbursement for the cost of the items destroyed. The insurance company permits Quality Chicken to use a joint-cost-allocation method. The splitoff point is assumed to be at the end of the production process. 1. Compute the cost of the special shipment destroyed using a. Sales value at the splitoff method Product Pounds of Produced Product Breasts 100 Wings 20 Thighs 40 Bones 80 Feathers 10 Total 250
SP per Sales Value Joint Cost Joint Cost Pound at Splitoff Pt. Allocated Per Pound \$0.55 \$55.00 \$33.74 \$0.3374 0.20 \$4.00 \$2.45 \$0.1227 0.35 \$14.00 \$8.59 \$0.2147 0.10 \$8.00 \$4.91 \$0.0613 0.05 \$0.50 \$0.31 \$0.0307 \$81.50 \$50.00
Cost of destroyed Product Product Joint Cost Produced Per Pound Breasts: \$0.3374 Wings \$0.1227 Total
Pounds Lost 40 15
Insurance Claim \$13.50 \$1.84 \$15.34
b. Physical measures method Product Pounds of Produced Product Breasts 100 Wings 20 Thighs 40 Bones 80
Joint Cost Allocated \$20.00 \$4.00 \$8.00 \$16.00
Joint Cost Per # \$0.2000 \$0.2000 \$0.2000 \$0.2000
Feathers Total
10 250
\$2.00 \$50.00
\$0.2000
Pounds Lost 40 15
Insurance Claim \$8.00 \$3.00 \$11.00
\$0.20
Cost of destroyed Product Product Joint Cost Produced Per Pound Breasts: \$0.2000 Wings \$0.2000 Total
2. What is joint-cost allocation would you recommend? Sales value at the splitoff method generates the higher insurance recovery value.
Problem 16-17 Joint products and byproducts (continuation of 16-16) Given: Quality Chicken grows and processes chickens. Each chicken is disassembled into five main parts. Information pertaining to production in July 2012 is: Parts Breasts Wings Thighs Bones Feathers
Pounds of Product Final Wholesale SP/Pound 100 \$0.55 20 0.20 40 0.35 80 0.10 10 0.05
Joint cost of production in July 2012 was \$50. Quality Chicken is computing the EI values for its July 31, 2012, balance sheet. EI amounts on July 31 are: Parts Pounds of Product Breasts 15 Wings 4 Thighs 6 Bones 5 Feathers 2 Quality Chicken's management wants to use the sales value at splitoff method. However, management wants you to explore the effect on ending inventory values of classifying one or more products as a byproduct rather than a joint product. 1. Compute the cost of the ending inventory if all products are accounted for as joint products. The sales values at the split-off point is used to assign joint manufacturing costs. Product Pounds of Produced Product Breasts 100 Wings 20 Thighs 40 Bones 80 Feathers 10 Total 250
SP per Sales Value Pound at Splitoff Pt. \$0.55 \$55.00 \$ 0.20 \$4.00 \$ 0.35 \$14.00 \$ 0.10 \$8.00 \$ 0.05 \$0.50 \$81.50
Joint Cost Joint Cost Allocated Per # \$33.7423 \$0.3374 \$2.4540 \$0.1227 \$8.5890 \$0.2147 \$4.9080 \$0.0613 \$0.3067 \$0.0307 \$50.0000
EI in Pounds 15 4 6 5 2
EI in Dollars \$5.0613 \$0.4908 \$1.2883 \$0.3067 \$0.0613 \$7.2086
2. Assume Quality Chicken uses the production method of accounting for byproducts. What are the EI values for each joint product on July 31, 2012, assuming breasts and thighs are the joint products and wings, bones, and feathers are byproducts? Product Pounds of Produced Product Breasts 100 Wings 20 Thighs 40 Bones 80
SP per Sales Value Pound at Splitoff Pt. \$0.55 \$55.0000 \$ 0.20 \$4.0000 \$ 0.35 \$14.0000 \$ 0.10 \$8.0000
Joint Cost Joint Cost Allocated Per # \$29.8913 \$0.2989 \$4.0000 \$0.2000 \$7.6087 \$0.1902 \$8.0000 \$0.1000
EI in Pounds 15 4 6 5
EI in Dollars \$4.4837 \$0.8000 \$1.1413 \$0.5000
Feathers Total
10 250
\$ 0.05
\$0.5000 \$81.5000
\$0.5000 \$50.0000
\$0.0500
2
\$0.1000 \$7.0250
3. Comment on the differences between the two methods.
Product Produced Breasts Wings Thighs Bones Feathers Total
EI in Pounds 15 4 6 5 2 32
Units Sold 85 16 34 75 8 218
(1) Joint Cost Per # \$0.3374 \$0.1227 \$0.2147 \$0.0613 \$0.0307
(1) Cost of Goods Sold \$28.680982 \$1.963190 \$7.300613 \$4.601227 \$0.245399 \$42.791411
(1) EI in Dollars \$5.0613 \$0.4908 \$1.2883 \$0.3067 \$0.0613 \$7.2086
(1) (2) Total Joint Cost Costs Per # \$33.7423 \$0.2989 \$2.4540 \$0.2000 \$8.5890 \$0.1902 \$4.9080 \$0.1000 \$0.3067 \$0.0500 \$50.0000
Both methods account for all of the \$50 of joint manufacturing costs as either COGS or EI. Both methods are arbitrary and acceptable under GAAP.
(2) Cost of Goods Sold \$25.407609 \$3.200000 \$6.467391 \$7.500000 \$0.400000 \$42.975000
(2) EI in Dollars \$4.4837 \$0.8000 \$1.1413 \$0.5000 \$0.1000 \$7.0250
(2) Total Costs \$29.8913 \$4.0000 \$7.6087 \$8.0000 \$0.5000 \$50.0000
Problem 16-20: Alternative methods of joint-cost allocation, ending inventories.
Given: The Evrett Company operates a simple chemical process to convert a single material into three separate items, referr at a single splitoff point. Product X and Y are ready for sale immediately upon splitoff without further processing or a being sold. There is no available market price for Z at the splitoff point.
The selling prices quoted here are expected to remain the same in the coming year. During 2012, the selling prices of X: 75 tons sold for \$1,800 per ton. Y: 225 tons sold for \$1,300 per ton Z: 280 tons sold for \$800 per ton
The total joint manufacturing costs for the year were \$328,000. Evrett spent an additional \$120,000 to finish product Z There were no beginning inventories of X, Y, or Z. At the end of the year, the following inventories of completed units X: 175 tons Y: 75 tons Z: 70 tons There was no beginning or ending work-in-process.
1. Compute the cost of inventories for X, Y, and Z for balance sheet purposes and the cost of goods sold for income s Relative Sales Value
Production (in Tons)
Separable Mfg. Costs
Final S. P. (per Ton)
\$450,000
X
250
\$0
\$1,800
\$390,000
Y
300
\$0
\$1,300
Estimated \$160,000
Z
350
\$120,000
\$800
Sales Price per Ton \$1,800 \$1,300 \$800
Relative SV at Split-off \$450,000 \$390,000 \$160,000 \$1,000,000
Joint Cost Allocated \$147,600 \$127,920 \$52,480 \$328,000
Joint Mfg. Costs \$328,000
a. NRV method of joint-cost allocation Products Produced X Y Z Total
Sales in Tons 75 225 280 580
EI in Tons 175 75 70 320
Tons Produced 250 300 350 900
b. Constant gross-margin % NRV method of joint-cost allocation
Products Produced X Y Z Total
Sales in Tons 75 225 280 580
EI in Tons 175 75 70 320
Tons Produced 250 300 350 900
Calculation of Overall Gross Margin: Production based Sales Joint Manufacturing Costs Separable Manufacturing Costs Gross Margin Gross Margin %
Sales Price per Ton \$1,800 \$1,300 \$800
\$1,120,000 (\$328,000) (\$120,000) \$672,000 60%
Production Based Sales \$450,000 \$390,000 \$280,000 \$1,120,000
Joint Cost Allocated (1) \$180,000 \$156,000 (\$8,000) \$328,000
(1) Note: The negative joint-cost alloca % NRV method. Some produ have the same gross-margin
Sales - GP - Separable Costs = Joi
o three separate items, referred to here as X, Y, and Z. All three end products are separated simultaneously thout further processing or any other additional costs. Product Z, however, is processed further before
ng 2012, the selling prices of the items and the total amounts sold were:
l \$120,000 to finish product Z.
ventories of completed units were on hand:
st of goods sold for income statement purposes as of 12/31/12, using
Final Sales Value \$450,000
\$390,000
\$280,000
Separable Mfg. Costs \$0 \$0 \$120,000 \$120,000
Total Assigned Mfg. Costs \$147,600 \$127,920 \$172,480 \$448,000
Cost per Ton \$590 \$426 \$493
Cost of Goods Sold \$44,280 \$95,940 \$137,984 \$278,204 \$448,000
Ending Inventory \$103,320 \$31,980 \$34,496 \$169,796
Actual Sales \$ \$135,000 \$292,500 \$224,000 \$651,500
Gross Margin \$90,720 \$196,560 \$86,016 \$373,296
Separable Mfg. Costs \$0 \$0 \$120,000 \$120,000
Total Assigned Mfg. Costs \$180,000 \$156,000 \$112,000 \$448,000
Costs per Ton \$720 \$520 \$320
Cost of Goods Sold \$54,000 \$117,000 \$89,600 \$260,600 \$448,000
Ending Inventory \$126,000 \$39,000 \$22,400 \$187,400
Actual Sales \$ \$135,000 \$292,500 \$224,000 \$651,500
Gross Margin \$81,000 \$175,500 \$134,400 \$390,900
e: The negative joint-cost allocation to Product Z illustrates one "unusual" feature of the constant gross-margin % NRV method. Some products may receive negative cost allocations in order that all individual products have the same gross-margin percentage.
s - GP - Separable Costs = Joint Cost Allocated
Gross Margin % 67.200% 67.200% 38.400%
Gross Margin % 60.000% 60.000% 60.000%
Problem 16-24: Accounting for a main product and a byproduct
Given: Tasty Inc. is a producer of potato chips. A single production process at Tasty Inc. yields potato chips as the main pro a byproduct that can also be sold as a snack. Both products are fully processed by the splitoff point, and there are no costs. For September 2012, the cost of operations is \$500,000. Production and sales data are as follows: Sales (#) 42,640 6,500
Production (#)
Main Product: Potato Chips Byproduct
52,000 8,500
Selling Price per # 16 10
There were no beginning inventories.
1. What is the gross margin for Tasty, Inc. under the production method and the sales method of byproduct accountin
Production Method: Byproduct revenue is recognized as a reduction of the joint mfg. costs at time of production. Product Produced Potato Byproduct Total
Pounds EI in Sold Pounds 42,640 9,360 6,500 2,000 49,140 11,360
Pounds SP per Sales Value Joint Cost Produced Pound at Splitoff Pt. Allocated 52,000 \$16 \$832,000 \$415,000 8,500 \$10 \$85,000 \$85,000 60,500 \$917,000 \$500,000
Joint Cost Per Pound \$7.98 \$10.00
Note: The byproduct sales of \$65,000 (6,500 X \$10) and equivalent assigned cost of \$65,000 are not displayed as e Sales Method: Byproduct revenue is recognized at the time of sale. Product Produced Potato Byproduct Total
Pounds EI in Sold Pounds 42,640 9,360 6,500 2,000 49,140 11,360
Pounds SP per Sales Value Joint Cost Produced Pound at Splitoff Pt. Allocated 52,000 \$16 \$832,000 \$500,000 8,500 \$10 \$85,000 \$0 60,500 \$917,000 \$500,000 Production Method--Y1
Revenues Main Product: Potato By-product: Snacks Total Revenues Cost of Goods Sold Total mfg. Costs Deduct byproduct revenue Net mfg. Costs Deduct main product inventory (EI) Total Cost of Goods Sold Gross Margin
42,640 6,500
8,500
\$16 \$10
\$10
\$682,240 \$682,240 \$500,000 (85,000) \$415,000 (74,700) \$340,300 \$341,940
Joint Cost Per Pound \$9.62 \$0.00
Sales Method-- Y1
\$
682,240 65,000 \$747,240 \$500,000 \$500,000 (90,000) \$410,000 \$337,240
2. What are the inventory values that should be reported in the balance sheets under each method of byproduct acco Year 1: Production Method--Y1 Sales Method-- Y1
EI of Main Product (Potato) EI of By-product
9,360 Pounds 2,000 Pounds
\$74,700 \$20,000
\$90,000 0
Overall proof: Production Method
GP - Y1 Sales - Y2 Main 9,360 ByPd 2,000 Byprod costs Total Revenues COGS GP - Y2 Total Gross Profit
\$16 \$10
Proof: Both Years Together Sales Main 52,000 \$16 ByPd 8,500 \$10 Jt. Mfg. Costs Total Gross Profit
Sales Method
\$341,940
\$337,240
\$149,760 \$20,000 (20,000) \$149,760 (74,700) \$75,060 \$417,000
\$149,760 \$20,000 0 \$169,760 (90,000) \$79,760 \$417,000
\$832,000 \$85,000 (\$500,000) \$417,000
ato chips as the main product and off point, and there are no separable re as follows:
\$500,000
elling Price per #
d of byproduct accounting?
sts at time of production.
Cost of Gross Goods Sold Margin \$340,300 \$ 341,940 \$ \$ \$340,300 \$ 341,940 \$435,000 \$65,000 ?? Under reported COGS 000 are not displayed as either Sales or COGS.
es Method-- Y1
EI in Sales Dollars Dollars \$74,700 \$ 682,240 \$20,000 \$ \$94,700 \$ 682,240
EI in Dollars \$90,000 \$0 \$90,000
Sales Dollars \$ 682,240 \$ 65,000 \$ 747,240
Difference (SM - PM)
\$
65,000 \$65,000
\$
85,000 \$ 85,000 (15,300) \$69,700 (\$4,700)
method of byproduct accounting.
es Method-- Y1
Cost of Goods Sold \$410,000 \$0 \$410,000 \$500,000
Gross Margin \$ 272,240 \$ 65,000 \$ 337,240
Difference (SM - PM)
(\$4,700)
\$20,000 (15,300) \$4,700 \$0
Problem 16-25: Joint Costs & Byproducts Given: Royston, Inc. is a large food processing company. It processes 150,000 pounds of peanuts in the Peanuts Department at a cost of \$180,000 to yield 12,000 pounds of product A, 65,000 pounds of product B, and 16,000 pounds of product C. Product A is processed further in the Salting Department to yield 12,000 pounds of salted peanuts at a cost of \$27,000 and sold for \$12 per pound. Product B (Raw Peanuts) is sold without further processing at \$3 per pound. Product C is considered a byproduct and is processed further in the Paste Department to yield 16,000 pounds of peanut butter at a cost of \$12,000 and sold for \$6 per pound. The company wants to make a gross margin of 10% of revenues on Product C and needs to allow 20% of revenues for marketing costs on product C. 1. Compute unit costs per pound for products A, B, and C, treating C as a byproduct. Use the NRV method for allocating joint costs. Deduct the NRV of the byproduct produced from the joint cost of products A and B (Production method).
Salting Department Processed to Af A 12,000 Pounds Peanuts Department Processing \$180,000
Product B Complete B 65,000 Pounds
Final Selling Price
Final Sales Value
Separable Separable Mfg. Marketing Costs Costs (20%)
Desired Gross Profit (10%)
Sales Value at Splitoff
Allocation Separable of Joint Mfg. Mfg. Costs Costs
Total Mfg. Costs
Total Mfg. Costs/#
\$27,000 \$12
\$144,000
\$27,000
\$0
\$117,000
\$46,800
\$27,000
\$73,800
\$6.15
49%
\$3
\$195,000
\$0
\$0
\$195,000
\$78,000
\$0
\$78,000
\$1.20
60%
\$6
\$96,000
\$12,000
\$19,200
\$55,200
\$55,200
\$12,000
\$67,200
\$4.20 \$4.20
\$367,200
\$180,000
\$39,000
\$219,000 \$219,000
Total Mfg. Costs
Total Mfg. Costs/#
\$0
150,000 Pounds Paste Department Processed to Cf C (Byproduct) 16,000 Pounds
\$12,000 \$9,600
2. Compute unit costs per pound for products A, B, and C, treating all three as joint products and allocating costs by the NRV method.
Salting Department Processed to Af A 12,000 Pounds
Peanuts Department Processing \$180,000
Product B Complete B 65,000 Pounds
Final Selling Price
Sales Value
Separable Separable Mfg. Marketing Costs Costs (20%)
Sales Value at Splitoff
Allocation of Joint Mfg. Costs
Separable Mfg. Costs
\$27,000 \$12
\$144,000
\$27,000
\$0
\$117,000
\$55,892
\$27,000
\$82,892
\$6.91
\$3
\$195,000
\$0
\$0
\$195,000
\$93,153
\$0
\$93,153
\$1.43
\$6
\$96,000
\$12,000
\$19,200
\$64,800
\$30,955
\$12,000
\$42,955
\$2.68
\$376,800
\$180,000
\$39,000
\$219,000
\$0
150,000 Pounds
Paste Department Processed to Cf C 16,000 Pounds
\$12,000
Note: Costs of individual products depend heavily on which assumptions are made and which accounting methods and techniques are used.
30% 10%?? Company wants to make a profit of 10%.
Problem 16-28: Comparison of alternative joint-cost-allocation methods, further-processing decision Given: The Chocolate Factory manufactures and distributes chocolate products. It purchases cocoa beans and processes them into two intermediate products: Chocolate-powder liquor base and Milkchocolate liquor base. These two intermediate products become separately identifiable at a single splitoff point. Every 1,500 pounds of cocoa beans yields 60 gallons of chocolate-powder liquor base and 90 gallons of milk-chocolate liquor base. The chocolate-powder liquor base is further processed into chocolate powder. Every 60 gallons of chocolate-powder liquor base yield 600 pounds of chocolate powder. The milk-chocolate liquor base is further processed into milk chocolate. Every 90 gallons of milk-chocolate liquor base yield 1,020 pounds of milk chocolate. Production and sales data for August 2012 are (assume no beginning inventory): Coco beans processed, 15,000 pounds Costs of processing cocoa beans to splitoff point (including purchase of beans), \$30,000
Chocolate powder Milk chocolate
Production 6,000 pounds 10,200 pounds
Sales 6,000 pounds 10,200 pounds
Selling Price \$4 per pound \$5 per pound
Chocolate Factory fully processes both of its intermediate products into chocolate powder or milk chocolate. There is an active market for these intermediate products. In August 2012, Chocolate Factory could have sold the chocolate-powder liquor base for \$21 a gallon and the milk-chocolate liquor base for \$26 a gallon.
1. Calculate how the joint costs of \$30,000 would be allocated between the chocolate-powder and milk-chocolate liqu Joint Processing Costs Allocated Using the Sales Value at the Splitoff Point Quantity of Split-off Product at Selling Splitoff Point Price
Chocolate Powder Liquor Processed to Chocolate Powder SP @ Splitoff
\$21 600 Gallons
Final SP
\$12,750 Separable Mfg. Costs
\$4 6,000 Pounds
Cocoa Beans processing \$30,000 Joint Mfg. Costs
15,000 Pounds Milk-Chocolate Liquor Base Processed to Milk Chocolate SP @ Splitoff
Final SP
600
21
\$26 900 Gallons
\$26,250 Separable Mfg. Costs
\$5 10,200
900
26
1,500
Pounds
2. What are the gross-margin percentages of chocolate powder and milk chocolate under each of the above methods
Final Sales Joint manufacturing costs Separable manufacturing costs Gross Margin Gross Margin %
Overall Gross Margin \$75,000 (30,000) (39,000) \$6,000 8.00%
Relative Sales Value at Spiltoff Chocolate Milk Powder Chocolate \$24,000 \$51,000 (10,500) (19,500) (12,750) (26,250) \$750 \$5,250 3.125% 10.294%
3. Could Chocolate Factory have increased its operating income by a change in its decision to fully process both of i
Final Sales Value Sales Value at Splitoff Increase in Sales Value Separable manufacturing costs Incremental operating income Process further
Chocolate Milk Powder Chocolate \$24,000 \$51,000 12,600 23,400 \$11,400 \$27,600 (12,750) (26,250) (\$1,350) \$1,350 Yes Yes Chocolate-powder liquor base should be sold at splitoff point. Operating income would increase by \$1,350 to \$7,350. Note: Earlier calculations would change. (Change G77 to "No" to see changes.) No
rocessing decision
ases cocoa beans e and Milkfiable at a single owder liquor base
ery 60 gallons of ocolate liquor liquor base yield
Separable Processing \$12,750 \$26,250
powder or milk 12, Chocolate milk-chocolate
ate-powder and milk-chocolate liquor bases:
ssing Costs Allocated Using the Joint Processing Costs Allocated Using the Value at the Splitoff Point Net Realizable Value at the Splitoff Point Sales Allocation Quantity of Final Final Separable NRV Value of Joint Final Product Selling Sales Mfg. at at Splitoff Mfg. Costs Produced Price Value Costs Splitoff \$12,600
\$10,500
6,000
\$4
\$24,000
\$12,750
\$11,250
Allocation of Joint Mfg. Costs \$9,375
\$23,400
\$19,500
10,200
\$36,000
\$30,000
16,200
\$5
\$51,000
\$26,250
\$24,750
\$20,625
\$75,000
\$39,000
\$36,000
\$30,000
e under each of the above methods?
les Value at Spiltoff Total \$75,000 (30,000) (39,000) \$6,000 8.000%
NRV at the Splitoff Point Quantity Method Chocolate Milk Chocolate Milk Powder Chocolate Total Powder Chocolate \$24,000 \$51,000 \$75,000 \$24,000 \$51,000 (9,375) (20,625) (30,000) (12,000) (18,000) (12,750) (26,250) (39,000) (12,750) (26,250) \$1,875 \$4,125 \$6,000 (\$750) \$6,750 7.813% 8.088% 8.000% -3.125% 13.235%
s decision to fully process both of its intermediate products? Both Products \$75,000 36,000 \$39,000 (39,000) \$0
der liquor base should be oint. Operating income by \$1,350 to \$7,350. alculations would change. o "No" to see changes.)
Total \$75,000 (30,000) (39,000) \$6,000 8.000%
Constant Gross Margin Meth Chocolate Powder \$24,000 (9,330) (12,750) \$1,920 8.000%
Quantity Joint Processing Costs Allocated Using the Method Constant Gross Margin Percentage (8%) NRV Method Allocation Quantity of Final Final Joint Cost Separable of Joint Final Product Selling Sales 92% of Sales Mfg. Mfg. Costs Produced Price Value Less Separable Costs \$12,000
6,000
\$4
\$24,000
\$9,330
\$12,750
\$9,330
\$18,000
10,200
\$30,000
16,200
Constant Gross Margin Method Milk Chocolate Total \$51,000 \$75,000 (20,670) (30,000) (26,250) (39,000) \$4,080 \$6,000 8.000% 8.000%
\$5
\$51,000
\$20,670
\$26,250
\$75,000
\$30,000
\$39,000
\$20,670
Problem 16-31 (12th edition): Joint and byproducts, NRV method (CPA) Given: The Harrison Corporation produces three products: Alpha, Beta, and Gamma. Alpha and Gamma are joint products, and Beta is a byproduct of Alpha. No joint costs are to be allocated to the byproduct. The production processes for a given year are as follows:
a. In Department 1, 110,000 pounds of DM, Rho, are processed at a total cost of \$120,000. After processing in Dept. 1 60% of the pounds are transferred to Dept. 2, and 40% of the pounds (now Gamma) are transferred to Depart. #3. b. In Department 2, the material is further processed at a total additional cost of \$38,000. Then 70% of the pounds (now Alpha) are transferred to Department 4; and 30% emerge as Beta, the byproduct, to be sold at \$1.20 per pound. Separable marketing costs for Beta are \$8,100. c. In Department 4, Alpha is processed at a total additional cost of \$23,660. After this processing, Alpha is ready for sale at \$5 per pound. d. In Department 3, Gamma is processed at a total additional cost of \$165,000. In this department, a nornal loss of Gamma occurs, which equals 10% of the good pounds of output. The remaining good pounds of output are then sold for \$12 per pound.
Department # 4 A
Department # 2 \$38,000 D
C 66,000 Pounds 60%
Department #1
B \$120,000 DM Rho is processed
Beta is a byproduct of Alpha
110,000 Pounds Department #3 \$165,000 E 44,000 Pounds 40%
Waste = 10% of Good Output Let X = good output 44,000 - 0.1X = X
44,000 = 1.1X X= 44,000/1.1 X = 40,000 1. Prepare a schedule showing the allocation of the \$120,000 joint manufacturing costs between Alpha and Gamma using the NRV method. The NRV of Beta should be treated as an addition to the sales value of Alpha. Relative sales value at point A: Relative sales value at point B Relative sales value at point C Less Department # 2 separable costs Relative sales value at point D Relative sales value at E Total sales value at 1st splitoff point Allocation of \$120,000 joint mfg. Costs: To Alpha/Beta Path To Gamma Path
\$207,340 15,660 \$223,000 (38,000) \$185,000 315,000 \$500,000
\$5 X 46,200 - \$23,660 \$1.20 X 19,800 - \$8,100
\$44,400 75,600 \$120,000
\$185,000/\$500,000 * \$120,000 \$315,000/\$500,000 * \$120,000
\$12 X 40,000 - \$165,000
2. Independent of your answer to requirement 1, assume that \$102,000 of total joint costs were appropriately allocate Alpha. Assume also that there were 48,000 pounds of Alpha and 20,000 pounds of Beta available to sell. Prepare a income statement through the gross-margin line item for Alpha using the following facts. a. During the year sales of Alpha were 80% of the pounds available for sale. There was no beginning inventory of b. The NRV of Beta available for sale is to be deducted from the cost of producing Alpha. The ending inventory of Alpha is to be based on the net cost of production. c. All other costs and selling-price data are listed in A through D above. Department # 4 A
Department # 2 \$38,000 D
Department #1
C 66,000 Pounds 60% B
\$120,000 DM Rho is processed 110,000 Pounds
Beta is a byproduct of Alpha
Department #3 \$165,000 E 44,000 Pounds 40% Sales of Alpha Less: Cost of Goods Sold Joint mfg. costs assigned Separable mfg. costs: Depart. # 4 Separable mfg. costs: Depart. # 2 plus: sales of Byproduct beta Net mfg. costs Less ending inventory (20%) Cost of Goods Sold (80%) Gross Margin from sales of 80% of Alpha
\$192,000 \$102,000 \$23,660 38,000 (15,900) \$147,760 (29,552) \$118,208 \$73,792
.80 X 48,000 X \$5
\$1.20 X 20,000 - \$8,100
ha and Gamma are joint products, ct. The production processes for
20,000. After processing in Dept. 1, ma) are transferred to Depart. #3.
8,000. Then 70% of the pounds oduct, to be sold at \$1.20 per
his processing, Alpha is ready for
his department, a nornal loss of good pounds of output are then
Department # 4 \$23,660
Alpha: \$5 46,200 Pounds 70%
Separable Marketing Costs \$8,100
Beta: \$1.20
Beta is a byproduct of Alpha
10% of Good Output X = good output 000 - 0.1X = X
19,800 Pounds 30%
Gamma: \$12 40,000 Pounds
4,000 = 1.1X = 44,000/1.1
osts between Alpha and n to the sales value of Alpha.
\$5 X 46,200 - \$23,660 \$1.20 X 19,800 - \$8,100
\$12 X 40,000 - \$165,000
\$185,000/\$500,000 * \$120,000 \$315,000/\$500,000 * \$120,000
t costs were appropriately allocated to of Beta available to sell. Prepare an
re was no beginning inventory of Alpha g Alpha. The ending inventory of
Department # 4 \$23,660
Alpha: \$5 48,000 Pounds 70%
Separable Marketing Costs \$8,100
Beta: \$1.20
Beta is a byproduct of Alpha 20,000 Pounds 30%
80 X 48,000 X \$5
\$1.20 X 20,000 - \$8,100
Gamma: \$12 40,000 Pounds
Chapter 16 Problem: Accounting for A Joint Manufacturing Process Doe Corporation grows, processes, cans, and sells three main pineapple products -- sliced pineapple, crushed pineapple, and pineapple juice. The outside skin is cut off in the Cutting Department and processed as animal feed. The skin is treated as a by-product. Doe's production process is as follows: 1. Pineapples first are processed in the Cutting Department. The pineapples are washed and the outside skin is cut away. Then the pineapples are cored and trimmed for slicing. The three main products and the by-product are recognized after processing in the Cutting Department. Each product is then transferred to a separate department for final processing. 2. The trimmed pineapples are forwarded to the Slicing Department where the pineapples are sliced and canned. Any juice generated during the slicing operation is packed in the cans with the slices. 3. The pieces of pineapple trimmed from the fruit are diced and canned in the Crushing Department Again, the juice generated during this operation is packed in the can with the crushed pineapple. 4. The core and surplus pineapple generated from the Cutting Department are pulverized into a liquid in the Juicing Department. There is an evaporation loss equal to 8% of the weight of the good output produced in this department which occurs as the juices are heated. 5. The outside skin is chopped into animal feed in the Feed Department. The Doe Corporation uses the net realizable value method (relative sales value method) to assign costs of the joint process to its main products. The by-product is inventoried at its market value. A total of 270,000 pounds were entered into the Cutting Department during May. The schedule presented below shows the costs incurred in each department, the proportion by weight transferred to the four final processing departments, and the selling price of each end product. Processing Data and Costs for the Month of May
Department Cutting Slicing Crushing Juicing Animal Feed
Costs Proportion of Product by Weight Incurred Transferred to Departments \$60,000 --\$4,700 35% \$10,580 28% \$3,250 27% \$700 10% \$79,230 100%
Selling Price Per Pound of Final Product --\$0.60 \$0.55 \$0.30 \$0.10
Required: 1. The Doe Corporation uses the NRV method to determine inventory values for its main products and by-product. Calculate: a. the pounds of pineapple that result as output for pineapple slices, crushed pineapple, pineapple juice, and animal feed. b. the net realizable value at the split-off point of the three main products. c. the amount of cost of the Cutting Department assigned to each of the main products and to the by-product in accordance with company policy. d. The total budgeted gross margins for each of the three main products. e. The anticipated gross margin % for each of the four products f. The amount of ending inventory valuation for the following units of ending inventory 1. Canned Sliced Pineapples: 4,000 2. Canned Crushed Pineapples: 3,000 3. Canned Pineapple Juice: 2,000 4. Animal Feed: 1,000
2. Comment on the significance to management of the gross margin information calculated above. 3. Discuss how a managers determine whether it is profitable to process a product post split-off. Solution: Question 1 \$4,700 35%
Slicing Department
Separable Mfg. Costs
Final SP \$0.60
94,500 Pounds
\$10,580 28%
Crushing Department
Separable Mfg. Costs
\$0.55
75,600 Pounds Cutting Department \$60,000 270,000 pounds \$3,250 27%
Juicing Department
72,900 Pounds 8% Evaporation Loss
Separable Mfg. Costs
\$0.30
X = 72,900 - .08 X 1.08 X = 72,9000 X = 72,900/1.08 67,500 Pounds \$700
10%
Feed Department
Separable Mfg. Costs
\$0.10
27,000 Pounds Products Produced Sliced Crushed Juice Animal Feed Total
Pounds Sales Price Produced per Ton 94,500 \$0.60 75,600 \$0.55 67,500 \$0.30 27,000 \$0.10 264,600 (a)
Relative SV at Split-off \$52,000 \$31,000 \$17,000 \$2,000 \$102,000 (b)
Joint Cost Allocated \$30,160 \$17,980 \$9,860 \$2,000 \$60,000 (c)
Separable Total Assigned Mfg. Costs Mfg. Costs \$4,700 \$34,860 \$10,580 \$28,560 \$3,250 \$13,110 \$700 \$2,700 \$19,230 \$79,230
Solution: Question 2
The cost figures determined are based upon an arbitrary cost assignment process. Therefore, these cost figures have no Such arbitrary cost assignment is used for the financial accounting purposes of inventory valuation and income determinat Solution: Question 3
Managers must compare incremental revenues with incremental costs. If incremental revenues are greater than incremen
Product Produced Canned Sliced Pineapples
Canned Crushed Pineapples
Canned Pineapple Juice
Cost per Ton \$0.36889 \$0.37778 \$0.19422 \$0.10000
Total Cost of Sales Goods Sold \$56,700 \$34,860 \$41,580 \$28,560 \$20,250 \$13,110 \$2,700 \$2,700 \$121,230 \$79,230
Gross Gross Margin Margin % \$21,840 38.519% \$13,020 31.313% \$7,140 35.259% \$0 0.000% \$42,000 34.645% (d) (e)
Ending Inventory \$1,476 \$1,133 \$388 \$100 \$3,097 (f)
refore, these cost figures have no real value for planning and control purposes. y valuation and income determination.
venues are greater than incremental costs, then the products should be processes further.
Problem 16-30: Joint-cost allocation. Elsie Diary Products Corp buys one input, full-cream milk, and refines it. This churning process from each gallon of milk produces 2 cups (1 pound) of butter and 2 quarts (8 cups) of buttermilk. During May 2011, Elsie bought 10,000 gallons of full-cream milk for \$15,000. Elsie spent another \$5,000 on the churning process to separate the milk into butter and buttermilk. Butter could be sold immediately for \$2 per pound and buttermilk could be sold immediately for \$1.50 per quart. Elisie chooses to process the butter further into spreadable butter by mixing it with canola oil, incurring an additional cost of \$.50 per pound. This process results in 2 tubs of spreadable butter for each pound of butter processed. Each tub of spreadable butter sells for \$2.50. Required: 1. Diagram the churning production process. Churning Process
Selling Price per Processing
Spreadable
Selling price
Cost per pound of butter
Butter
per pound of
Mixing with Canola Oil
Tub
Butter
\$0.50
\$2.50
\$2.00 \$20,000
20,000
Cups
10,000
Pounds
80,000
Cups
20,000
Quarts
\$45,000 \$15,000 + \$5,000 full-cream milk 10,000 Gallons \$30,000 \$1.50 Buttermilk per quart of Selling price
2. Allocate the \$20,000 joint manufacturing cost to the spreadable butter and the buttermilk using the a. Physical-measure method (using cups) of joint cost allocation (20,000/100,000) X \$20,000 =
\$4,000
Butter
(80,000/100,000) X \$20,000 =
\$16,000
Buttermilk
b. Sales value at splitoff method of joint cost allocation (20,000/50,000) X \$20,000 =
\$8,000
Butter
(30,000/50,000) X \$20,000 =
\$12,000
Buttermilk
c. NRV method of joint cost allocation (45,000/75,000) X \$20,000 =
\$12,000
Butter
(30,000/75,000) X \$20,000 =
\$8,000
Buttermilk
d. Constant gross margin percentage method of joint cost allocation Sales:
Butter
\$50,000
Buttermilk
\$30,000
Costs Joint mfg. costs
(\$20,000)
Separable mfg. costs
(\$5,000)
Gross Margin
\$55,000
GM Percentage
68.75%
\$50,000 - (.6875 X \$50,000) - (\$.50 X 10,000) = \$50,000 - \$34,375 - \$5,000 =
\$10,625
Butter
\$10,625
Butter
\$9,375
Buttermilk
\$9,375
Buttermilk
Cost/Tub \$2.50 - (.6875 X \$2.50) - (\$.50/2) =
\$0.5313
\$30,000 - (.6875 X \$30,000) = Cost/Quart \$1.50 - (.6875 X \$1.50) =
\$0.4688
Problem 16-31: Further processing decision (continuation of 16-30) Elsie has decided that buttermilk may sell better if it was marketed for baking and sold in pints. This would involve additional packaging at an incremental cost of \$.25 per pint. Each pint could be sold for \$.90 (Note: 1 quart = 2 pints). 1. If Elsie uses the sales value at splitoff method, what combination of products should Elsie sell to maximize profits? Incremental sales value
Buttermilk
Final sales value
\$.90 X 20,000 X 2 =
\$36,000
Current sales value
\$1.50 X 20,000 =
\$30,000
Incremental costs
\$.25 X 20,000 X 2 =
Disadvantage of further processing Incremental sales value
(\$4,000) Do not process further
Butter
Final sales value
\$2.50 X 10000 X 2 =
\$50,000
Current sales value
\$2.00 X 10,000 =
\$20,000
Incremental costs
\$6,000 \$10,000
\$.50 X 10,000 =
Benefit of further processing
\$30,000 \$5,000 \$25,000 Process further
2. If Elsie uses the physical- measure method, what combination of products should Elsie sell to maximize profits? Same as question #1
3. Explain the effect that the different cost allocation methods have on the decision
to sell the products at split off or to process them further. The method of joint manufacturing cost allocation has no effect on the process further decision.
Problem 16-35: Accounting for a byproduct Given: Sanjana's Silk Shirts (SSS) hand-makes blouses and sells them to high-end department stores. SSS buys bolts of silk for \$300 each. Out of each bolt it gets 30 blouses, which it sells for \$90 each. SSS's new manager has suggested taking the scraps left after cutting out the blouses and using them to make scarves. By carefully cutting the blouses, SSS can produce 6 scarves from each bolt, which it can sell for \$25 each. During September, SSS buys 50 bolts of silk and spends an additional \$10,500 on the cutting and sewing process. By the end of the month, SSS sells 1,200 blouses and 260 scarves made from these bolts. Because the scarves are lower in value than blouses, SSS decides to treat the scarves as a byproduct.
Blouses Selling price @ split off
\$90
Production
Sales
1,500
1,200
\$135,000
\$108,000
Joint Mfg. Costs conversion costs added = silk bolts =
50
\$10,500 \$300
\$15,000 \$25,500 Production
Selling price @ split off
Sales
\$25
300
260
Scarves
\$7,500
\$6,500
Required: 1. Assuming SSS accounts for the byproduct using the production method, complete the chart below. Production method:
Byproduct revenue is recognized as a reduction of the jt. mfg. costs at the time of production.
Products
Units
EI
Units
SP per
Sales Value
Joint Cost
Joint Cost
EI in
Sales
Cost of
Gross
Produced
Sold
Units
Produced
Unit
at Splitoff Pt.
Allocated
Per Unit
Dollars
Dollars
Goods Sold
Margin
Blouses
1,200
300
1,500
\$90
Scarves
260
40
300
\$25
\$135,000
Total
\$18,000
\$12.00
\$3,600
\$
\$7,500
\$7,500
\$25.00
\$1,000
\$
\$142,500
\$25,500
\$4,600
\$
108,000 108,000
\$14,400 \$
\$14,400
\$
93,600
\$ \$
93,600
2. Assuming SSS accounts for the byproduct using the sales method, complete the chart below. Sales Method: Byproduct revenue is recognized at the time of sale. Products
Units
EI
Units
SP per
Sales Value
Joint Cost
Joint Cost
EI in
Sales
Cost of
Gross
Produced
Sold
Units
Produced
Unit
at Splitoff Pt.
Allocated
Per Unit
Dollars
Dollars
Goods Sold
Margin
Blouses
1,200
300
1,500
\$90
Scarves
260
40
300
\$25
Total
\$135,000
\$25,500
\$17.00
\$7,500
\$0
\$0.00
\$142,500
\$25,500
\$5,100
\$
\$0
\$
\$5,100
\$
108,000
\$20,400
\$
87,600
6,500
\$0
\$
6,500
114,500
\$20,400
\$
94,100
3. Complete the following journal entries for September assuming SSS accounts for the byproduct using (a) the production method and (b) the sales method. If no entry is needed, indicate that that is the case by putting "NE" in the amount column. Use the following accounts: Account #
Account name
Account #
Account name
Account #
Account name
4
Cash (or A/R, or A/P)
12
FG, Inventory -- Blouses
18
Sales Revenue -- Blouses
6
Cost of Goods Sold -- Blouses
14
FG, Inventory -- Scarves
20
Sales Revenue -- Scarves
8
Cost of Goods Sold -- Scarves
16
Miscellaneous Accounts
22
WIP, Inventory
10
Direct Materials Inventory Production Method
Always select the best possible account.
Sales Method
Debit
Credit
Debit
Credit
Debit
Credit
Account #
Account #
Amount (\$)
Amount (\$)
Amount (\$)
Amount (\$)
a. Give the journal entry to record the purchase of silk bolts Direct Materials Inventory
10
Cash (or A/R, or A/P)
15,000 4
15,000 15,000
15,000
b. Give the journal entry to record the use of the silk bolts WIP, Inventory
22
Direct Materials Inventory
15,000 10
15,000 15,000
15,000
c. Give the journal entry to assign conversion costs WIP, Inventory
22
Miscellaneous Accounts
10,500 16
10,500 10,500
10,500
d. Give the JE to record the cost of goods manufactured FG, Inventory -- Blouses
12
FG, Inventory -- Scarves
14
WIP, Inventory
18,000
25,500
7,500 22
25,500
25,500
e. Give the JE to record the COGS associated with sales of blouses Cost of Goods Sold -- Blouses
6
FG, Inventory -- Blouses
14,400 12
20,400 14,400
20,400
f. Give the JE to record the sales of blouses Cash (or A/R, or A/P)
4
Sales Revenue -- Blouses
108,000 18
108,000 108,000
108,000
e. Give the JE to record the COGS associated with sales of scarves Cost of Goods Sold -- Scarves
8
FG, Inventory -- Scarves
NE 20
NE NE
NE
f. Give the JE to record the sales of scarves Cash (or A/R, or A/P)
4
6,500
FG, Inventory -- Scarves
14
Sales Revenue -- Scarves
20
6,500 6,500 6,500
#### Comments
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1. What's the shortest function that implements the spec? We found one with four instructions, but didn't rule out the possibility of one with three.
2. In general, how does the minimum solution length relate to the dimensions of the problem (the distribution of 221 values into 21 groups, and 8 bits)? When does the problem become impossible?
This is a combinatorics problem. Fifteen years ago I would have asked on a Usenet math newsgroup.
A reason not to use this technique is that the numbers will change as the code evolves, and I don't necessarily want to do a new search every time.
1. Can we optimize the search? I noticed symmetries in the results — i.e. there were 12 functions that had the same output distribution as `(id & 230) & (id + 13)`. Most search spaces for superoptimization are so big that they need to use stochastic techniques, but we could probably do some math for this specific problem to prune the search space.
2. What if the enum labeling was constrained by two or more lookup tables? That is, how would we find two expressions to replace two lookup tables on the same set of identifiers? This will probably happen in oil, and that's another reason I won't use this solution right away.
3. Our problem was to map 221 arbitrary IDs to 21 arbitrary IDs, which is an easier problem than mapping 221 integers to 21 integers. So easy that we can solve it with single-threaded Python.
What other kinds of functions can be easily superoptimized?
1. How does the solution compare with perfect hashing? I believe perfect hash functions would be longer because the problem is more constrained. And some algorithms generate hashes that require lookup tables, which defeats the purpose of this optimization. (The Python code in this blog post may be useful.)
2. Do any compilers do this? They might do other kinds of superoptimization, but this problem appears unique because allowing the enum labels to vary makes the search exponentially easier.
C and C++ specify that unlabeled enums are assigned values from 0 to N-1, but the compiler should be able to detect if the real value is actually "observed" by any code. Strongly-typed enums in C++ 11 require an explicit `static_cast<int>` to be used as an integer.
1. What other techniques are there for replacing lookup tables with functions? Several google searches yielded nothing.
2. Finally, what's the difference between this code generated by GCC versus Clang? I wanted to verify that we're actually removing a memory access, so I compiled the code with `-O3` with both compilers.
The Clang code produces a sequence of four instructions as you might expect (`xor add and and`), but GCC produces a sequence with a `movzx` instruction in the middle.
I suspect there's no real difference: the `movzx` is a register-to-register operation, and it happens at the end of the Clang code too. So we have indeed removed a memory access with our algorithm.
This function:
```kind = 175 & id & ((id ^ 173) + 11)
```
compiles to this:
```blog-code/id-kind-func\$ ./run compare-gcc-clang
...
```
```GCC
0000000000400580 <_Z10LookupKind2Id>:
400580: 89 f8 mov eax,edi
400582: 81 e7 af 00 00 00 and edi,0xaf # 0xaf is 175
<b>40058b: 0f b6 c0 movzx eax,al</b>
40058e: 83 c0 0b add eax,0xb # 0xb is 11
400591: 21 f8 and eax,edi
400593: c3 ret
400594: 66 2e 0f 1f 84 00 00 nop WORD PTR cs:[rax+rax*1+0x0]
40059b: 00 00 00
40059e: 66 90 xchg ax,ax
...
Clang
00000000004005f0 <_Z10LookupKind2Id>:
4005f0: 40 88 f8 mov al,dil
4005f5: 04 0b add al,0xb # 0xb is 11
4005f7: 40 20 f8 and al,dil
4005fa: 24 af and al,0xaf # 0xaf is 175
4005fc: 0f b6 c0 movzx eax,al
4005ff: c3 ret
```
Versions:
```blog-code/id-kind-func\$ ./run.sh show-versions
c++ (Ubuntu 4.8.4-2ubuntu1~14.04.1) 4.8.4
...
clang version 3.8.0 (tags/RELEASE_380/final)
Target: x86_64-unknown-linux-gnu
Is there any other difference between these two compilations? I'm not sure what the `nop` and `xchg` are for in the GCC listing -- they both look like no-op instructions. | 1,150 | 4,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-06 | latest | en | 0.944633 |
http://excel.bigresource.com/Calculating-hours-worked-for-3rd-shift-JtzqBR.html | 1,555,642,537,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526966.26/warc/CC-MAIN-20190419021416-20190419043416-00338.warc.gz | 58,994,747 | 12,791 | # Calculating Hours Worked For 3rd Shift
Nov 27, 2012
Weekly Timesheet.xlsx
This spreadsheet calculates hours worked great for first and second shift but when you enter times for third shift it goes all whacky with the outcome.
## Calculating Actual Hours Worked Only In Core Hours?
Aug 11, 2010
I have been working on a timesheet but the problem I have come across is calculating actual hours worked only in the core hours and any work outside the core hours is calculated in the outside hours column. A standard work day is 7.6 hours working between 8.30am and 5.00pm. However if someone was to commence work either before 6am or after 8pm this is outside of core hours. I have attached an example of my timesheet for you to see what I am talking about.
## Operational Model - Shift Start Time Plus Amount Of Hours Worked In One Cell.
Feb 19, 2009
Is it possible that a cell contains both numeric and alphanumeric data and to do calculations on that?
For example: if a cell conatain the value "10a" or "8.5b" etc. Would it be possible to have a column that gives me the hours worked (the numeric value in the cell) and a line that gives me the amount of people that are working on shift "a" (the alphanumeric value in the cell).
Is this at all possible? Or does that require VBA/Macros and stuff (in which case this is posted in the wrong part of the forum )
## Calculating Hours Worked
Jan 1, 1970
I can do the timesheet formula for adding the hours worked as follows:
Start Finish Total
08:45 17:15 8.5
However, it doesn't work when I fill in a whole week work of hours in this format:
Start Finish Total
08:45 17:15 08:30
It works on a daily basis, but when total hours exceeds 24, the formula get's all mixed up - how to I format the total column to account for every 5 minutes worked, which you can't do when converting to decimal??
## Calculating Hours Worked With Overtime Clause
Jul 23, 2007
I need to worked out Hours worked in a timesheet. This was the easy part, the hard part is the clause tha HR threw in, which is:
If you have worked and 8 hour day WITH 1 hr lunch then you qualify for overtime.
if you work a 8 hr day and work through your lunch (1 hour) (so equivelant to 9 hrs) you still do NOT qualify for overtime there for Overtime = 0.
This is cause some people work though their lunch to get overtime, but legally they have to have a break so we are not paying overtime for it. I have basically tried in a formula to replicate this but it works with some data and not with all.
attached is an example, as you'll see the formula works in some cells, but not others.
## Finding All Occurrences Of Person's Name And Calculating Hours Worked
Jan 28, 2014
I'm trying to create a macro that will take name values from a list (Last name, First name) and then find these particular names in a long list of names in another sheet. Once all the occurrences of the name are found, the macro would need to add up all the hours that the person has worked during a given month. For example, I have Jane Doe and John Smith on my shortlist, the macro would have to take the name of the first person, Jane Doe, find all the occurrences of that name in the second sheet that contains names, dates and hours worked, then add up all the hours worked by that person and return the total hours worked value for that parson. Then take the name of the second person, John Smith, and do the same, and so on. Is it possible to do it with macros?
## Calculating Hours Worked Based On Elapsed Time Over 24H
Aug 6, 2007
If a Rescue Officer is called out at 23:00 and is back at 04:00, this should equate to 5 hours worked.
It seems that if my times are all on one side or the other of a 24 hour cycle, my calculation work fine but it it breaks across the 24 hour (as above, it doesn't work.
A2=04:00
A1=23:00
Using (A2-A1)*24 give me -19.00 hours
My SS macro has a line:
s = (wks.Cells(c, 3) - wks.Cells(c, 2)) * 24 'calculates the duration of time worked
Is there any way of getting excel to calculate an elapsed time in hours when the start and end times roll over from one day to the next?
## Calculating Night Shift Hours
Jan 15, 2009
I'm trying to calculate the hours worked for both my day shift and my night shift.
Day shift (thanks to search ) I have managed to figure out and worked quite well.
=ROUND((E7-D7)*96,0)/4
It totals adds up the time and converts it into a decimal of hours worked.
For example Joes starts at 1100 and finishes at 1330 it returns a total of 2.5 hours worked.
However I strike a problem with nightshift.
They start in the late afternnon and work thoguh into the am.
I have used the same formula but it doesn't seem to work:
=ROUND((K7-L7)*96,0)/4
I assume because once the clock strikes 12 it's a new day and it can't work out the maths.
Lets use the example form about but make it pm.
Joe starts at 2300 and finishes at 0130 it should give me a total of 2.5 hours instead it gives me 21.5 hours
## Calculating Total Hours Worked By Group Of Employees For Certain Hour
Mar 23, 2013
I need creating a formula that will tell me the total number of employee hours worked during a certain hour (6:00 AM) for a particular department. Some employees clock in at 6:00 AM, some clock in at 6:30. Here is the current formula I'm using which only tells me the amount of employees in the department not how many hours worked . I have three employees that clock in at 6:00AM and one at 6:30AM, the total I'm looking for is 3.5 but my formula gives me the result of 4.
C5:C1446 is a list of departments
T1503 is the 6:00 AM
D5:D1446 are the Clock In times
K5:K1446 are the Clock Out Times
=SUMPRODUCT(--(\$C\$5:\$C\$1446="Shipping"),(\$T\$1503>=\$D\$5:\$D\$1446)*(\$T\$1503
## Calculating Staff Meal Deductions Based On Hours Worked?
Jun 30, 2014
I would like to calculate staff meal deductions for each employee based on the number of hours they worked for that day.
I have the hours listed all on one page, per day, per employee.
I have the staff meal deductions range listed on another page in the workbook.
The range is as follows: *Please note anyone working over 8.00 hours is deducted a maximum of \$2.40 for that shift/day.
Hours
Deduction
1.00
\$0.30
[Code].....
## Calculating Day And Night Shift Hours From Date / Time
Dec 22, 2009
I want to find the hours worked in day shift and night shift. Day shift is from 06:30 to 18:30 and night shift is from 18:30 to 06:30. Listed below is an example of my date/times.
Note that the night shift carries over to the next day.
Start/Finish
21/12/09 07:00 to21/12/09 11:09
21/12/09 07:46 to21/12/09 14:41
21/12/09 12:13 to21/12/09 22:08
21/12/09 16:40 to21/12/09 18:05
21/12/09 19:40 to22/12/09 02:34
21/12/09 23:20 to22/12/09 04:39
22/12/09 02:06 to22/12/09 06:15
## Convert Hours Worked In Week To Hours Worked In Month With Formula?
Feb 21, 2013
Is there a way to conver a persons time spent (given in weeks) to adjust/convert to show per month. Attached is the sheet. Do note that week 2/25 - 3/1 is a combination of Jan and Feb so hours should be logically divided into jan and feb...
Name 2/18 - 2/22 2/25 - 3/1 3/4 - 3/8 3/11 - 3/15 Feb mar
Tom 40 10 0 20 ?? ??
name
2/18-2/22
2/25 - 3/1
3/4 - 3/8
3/11-3/15
Feb
Mar
tom
40
10
0
20
??
??
## Calculate Pay For Shift Work With Different Rates Based On Shift Hours
Apr 11, 2008
a person works for certain hours and get paid according to the hours worked either by day or by night or a mix of both. Day payment is \$8 when worked between 08:00 and 19:59 , night payment is \$12 when worked between 20:00 and 07:59. The excel cell are formatted as datetime with yyyy-mm-dd hh:mm , the function works fine in getting the time information and checking whether the whole work is all day or all night , yet the if-then-else statements for calculation seems to be wrong!!
examples:
start = 2008-01-01 09:15 , end = 2008-01-01 11:40 , all day as it is between 08:00 and 20:00 and cost = 8/hr = 19.333
start = 2008-01-03 21:05 , end = 2008-01-04 02:05 , all night as it is between 20:00 and 08:00 and cost = 12/hr = 60.000
start = 2008-02-02 19:00 , end = 2008-02-02 20:05 , cost = 9.000 as 1 hour day = 8.000 plus 5minutes night = 1.000
Function prod(st As Date, en As Date) As Double
Dim shour As Integer
Dim smin As Integer
Dim ehour As Integer
Dim emin As Integer
Dim stod As String
Dim etod As String
pday = 8
pnight = 12
shour = Hour(st)
smin = Minute(st) + shour * 60
If (shour >= 8 & shour < 20) Then
stod = "day"
Else
stod = "night"
End If
ehour = Hour(en)
emin = Minute(en) + ehour * 60
If (ehour >= 8 & ehour < 20) Then.................
## Calculate Hours Worked With Lunch / Overtime / Holiday / Vacation And Saturday Hours?
Jan 13, 2014
Lunch is not paid. Holiday and vacation hours get calculated at the regular pay rate. Overtime is anything in excess of 8 hours per day and/or in excess of 40 hours per week and/or over 5 working days per week. Saturdays for most the employees will be overtime because it will be their 6th workday of the week; but it will be regular time for one employee as it will only be his 5th workday of the week.
For accounting and payroll purposes, we need the totals to display in both hour and decimal format.
So far, I have Lunch, Regular and Overtime hours figured out, but I still need to work with Saturday, Vacation and Holiday hours. Also, currently, the time in and out has to be typed in with the colon and AM or PM. Is there another way to input the info without having to type in those items? I'm trying to make it as user friendly as possible.
## Adding Hours Worked Minus Lunch Break IF Over 5 Hours?
Dec 15, 2013
I'm attempting to make a simple time sheet for a handful of employees. I'd like to enter the clock in time and clock out time for each day. The end cell should be the running total for the week. The tricky part for me is having the formula subtract an hour for each day that is over 5 hours.
## Time Scheduling: Take Out A 30 Min Break If The Hours Worked Is Over 6 Hours
Jan 13, 2009
I am making a schedule and I would like it to take out a 30 min break if the hours worked is over 6 hours.
I have so far
A B
1 11:00 7:30
=24*(B1-A1)
Gives me 8 hours, I would like it to subtract the 30 minutes only ifthe sum is over 6 and not alter the sum if it is under 6.
## Looking For A Formula That Can Separate Day Shift Hours From Night Hours?
Aug 6, 2013
I have got a formula that can separate day shift hours from night hours, in this case night begins at 7pm to 7am, however the problem is after 12am we get into negative numbers, what formula would fix this and can be combined with the formulas below?
E10 = 19:00 or start of night time hours
B3 = start time
C3 = end time
D3 = day hours workeds =24*IF(E10
## Pivot For Hours Worked Sum
Sep 24, 2012
Timekeeper to tally total hours worked by employee. When doing a pivot for sum total hours worked for FY13 it does not calculate correctly. I understand they formatted that column/custom h:mm but when I change the 8:00 hours to a number I come up with 0.33 .
I am attaching a sample file : sample time.xlsx
I just want them to get a running total of hours worked/pay.
## Calculation Hours Worked?
Jul 3, 2014
I'm trying to make a way to track if I've worked more or less than the 39 hours/week I'm paid for. At the end of each week, I have a total of how many hours and minutes that I've worked .
On column C I have what I should work.
On column D I have what I did actually work
On column E I'd like to convert automatically Columd D to minutes for calculation purposes
Column F to know if I worked more or less than what I should've subtracting C and E
Column G to have an ongoing tally to know if I need to work more or less
Column H and I could probably be the same thing. Ideally what I'd like is to have a formula pull the information from column G and put it into workdays, hours, and minutes with 1 workday being 7 hours and 48 minutes.
Since I tend to work too much, I'd like to know if I've worked 3 days too much during 1 month, I can take 3 days off the next month to get everything zeroed back to where I don't owe the company anything and vice versa.
## Hours Worked Calculation ....
Dec 2, 2008
I am trying to create a spreadsheet that auto calculates my emp. time.
However I do not want to use military time. I can get it to work by =a2-a1 but only if it is 8.5 and 17.5. Any ideas how I can do clock in 8:30 clock out 4:30 = 8 hours?
## Worked Hours Between Two Periods
Feb 10, 2010
I'm trying to calculate the total hours worked for two given periods over a shift , which can span two consecutive days ie. start 15:45 and finish at 00:15 the next day. Hours worked between 6am and 6pm are paid at standard rate, whilst hours worked between 6pm and 6am attract penalty rates. Hours are cacluated in 24hr time
I have attached a copy of the timesheet that we use so you can see exactly what I'm trying to achieve, and included most of the shifts that we have.
## Track Hours Of Overtime Worked
May 9, 2008
I am creating a spreadsheet that will track hours of overtime worked and within the spreadsheet there are several separate departments listed.
I have made it dynamic so that the summary spreadsheet will update as employees are added. I’m using a macro and some complicated helper cells to be able to sort the employees based on their total OT hours worked.
## Converting Time To Hours Worked
Mar 15, 2012
Is there a function or a macro to calculate number of hours worked from a single cell value.
For example, cell A1 has "1600 - 1715" and need it to convert to "1.25" on cell B1
## Formula To Calculate Hours Worked
Feb 24, 2009
I have this spreadsheet and in it the time is changed from military time to regular and then I use a formula to calculate hours worked. On some of these the total is off by one minute. Does anyone know how to fix this?
I don't know how to paste the spreadsheet so you can see formulas,
## Macro To Calculate Hours Worked
Feb 1, 2007
I need to develop a work sheet for agency booked people to count the number of hours worked by them on daily basis. Agency can clock in at three different times and clock out at 6 different times. i tried but could not even develop logic to calculate the total earned hours. i attached the sheet for reference.
## Formula To Add Total Hours Employee Worked For Each Day?
Dec 21, 2013
Refer to the attached sheet which is Daily Schedule for employee. I need a formula to add hours worked on single day in cell C2 for Monday, E2 for Tuesday, G2 for Wednesday, and so on.
Every day we have Clockin_Clockout info for each employee as shown for employee a & b.
FYI : I am using below formula to add employee hours for the week as (formula in cell R4).
[Code] .....
## How To Calculate Hours Worked Using 24hr Clock
Jan 28, 2014
Attached sheet, I am struggling with the formula that will add up the hours overtime worked per day when I enter start and finish times.
Standard working hours are :
mon to thurs 8 hrs per day 8m to 16.30 (with 30 mins unpaid break)
Friday 6 hrs per day
Saturday all hours are overtime
Hours Commited sheet.xlsx‎
## Subtracting Break From Total Hours Worked
May 21, 2009
This is probably a very simple problem that has me going around in circles. I am attempting to set up a time roster, where I simply want to check:
If "end-time" minus "start-time" is greater than 4:00 (hrs), then deduct 00:30 (minutes) and place that result in another cell.
If it is not greater than 4:00 (hrs) then leave unchanged. I have read thru countless examples on the Forum - but I think that such great learning is driving me mad. Although I do believe that I have the correct format [h].mm - but attempts with IF's have got me confused. This is one of those "Looking down the tunnel towards the flickering light" moments.
## Productivity Formula: Calculate Worked Hours
Nov 6, 2009
i m trying to work out the productivity of employees based on how many hours they work (Time in Back Office). How many pieces of work they complete(Back Office items Completed) if 1 piece of work should take 7 mins. the item in red is what i cant seem to figure out.
## Calculate Hours Worked, Minus Weekends
Nov 11, 2009
to calculate how long a ticket is open in our system before being resolved. I don't want to count weekends, and if the ticket is 'suspended', I don't want to count that either. There is also the factor that the ticket 'un-suspend' date may be later than the ticket 'closed' date. Which is the bit that's throwing me.
So, I have the following fields
Ticket Open, Ticket Closed, Ticket Suspended Date, Ticket Unsuspended Date
A sample ticket might be (using above fields)
02/11/09 09/11/09 04/11/09 30/11/09
That 'should' equal two days (16 hours) as the Unsuspend date falls after the close date so it was suspended from the 4th until closure.
Now I want to know, in hours (8 hour day) how long that ticket took to resolve (i.e close), remembering you can't count the time it was suspended, or any time that fell over a weekend. Also not all tickets are suspended. | 4,616 | 17,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | longest | en | 0.94312 |
https://mersenneforum.org/showthread.php?s=6f1c14444e6885dbb663427053b5c1a0&p=601838 | 1,670,164,290,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710974.36/warc/CC-MAIN-20221204140455-20221204170455-00505.warc.gz | 416,146,644 | 13,437 | mersenneforum.org mod x^4+1
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2022-03-16, 01:06 #12
Dr Sardonicus
Feb 2017
Nowhere
22·52·61 Posts
Quote:
Originally Posted by paulunderwood I did say "any k". Powers k never seem to map x+1 to 1-x. HTH.
So you did. I just read "n" all the way through. Oops.
No good ideas for this question, either. At least not yet. I'm pretty sure I know how to get closed formulas for the coefficients of (1+x)^k (mod x^4 + 1), but I'm not sure they'd help much with the question at hand.
2022-03-16, 05:45 #13 paulunderwood Sep 2002 Database er0rr 29×151 Posts The companion matrix of x^4+1 is: Code: ? cm=([0,0,0,-1;1,0,0,0;0,1,0,0;0,0,1,0]) [0 0 0 -1] [1 0 0 0] [0 1 0 0] [0 0 1 0] Then x+1 is equivalent to: Code: ? cm+1 [1 0 0 -1] [1 1 0 0] [0 1 1 0] [0 0 1 1] The determinant of this matrix is 2: Code: ? matdet(cm+1) 2 So is it safe to say a test of PRP test of x+1 over x^4+1 is necessarily a 2-PRP? Answer: Yes, according to https://en.wikipedia.org/wiki/Determ..._matrix_groups This makes verification to 2^64 easy. I/4 of odd n for n%8==5 and 8 Selfridges per test... Verified to 2^64 using Feitsma's list of 2-PSPs. Last fiddled with by paulunderwood on 2022-03-16 at 07:34
2022-03-16, 12:17 #14
Dr Sardonicus
Feb 2017
Nowhere
10111110101002 Posts
Quote:
Originally Posted by paulunderwood So is it safe to say a test of PRP test of x+1 over x^4+1 is necessarily a 2-PRP?
Quite safe. Here's another way to look at this:
If Mod(Mod(1,n)*x + 1, x^4 + 1)^n == Mod(1 - x, x^4 + 1)
taking the norm gives (Mod(2, n))^n = 2.
Quote:
Verified to 2^64 using Feitsma's list of 2-PSPs.
Well done, Sir!
2022-03-16, 18:07 #15 paulunderwood Sep 2002 Database er0rr 29×151 Posts FWIW, not a peep out of: Code: {forstep(n=5,1000000,8, if(!ispseudoprime(n), for(a=1,(n-1)/2, if(gcd(a,n)==1, Det=Mod(a,n)^4+1; if(Det^n==Det&&Mod(Mod(x+a,n),x^4+1)^n==a-x, print([n,a]))))));} Last fiddled with by paulunderwood on 2022-03-16 at 18:16
2022-03-17, 01:29 #16
Dr Sardonicus
Feb 2017
Nowhere
22×52×61 Posts
Quote:
Originally Posted by paulunderwood You may be aware that no one has found a counterexample to the test Mod(Mod(x+2,n),x^2+1)^(n+1)==5 for n%4==3.
BTW, similar to your observation with Mod(Mod(1,n)*x + 1, x^4 + 1),
Mod(Mod(1,n)*x + 2, x^2 + 1)^(n+1) == 5 implies 5^(n-1) == 1 (mod n) for n == 3 (mod 4) (assuming 5 does not divide n). That is, n is a Fermat pseudoprime to the base 5. I do not know if these have been as extensively computed as Fermat pseudoprimes to the base 2.
But my guess is, even if they haven't been, if you run the base 5 test first, and only run Mod(Mod(x+2,n),x^2+1)^(n+1)==5 on the composite survivors, that will speed things up.
2022-03-17, 13:22 #17
paulunderwood
Sep 2002
Database er0rr
104338 Posts
Quote:
Originally Posted by Dr Sardonicus BTW, similar to your observation with Mod(Mod(1,n)*x + 1, x^4 + 1), Mod(Mod(1,n)*x + 2, x^2 + 1)^(n+1) == 5 implies 5^(n-1) == 1 (mod n) for n == 3 (mod 4) (assuming 5 does not divide n). That is, n is a Fermat pseudoprime to the base 5. I do not know if these have been as extensively computed as Fermat pseudoprimes to the base 2. But my guess is, even if they haven't been, if you run the base 5 test first, and only run Mod(Mod(x+2,n),x^2+1)^(n+1)==5 on the composite survivors, that will speed things up.
Computing 5-PSPs up to 2^64 would be a mammoth task involving many dedicated years of running efficient code on big hardware. I think getting the list of 2-PSPs involved some mathematical tricks that I do not understand. All I can offer beyond my 2^50 check is a scan of Feitsma's sub-list of Carmichael numbers, which I just check up to 2^64 in less than a second, not counting the time to read the list into memory.
2022-03-17, 14:06 #18 Dr Sardonicus Feb 2017 Nowhere 137248 Posts Let Code: v=[Mod(1/4, x^4 - 4*x^3 + 6*x^2 - 4*x + 2), Mod(-1/4*x^3 + 3/4*x^2 - 3/4*x + 1/4, x^4 - 4*x^3 + 6*x^2 - 4*x + 2), Mod(-1/4*x^2 + 1/2*x - 1/4, x^4 - 4*x^3 + 6*x^2 - 4*x + 2), Mod(-1/4*x + 1/4, x^4 - 4*x^3 + 6*x^2 - 4*x + 2)]; Then for k = 0, 3 the coefficient of x^k in lift(Mod(1+x, 1+x^4)^n) is trace(v[k+1]*x^n). The only really "nice" expression is for the constant term, trace(Mod(1/4,x^4 - 4*x^3 + 6*x^2 - 4*x + 2)*x^n). This may be expressed as $\frac{1}{4}\sum_{m=1}^{4}$$1 + e^{\frac{2\pi i(2m-1)}{8}$$^{n}$ The coefficients all grow exponentially with n, with the ratio of consecutive terms approaching sqrt((2 + sqrt(2))). Scratch that. The polynomial has two (complex conjugate) zeroes of absolute value sqrt(2 + sqrt(2)) and two of absolute value sqrt(2 - sqrt(2)). The expression is zero for n == 4 (mod 8). What is true is that the n-th root of the absolute value of the n-th constant term has an upper limit of sqrt(2 + sqrt(2)). Note: The polynomial x^4 - 4*x^3 + 6*x^2 - 4*x + 2 is charpoly(Mod(x+1,x^4 + 1)). I wasn't advocating compiling a table of base-5 pseudoprimes. I was merely thinking that most composites would "flunk" a base-5 psp test, which (I am assuming) is much faster than the Mod(x+2,1+x^2) test. This would greatly reduce the number of slower Mod(2 + x, 1 + x^2) tests that needed to be done in searching for psp's for that test. Last fiddled with by Dr Sardonicus on 2022-03-18 at 01:50 Reason: Fix bad notation;correct misstatement
2022-03-18, 15:02 #19 Dr Sardonicus Feb 2017 Nowhere 10111110101002 Posts As you can see from the following, I have obtained closed-form expressions for the coefficients of the degree-less-that-4 lift of Mod(1 + x, 1 + x^4)^n. They can be written as 4-term sums similar to the one already given. Unlike the constant term, though, the coefficient is a root of unity of degree greater than 1; a 4-th or 8-th root of unity. The trace may be written as a sum of algebraic conjugates; the conjugates of Mod(x, x^4 + 1) are Mod(x^3, x^4 + 1), Mod(-x, x^4 + 1), and Mod(-x^3, x^4 + 1). This allows the traces to be expressed as explicit 4-term sums. These expressions are unlikely to be useful computationally, but may be helpful theoretically. Code: ? for(n=0,15,v=[lift(Mod(x + 1, x^4 + 1)^n),trace(Mod(-x, x^4 + 1)*Mod(1 + x, x^4+1)^n)/4, trace(Mod(-x^2, x^4 + 1)*Mod(x + 1, x^4 + 1)^n)/4, trace(Mod(-x^3, x^4 + 1)*Mod(x + 1, x^4 + 1)^n)/4, trace(Mod(x + 1, x^4 + 1)^n)/4];print(n" "v)) 0 [1, 0, 0, 0, 1] 1 [x + 1, 0, 0, 1, 1] 2 [x^2 + 2*x + 1, 0, 1, 2, 1] 3 [x^3 + 3*x^2 + 3*x + 1, 1, 3, 3, 1] 4 [4*x^3 + 6*x^2 + 4*x, 4, 6, 4, 0] 5 [10*x^3 + 10*x^2 + 4*x - 4, 10, 10, 4, -4] 6 [20*x^3 + 14*x^2 - 14, 20, 14, 0, -14] 7 [34*x^3 + 14*x^2 - 14*x - 34, 34, 14, -14, -34] 8 [48*x^3 - 48*x - 68, 48, 0, -48, -68] 9 [48*x^3 - 48*x^2 - 116*x - 116, 48, -48, -116, -116] 10 [-164*x^2 - 232*x - 164, 0, -164, -232, -164] 11 [-164*x^3 - 396*x^2 - 396*x - 164, -164, -396, -396, -164] 12 [-560*x^3 - 792*x^2 - 560*x, -560, -792, -560, 0] 13 [-1352*x^3 - 1352*x^2 - 560*x + 560, -1352, -1352, -560, 560] 14 [-2704*x^3 - 1912*x^2 + 1912, -2704, -1912, 0, 1912] 15 [-4616*x^3 - 1912*x^2 + 1912*x + 4616, -4616, -1912, 1912, 4616] If you want lift(lift(Mod(Mod(1,n)*x + 1, x^4 + 1)^k)) = (n-1)*x + 1, then trace(Mod(x + 1, x^4 + 1)^k)/4 has to be congruent to 1 (mod n); trace(Mod(-x^3, x^4 + 1)*Mod(x + 1, x^4 + 1)^k)/4 has to be congruent to -1 (mod n); and finally, trace(Mod(-x^2, x^4 + 1)*Mod(x + 1, x^4 + 1)^k)/4 and trace(Mod(-x, x^4 + 1)*Mod(1 + x, x^4+1)^k)/4 both have to be divisible by n. EDIT: From the above data, you can see that for k == 6 (mod 8), the coefficient of x in lift((Mod(x + 1, x^4 + 1)^k) is zero; and if k == 4 (mod 8) the constant term is 0. I'm sure there's a straightforward proof, but I'm too lazy to work out the details It is thus impossible to have lift(lift((Mod(Mod(1,n)*x + 1, x^4 + 1)^k)) = 1 - x for any n, if k is congruent to 4 or 6 (mod 8). Last fiddled with by Dr Sardonicus on 2022-03-18 at 16:50
2022-03-20, 15:08 #20 Dr Sardonicus Feb 2017 Nowhere 22·52·61 Posts "That's the stupidest proof I've ever seen in my life!" I finally thought of a very simple, straightforward proof of the "zero-ness" of the coefficients of x^3, x^2, x and 1 in lift(Mod(x+1, x^4 + 1)^k) for k == 2, 0, 6 and 4 (mod 8) respectively. All that is required is numerical verification that four consecutive values are 0. The rest being 0 then follows because for any given r, the subsequence lift(Mod(x+1,x^4 + 1))^(8*n + r) satisfies a linear recurrence with constant coefficients of order 4. Boom! Done. The characteristic polynomial is charpoly(Mod(x, x^4 - 4*x^3+6*x^2-4*x+2)^8)) = x^4 + 272*x^3 + 18528*x^2 + 4352*x + 256. The characteristic polynomial x^4 + 272*x^3 + 18528*x^2 + 4352*x + 256 factors as (x^2 + 136*x + 16)^2. I am too lazy to check whether any of the subsequences of terms in a residue class mod 8 actually satisfy a linear recurrence of order 2.
2022-03-20, 21:13 #21 paulunderwood Sep 2002 Database er0rr 10001000110112 Posts Powers of x over x^2-136*x+16 are interesting. Mod(Mod(x,n),x^2-136*x+16)^(n+1)==16 is good for n%8==3 and n%8==5 I can also check Mod(Mod(x,n),x^2-136*x+16)^((n+1)/4)==2 for n%8==3, and Mod(Mod(x,n),x^2-136*x+16)^((n+1)/2)==4 for n%8==5. The test Mod(Mod(x,n),x^2-136*x+16)^(n+1)==16 can be transformed into Mod(Mod(x,n),x^2-1154*x+1)^((n+1)/2)==1 for n%8={3, 5} for Fermat-2-PRP, making a quick check of Feitsma's 2^64 2-PRP possible. As ever n%8==1 has psudoprimes. Without checking Feitsma, we now have simple tests for n%8 = {3,5,7} Last fiddled with by paulunderwood on 2022-03-20 at 21:51
2022-03-21, 13:14 #22 Dr Sardonicus Feb 2017 Nowhere 22×52×61 Posts For every integer r = 0 to 7, the sequence an = lift(Mod(x+1,x^4+1)^(8*n + r)), n = 0, 1, 2, ... satisfies the recurrence an+2+136*an+1 + 16*an = 0. Thus, for any given r, the coefficients of 1, x, x^2 and x^3 in the lift are four sequences, each satisfying this recurrence. I have noted cases for even r where one of the four sequences is the zero sequence. I note that the discriminant of the quadratic x^2 + 136*x + 16 is 1362 - 4*16 = 128*144 = 2*82*122. Note that Mod(1 + x, 1 + x^2)^4 has the linear minimum polynomial x + 4 (that is, (1 + I)^4 = -4), and now we have that Mod(1 + x, 1 + x^4)^8 has the quadratic minimum polynomial x^2 + 136*x + 16. Hmm, that's funny... I checked, and - sure enough! Mod(1 + x, 1 + x^8)^16 has a degree-4 minimum polynomial; and Mod(1 + x, 1 + x^16)^32 has a degree-8 minimum polynomial. I'm sold. Mod(1 + x, 1 + x^(2k))^(2k+1) has minimum polynomial of degree 2k-1. Hmm, in every case so far, the zeroes of the minimum polynomial are real... EDIT: Of course they're all real. And there's nothing special about powers of two. We have 1 + exp(2*I*pi/n) = exp(2*I*pi/(2*n))*[exp(-2*I*pi/(2*n)) + exp(2*I*pi/(2*n))] = exp(2*I*pi/(2*n))*2*cos(2*pi/(2*n)). This is a 2n-th root of unity times a real number. Raising to the n-th power gives a real number. Last fiddled with by Dr Sardonicus on 2022-03-21 at 13:39
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https://questions.examside.com/past-years/gate/question/pconsider-the-following-context-free-grammar-where-the-set-gate-cse-theory-of-computation-finite-automata-and-regular-language-eofck2gswck1vnis | 1,708,632,590,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.45/warc/CC-MAIN-20240222193722-20240222223722-00645.warc.gz | 487,790,962 | 30,398 | 1
GATE CSE 2021 Set 1
MCQ (Single Correct Answer)
+2
-0.67
Consider the following context-free grammar where the set of terminals is {a, b, c, d, f}.
S → d a T | R f
T → a S | b a T | ϵ
R → c a T R | ϵ
The following is a partially-filled LL(1) parsing table.
Which one of the following choices represents the correct combination for the numbered cells in the parsing table ("blank" denotes that the corresponding cell is empty)?
A
(1) S → R f (2) S → R f (3) T → ϵ (4) T → ϵ
B
(1) blank (2) S → R f (3) blank (4) blank
C
(1) S → R f (2) blank (3) blank (4) T → ϵ
D
(1) blank (2) S → R f (3) T → ϵ (4) T → ϵ
2
GATE CSE 2021 Set 1
Numerical
+2
-0.67
In a pushdown automaton P = (Q, ∑, Γ, δ, q0, F), a transition of the form,
where p, q ∈ Q, a ∈ Σ ∪ {ϵ}, and X, Y ∈ Γ ∪ {ϵ}, represents
(q, Y) ∈ δ(p, a, X).
Consider the following pushdown automaton over the input alphabet ∑ = {a, b} and stack alphabet Γ = {#, A}.
The number of strings of length 100 accepted by the above pushdown automaton is ______
3
GATE CSE 2021 Set 1
MCQ (Single Correct Answer)
+2
-0.67
Let $$\left\langle M \right\rangle$$ denote an encoding of an automation M. Suppose that ∑ = {0, 1}. Which of the following languages is/are NOT recursive?
A
L = { $$\left\langle M \right\rangle$$ | M is a PDA such that L(M) = ∑*}
B
L = { $$\left\langle M \right\rangle$$ | M is a DFA such that L(M) = Φ}
C
L = { $$\left\langle M \right\rangle$$ | M is a PDA such that L(M) = Φ}
D
L = { $$\left\langle M \right\rangle$$ | M is a DFA such that L(M) = ∑*}
4
GATE CSE 2020
MCQ (Single Correct Answer)
+2
-0.67
Consider the following languages.
L1 = {wxyx | w, x, y ∈ (0 + 1)+}
L2 = {xy | x, y ∈ (a + b)*, |x| = |y|, x ≠ y}
Which one of the following is TRUE?
A
L1 is regular and L2 is context-free.
B
L1 is context-free but not regular and L2 is context-free.
C
Neither L1 nor L2 is context-free.
D
L1 is context-free but L2 is not context-free.
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Class 12 | 815 | 2,059 | {"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} | 2.515625 | 3 | CC-MAIN-2024-10 | latest | en | 0.746112 |
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By BYJU'S Exam Prep
Updated on: September 25th, 2023
As you must know that 1 Indian rupee can be split into 100 paise. Similarly, the US currency which is a dollar can be split into 100 cents. Here, we are going to find out the value of five cents in Indian rupees using the conversion formula.
## Answer: Five Cents Value in Indian Rupees is 3.898.
Let us understand its explanation by using the dollar to rupee conversion formula which is 1 USD = 77.96 INR.
Explanation:
The value of five cents in Indian rupees is 3.898.
### One Cent in Indian Rupees
As dated today, the value of 1 US Dollar is equal to 77.96 Indian Rupees. We can rewrite it as,
100 cents = 77.96 INR
⇒ 1 cent = (77.96 ÷ 100) INR
Therefore, 1 cent in Indian Rupees is 0.7796. Now, we have to find what is five cents value in Indian Rupees.
If 1 cent = 0.7796 INR
5 cents = (5 × 0.7796) INR
Hence, 5 cents = 3.898 INR
Here, it is important to note that the conversion formula of USD and INR keeps on changing based on various economic and social factors. Therefore, you must apply the current rate while calculating the value of five cents in Indian Rupees.
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1. Feb 18, 2012
### feyomi
If A = {(x,y) : x³ = y² = e, yx = xˉ¹y} and B = {(u,v) : u^4 = v² = e, vu = uˉ¹v}, how do I go about finding the orders of, say, (x,v) and (x,u²) in A x B?
Thanks.
2. Mar 28, 2012
### DonAntonio
Your writing seems to be wrong: as you put it, both A and B are sets of ordered pairs, but then again you write elements of A x B as pairs of elements, NOT of pairs...I think you meant to write that A, B are groups, A generated by x, y and B by u,v with the given relations.
Then in A x B, both elements (x,v), (x, u^2) clearly have order dividing 6. The only thing left to do is to actually research deeper what exactly x,u,v,y are within those groups...(for example, in A the element of order dividing 3 is conjugate to its inverse, whereas in B the element of order dividing 4 is conjugate to its inverse...what does this mean?) | 290 | 912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-47 | longest | en | 0.935273 |
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### Answer key with solutions to UCEED 2016 previous paper
you are here ....... UCEED >> Answer Key to 2016 UCEED official paper
-------------------------------------------------------------------------------------------
Hi,
It's been a while since I gave solutions since I myself have been busy with my works. Anyway I managed time once in every three/four days to compose the answers and finally it's here - the complete explanation/solutions to official UCEED 2016 previous paper.
You can download the question paper - UCEED 2016 question paper here
Hope you enjoy the explanations that I gave :)
#### Section A: NAT
1) 19
We know that a pyramid has a base and triangles connecting the base sides and a common point. Usually the base is square. In the given question, pyramid base size (7cm) is greater than the size of the cube (5cm). So, some part of the pyramid will protrude out. Picture below shows the pre step of adding the shapes.
Below image shows the 3D view of the solid formed after joining. Sorry for the poor drawing, but hope you can visualize that. Note that the top portion is also a small sized pyramid shape. As can be seen the shape is in symmetry.
No. of faces in front direction = 2 (not considering the slant portion)
No. of faces in other three directions (right, back and left) = 2*3 = 6
Surfaces from top = 9
Surfaces from bottom = 2
Total = 2+6+9+2 = 19
2) 26
In the below image, I've marked the different symbols with the letter '1' in red.
3) 8
Couldn't spend much on this coz of time and this one is really tough, have been scratching my head on how to solve. Will check this again during free time and update about the way of solving this question.
4) 11
Below figure shows one possibility. We can have similar possibilities.
5) 50
Before cloning the shapes, we will count the no. of triangles so that we can just multiply the number of triangles after cloning by the number of clones. Below image shows the same image in two separate forms. Number of triangles are 6. Now, since we are cloning to the left and then to the right, and finally the whole (3 patterns) to the bottom, we will have six such patterns.
Now, after cloning, we can count the following 4 triangles as shown below.
The following two images shows further number of triangles as 8.
Finally, the below image shows 2 triangles formed.
Total no. of triangles = 6*6+4+8+2 = 50
6) 26.17
Net profit of the company = 68799-2456-4789 = 61554
Now, operating Profit Ratio = 61554/Sales = 61554/235126 = 0.2617
So, 26.17 %
7) 12
The several possibility ways are
A-D-G-H-I
A-D-G-H-E-B-C-F-I
A-D-G-H-E-F-I
A-D-E-H-I
A-D-E-F-I
A-D-E-B-C-F-I
A-B-E-H-I
A-B-E-F-I
A-B-E-D-G-H-I
A-B-C-F-I
A-B-C-F-E-H-I
A-B-C-F-E-D-G-H-I
8) 85
The problem can be solved as follows.
Given:
8 men and four women shook hands with same gender. So, it mean there are 4 groups of 2 men each, whom shook their hands with each other, totaling to 4 hand shook consisting of 8 men. Similarly since 4 women found that they shook with same gender, it mean there are 2 groups of 2 women each which counts to two hand shakes. Got it.
So, outting it like
MM + MM + MM + MM + MM = 8 men and 4 shakes
WW + WW = 4 women and 2 shakes
counting to total 4+2 =6 hand shakes,
Given, total hand shakes is 87, so the left (87-4-2 = 81) hand shakes has to be done by a group consisting of one men and one women each that is MW.
Adding these 81W from the 81 groups to the 4 women earlier we get 85 women as total.
9) 65*(25/60) minutes
It's obvious,
For every 60 ticks of the minute hand, the hour hand would have ticked 5 times.
put it like
60 ticks of minute hand = 5 ticks of hour hand
So,
1 tick of minute hand = 5/60 ticks of hour hand
5 ticks of minute hand = 5*5/60 ticks of hour hand
Now, after the 12 o' clock, they will meet somewhere after an hour, i.e. when the hour hand is near to 1 hr and the minute hand also at the same location.
That is, the minute hand had to travel an whole round (60 ticks) when the hour hand travels to 1'o clock. But still they haven't met. The minute hand has to travel further to 5 minutes, adding to a total ticks of 60+5 = 65 minutes.
But by the time the minute hand reaches the 5 min zone (number 1 on the clock), the hour hand would have moved 5*5/60 = 25/60 past number 1 (5 min) - check above for how 5*5/60 came). So, approax, they meet at 65 + 25/60 minutes
10) 12
Similar question has been solved already in my solutions to previous papers.
If it's just circles, then around a coin/circle, we can place 6 circles, but now the situation has spheres, so apart from this 6 spheres on the same horizontal plane, 3 spheres can be placed above and 3 spheres below. So, total 12 spheres. Hope you got it. If not, I can draw and explain you!
11) 99
Given Tank dimensions = 20X5X10
Volume of water that can fill the tank = 20*5*10 = 1000 cc
Given volume of water already in it = 307 cc
So, balance volume to be filled with bricks = 1000-307 = 693
Volume of brick = 5*2*1 = 10 cc
but, given that the brick absorbs 30% of its own volume, which will be = 10*30/100 = 3 cc
so, it mean if we place one brick in the water, it will consume 3 cc of water apart from displacing the water.
So, it's better t oassume that the volume of water displaced by each brick will be = it's own volume - volume consumed by it
= 10 - 3 = 7 cc
So, the required no. of bricks = 693/7 = 99 no's
12) 12
check the below image
13) 1068
Given dimension is 48m X 1.2m,
converting to centimeters, we observe dimension as 4800 X 120 cm
Now, the below picture shows one best possible arrangement. As can be seen, arrange two sheets in a row, such that 25 side is vertical and 21cm side is horizontal (shown as two sheets) in the left bottom. And also arrange three other sheets such that 25cm side is horizontal and 21cm side is vertical. Three of such sheets can be placed.
Now consider the bottom left most sheet, similar sheets when placed vertical will occupy 4800/25 = 192 sheets.
Since there are two such sheets, we have 192*2 = 384
Now, consider the sheet placed such that 21cm side is vertical, placing such sheets one above other, we can get 4800/21 = 228
Three such columns are there, so, 228*3 = 684
So, total no. = 684+384 = 1068
14) 22
Most of you might get 21 as answer, but let me show the hidden tree branch which you might count as 2 instead of 3. This is the right most part of the given image.
15) 33
Instead of drawing the whole image, let me show the part given by them with illustration of what happens when we join sides. As can be seen, the semi circle at 1 will form a circle, similarly the triangle at 2 forms a rhombus. Similarly the cut at 3 forma a square. Neglecting the shapes at 1,2,3, let us count the number of shapes as 6. Since there are four such patterns. So, total number neglecting 1,2,3 is 6*4 = 24.
Now, total no. of 1's = 2*2
total no. of 2's = 2*2
total no. of 3's = 1
SO, total no. of holes = 24+4+4+1 = 33
16) 26
Below I've identified the faces with glasses for your benefit. But, remember, during exam, you have the face questions on the computer and not on paper, so, you need to do this mentally. So, use your concentration for figuring out the required.
17) 810000
The best possible combination is in order below
1) press 9
2) press 00
3) press X
4) press 9
5) press 00
6) Press =
So, the largest number = 900*900 = 810000
18) 10
I'm not pretty confident about this. But I tried my best to figure out. Below image shows my findings. Correct me if i'm wrong.
19) 451
20) 0.82
Probability that only one of them finds a parking space = prob. that either one of them finds a park
= (prob that car finds a space AND prob. that bike doesn't finds a space) OR (prob that car doesn't finds a space AND prob. that bike finds a space)
In probability, AND is denoted by '*' and OR is denoted by '+'
So, probability = 0.1*0.1 + (1-0.1)*(1-0.1) = 0.01 + 0.81 = 0.82
#### Scetion B: MSQ
21) B,C
(I don't think explanation is required for paragraph related questions)
For more on paragraph related questions, check Resource page of UCEED
22) A,D
23) A,D
24) B,D
25) A,B,D
26) A,B,C,D
27) C
Check this on how to deal with unfolding of solids - 3D Manipulation
28) A,B,D
By observation
29) B,C
30) A,C
31) A,B,C
32) B,C
Check the below images for explanation. I followed the following numbering for explanation
33) A,D
Check this on how to deal with unfolding of solids - 3D Manipulation
34) D
Unless there are sliding joint in the first three mechanisms, it cannot be folded. Mechanism in figure 4 is the standard four bar mechanism, and it can easily be folded.
35) A,C,D
By observation
36) A,B,C,D
In figure one, curve is bulging at the top, it means more aged people are there compared to infants and teens. Second graph has steady rate for all age groups. In graph 3, infants are more in no. than aged people. Remember that in developed nations, population of infants has been controlled, compared to adult and age groups or at least equal.
So, all four options apply.
37) C
The answer is simple. Find the number of alphabets and repetitions in it. There are total 7 alphabets with two repeated (A is appearing twice). In the options check the pattern that has this feature - that is two repetitions and total 7 items. Option C has that feature by using colors.
38) D
The questions seems to be tricky with no clue. But, the first and best check that one can do is check the number boxes in the five segments in each option, since they have mentioned that they should be continuous and have equal no. of families. Fortunately, only option D satisfies!
Update:
Checking that with the above simple way, we get two options that satisfy it. I therefore discuss with you the actual procedure for solving this problem. Considering the five patterns in option D, we try to arrange it such that the whole combination forms the rectangular grid shape as given in the actual picture, which when joined will be as shown below.
As you can see, I've marked the separate parts with numbers from 1 to 5. I've also marked the parts which according to the ques are blue (to the left of section line) and red (to the right of the section line).
Now the next image shows my explanation, showing the parts sepearately and the count of Blue and red parties according to the above image division.
As you can see, for the first part (of 10 families/boxes) 6 families/boxes are blue while the rest 4 are red, so blue won, likewise
II pattern - red won
III - Blue won
IV - Red won
V - Blue won
So, overall, blue won the election in three divisions while red only in two, so that will be the answer, Now, I hope you can check for others too.
39) A,B,C
Option A is the top view of the given image in the box. You don't have to check the whole lines initially, start with the bottom two squares formed by four different colors, observe the order in which they are arranged. Like in the below image (left pattern in the box), one of the square base has been numbered as 1,2,3,4 for red, yellow, blue and green colors respectively. Observe similar arrangement for the other base square. Now, in the options, observe these 1,2,3,4 sequence for coincidence (for both the base squares). Once you screened the appropriate options, you can now check the triangles joining the base to the top vertex for the pattern.
Note that option C is the bottom view of the given pattern n the box. So, all the color order are just reversed.
40) Water flows out through A,B,D holes
It's bit difficult to explain this in paper. But the only clue I can give is, cheek for the tunnels which are either in the same row from top or a row adjacent to the water flowing row. Because, water will find way to flow in both cases. Other than the tunnel numbered 8 in front face (tunnel numbered 8 in the front face middle, 4th row from top), water will flow in all other holes.
#### Section C: Multiple Choice Questions
41) B
The answer is simple. Consider the donut shaped circle with plus/cross (+) sign as highlihted in the beliw first picture. Now, the circle has moved to the top of the center circle as shown in the second image. Let's consider it is turning clockwise and the rotation is thus 180 degrees. Now i nthe third image, the same circle is rotate di nthe same direction (clockwise) but with angle 180+90 = 270. In the fourth image, rotate the ciecle in the same direction (clockwise) but with angle 180+90+90 = 360 degree. So, obviously in the firth sequence, the ciecle has to be rotated clockwise in 180+90+90+90 = 450 degree. Got it ? all the four circled symbols inside the pattern follow the same proceedure. Now, check the option that represnets 450 degree rotated view (of inner 4 circles) rotated from the fourth sequences. Option B and D matches. Now, observe the center circle. In the first image it's a circle with dot inside, in the second image it's a donut (empty shape inside a black circle). follow the pattern
first - circle with inner dot
second - donut
third - circle with inner dot
fourth - donut
So, obviously, the next pattern should have 'circle with inner dot'. Thus option B suits
For more on how to solve this question, check the mental ability part of the resource page.
42) D
All the options have same color, so let's forget about the color criteria given in the conditions.
As given in the grid form with numbers, and following the arrow-ed directions, we get
(Grid No.) - (no. of sides of polygon)
start - 5 sides (since it's pentagon shown in red color)
4 -> 5 + 1(rule i) +1 (rule iii) = 7
9 -> 7 - 2(rule ii) - 1 (rule iv) = 4
8 -> 4 + 1(rule i) +1 (rule iii) = 6
5 -> 6 - 2(rule ii) = 4
2 -> 4 + 1(rule i) +1 (rule iii) = 6
7 -> 6 - 2(rule ii) = 4
6 -> 4 + 1(rule i) +1 (rule iii) - 1 (rule iv) = 5
1 -> 5 - 2(rule ii) = 3
2 -> 3 + 1(rule i) +1 (rule iii) = 5
So, the final shape is again five sided, i.e. pentagon
43) C
By careful observation, this can be solved. If you're not confident, then better draw the pattern by linking similar sides.
44) C
It's obvious that the other shoe sole should be the mirror image of the given shoe. All the options might look similar but there's difference in the wave like lines at the c=middle of all the soles. Only option C is proper.
45) D
The colors are analysed as follows
For the top part of the circles
Yellow + Blue (yields) -> Dark Green
Dark green (yields) -> light green
For the left part of the circles
Blue + Red -> bright violet
bright violet + yellow -> dark brown
For the right part of the circles
Red + yellow -> dark orange
orange + yellow -> light orange
For more about color combination - check Resource page of UCEED
46) A
Obviously it should rotate in the opposite direction as the rotation when the water is sprinkled through the nozzles, which will be in clockwise direction for the given shape of the sprinkler.
47) C
It will not be sphere, since the question needs the portion that is common to all the four spheres. So, the sphere part is cut a bit to a shape something similar to C.
48) A
The first impression that I got after reading the question gave me a shock! was wondering whether something like hammer is used for any instrument that too musical ? then later realized that it's the kind of mechanism that they wanted. Hammering here they mean the technique of hitting the string rather than sliding (as in violin) or picking (like in guitar).
49)
50) A
I don't know whether there is any such mathematics existing to use parameters to construct a shape or not (I think there isn't any standard in math literature), but I do see some pattern from the given images. The values of R and r will give us an idea of how many sides the shape is having. Like R/r is the no. of sides of the shape.
For the first image, R/r = 5, so, it has five sides, (all formed by 5 circles intersecting one another)
For the second image, R/r = 5, so therea re 5 sides
For the third image, R/r = 16/4 = 4, so there are 4 sides
For the second image, R/r = 108/4 = 27, so there are 27 sides, all formed with circles.
Now, 'c' might represent the amount of bulge of the circle used to construct the shape, in other words the curvature or radius of the circle used for constructing.
As can be observed, when c i more, the intersection is more, as compared to when c is less.
So, in the required shape, which ahd three sides, the best combination is R/r = 3 (24/8 in option) and since the shapes are bulges out, the value of c might be the least. So, option A fits.
51) C
I don't think explanation is required for this
You can check more paragraph related question here - check Resource page of UCEED
52) D
Obtained by following the procedure that they gave.
Below image shows the step by step change made for the given sequence. Green symbols (and arrows) indicate the change that has been made for the given pair of symbol.
53) D
Solution is simple, but bit of observation is needed.
As can be seen in the below picture, it's obvious that we don;t have to bother about circles as all the squares have two circles each. Now, observe the first and third numbered square in the below image. As can be seen. Shape A of 4 sides/peaks has been transformed to A' of three sides/peaks (with difference 1). Similarly shape B of 9 sides/peaks has been transformed to shape B' of 10 sides (with difference 1). And note that A+B = 13, as well as A'+B' = 13.
Keeping this pattern in mind, shapes in square numbered 2, viz. C and D (C+D = 13 sides) should be transformed to shapes C' and D', such that C'+D' = 13, and difference should only be 1 between them, the possible combinations as are as shown in image, i.e C' can be either 6 or 5, and D' can be either 7 or 8. Now, check the options which best matches this pattern. Option D suits I guess.
54) B
It's really tough (at least for me) to exactly figure out the pattern. However, as always, we can derive some pattern out of given sequence. So, here is what I observed. As can be seen in the below image, wherever circles are appearing in the main image (highlighted in red color), then we can see a dark water drop symbol accompanying it (highlighted in blue color).
Similarly when there is unbound lines (lines that are not closed to form a shape) as highlighted in the below image, we see the shape of two zigzag short lines accompanying that as highlighted in blue color.
Now, in Fig.1. since we can see both circles as well as lines (not rectangles), we can assume that the two symbols (water drop and zigzag) must also exist. So, option B seems to be correct. Although this might not be a correct interpretation, but this is how I solve.
55) C
Below image shows the construction. I've numbered the parts for your convenience. Note that the parts numbered 3 and 4 has to be rotated 90 degree clockwise for the final assembly and the final assembly is a flipped E. We can just flip it to get the actual shape as in Fig.1.
56) D
57) C
Dip your toe in the water -> tentative step
Come hell or high water -> Persistent in the face of difficulty
Mout hwatering -> tasty
58) B
In downward to-and-fro position, we tend to bend bit and try to push and pull the hack saw. So, keeping the handle either horizontally or vertically will only cause strain to hands and will not function well as there will be bending possibly between every to-and-fro motion . Posture in option D will separate the blade frame from the material in downward f=direction (because of handle direction being 45 degree outer), which is one inefficient way. Keeping handle as in option B works fine, and will always ensure that the tool (hack saw blade) is in contact with the material all the time.
59)
I'm not pretty sure. Probably if I come across a pilot, I will ask him ;)
60) A
Tough to explain this in paper, but what you can do is speak yourself "welcome to uceed" very slowly and observe how your mouth is moving.
Tutorial 1
Tutorial 2
61) B
Okay, here goes the convention :
1. Whenever tail is included with the bubble, then it means converssation is happening either directly or through whisphering
2. When the bubble is dashed with a tail, then it indocates whispering
3. When the bubble is made of continuous line with a tail, then it is a direct conversation (speech)
4. When the bubble is made of zigzag lines, it mean shouting/screeming etc
5. When the bubble is followed by a circled dots at bottom as shown in the below picture, it indicates thinking process.
6. When instead you see box, then it is just explanation of the situation (past or present or future) which is neither been spoken by anyone nor indicates thinking process but is just a narration of the situation.
7. Below image should give you an idea.
For more details refer to - Link 1
wikipedia guide
62) A
63) C
64) A
For more on animation, refer this - Animation study
65) D
I was able to spot 6 differences
Below image shows the difference. Note that difference numbered 2 and 3 indicates the color change.
I think, the options that they showed are wrong! They showed like the front two legs will go crossed while running which I suspect whill not happen, and the back hind legs will not be on the same line as far as I observed.
67) D
You can play with these online tool and learn about different combinations -
68) C
69) A
70) C
71) B
two reflections one each on the mirror surface and one reflection at the intersection of the two mirrors. If the ball was placed in between two parallel mirrors, then there would be infinity reflections!
72) D
COG lies at the mid way of the mass and not at the mid if the object. So, the top surface of the T shape contributes more weight and hence the COG should be somewhere around the border of the two surface intersections, but placed little distance away from the horizontal top T surface as shown in option D.
73) C
Option C is the correct rotated view
74) B
By observation
75) D
The Kyoto Protocol is an international treaty which extends the 1992 United Nations Framework Convention on Climate Change (UNFCCC) that commits State Parties to reduce greenhouse gases emissions, based on the premise that (a) global warming exists and (b) man-made CO2 emissions have caused it.
The Ramsar Convention 1971 is an international treaty for the conservation and sustainable use of wetlands.
The Vienna Convention 1981 on the Law of Treaties (VCLT) is a treaty concerning the international law on treaties between states.
Montreal Protocol 1989 on Substances that Deplete the Ozone Layer (a protocol to the Vienna Convention for the Protection of the Ozone Layer) is an international treaty designed to protect the ozone layer by phasing out the production of numerous substances that are responsible for ozone depletion
i indicates ecofriendly disposable
ii indicates Hazard/warning/dangerous etc
iii indicates recyclable products
iv indicates caution
For more list of symbols check the following pages - wikipedia page
The best option could be B, if X-iii was used instead X-iV
77) D
78) A
I think so,
79) A,B
Pellagra is a vitamin deficiency disease most frequently caused by a chronic lack of niacin (vitamin B3 or synonym: vitamin PP (from: Pellagra Preventing factor)) in the diet. It can be caused by decreased intake of niacin or tryptophan, and possibly by excessive intake of leucine.
Anorexia - Eating disorders
(source: internet)
80) B
It's clearly identifiable by observation.
1. can you give me explaination of ques 3
1. I spent 15 min but couldn't solve, probably have to spend more time and figure it out. Really a tough one, with no pattern ?
Will update if I solve.
2. Ya it is very hard to solve. Hats off to the question maker.
3. I did come up with a logic to solve this.
It is kind of arbitrary,but it fits the pattern
4. that's really nice of you, thanks for sharing your solutions with everyone :)
5. Its the least that I can do :)
6. Hi Bhanuchander, My daughter have worked out solution for Ques number 3 of Uceed 2016.
The solution is
Ist row + 3rd row - 2nd row = 4th row. eg: 6849+4675-8659 = 2865. Similarly the last row if you apply the same rule the answer is 2nd row + 4th row - first row = 3rd row. ie., 4492+6844-3284 = '8'052. Hence answer is 8.
7. That's really great of her sir. Thank you sharing with everyone.
8. Awesome solution by Murali Bharatwaj, thank you for this sir and thank for your daughter too.
9. Thank you very much
2. I think in that ques abt counting glasses.. u answered 26.. and the red squares are 27.. bt i saw some glasses on there head.. it should b concedard or not..??
1. what's Q no. ?
2. red rectangles in the picture are 26 in no. and not 27! pls count again n since the ques asked "how many are wearing" we should be bothered about those wearing the glass and not others.
3. Q NO.-1 DRAWING???
SIR NOT ABLE TO VISUALIZE THE DRAWING.
1. It's been updated
4. This comment has been removed by the author.
1. Pl don't look at their size that I drew, look at the dimensions (length) I mentioned, I have drawn that by hand without minding the actual size, but as I showed in the picture above, it looks like that.
2. sorry sir, i get confuse in case of centroid...i just understand now! thanks
3. sorry sir , i am not very use to with internet so by mistake comment was deleted . so i wanna ask that like in question 31....we must know all rivers and sort of things like mountains and all ?
5. 71st q "one reflection at the intersection of the two mirrors"???
6. Hello sir,
Thankyou so much for such a helpful blog
While doing q.4 i got an ans of 16
I wanted to ask you that why cant we have 7 triangles in the last row instead of one
If we divide the length of one small triangle with 11.2 we get an approximate ans of 4 and so we can have 4 rows of trianles with 1,3,5 & 7 triangles resp.
1. That's the thing, if we divide 11.2 with 3, we will get <4 and not 4, had it been 4, then we can go with the counting like u did, but
(11.2 - 3*3) = 2.2
is the leftiver row height at the bottom as shown in the above and how can we fit a equilateral triangle of size 3 perfectly ? unless we tilt it and try to fix, which if we do, would allow us to fit only 2 triangles as shown in the pic. So the end row cannot accommodate 7 triangles but instead only 2 triangles.
2. But sir why would we divide 11.2 by 3
The length of an equilateral triangle with side 3 cm would be approx 2.6 by pythagoras theorem right?
N when we divide 11.2 by 2.6 we get an ans slightly more than 4 which means we would have 4 rows.
Sir thanks for such a responsive blog
3. That's great Neelu,
Yeah, 3cm is the side and not the height of the equilateral triangle, as I thought. I couldn't check that. So, you are right in saying that 3cm sided equilateral triangle will be approx. 2.6cm.
But, we need to apply the same principle to the outer triangle of length/side 11.2 also, right? So the altitude or height of the big outer equi triangle will be (sqrt(3)/2)*side = (1.732/2)*11.2 = 9.7
Now the above figure of 9.7 should justify u why only 11 is possible. 9.7/2.6 = 3.7 (approx) and so we could only fit 3 row of triangles with the fourth row having only two triangles again.
That was an interesting coincidence from my side :D
Probably you missed the length 11.2 part thinking that as height rather than side. I know it's not ur mistake as the ques should have been clear.
Thanks for bringing this up and finding the mistake.
4. Thank YOU for clearing the doubt sir. You are really humble:)
5. sorry to say ...ur given figure is wrong for question 4....as if u say 7,10,11 are equilateral triangle then the triangle in between must be an equilateral triangle as all sides are of 3cm....btw nice help for other solutions
7. Q 4 solution is wrong , only 9 are possible. Becoz if u turn the triangle in any other alignment the height will be greater than the height of the equilateral triangle (ie 2.6 cm ) . I also tried it practically.
1. yes, If we tilt the triangles, height should increase. I think I've to recheck that in depth.
8. 65. Other difference is dot on her nose.
9. Sir diagram for 13th sum sir
1. it's updated, thanks for reminding me
10. qNo.. 19. Please explain. It.
1. it's GA question, nothing to explain
2. on googling, i found out that Ray Bradbury's most famous novel is "Fahrenheit 451".
now get it?
1. will look into it and reply u later
2. Yup..thanks sir!
I've wasted more than 1 hour on this question..still haven't been able to figure it out
3. The asnwer has been given in a link in the first comment by Sourav, you can chk that.
12. Hello sir , could you please clear how 52nd one is solved ?
1. It's updated with an image above. U can chk now
13. Hi. Can you please elaborate on the 35 question about the tyre mark.
1. I will make a pic n update there soon
14. sir i made in graph paper q4 and only 9 triangle fit not more if i tilt than triangle get disorder
1. hmm, seems like I too need to do the exact measurement for checking. Thnk u, will chk that soon
15. question 2 there are 27 symbols sir, if you see closely 3rd box in 3rd row and 3rd box in 4th row are two different symbols.
1. I think both r same.
16. and in q4 if we calcute area of larger triangle and divide it by area of smaller triangle then the answer comes out to be 12 triangles.
17. Q 80. can you describe elaborately. i think it is meta.
18. bhanu bhaiya in 38th ques. i think you missed to check option B. It also satisfies acc. to your explanation. it is also continuous and have equal no. of families. i have a diff. logic , but still please justify if possible.
1. It's explained above with pictures bro. You can check now for the answer.
2. thankyou so much. :)
19. Hey could You please suggest some sites or books from where I can get the required GK?
Coz I am not able to get the source for improving my GK in the fields required by UCEED
1. It's already available in UCEED resource page, chk there.
20. Figure image is not shown ......plz help to resolve this
1. which figures ?
21. How are we supposed to solve it in exam hours if we even can't solve in with two days..
1. It's all about math apti and visualizing. If u have some prior knowledge, then u will feel it as easy. Try ur best till the last day of ur practice and in exam, if u r not able to answer them i nshort time, JUST IGNORE THEM!
coz it's better to spend time for some answerable ques rather than pondering on something which eats ur time leaving a 50-50 chance of being correct. So, just ignore those type of ques during exam, but make sure that u try ur best solving n understandung them until ur last day of prep!
22. Like .... question1 is not shown and others also have same problem
23. sir i couldn't understand question number 9.WHY DID YOU DO 5*5/60?.
1. For every 60 ticks of the minute hand, the hour hand would tick 5 times right? in other words,
60 ticks minute hand -> 5 ticks hour hand
1 tick minute hand -> (5/60) ticks hour hand
5*1 = 5 tick minute hand -> 5*(5/60) ticks hour hand
2. my answer for this question came 720/11 minutes (don't know if it is correct or not). Here's the method:
the minute hand rotates 360° in 60 minutes. so it will rotate 6° in one minute.
the hour hand rotates 360° in 12 hours. so it will rotate 0.5° in one minute (simple math)
now let us assume they meet after 't' minutes. so rotation by hour hand = 0.5t and by minute hand = 6t.
it is easy to understand that the minute hand would complete one revolution by the time the hour hand rotates 30°. so they will meet sometime after an hour.
6t = 0.5t + 360. on solving we get t = 360/5.5
let me know if you find any errors in my method.
24. I don't think than paper and pen is provided in uceed exam as its totally online. How are we supposed to do such tedious calculations!???????
1. Even if it's online, I think u will be provided with a scribble pad by the invigilators, n u have to carry pens/pencils.
25. hello sir,
would you please explain how come the answer to Q.12 is 12 only. ( I found more than 12 diamonds in the figure. )
1. I'm getting 12 again, I counted the four sided rhombus, can u tell how u got > 12 ?
Here it is, please correct me if i'm wrong.
1. good try,
I've updated my sol. image above for reference.
what I understood from ur sol is u r considering trapezium shapes too, apart from diamond shapes (rhombus) while u were asked to count black shapes in diamond form, so, although diamond shape assumes as a part of the trapezium shape, still they are doesn't in parts and represent a single shape. They shouldn't be counted.
27. ohh, got it! thanks a lot for your time. :)
and lastly i have been pondering over q.59 from past one hour and i think the answer should be blue or brown goggles as black and red would not be suitable for night vision.what's your opinion?
1. Even I was not sure, that's y I didn't give correct option above. Ya I agree with ur opinion, probably red could also be considered.
2. ohh,yes you're right. i just asked an air officer . the answer is ombre(red).
28. sir please explain q33 its abit complicated then the link you have provided . if you have time. thanks :)
1. will chk when i get time, but logic is same, it's just unwrapping in an order minding the pattern in option, probably u can give it a shot.
29. Sir, shouldn't the question 49 have the answer as 360?
30. in question number 16 what about the right side top corner fellow isnt he wearing spectacles?
1. It's not spects I suppose
31. i have got the solutions of Q.3...the number puzzle....the soln is as follows......first add the numbers from 1st and 3rd row and subtract the number from 2nd row...you will get answer as 4th row...this goes with every box,,,,...e.g....... (6 8 4 9 + 8 6 5 9)-4 6 7 4 = 2 8 6 5
1. Thank you Bablu, already the same answer was given in the first comment by someone :)
32. Sir are there any easy tricks for these for these diagrmatic questions?
1. yes, they r available, chk the resource pg of uceed or ceed, u will find the links.
2. Sir can u prepare a list of the main logos and their designers? i checked in net and iam not able to find.Also tell me a site where i can study about photography concepts..plzz sir :D
3. sir i didnt get you resource page in the sense....?
4. Bro, i think u r not exploring the blog,
u will find everything that u asked, chk the three links below
http://stuffyoulook.blogspot.in/2014/06/logos-of-Indian-institutes-and-corporations-1.html
http://stuffyoulook.blogspot.in/2015/05/books-and-study-materials-for-uceed.html
https://stuffyoulook.blogspot.in/p/uceed.html
pls explore the blog for further topics since i cannot spend time sharing individual links with everyone.
33. sir in question no. 9 how it is 25/60min sir i am not able to understand. please help me
34. SIR SAME QUESTION NO-9. I ALSO DID NOT UNDERSTAND.WHY 65*(25/60) ??PLEASE GIVE THE EXPLANATION FOR 25/60.
35. Ok, guys, I've rewritten the answer for Q9, which I hope will be clear. Hope u r doubts will get cleared.
36. THANK YOU VERY MUCH SIR FOR YOUR REPLY.SIR I WANTED TO KNOW WHAT IS MECHANICAL ABILITY??.IS IT ABOUT THE PULLEY,GEAR RATIO ETC.ACTUALLY I WANTED TO KNOW THE TOPICS THAT COMES UNDER MECHANICAL ABILITY.
1. yes, it is,
anything related to mechanical like mechanisms, mirroring, functioning of products, gears etc are comes in mech ability. Related material PDFs r available in uceed resource pg, u can chk them
37. Sir can you explain Q.55, i think the answer can be option D too and here is how i got that.Moving from top to bottom pieces,piece 1 and 2 can be combined to make first part of the letter E,then 3rd part can be rotated 90degree to make the centre line of E.And pieces 4aand5 can be flipped to the left and merged to form the lower half.
1. No, option D cannot come, actually their thickness are less and so, even if use all those parts, u will not get the complete E, but only something like C, with the ceeter extension - missing
38. Hai sir their is NO IMAGE for Q55.Plz uplod the picture
1. it's there, visible, chk there
39. Sir in question 66 i found A be the most appropriate answer as when we run fast our opp. leg always land ahead of other
40. Sir I have a query
in Ques 33
Option D can only be folded into the shape if the part in blue is joined to the part in red and orange (in the image I have linked)
But from the image in the question paper it is not obvious that it is not joined :(
41. Please draw and explain for question no 10.
1. It will be no use even if I draw in 3D for that ques, as most spheres will be hidden, u have to imagine the situation,
consider a cubic box of same dimensions all sides. Place a sphere in the center of the box. six spheres around the center spheres horizontally will surround the center sphere, 3 spheres above n 3 spheres below will cover the top and bottom portion. So, altogether 12
2. actually, could you please link the page where you explain how six circles will completely surround a circle, as i dont quite get it, neither do i get how 3 spheres will cover the top and bottom
42. I am a drop out 12thpassed.. for uceed exam verification id question is
whether aadhar id card of me is enough for the id proof ..or i need to submit any thing extra?
1. usually 1 id is enough, aadhar should do the job, anyway chk the brochure if they have any specific requirement of college id's, i don't think it's req, but still as safety chk the brochure
43. Sir which book should be refered for the kind of question no's:21,70,75
1. hello Akthar,
thankfully this time u commented on the right page instead of asking somewhere else :D
Well, Q21 is paragraph comprehension type and Q70 and Q71 are GA/GK which unfortunately are not available in any single as far as i know. I think manorama book mightcontain but u will be lost wondering how to cover the whole book, u can check those materials in the resource page. Don't roam around only on mock tests and previous papers, chk theory part, resources of which r available in
http://stuffyoulook.blogspot.in/p/uceed.html
44. Sir plz explain Q19
45. In ques 18 i think that. .. worried and demarcate are of same fonts..... Pls hell me... With such ques..
1. font related materials r available here - http://stuffyoulook.blogspot.com/2015/12/useful-webpages-for-uceed-exam.html
and
http://stuffyoulook.blogspot.com/2016/11/nid-dat-resources-update-new-syllabus.html
46. plz further explain question 80, I don't understand what to observe
47. hello sir,
how u draw 11 triangles of side 3 cm in equi.triangle of 11.2 cm
1. they're tilted so that they can be placed in the available space, check the earlier comments (at the beginning) for further discussion already happened in this page.
48. Sir I got question no 3
1. we can identify font type based on certain pattern that each follow. Once u understand that pattern, then it's easy, there are some 4-5 basic font styles that one has to recognize, u can study more here -
http://stuffyoulook.blogspot.in/2015/12/useful-webpages-for-uceed-exam.html
50. Sir please explain Q19 and Q03.I think this is the right time for Q03.Only FOUR DAYS are left over.PLZ PLZ explain these questions
1. Q19 is GA type related to web design, i will try to share related matl. if possible very soon, nothing else to help in that coz they r standard things to be memorized and no shortcut. Solution for Q3 is given in first comment of this page by Mr.Muralidaran, pl chk abova
51. Sir, can you please explain Q.28 and how did we observe such a pattern????
52. In Q.no 16 you have circled a person who is not seeing through the spectaclesso he is not wearing them....also on the right top corner there is a person who is wearing spectacles but you have not circled him.
53. In the 80th question option D looks similar to the word. can you explain why it is B ?
1. for the given height of the alphabets in the given word, observe the width of the lines, especially vertical lines. In option D, the width for that small height seems comparatively more, if we zoom in to get to the same height as in the questioned word
54. sir... in qno.18, the words u underlined.."worried" & "demarcate" are of same font...
i spent observe all words much time.. and found that IF THE RIGHT ANSWER IS 10,
then the following words are of same font..
1. amazing, sandalwood, greetings
2. discussion, patient, encourage
3. worried, demarcate, lavender, fantastic
4. sampled, happily, estate, temporary
5. entrance, transfer, relinquish, delightful
6. deployed, metal, wavelength
7. canine
8. cautiously, workable, safety
9. karmic
10.achievement
once more go through it and hope u wil improve the answer 4 evryone............ :)
55. Bhanu chander bhai what is the minimum mark a person should get to qualify into uceed from the experience that you have
56. Plz sir help me understand y not option C in q28?
57. Hello sir I want to thank you for the solutions and papers you post for people like me preparing for different papers.. So thank you. ☺️ sir it would be kind if you could solve ceed 2006 paper to 2017's paper also.. Thank you waiting for the solutions..
1. they r already available in this blog, check the solutions or CEED page.
58. Plz sir help me in last question ( Q 80)?
59. Here a solution of 49th question -
First find the minimum length of the chord (perpendicularbto the centre) which is 24.18, then the maximum length that can pass through the point which is the diameter itself (34). Hence the lengths of chords are 25,26,27,28,29,30,31,32,33,34.
Therefore the number of chords would be 9 + diameter + 9(on the other half of the circle)
61. I think question 4 diagram is wrong
62. Ans no. 4 is definitely wrong coz I have tried a lot and failed to mention 11 triangles in the figure. Plz check it out and give the correct way of doing it.
63. Tysm sir your explanation are really Amazing Tysm once again...
A small request can you send me the link of your blog. ...
1. u r already in my blog page bro :)
64. In question 65 one difference is in nose
65. my answer in Q18 is amazing discussion worried sampled entrance deployed relinquish cautiously karmic greetings
66. Can you draw the soultion of question no. 27
67. 1) right anwser is 22
68. Sir! There is a 7 differences in question no. 65. There is a difference in nose pin. You can check that...
69. Sir can you please explain question no. 28? I can’t seem to distinguish the pattern of C. Where as the answer is ABD. Thank you sir. )
70. hello sir!My name is srujan. Thanq for these solutions. Q.18 ANSWER IS 11. U MISSED THE WORD "greetings".
71. helllooooo? anybody there. bhanu sir pls respond.my exam is tomorrow.
Any of your comments for this post will be sent to Bhanu for his approval. It will get published only after his approval. You will receive mail for all the replies to your comment. | 11,139 | 42,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-43 | longest | en | 0.944549 |
https://byjus.com/maths/hex-to-decimal/ | 1,632,386,192,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00447.warc.gz | 200,794,110 | 148,842 | # Hex to Decimal
The number system is an important concept in maths as it is used to represent the numbers and classify them based on their base numbers. There are various types of number systems namely binary, octal, decimal, hexadecimal and so on. We can easily convert one base system to another using certain rules of conversion. In this article, you will learn about hexadecimal and decimal systems, conversion from hex to decimal, charts of conversion, tables along with the examples.
The base of an hexadecimal system is 16. The 16 symbols involved in this system include 10 decimal digits and the first six letters of the English alphabet, i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F. Here, the alphabets can be treated 10, 11, 12, 13, 14 and 15, respectively. Learn more about the hexadecimal system here.
### Decimal Number System
A number system that uses digits from 0 to 9 to represent a number with base 10 is called the decimal number system. The number is expressed in base-10, where each value is denoted by 0 or the first nine positive integers. Each value in this number system has the place value of power 10. It means the digit at the tens place is ten times greater than the digit at the unit place.
## Conversion from hex to decimal
As we know, number systems can be converted from one base to another. Thus, we can convert hexadecimal numbers to decimal easily. This number system conversion can be done as explained in the example given below:
Example: Convert 7CF (hex) to decimal.
Solution:
7 = 7
C = 12
F = 15
To convert this into a decimal number system, multiply each digit with the powers of 16 starting from units place of the number.
7CF = (7 × 162) + (12 × 161) + (15 × 160)
= (7 × 256) + (12 × 16) + (15 × 1)
= 1792 + 192 + 15
= 1999
From this the rule can be defined for the conversion from hex numbers to decimal numbers.
Suppose below is the hex number with n digits:
dn-1 … d3 d2 d1 d0
Multiply each digit of the hex number with its corresponding powers of 16 and add them such as:
dn-1 × 16n-1 + … + d3 × 163 + d2 × 162 + d1 × 161 + d0 × 160
Thus, the resultant number will be taken as base 10 or decimal system of number.
dn-1 … d3 d2 d1 d0 (hex) = dn-1 × 16n-1 + … + d3 × 163 + d2 × 162 + d1 × 161 + d0 × 160 (decimal)
### Hex to Decimal Converter
There is a tool available to convert the numbers from hexadecimal to decimal number system.
### Hex to Decimal Table
The conversion table for the numbers from hexadecimal to decimal is given below:
Hexadecimal 0 1 2 3 4 5 6 7 8 9 A B C D E F Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
This table will help in representing the digits and letters individually in the large numbers in base 16 system as explained above.
Also, learn about decimal to hex conversion here.
### Solved Examples
Let us have a look at the examples of conversion of numbers from hexadecimal (base 16) to the base 10 number system, with detailed explanations.
Example 1: Convert (1DA6)16 to decimal.
Solution:
(1DA6)16
Here,
1 = 1
D = 13
A = 10
6 = 6
Thus,
(1DA6)16 = (1 × 163) + (13 × 162) + (10 × 161) + (6 × 160)
= (1 × 4096) + (13 × 256) + (10 × 16) + (6 × 1)
= 4096 + 3328 + 160 + 6
= 7590
Therefore, (1DA6)16 = (7590)10
Example 2: Convert (E8B)16 to decimal system.
Solution:
(E8B)16
Here,
E = 14
8 = 8
B = 11
Thus,
(E8B)16 = (14 × 162) + (8 × 161) + (11 × 160)
= (14 × 256) + (8 × 16) + (11 × 1)
= 3584 + 128 + 11
= 3723
Therefore, (E8B)16 = (3723)10
Test your knowledge on hex to decimal conversion | 1,082 | 3,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2021-39 | latest | en | 0.87287 |
https://www.coursehero.com/file/p5mnmukd/ALGORITHM-Step-1-Input-VALUE1-VALUE2-Step-2-if-VALUE1-VALUE2-then-MAX-VALUE1/ | 1,606,905,041,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141706569.64/warc/CC-MAIN-20201202083021-20201202113021-00453.warc.gz | 619,838,797 | 221,149 | ALGORITHM Step 1 Input VALUE1 VALUE2 Step 2 if VALUE1 VALUE2 then MAX VALUE1
# Algorithm step 1 input value1 value2 step 2 if value1
This preview shows page 27 - 36 out of 36 pages.
ALGORITHM Step 1: Input VALUE1, VALUE2 Step 2: if ( VALUE1 > VALUE2) then MAX VALUE1 else MAX VALUE2 endif Step 3: Print “The largest value is”, MAX
Example 5 MAX VALUE1 Print “The largest value is”, MAX STOP Y N START Input VALUE1,VALUE2 MAX VALUE2 is VALUE1>VALUE2
NESTED IFS One of the alternatives within an IF– THEN–ELSE statement may involve further IF–THEN–ELSE statement
Example 6Write an algorithm that reads threenumbers and prints the value of the largest number.
Example 6 Step 1: Input N1, N2, N3 Step 2: if ( N1>N2) then if ( N1>N3) then MAX N1 [N1>N2, N1>N3] else MAX N3 [N3>N1>N2] endif else if ( N2>N3) then MAX N2 [N2>N1, N2>N3] else MAX N3 [N3>N2>N1] endif endif Step 3: Print “The largest number is”, MAX
Example 6 Flowchart: Draw the flowchart of the above Algorithm.
Example 7Write and algorithm and draw a flowchart to a)read an employee name (NAME), overtime hours worked (OVERTIME), hours absent (ABSENT) andb)determine the bonus payment (PAYMENT).
Example 7 Bonus Schedule OVERTIME – (2/3)*ABSENT Bonus Paid >40 hours >30 but 40 hours >20 but 30 hours >10 but 20 hours 10 hours \$50 \$40 \$30 \$20 \$10
Step 1: Input NAME,OVERTIME,ABSENT Step 2: if (OVERTIME–(2/3)*ABSENT > 40) then PAYMENT 50 else if ( OVERTIME–(2/3)*ABSENT > 30) then PAYMENT 40 else if ( OVERTIME–(2/3)*ABSENT > 20) then PAYMENT 30 else if ( OVERTIME–(2/3)*ABSENT > 10) then PAYMENT 20 else PAYMENT 10 endif Step 3: Print “Bonus for”, NAME “is \$”, PAYMENT
Example 7Flowchart: Draw the flowchart of the above algorithm?
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• Fall '18
• Abdul Umar
• Pseudocode & Algorithm, Input | 596 | 1,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-50 | latest | en | 0.585197 |
https://www.pcreview.co.uk/threads/if-function-help.3845120/ | 1,660,961,784,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573876.92/warc/CC-MAIN-20220820012448-20220820042448-00373.warc.gz | 798,206,640 | 14,609 | # if function help
H
#### Haz
Hi,
I would like to return 'no result' if the cells in the sum below are empty.
=sum(a5:a9)+(a4*3)
I think I need an if function in the sum above so if all cells
A4,A5,A6,A7,A8 & A9 have no values in then give 'no result''. If any of these
cells have a value which may come to 0 then give 0 or any other value of the
sum.
hope this clear I can't get this to work
Thanks
L
#### Luke M
If you meant display "no result" if ALL cells are empty:
=IF(SUMPRODUCT(--ISBLANK(A4:A9))=5,"no result",SUM(A5:A9)+A4*3)
If you want to display "no result" if ANY of the cells are empty:
=IF(SUMPRODUCT(--ISBLANK(A4:A9))>0,"no result",SUM(A5:A9)+A4*3)
T
#### T. Valko
If any of these cells have a value which may
come to 0 then give 0
Not real sure what "which may come to 0" means.
See if this does what you want:
=IF(COUNT(A5:A9)=0,"No Result",IF(COUNTIF(A5:A9,0),0,SUM(A5:A9,A4*3)))
H
#### Haz
Hi Thanks for coming back to me,
The first Sumproduct function works perfectly but only if A4:A9 are blank. I
have formulas in a5:a9 and they display £0.00 if nothing is entered or 0.00.
Can this be adapted to except £0.00 as 0 values to display 'no result'?
Thanks again | 390 | 1,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.788275 |
http://blogs.msdn.com/b/ericlippert/archive/2006/01/20/regular-expressions-from-scratch-part-twelve-superposition-of-states.aspx | 1,448,592,574,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447906.82/warc/CC-MAIN-20151124205407-00068-ip-10-71-132-137.ec2.internal.warc.gz | 30,141,540 | 19,928 | Regular Expressions From Scratch, Part Twelve: Superposition of States
Regular Expressions From Scratch, Part Twelve: Superposition of States
Happy New Year everyone. Over the break I had a wonderful time reconnecting with my friends and family. And of course I came back to a huge pile of work! We're going through the flaws that were discovered in C# 2.0 too late in the cycle to risk fixing, and some of them illustrate interesting corner cases in the language. But those will have to wait, as we've still got a lot of ground to cover before I'm done this crazy long series on regular expressions.
To summarize the story so far: we've defined alphabets and languages over those alphabets. We've shown how to create a special "regular expression" alphabet and language for any given alphabet. We've also come up with a rule that associates a language with a regular expression. This language is the set of strings which "matches" the regular expression.
We've determined that there are deterministic finite state computers which consume one character of a string at a time that can "recognize" strings from some regular languages. We've also determined that there are "nondeterministic" finite automata that somehow "magically" figure out which rules to apply at any time to match a string.
We've also shown that any NFA that has rules that act on more than one character can be turned into an equivalent NFA which only has single-character or no-character transitions.
What we would like to do is show the following three facts: first, that every NFA is equivalent to a DFA. Second, that every DFA/NFA recognizes a regular language. And third, that every regular language has a DFA/NFA that recognizes it.
We've still got a ways to go to show that first result. Rather than prove the general result, which will be tedious, I'll just go through an example and hope that it is illustrative enough to convince you that we could give this treatment to any NFA.
Let M be an NFA with alphabet {a, b}, states {1, 2, 3, 4}, start state is 1, and the set of acceptable states is {4}. We'll assume that we've already eliminated all the multi-character rules and have a set of rules as follows:
(1, e) → 2
(1, b) → 3
(2, a) → 1
(2, e) → 3
(2, a) → 4
(3, b) → 4
This NFA accepts the language a*( bbba ), and as you can see, we've got lots of empty and ambiguous rules.
Here's the trick: given an NFA with 4 states like this one, we can find a DFA which is equivalent to this guy that has 24 = 16 states or fewer. We can think of the NFA as at any time living in a "superstate" that consists of all of the states that it COULD be in right now.
Let me try to explain that a little better. We start in state 1, right? But since (1, e) → 2 and (2, e) → 3, we could also be starting in states 2 or 3 before we process any input. Let's create a new automaton with a start state that represents the concept "right now M could be in states 1, 2, or 3." We'll call that state 123x, which is one big symbol, not a string of four symbols. In our new automaton, the start state is 123x.
We're trying to write a DFA here, so there needs to be a rule for every input and every state:
(123x, a) → ?
(123x, b) → ?
Look at the original automaton. When we were in states 1, 2 or 3, what were the possible state transitions for a? The only ones were (2, a) → 1 and (2, a) → 4. But we're not finished! Again, we need to consider what could happen from states 1 and 4 if we processed the "empty string" transition rules. Since (1, e) → 2 and (2, e) → 3, we could end up in state 1, 2, 3 or 4. Let's create a new state called 1234 for our DFA and finish off that rule:
(123x, a) → 1234
Now do the same analysis for (123x, b) and we'll see that in the original automaton, the only possible resulting states starting in 1, 2 or 3, and processing a b are 3 and 4. Add another new state:
(123x, b) → xx34
Are we done? No. We've added two more states, 1234 and xx34, so we need to figure out the state transitions for them too, which we do by the same process. We discover that
(1234, a) → 1234
(1234, b) → xx34
(xx34, a) → ?
(xx34, b) → xxx4
Uh oh. There are no state transitions in the original NFA that start in states 3 or 4 and take an a. Remember that in that case we assume that the NFA goes into a special "reject" state. Let's call the reject state xxxx.
We now have two more new states to work out the transitions for. If we're in the reject state, we stay in the reject state, and we see from the original NFA that there are no transitions out of state 4, so we can round out our list with:
(xx34, a) → xxxx
(xxxx, a) → xxxx
(xxxx, b) → xxxx
(xxx4, a) → xxxx
(xxx4, b) → xxxx
Since 4 was the acceptable state in the original NFA, any "superstate" that contains 4 must be an acceptable state. So our new DFA is a machine M2 with alphabet {a}, b}, states {1234, 123x, xx34, xxx4, xxxx}, start state is 123x, and the set of acceptable states is {1234, xx34, xxx4}. The rules are
(1234, a) → 1234
(1234, b) → xx34
(123x, a) → 1234
(123x, b) → xx34
(xx34, a) → xxxx
(xx34, b) → xxx4
(xxx4, a) → xxxx
(xxx4, b) → xxxx
(xxxx, a) → xxxx
(xxxx, b) → xxxx
This DFA accepts the language a*( bbba ), so we've found an equivalent DFA to our NFA. We've turned an NFA with four states and five rules into an equivalent DFA with five states and ten rules. We did well -- you can find NFAs that require 2n new states, where n is the number of NFA states. An NFA with a thousand states could require 21000 states to represent as a DFA! 21000 is a finite, albeit rather large number. But it's not that big. Clearly we could build a 21000 state machine with only 1000 bits to store the state information. It's the state transition rules that are the pain to work out!
Using the techniques from this and the previous entry we can turn any NFA into an equivalent DFA, so NFAs are really not magic at all. They're just a convenient way to talk about DFAs.
Now, obviously what I've done here isn't a proof; one example does not a proof make. But the proof is both tedious and complicated, so I'm going to skip it. The key result here is that we can stop saying "deterministic finite automaton" and "nondeterministic finite automaton" and just start saying "finite automaton", because they're essentially the same thing.
Now that we know that we can use NFAs with impunity, we can try to answer questions such as:
Is every regular language recognizable by a finite automaton?
Are there any FAs that recognize irregular languages?
Tune in next time to find out!
• You should create a category for Regular Expressions!
• When it comes to validating input regular expression becomes a very important part of your security plan.&nbsp;...
• Woo! Theory of computation! (Fresh in my mind since I just graduated within the year. :D)
Looking forward to your posts so that I may brush up on Finite Automata. Will you be covering GFNAs and the Pumping Lemma or content only related to programming concepts (Regular Expressions and the like)?
I think it's time I dig out my math text again... :(
To think I liked this math course.
-- Chris
• Hey, what happened to the rest of this series? You can't just leave us hanging!
• Indeed, I would love to read more if you have the time to write it.
Page 1 of 1 (7 items) | 1,931 | 7,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2015-48 | latest | en | 0.967924 |
http://www-cs.engr.ccny.cuny.edu/~zhu/CSCI6716-2009/Homework/Fall2009-Assignment4.html | 1,638,591,457,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00529.warc.gz | 77,984,193 | 2,925 | Computer Science - The City College of New York
CSC I6716 - Fall 2009 3D Computer Vision
Assignment 4.
Stereo and Motion ( Deadline: Nov 30 before class)
(Those marked with * are optional for extra credits)
1. (Stereo- 15 points ) Estimate the accuracy of the simple stereo system (Figure 7.4 in Trucco & Verri’s book) assuming that the only source of noise is the localization of corresponding points in the two images. Discuss the dependence of the error in depth estimation as a function of the baseline width and the focal length.
Hint: Take the partial derivatives of Z with respect to x, T, f respectively.
2. (Motion- 20 points) Could you obtain 3D information of a scene by viewing the scene by a camera rotating around its optical center? Show why or why not. What about moving the camera along its optical axis?
3. (Motion- 15 points) Show that the aperture problem can be solved if a corner is visible through the aperture.
4. (Stereo Programming - 50 basic points + 20 extra points ) Use the image pair ( Image 1, Image 2) for the following exercises.
(1). Fundamental Matrix. - Design and implement a program that , given a stereo pair, determines at least eight point matches, then recovers the fundamental matrix (15 points ) and the location of the epipoles (5 points). Check the accuracy of the result by measuring the distance between the estimated epipolar lines and image points not used by the matrix estimation (5 points). Also, overlay the epipolar lines of control points and test points on one of the images (say Image 1- I already did this in the starting code below). Control points are the correspondences (matches) used in computing the fundamental matrix, and test points are those used to check the accuracy of the computation.
Hint: As a first step, you can pick up the matches of both the control points and the test points manually. You may use my matlab code (FmatGUI.m) as a starting point - where I provided an interface to pick up point matches by mouse clicks. The epipolar lines should be (almost) parallel in this stereo pair. If not, something is wrong either with your code or the point matches. Make sure this is achieved before you move to the second step* - that is to try to search for point matches automatically by your program. However the second step is optional (for extra 5 points)
(2). Feature-based matching. - Design a stereo vision system to do "feature-based matching" and explain your algorithm in writing (10). The system should have a user interface that allows a user to select a point on the first image, say by a mouse click (5 points). The system should then find and highlight the corresponding point on the second image, say using a cross hair (5 points). Try to use the epipolar geometry derived from (1) in searching correspondences along epipolar lines (5 points).
Hint : You may use a similar interface as I did for question (1). You may use the point match searching algorithm in (1) (if you have done so), but this time you need to constrain your search windows along the epipolar lines.
(3) *Discussions. Show your results on points with different properties like those in corners, edges, smooth regions, textured regions, and occluded regions that are visible only in one of the images (5 points). Discuss for each case, why your vision system succeeds or fails in finding the correct matches (5 points). Compare the performance of your system against a human user (e.g. yourself) who marks the corresponding matches on the second image by a mouse click (5 points). | 801 | 3,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-49 | latest | en | 0.848073 |
https://stats.stackexchange.com/questions/linked/66448?sort=hot&pagesize=30 | 1,596,952,343,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738425.43/warc/CC-MAIN-20200809043422-20200809073422-00045.warc.gz | 513,039,396 | 32,043 | 428 views
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571 views | 1,256 | 6,042 | {"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} | 2.78125 | 3 | CC-MAIN-2020-34 | latest | en | 0.927079 |
https://www.jiskha.com/questions/1461662/a-block-slides-down-a-frictionless-plane-having-an-inclination-of-13-2-the-block | 1,603,822,697,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894426.63/warc/CC-MAIN-20201027170516-20201027200516-00493.warc.gz | 756,723,328 | 5,268 | Physics
A block slides down a frictionless plane having an inclination of θ = 13.2°. The block starts from rest at the top, and the length of the incline is 1.70 m.
-> Find the acceleration of the block.
(m/s2 down the incline)
-> Find its speed when it reaches the bottom of the incline.
m/s)
1. 👍 0
2. 👎 0
3. 👁 1,293
1. weight ccomponent down slope
F = m g sin 13.2
so
m g sin 13.2 = m a
so
a = 9.81 sin 13.2
( what else is new?)
d = (1/2) a t^2
1.7 = (1/2) a t^2
solve for t
then
v = a t
1. 👍 0
2. 👎 1
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http://mathhelpforum.com/calculus/88340-yet-another-series-question.html | 1,481,144,906,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542246.21/warc/CC-MAIN-20161202170902-00218-ip-10-31-129-80.ec2.internal.warc.gz | 186,668,975 | 13,252 | # Thread: Yet another series question
1. ## Yet another series question
Determine whether the following diverges or converges:
$\sum_{k=1}^{\infty}\frac{k!}{k^{k}}$
The ratio test is inconclusive, and it seems like I'll need to use some form of the comparison test, but I'm stuck on where to begin.
2. Originally Posted by Pinkk
Determine whether the following diverges or converges:
$\sum_{k=1}^{\infty}\frac{k!}{k^{k}}$
The ratio test is inconclusive, and it seems like I'll need to use some of the comparison test, but I'm stuck one where to begin.
Note that for $k > 2$ that
$k! < k^k$
check the case k =4
$1\cdot 2\cdot 3 \cdot 4 < 4\cdot 4 \cdot 4 \cdot 4$
3. Originally Posted by Pinkk
Determine whether the following diverges or converges:
$\sum_{k=1}^{\infty}\frac{k!}{k^{k}}$
The ratio test is inconclusive, and it seems like I'll need to use some form of the comparison test, but I'm stuck on where to begin.
actually the ratio test works very nicely. you'll get $\lim_{k\to\infty} \frac{a_{k+1}}{a_k} = \frac{1}{e} < 1,$ and so the series is convergent.
4. For the ratio test I got:
$\lim_{k\to \infty}\frac{(k+1)k!}{k\cdot k^{k}}\cdot \frac{k^{k}}{k!}=\lim_{k\to \infty} \frac{k+1}{k}=1$
And I understand that $k! \le k^{k}$, but I'm not sure how to apply that using a comparison; I always end up setting up a comparison where the larger series is divergent, which doesn't help.
5. Originally Posted by Pinkk
Determine whether the following diverges or converges:
$\sum_{k=1}^{\infty}\frac{k!}{k^{k}}$
The ratio test is inconclusive, and it seems like I'll need to use some form of the comparison test, but I'm stuck on where to begin.
Use NoncomAlg observation
$\lim_{k \to \infty} \bigg|\frac{(k+1)!}{(k+1)^{k+1}}\cdot \frac{k^k}{k!} \bigg|$
$\lim_{k \to \infty} \bigg|\frac{(k+1)!}{(k+1)^{k}(k+1)}\cdot \frac{k^k}{k!} \bigg|$
$\lim_{k \to \infty} \bigg| \frac{k^k}{(k+1)^k}\bigg|$
$\lim_{k \to \infty} \bigg| \left( \frac{k}{k+1}\right)^k\bigg|=\lim_{k \to \infty} \bigg| \left( \frac{1}{1+\frac{1}{k}}\right)^k\bigg|=\frac{1}{e}$
6. Bah, I'm an idiot; forgot to change the base of $k^{k}$. Thanks.
7. here's a related question: what is $\lim_{k\to\infty} \frac{k}{\sqrt[k]{k!}}$ ?
8. $\sum_{k=1}^{\infty}\frac{k^{k}}{k!}$
$\lim_{k\to \infty}\frac{a_{k+1}}{a_{k}}=\lim_{k\to \infty}\frac{(k+1)^{k+1}}{(k+1)k!}\cdot \frac{k!}{k^{k}}=\lim_{k\to \infty}(\frac{k+1}{k})^{k} = e$
And since $e > 1$, by the ratio test, the series diverges.
I guess you can see that it diverges that since $k^{k}\ge k!$, the sequence is increasing, which means $\lim_{k\to \infty} a_{k} \ne 0$, and the series must therefore diverge.
9. Originally Posted by Pinkk
$\sum_{k=1}^{\infty}\frac{k^{k}}{k!}$
$\lim_{k\to \infty}\frac{a_{k+1}}{a_{k}}=\lim_{k\to \infty}\frac{(k+1)^{k+1}}{(k+1)k!}\cdot \frac{k!}{k^{k}}=\lim_{k\to \infty}(\frac{k+1}{k})^{k} = e$
And since $e > 1$, by the ratio test, the series diverges.
I guess you can see that it diverges that since $k^{k}\ge k!$, the sequence is increasing, which means $\lim_{k\to \infty} a_{k} \ne 0$, and the series must therefore diverge.
my question was to find $\lim_{k\to\infty} \frac{k}{\sqrt[k]{k!}}.$ the point here's that $\lim_{k\to\infty} \left| \frac{a_{k+1}}{a_k} \right|=\lim_{k\to\infty} \sqrt[k]{|a_k|},$ if one of the limits exists of course. that's why the ratio test is inconclusive if and only if the root test is
inconclusive. now let $a_k=\frac{k^k}{k!}.$ you showed that $\lim_{k\to\infty} \frac{a_{k+1}}{a_k}=e.$ thus we must also have $e=\lim_{k\to\infty} \sqrt[k]{a_k}=\lim_{k\to\infty} \frac{k}{\sqrt[k]{k!}}.$
10. You seem to like series. Try this one
$
\sum_{k=1}^{\infty} 2^{n} -1
$
11. My mistake for misreading your question. Also, for some reason our Calculus 3 class omits teaching the root test, so I'll have to read up on that on my own when time permits. Right now, I have to focus on the material that's gonna be on my upcoming exam.
And I'll try that one another time, my brain is dead for the day.
12. Give it a try sometime.
It's a sneaky series that find it's way into my first test on infinite series last semester. It took 15 minutes for me to figure it out.
13. Well, I'm pretty sure it'll diverge because if I recall correctly from class, since you can express that series as:
$\sum_{k=1}^{\infty}2^{k} -\sum_{k=1}^{\infty}1$
It is pretty obvious $\sum_{k=1}^{\infty}1$ will diverge, so the whole series diverges. Correct me if I'm wrong though.
14. It is necessary to use ratio test if the given sequence is a decreasing sequence so i d wish to know how to prove k!/k^k is decreasing everywhere ......
15. Originally Posted by Pinkk
Well, I'm pretty sure it'll diverge because if I recall correctly from class, since you can express that series as:
$\sum_{k=1}^{\infty}2^{k} -\sum_{k=1}^{\infty}1$
It is pretty obvious $\sum_{k=1}^{\infty}1$ will diverge, so the whole series diverges. Correct me if I'm wrong though.
I made a mistake with the question.
it is supposed to be
$\sum_{k=1}^{\infty}2^{\frac{1}{k}} -1$
Other than that it would be trivial | 1,793 | 5,088 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 37, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2016-50 | longest | en | 0.799165 |
http://openstudy.com/updates/511d881ce4b06821731bac9c | 1,448,832,308,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398459875.44/warc/CC-MAIN-20151124205419-00153-ip-10-71-132-137.ec2.internal.warc.gz | 162,645,352 | 13,557 | ## love_jessika15 2 years ago Given: line BD is a diameter m angle 1 = 100° m semi circle BC = 30° m semi circle AC= a. 60 b. 110 c. 200
1. love_jessika15
2. hartnn
have you tried this ? which step are you stuck ?
3. love_jessika15
All of the steps haha
4. hartnn
do you know that measure of central angle equals measure of arc. |dw:1360890234039:dw|
5. love_jessika15
Yes I knew that.
6. hartnn
now , measure of semicircle =180 you knew this ? |dw:1360890434105:dw|
7. love_jessika15
I know that I also know that a whole circle equals 360 degrees.
8. love_jessika15
I just need to know how to find out AC
9. hartnn
okkk, AC = AB+BC = angle 2+ angle 3, right ?
10. love_jessika15
Yupp.
11. love_jessika15
The answer is 110! Thank youu (:
12. hartnn
and since BC =30, you know angle 3=30 ... oh, you got it ! good :) | 284 | 838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2015-48 | longest | en | 0.802411 |
https://www.exactlywhatistime.com/days-before-date/december-4/8-days | 1,695,422,651,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506423.70/warc/CC-MAIN-20230922202444-20230922232444-00882.warc.gz | 845,232,442 | 5,645 | # What date is 8 days before Monday December 04, 2023?
## Calculating 8 days before Monday December 04, 2023 by hand
This page helps you figure out the date that is 8 days before Monday December 04, 2023. We've made a calculator to find the date before a certain number of days before a specific date. In this example, we want to know the date 8 days before Monday December 04, 2023.
Trying to do this in your head can be really hard and take a long time. An easier way is to use a calendar, either a paper one or an app on your phone or computer, to look at the days before the date you're interested in. But the best and quickest way to find the answer is by using our days before specific date calculator, which you can find here.
If you want to change the question on this page, you have two choices: you can change the URL in your browser's address bar, or go to our days before specific date calculator to type in a new question. Remember, figuring out these types of calculations in your head can be really tough, so we made this calculator to help make it much easier for you.
## Sunday November 26, 2023 Stats
• Day of the week: Sunday
• Month: November
• Day of the year: 330
## Counting 8 days backward from Monday December 04, 2023
Counting backward from today, Sunday November 26, 2023 is 8 before now using our current calendar. 8 days is equivalent to:
8 days is also 192 hours. Sunday November 26, 2023 is 90% of the year completed.
## Within 8 days there are 192 hours, 11520 minutes, or 691200 seconds
Sunday Sunday November 26, 2023 is day number 330 of the year. At that time, we will be 90% through 2023.
## In 8 days, the Average Person Spent...
• 1718.4 hours Sleeping
• 228.48 hours Eating and drinking
• 374.4 hours Household activities
• 111.36 hours Housework
• 122.88 hours Food preparation and cleanup
• 38.4 hours Lawn and garden care
• 672.0 hours Working and work-related activities
• 618.24 hours Working
• 1011.84 hours Leisure and sports
• 549.12 hours Watching television
## Famous Sporting and Music Events on November 26
• 1942 "Casablanca" directed by Michael Curtiz and starring Humphrey Bogart and Ingrid Bergman premieres at Hollywood Theater, NYC (Academy Awards Best Picture 1943)
• 1917 NHL forms with Montreal Canadiens, Montreal Maroons, Toronto Arenas, Ottawa Senators & Quebec Bulldogs; National Hockey Association disbands | 610 | 2,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-40 | latest | en | 0.925084 |
http://www.chegg.com/homework-help/questions-and-answers/92-mapping-z-plane-s-planeconsider-inverse-relation-given-z-e-s-ts-map-z-plane-s-planea-ex-q1887705 | 1,387,699,913,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345777253/warc/CC-MAIN-20131218054937-00075-ip-10-33-133-15.ec2.internal.warc.gz | 329,433,679 | 7,469 | # 9.2- Mapping of the z-plane into the s-plane
0 pts pending
9.2- Mapping of the z-plane into the s-plane
Consider the inverse relation given by z=e^(s*Ts)- That is, how to map into the z-plane into the s-plane
a) Find an expression for the s in terms of z from the relation z=e^(s*Ts) | 85 | 286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.824434 |
https://oeis.org/A161620 | 1,586,141,964,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371612531.68/warc/CC-MAIN-20200406004220-20200406034720-00208.warc.gz | 603,740,357 | 4,655 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A161620 Primorial numbers of the form n^2 + n for some integer n. 2
2, 6, 30, 210, 510510 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Primorial numbers m such that 4m+1 is a square. Intersection of the sequences A002110 and A002378. If it exists, a(6) > A034386(10^11). - Max Alekseyev, Oct 23 2011 The form is n^2 + n = n(n + 1), and the values n + 1 constitute A215659. - Jeppe Stig Nielsen, Mar 27 2018 LINKS C. Nelson, D. E. Penney, and C. Pomerance, 714 and 715, J. Recreational Mathematics (1974) 7(2), 87-89. [Warning: As of March 2018 this site appears to have been hacked. Proceed with great caution. The original content should be retrieved from the Wayback machine and added here. - N. J. A. Sloane, Mar 29 2018] FORMULA a(n) = A034386(A215658(n)). - Jeppe Stig Nielsen, Mar 27 2018 EXAMPLE 2 = 1*2 = 2 2*3 = 2*3 = 6 2*3*5 = 5*6 = 30 2*3*5*7 = 14*15 = 210 2*3*5*7*11*13*17 = 714*715 = 510510 MATHEMATICA p=1; Do[p=p*Prime[c]; f=Floor[Sqrt[p]]; If[p==f*(f+1), Print[p]], {c, 1000}] PROG (PARI) N=10^8; si=30; q=vector(si, i, nextprime(i*N)); a=vector(si, i, 1); forprime(p=2, N, for(i=1, si, a[i]=(a[i]*p)%q[i]); v=1; for(i=1, si, if(kronecker(4*a[i]+1, q[i])==-1, v=0; break)); if(v, T=1; forprime(r=2, p, T*=r); print1(T", "))) (PARI) pr=1; forprime(p=2, , pr*=p; s=sqrtint(pr); s*(s+1)==pr&&print1(pr, ", ")) \\ Jeppe Stig Nielsen, Mar 27 2018 CROSSREFS Cf. A002110, A002378, A215658, A215659. Sequence in context: A294925 A091456 A293756 * A333508 A205569 A108204 Adjacent sequences: A161617 A161618 A161619 * A161621 A161622 A161623 KEYWORD nonn,hard,more AUTHOR Daniel Tisdale, Jun 14 2009 EXTENSIONS Edited by Hans Havermann, Dec 02 2010 Edited by Max Alekseyev, Dec 03 2010 Edited by Robert Gerbicz, Dec 04 2010 STATUS approved
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Last modified April 5 22:58 EDT 2020. Contains 333260 sequences. (Running on oeis4.) | 837 | 2,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-16 | latest | en | 0.609707 |
http://aplusphysics.com/wordpress/regents/tag/regents-exam/ | 1,369,124,112,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699798457/warc/CC-MAIN-20130516102318-00082-ip-10-60-113-184.ec2.internal.warc.gz | 10,537,218 | 34,140 | # Posts tagged Regents Exam
## Castle Learning Review Assignments
0
As we prepare for our comprehensive Regents examination in June, it is important to make sure we are up to speed on material covered throughout the entire year. Toward that end, we will undertake a series of seven Castle Learning review assignments consisting of 50-60 multiple choice questions on each of the major topics we have covered this year. These topics correspond to the short review podcasts available on iTunes and on our course videos page. I would recommend viewing the appropriate review lessons before tackling the Castle Learning assignments. Then, take the Castle Learning assignments with your reference table, the calculator you will use on the Regents exam, and your notebook handy.
Each assignment is worth 50 to 60 points, with second chance correct scores counted for full credit! These are weighty assignments, with distinct opening and closing dates. Because these are being provided well in advance of due dates, you should have opportunity to plan your time accordingly. No credit will be given for late assignments or submissions, regardless of attendance or illness issues. Assignments and Review Schedule is as follows:
Please take these assignments seriously, and be diligent in your planning and submissions. This is a large portion of our fourth quarter grading, and is an excellent opportunity to put yourself in position for achieving an optimal score on the Regents Physics Exam!
## 10 Quick Tips to Maximize your Regents Physics Score
0
Although by no means an exhaustive list, these 10 quick tips may help you secure that extra point or two on your upcoming Regents Physics exam.
1. Mass and inertia are the same thing.
2. To find the resultant, line your vectors up tip-to-tail, and draw a line from the starting point of the first vector to the ending point of the last vector.
3. Any object moving in a circular path is accelerating toward the center of the circle.
4. Acceleration of an object is equal to the net force on the object divided by the object’s mass.
5. The normal force always points at an angle of 90° from the surface.
6. Opposite charges and magnetic poles attract, likes repel.
7. Gravitational forces and electrostatic forces both follow an inverse square law relationship, where the strength of the force is related to one divided by the square of the distance between the charges/masses.
8. The force of gravity on an object, commonly referred to as weight, is equal to mg, where g is the gravitational field strength (also referred to as the acceleration due to gravity).
9. The mass-energy equivalence can be calculated using E=mc^2. If a mass is given in universal mass units, however, you can do a straight unit conversion using 1u = 931 MeV.
10. Protons and neutrons fall into the category of baryons, which are hadrons. Smaller particles, such as electrons, fall into the category of leptons. Mesons are rare, weird particles you probably haven’t heard of.
Most importantly, use your reference table. When in doubt, write down the information you’re asked to find, what you’re given, and use your reference table to help you narrow down what you should be doing. In the free response part of the test, make sure to show your work in detail with a formula, substitution with units, and an answer with units.
Find these and many more tips for success at APlusPhysics.com.
## 87 “Phacts” for the Physics Regents Exam
0
(As adapted from several sources, beginning with Jim Davidson, Physics Teacher)
I. Mechanics
1. Mass and inertia are the same thing. (Mass actually measures inertia – in kilograms… Much as monetary resources measures financial wealth – in dollars.)
2. Weight (force of gravity) decreases as you move away from the earth by distance squared. (It decreases, but only approaches zero, never reaching it, even far beyond the solar system.)
3. Weight (in newtons) is mass * acceleration (w = mg). Mass is not Weight! Mass is a scalar and measured in kilograms, weight is a force and a vector and measured in Newtons.
4. Velocity can only be constant when the net force (and acceleration) is zero. (The velocity can be zero and not constant – for example when a ball, thrown vertically, is at the top of its trajectory.)
5. Velocity, displacement [s], momentum, force (weight), torque, and acceleration are vectors.
6. Speed, distance [d], time, length, mass, temperature, charge, power and energy (joules) are scalar quantities.
7. The slope of the distance-time graph is velocity.
8. The slope of the velocity-time graph is acceleration.
9. The area under a velocity-time graph is distance.
10. Magnitude is a term used to state how large a vector quantity is.
11. At zero (0) degrees two vectors have a resultant equal to their sum. At 180 degrees two vectors have a resultant equal to their difference. From the minimum value (at 180) to the maximum value (at zero) is the total range of all the possible resultants of any two vectors.
12. An unbalanced force must produce an acceleration and the object cannot be in equilibrium.
13. If an object is not accelerating, it is in equilibrium and no unbalanced forces are acting.
14. The equilibrant force is equal in magnitude but opposite in direction to the resultant vector.
15. Momentum is conserved in all collision systems. Energy is conserved (in the KE of the objects) only if a collision is perfectly elastic.
II. Energy
16. Mechanical energy is the sum of the potential and kinetic energy.
17. UNITS: a = [m/sec2]; F = [kg•m/sec2] = Newton; work = PE = KE = [kg•m2/sec2] = Joule; Power = [kg•m2/sec3] = [Joules/sec] = Watt
18. 1ev is a very small energy unit equal to 1.6 x 10-19 joules – used for small objects like electrons. This is on the Reference Table.
19. Gravitational potential energy increases as height increases.
20. Kinetic energy changes only if mass or velocity changes.
21. Mechanical energy (PE + KE) does not change for a free falling mass or a swinging pendulum. (when ignoring air friction)
III. Electricity and Magnetism
22. A coulomb is charge, an amp is current [coulomb/sec] and a volt is potential difference [joule/coulomb].
23. Short, fat, cold wires make the best conductors.
24. Electrons and protons have equal amounts of charge (1.6 x 10-19 coulombs each – known as one elementary charge). This is on the Reference Chart.
25. Adding a resistor in series increases the total resistance of a circuit.
26. Adding a resistor in parallel decreases the total resistance of a circuit.
27. All resistors in series have equal current (I).
28. All resistors in parallel have equal voltage (V).
29. If two similar charged spheres touch each other add the charges and divide by two to find the final charge on each sphere after they are separated.
30. Insulators contain no electrons free to move.
31. Ionized gases conduct electric current using positive ions, negative ions and electrons.
32. Electric fields all point in the direction of the force on a positive test charge.
33. Electric fields between two parallel plates are uniform in strength except at the edges.
34. Millikan determined the charge on a single electron using his famous oil-drop experiment.
35. All charge changes result from the movement of electrons not protons. (an object becomes positive by losing electrons)
36. The direction of a magnetic field is defined by the direction a compass needle points. (The direction an isolated north pole would feel.)
37. Magnetic fields point from the north to the south outside the magnet and south to north inside the magnet.
38. Magnetic flux is measured in webers.
39. Left hands are for negative charges and reverse answer for positive charges.
40. The first hand rule deals with the B-field around a current bearing wire, the second hand rule deals with the magnetic field from a wire around a solenoid, and the third hand rule looks at the force on charges moving in a B-field.
41. Solenoids are stronger with more current or more wire turns or adding a soft iron core.
IV. Wave Phenomena
42. Sound waves are longitudinal and mechanical.
43. Light slows down, bends toward the normal and has a shorter wavelength when it enters a medium with a higher index of refraction (n).
44. All angles in wave theory problems are measured to the normal.
45. Blue light has more energy, a shorter wavelength and a higher frequency than red light (remember- ROYGBIV).
46. The electromagnetic spectrum are listed highest energy (on left) to lowest (on right). They are all electromagnetic waves and travel at the speed of light (c = f *l ).
47. The speed (c) of all types of electromagnetic waves is 3.0 x 108 m/sec in a vacuum.
48. As the frequency of an electromagnetic wave increases its energy increases (E = h * f) and its wavelength decreases and its velocity remains constant as long as it doesn’t enter a medium with a different refractive index (i.e. optical density).
49. A prism produces a rainbow from white light by dispersion. (red bends the least because it slows the least).
50. Transverse wave particles vibrate back and forth perpendicular to the direction of the wave’s velocity. Longitudinal wave particles vibrate back and forth parallel to the direction of the wave’s velocity.
51. Light wave are transverse (they, and all (and only)transverse waves can be polarized).
52. The amplitude of a non-electromagnetic wave (i.e. water, string and sound waves) determines its energy. The frequency determines the pitch of a sound wave. Their wavelength is a function of its frequency and speed (v = f * l ). Their speed depends on the medium they are traveling in.
53. Constructive interference occurs when two waves are zero (0) degrees out of phase or a whole number of wavelengths (360 degrees.) out of phase.
54. At the critical angle a wave will be refracted to 90 degrees. At angles larger than the critical angle, light is reflected not refracted.
55. Doppler effect: when a wave source moves toward you, you will perceive waves with a shorter wavelength and higher frequency than the waves emitted by the source. When a wave source moves away from you, you will perceive waves with a longer wavelength and lower frequency.
56. Double slit diffraction works because of diffraction and interference.
57. Single slit diffraction produces a much wider central maximum than double slit.
58. Diffuse reflection occurs from dull surfaces while regular (spectacular) reflection occurs from smooth (mirror-like) surfaces.
59. Only waves show diffraction, interference and the polarization.
60. The period of a wave is the inverse of its frequency (T = 1/f ). So waves with higher frequencies have shorter periods.
61. Monochromatic light has one frequency.
62. Coherent light waves are all in phase.
V. Modern Physics
63. In order to explain the photoelectric effect, Einstein proposed particle behavior for light (and all electromagnetic waves) with E = h* f and KEmax = hf – Wo, where W is the work function.
64. A photon is a particle of light (wave packet).
65. To preserve the symmetry of the universe, DeBroglie proposed wave behavior for particles ( l = h/mv). Therefore large fast moving objects (baseballs, rockets) have very short wavelengths (that are unobservable) but very small objects, particularly when moving slowly have wavelengths that can be detected in the behavior of the objects.
66. Whenever charged particles are accelerated, electromagnetic waves are produced.
67. The lowest energy state of a atom is called the ground state.
68. Increasing light frequency increases the kinetic energy of the emitted photo-electrons in the photo-electric effect (KEmax = hf – Wo).
69. As the threshold frequency increases for a photo-cell (photo emissive material) the work function also increases (Wo = h fo)
70. Increasing light intensity increases the number of emitted photo-electrons in the photo-electric effect but not their KE (i.e. more intensity>more photons>more electrons emitted). This is the particle nature shown by light.
VI. Motion in a plane
71. Key to understanding trajectories is to separate the motion into two independent components in different dimensions – normally horizontal and vertical. Usually the velocity in the horizontal dimension is constant (not accelerated) and the motion in the vertical dimension is changing (usually with acceleration of g).
72. Centripetal force and centripetal acceleration vectors are toward the center of the circle- while the velocity vector is tangent to the circle. (Centripetal means towards the center!)
73. An object in orbit is not weightless – it is its weight that keeps it moving in a circle around the astronomical mass it is orbiting. In other words, its weight is the centripetal force keeping it moving in a circle.
74. An object in orbit is in free fall – it is falling freely in response to its own weight. Any object inside a freely falling object will appear to be weightless.
75. Rutherford discovered the positive nucleus using his famous gold-foil experiment.
76. Fusion is the process in which hydrogen is combined to make helium.
77. Fission requires that a neutron causes uranium to be split into middle size atoms and produce extra neutrons, which, in turn, can go on and cause more fissions.
78. Radioactive half-lives are not effected by any changes in temperature or pressure (or anything else for that matter).
79. One AMU of mass is equal to 931 MeV of energy. (E = mc2). This is on the Reference Charts!
80. Nuclear forces are very strong and very short-ranged.
81. There are two basic types of elementary particles: Hadrons & Leptons (see Chart).
82. There are two types of Hadrons: Baryons and Mesons (see Chart).
83. The two types of Hadrons are different because they are made up of different numbers of quarks. Baryons are made up of 3 quarks, and Mesons of a quark and antiquark.
84. Notice that to make long-lived Hadron particles quarks must combine in such a way as to give the charge of particle formed a multiple of the elementary charge.
85. For every particle in the "Standard Model" there is an antiparticle. The major difference of an antipartcle is that its charge is opposite in sign. All antiparticles will anhililate as soon as they come in contact with matter and will release a great amount of energy.
85. Notice that to make long-lived Hadron particles quarks must combine in such a way as to give the charge of particle formed a multiple of the elementary charge.
86. Notice that the retention of the Energy Level Diagrams on the new charts implies that there will be questions on it. The units (eV) can be converted to Joules with the coversion given on the first Chart of the Regents Reference tables. And can be used with the formula (given under Modern Physics formulas) to calculate the energy absorbed or released when the electron changes levels.
And by using another formula (given under Modern Physics formulas) you can calculate the frequency of electromagnetic radiation absorbed or released. AND using the Electro-magnetic spectrom given on the charts you can find out what kind of electromagnetic radiation it is (infrared, visible light, UV light, etc.)
87. Physics is phun!! (This is key. Honest!)
## Castle Learning Review Assignments
0
As we prepare for our comprehensive Regents examination in June, it is important to make sure we are up to speed on material covered throughout the entire year. Toward that end, we will undertake a series of seven Castle Learning review assignments consisting of 50-60 multiple choice questions on each of the major topics we have covered this year. These topics correspond to the short review podcasts available on iTunes. I would recommend viewing the appropriate review lessons before tackling the Castle Learning assignments. Then, take the Castle Learning assignments with your reference table, the calculator you will use on the Regents exam, and your notebook handy.
Each assignment is worth 50 to 60 points, with second chance correct scores counted for full credit! These are weighty assignments, with distinct opening and closing dates. Because these are being provided well in advance of due dates, you should have opportunity to plan your time accordingly. No credit will be given for late assignments or submissions, regardless of attendance or illness issues.
Assignments and Review Schedule is as follows:
Assignment Podcasts Open Close Units Vectors Scalars R01 4/29/2010 5/6/2010 Kinematics R02,R03 5/6/2010 5/13/2010 Dynamics R04,R05 5/13/2010 5/20/2010 Momentum and WEP R06,R07,R08 5/20/2010 5/27/2010 Electricity and Magnetism R09-R12 5/27/2010 6/3/2010 Waves and Optics R13,R14 6/3/2010 6/8/2010 Modern Physics R15 6/8/2010 6/14/2010
Please take these assignments seriously, and be diligent in your planning and submissions. This is a large portion of our fourth quarter grading, and is an excellent opportunity to put yourself in position for achieving an optimal score on the Regents Physics Exam!
## Mid-Term Exam Results 2010-2011
0
Mid-Term results have been compiled and published. In general, scores are right in line with performance within the last few years. The exam itself was styled after the Regents Exam which is given at the end of the school year, and included topics from all aspects of Newtonian Mechanics (roughly half of the Regents Exam). This included 46 multiple choice questions, and 25 open-ended, or “free response,” type questions. Students were given 180 minutes to complete the exam, as well as a formula sheet (Regents Physics Reference Table), and a ruler. Students were permitted the use of a scientific calculator.
Summary Data
Summary data from this year’s exam, along with the previous two years of data, are shown at right. Average score on the exam, with a fairly generous curve, was 76%, compared to 73% in 2010 and 74% in 2009. Median score was 78%, two points higher than last year’s exam, and right in line with 2009 scores. Percentage of students scoring above 85%, considered “Mastery Level” by Regents Physics standards, was 25%, slightly lower than in 2010 and 2009, but passing percentage was 83%, several points higher than in past years. Even more promising is a glance at a histogram breaking down student scores further. This data indicates that we have many students in the 80-85% range. Past history has shown me that it is quite possible for students to improve their scores from 10-15% points between now and the actual Regents Exam in June with focused effort, study, and motivation. This indicates the potential for outstanding standardized exam performance in June if students buckle down and focus their efforts. Unfortunately, the Physics Regents Exam is usually one of the last tests offered (sometimes even after official graduation), so obtaining and maintaining student focus at this point of the year can be a challenge.
Cluster Data
Breaking down exam performance by topic, we observe students demonstrating a very strong understanding of Newton’s Laws of Motion, Hooke’s Law (springs), and Work and Energy. These have been focus areas during the year, although there is room for further improvement in understanding Newton’s 2nd Law as well as momentum and impulse.
Cluster data indicates weaknesses in Newton’s Law of Universal Gravitation, although a deeper look at the questions and errors themselves indicates that most of the confusion here is a result of mathematical errors rather than conceptual misunderstandings. This is being re-emphasized as we work through Coulomb’s Law / Electrical Force problems, which utilize almost the exact same mathematical model. Other areas for improvement include 2-D Kinematics (projectile motion), a perennial challenge, and utilizing and understanding the metric system, a topic that has been emphasized and will continue to be reviewed throughout the year. Performance on momentum problems was, unsurprisingly, problematic. This relates directly to a “class as a whole” issue with independent practice and motivation throughout that unit, and will be our top priority for review in mid-June.
Conclusions
Student performance is in line with, if not slightly above, expectations for this point in the year. With a focused effort, students can fairly routinely increase their scores 10 to 15 percentage points by June. For struggling students, a variety of remediation protocols are being put in place, including the entire course content and sample quizzes available on the web, tailored directly to this course’s requirements (APlusPhysics.com), video lectures on topics corresponding directly to our textbook (Hippocampus), and even review sessions which can be viewed from any computer supporting iTunes (Windows, Mac, Linux, etc.) as well as iPods, iPhones, Blackberries, Android devices, etc. All of these resources are available both in the classroom as well as the school library for those who don’t have ready access to these resources outside of school.
Data has also been broken down to the same level for individual students, and will be utilized to develop both entire-class and individual review plans to allow students to focus on areas that will provide the biggest “bang for the buck” in their studies.
Go to Top | 4,610 | 21,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2013-20 | latest | en | 0.925672 |
https://www.physicsforums.com/threads/sequences-and-series.747572/ | 1,597,059,779,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738674.42/warc/CC-MAIN-20200810102345-20200810132345-00155.warc.gz | 768,161,308 | 18,949 | # Sequences and Series
## Homework Statement
So, I actually have a bunch of these problems and I cannot do any of them. I don't think I'm really understanding it. Here is the question: (one of them) The way I wrote them, a_n means a sub n
For each sequence a_n find a number k such that n^k a_n
has a finite non-zero limit.
(This is of use, because by the limit comparison test the series ∑from 1 to infinity (a_n) and ∑from 1 to infinity (n^-k) both converge or both diverge.)
The question:
a_n= (6+3n)^-7
What does k equal?
N/A
## The Attempt at a Solution
I'm actually 100% lost on this problem. I don't even know where to start :(
Here's my attempt. It may not make any sense because I don't understand this. But here it is:
(6+3n)^-7>0
1/(6+3n)^7>0
1>0
Ans: 1
But that is incorrect.
Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
## Homework Statement
So, I actually have a bunch of these problems and I cannot do any of them. I don't think I'm really understanding it. Here is the question: (one of them) The way I wrote them, a_n means a sub n
For each sequence a_n find a number k such that n^k a_n
has a finite non-zero limit.
(This is of use, because by the limit comparison test the series ∑from 1 to infinity (a_n) and ∑from 1 to infinity (n^-k) both converge or both diverge.)
The question:
a_n= (6+3n)^-7
What does k equal?
N/A
## The Attempt at a Solution
I'm actually 100% lost on this problem. I don't even know where to start :(
Here's my attempt. It may not make any sense because I don't understand this. But here it is:
(6+3n)^-7>0
1/(6+3n)^7>0
1>0
Ans: 1
But that is incorrect.
Here's a hint. (6+3n)=n(6/n+3). Try and do something with that.
Would you mind explaining the meaning? How would I find k from it?
Dick
Homework Helper
Would you mind explaining the meaning? How would I find k from it?
Ok, so (6+3n)^(-7)=(n(6/n+3))^(-7)=n^(-7)*(6/n+3)^(-7). What's the limit n->infinity of (6/n+3)^(-7)?
isn't it infinity?
Dick
Homework Helper
isn't it infinity?
I don't think so. Why do you? What's limit 6/n as n->infinity?
because isnt it the same as saying ((n+3)/6)^7 so as n approaches infinity, it approaches infinity?
but how would doing that help me find what k is?
Dick
Homework Helper
because isnt it the same as saying ((n+3)/6)^7 so as n approaches infinity, it approaches infinity?
No, it isn't. ((6/n)+3)^(-7)=1/((6/n)+3)^7. It's not 6/(n+3). It's (6/n)+3. They are very different.
and using this new limit (6/n)+3, how would it lead me to solve for k?
Dick
Homework Helper
and using this new limit (6/n)+3, how would it lead me to solve for k?
I think you should stop asking that until you do the algebra correctly and tell me what limit n->infinity ((6/n)+3) is. Then take a breath and think about it.
the limit is 3 because the 6/n goes to 0
Dick
Homework Helper
the limit is 3 because the 6/n goes to 0
Ok, so what's limit ((6/n)+3)^(-7). Have you thought about what that might mean for your question of what k is?
1/3^7= 1/2187.
I still dont know where k comes into play.
Is k 7?
Here's a hint. (6+3n)=n(6/n+3). Try and do something with that.
Where did you get the n(6/n+3) here? Why was it necessary to factor out the n?
Dick
Homework Helper
Is k 7?
Ok, so (6+3n)^(-7)=(n(6/n+3))^(-7)=n^(-7)*(6/n+3)^(-7). As n->infinity (6/n+3)^(-7)->1/3^7. Does it make sense to you that k=7 works??
Dick
Homework Helper
Where did you get the n(6/n+3) here? Why was it necessary to factor out the n?
Because you want the series to look like n^(-k) times something that approaches a nonzero constant. It's a good idea to factor out the n and see what's left.
yes it does. so just to recap, i had to factor out an n to create the n^k? what if n were in the demonimator instead?
Dick
Homework Helper
yes it does. so just to recap, i had to factor out an n to create the n^k? what if n were in the demonimator instead?
Then you would factor out a 1/n^(k)=n^(-k). It just changes the sign of k. Doesn't it?
i have a problem that says 7/(n^3+n) you dont have to solve this, but i just want to know, would i have to factor out n? and i would have 7n^-3?
Dick
Homework Helper
i have a problem that says 7/(n^3+n) you dont have to solve this, but i just want to know, would i have to factor out n? and i would have 7n^-3?
That will work. What's the limit of what's left after you factor that out?
is it 0?
Dick | 1,355 | 4,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 4 | CC-MAIN-2020-34 | latest | en | 0.963982 |
https://www.bearnaiserestaurant.com/review/whats-the-difference-between-mass-and-weight/ | 1,723,272,049,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00138.warc.gz | 509,095,195 | 11,210 | # Whats the difference between mass and weight?
## Whats the difference between mass and weight?
Mass is a measure of how much matter there is in an object, while weight is a measure of the size of the pull of gravity on the object. Mass is the amount of matter in an object. Mass is measured in kilograms (kg). Weight is a force due to the pull of gravity on an object.
## What is difference between mass and weight with example?
It refers to the measure of the amount of force that acts on mass because of the pull of gravity. Mass is the measure of inertia. Weigh is the measure of force. It stays the same everywhere, irrespective of location.
What is mass in simple words?
mass, in physics, quantitative measure of inertia, a fundamental property of all matter. It is, in effect, the resistance that a body of matter offers to a change in its speed or position upon the application of a force. The greater the mass of a body, the smaller the change produced by an applied force.
What is the 5 difference between mass and weight?
The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass. Mass is the measure of the amount of matter in a body. Mass is denoted using m or M.
### How do you teach preschoolers about weight?
Weigh the Objects Turn on kitchen scale, make sure it is tared and set to grams or ounces. Show your child how to weigh one of the boxes and read the scale. You might like to record the amount on a piece of paper or sticky note to keep track for sorting.
### Which statement summarizes the difference between mass and weight?
Which statement summarizes the difference between mass and weight? Mass is the amount of matter in an object, and weight is the downward pull on an object due to gravity.
What is the difference between mass and weight essay?
Mass is the quantity of matter in a body regardless of its volume or of any forces acting on it. Weight is a measurement of the gravitational force acting on an object. Mass is expressed in kilogram (kg), grams (g), and milligram (mg).
What is mass for kindergarten?
Mass is the amount of matter or substance that makes up an object. It is measured in units called kilograms, which can be abbreviated kg. It’s important to remember that mass is different from weight. Mass always stays the same, while weight changes with changes in gravity.
## What is the difference between weight and mass in science?
In scientific contexts, mass is the amount of “matter” in an object (though “matter” may be difficult to define), whereas weight is the force exerted on an object by gravity.
## What is the difference between mass and weight Wikipedia?
What is mass in simple language?
• August 17, 2022 | 589 | 2,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.968806 |
https://www.thestudentroom.co.uk/showthread.php?t=3504133 | 1,656,661,667,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103922377.50/warc/CC-MAIN-20220701064920-20220701094920-00077.warc.gz | 1,089,848,503 | 32,969 | # Edexcel Maths/Further Maths Grade Check
#1
Hello,
These are my results for the 12 Edexcel modules for Maths/Further Maths:
C1-93, C2-88, C3-93, C4-96
FP1-85, FP2-80
S1-79, S2-80
M1-72, M2-70
D1-82, D2-77
On my results sheet i was given an A* in Maths and B in Further Maths,
But i think i should be given an A in Further Maths, or i could be wrong.
This combination:
Maths:
C1,C2,C3.C4,M1,M2, Total=512 Ums
Further Maths:
FP1,FP2,S1,S2,D1,D2 Total=483 Ums
I may be wrong but 480 ums is an A and 90% overall in c3 and c4 gives you an A*.
So why was i not given this combination, rather S1 and S2 went to my maths grade and m1 and m2 went to my furthe rmaths grade.
Any help will be grateful.
Thank You
0
6 years ago
#2
(Original post by Uthman)
Hello,
These are my results for the 12 Edexcel modules for Maths/Further Maths:
C1-93, C2-88, C3-93, C4-96
FP1-85, FP2-80
S1-79, S2-80
M1-72, M2-70
D1-82, D2-77
On my results sheet i was given an A* in Maths and B in Further Maths,
But i think i should be given an A in Further Maths, or i could be wrong.
This combination:
Maths:
C1,C2,C3.C4,M1,M2, Total=512 Ums
Further Maths:
FP1,FP2,S1,S2,D1,D2 Total=483 Ums
I may be wrong but 480 ums is an A and 90% overall in c3 and c4 gives you an A*.
So why was i not given this combination, rather S1 and S2 went to my maths grade and m1 and m2 went to my furthe rmaths grade.
Any help will be grateful.
Thank You
sorry i cant help much just going to say you should speak to your school/college and definitely your maths teacher, try emailing them or calling up the school
0
6 years ago
#3
(Original post by Uthman)
Hello,
These are my results for the 12 Edexcel modules for Maths/Further Maths:
C1-93, C2-88, C3-93, C4-96
FP1-85, FP2-80
S1-79, S2-80
M1-72, M2-70
D1-82, D2-77
On my results sheet i was given an A* in Maths and B in Further Maths,
But i think i should be given an A in Further Maths, or i could be wrong.
This combination:
Maths:
C1,C2,C3.C4,M1,M2, Total=512 Ums
Further Maths:
FP1,FP2,S1,S2,D1,D2 Total=483 Ums
I may be wrong but 480 ums is an A and 90% overall in c3 and c4 gives you an A*.
So why was i not given this combination, rather S1 and S2 went to my maths grade and m1 and m2 went to my furthe rmaths grade.
Any help will be grateful.
Thank You
Are you down for pure maths as opposed to normal maths and further maths?
0
#4
(Original post by Slowbro93)
Are you down for pure maths as opposed to normal maths and further maths?
Hi,
On the results sheet it says: award 9371 - Mathematics, Why does that matter?
0
6 years ago
#5
You're not wrong, you should talk to your teachers, it might be that your AS results were cashed-in so they couldn't be rearranged.
0
6 years ago
#6
(Original post by Uthman)
Hi,
On the results sheet it says: award 9371 - Mathematics, Why does that matter?
Because what it looks like is that you've actually been given a different qualification.
There's two more A Level combos that you can do from your marks called Pure maths and Further Additional Maths. Should Edexcel have done that then you would have an A* in Pure maths and a B in Additional.
Given that hasn't happened, it looks like there has been a calculation error and that can be sorted by the exam board quite easily. I'd recommend seeing your exam officer with your maths teacher ASAP so this can be sorted
Hope this gets sorted quickly
1
#7
(Original post by Drakton15)
You're not wrong, you should talk to your teachers, it might be that your AS results were cashed-in so they couldn't be rearranged.
Hi,
I payed before my exams to get cashed in for both A level Maths and Further Maths, so don't all option units become available once i do that?
0
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https://file.scirp.org/Html/68116_68116.htm | 1,723,678,265,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00464.warc.gz | 201,355,727 | 17,598 | Fireman Numerical Solution of Some Existing Optimal Control Problems
Open Access Library Journal
Vol.02 No.03(2015), Article ID:68116,14 pages
10.4236/oalib.1101352
Fireman Numerical Solution of Some Existing Optimal Control Problems
J. O. Olademo, A. A. Ganiyu, M. F. Akimuyise
Mathematics Department, Adeyemi College of Education, Ondo, Nigeria
Received 18 February 2015; accepted 6 March 2015; published 10 March 2015
ABSTRACT
This study employs the Fireman Method in the solutions of optimal control problems of the form, under an admi- ssible control, which causes to follow admissible trajectory. The Hamiltonian Principle was employed for the analytical solutions of the given optimal control problems. It has been observed that this method converges close to the analytical solution for some class of problems.
Keywords:
Fireman Method, Optimal Control, Hamiltonian Principle, Linear Quadratic Regulator
Subject Areas: Mathematical Analysis, Mathematical Statistics, Operational Research
1. Introduction
A well known approach to the principle of optimization was first scribbled centuries ago on the walls of an ancient Roman bathhouse in connection with a choice between two aspirants for emperorship of Rome. Out of two evils, we always choose the lesser. In everyday life, decisions are made to accomplish certain tasks. Normally, there exist several possible ways or methods by which a certain task can be accomplished. Some of these methods may be more efficient or reliable than others and the presence of physical constraints implies that, not just any method can be used. It thus becomes necessary to consciously determine the “best” or “optimal” way to accomplish the task [1] .
Optimization is the act of obtaining the best policies to satisfy certain objectives while at the same time satisfying some fixed requirements or constraints. It involves the study of optimality criteria for problems, the determination of algorithmic methods of solution, the study of the structure of such methods and computer implementation of the methods under both trial conditions and real life problems [2] .
According to Wikipedia, optimization can be defined as a process of finding an alternative with the most costeffective or highest achievable performance under given constraints, by maximizing desired factors and minimizing undesired ones. In comparison, maximization means trying to attain the highest or maximum result or outcome without regard to cost or expense. The Practice of optimization is restricted by lack of full information and lack of time to evaluate available information. In computer simulation (modeling) of business problems, optimization is achieved usually by usinglinear programming techniques of operations research.
Optimization takes its root from the word optimizes, which is to make as perfect, effective, or functional as possible. In engineering, optimization is a collection of methods and techniques to design and make use of engineering systems as perfectly as possible with respect to specific parameters. In industrial engineering, one typical optimization problem is in inventory control. For this problem, we want to reduce the costs associated with item stocking and handling in a warehouse. In the simplest form of this problem, the parameters to be optimized are the quantity of inventory required to fill existing and anticipated orders, when that inventory has to be available and the physical capacity of the warehouse. Optimization requires the representation of the problem in a mathematical model where the decision variables are the parameters of the problem [3] .
Optimization is an act of finding an alternative with the most cost effective of highest achieveable perfor- mances under the given constraints by a maximizing desire factors and minimizing undesire ones. In comparison, maximization means trying to attain the highest of mininum result or outcome without regard to cost or expense. Practice of optimization is restricted by the lack of full information and the lack of time to evaluate what information is available. The essence of an optimization problem is like catching a black cat in a dark room in minimal time. (A constrained optimization problem corresponds to such a roomful of furniture). A light, even dim, is needed. Hence optimization methods explore assumptions about the character of response of the goal function to varying parameters and suggest the best way to change them. The variety of a priori assumptions corresponds to the variety of optimization methods [4] .
Optimization is a process, or methodology of making something (as a design, system, or decision) as fully perfect, functional, or effective as possible; specifically: the mathematical procedures (as finding the maximum of a function) involved in this [5] .
1.1. Optimal Control
In many areas of the empirical sciences such as Mathematics, Physics, Biology, and Chemistry, as well as in Economics, we study the time development of systems. Certain essential properties or characteristics of the system, called the state of the system, change over time. If there are n state variables we denote them by The rate of change of the state variables with respect to time is usually subject to an error. This is due to many factors, including the actual values of the and on certain parameters that can be controlled from the outside. These parameters are called control variables and are denoted by. We shall assume throughout this work that the laws governing the behavior of the system over time are given by systems of ordinary differential equations. The control variables can be freely chosen within certain bounds. If the system is in some state x0 at time, t0, and if we choose a certain control function u(t), then the system of differential equations will, usually, have a unique solution, x(t). If there are a priori bounds on the values of the state variables, only those control functions will be admissible which give rise to state functions x(t) satisfying the bounds. In general, there will be many admissible control functions, each giving rise to a specific evolution of the system. In an optimal control problem, an optimality criterion is given which assigns a certain number (a “utility”) to each evolution of the system. The problem is then to find an admissible control function which minimizes the optimality criterion in the class of all admissible control functions. The tools available for solving optimal control problems are analogous to those used in static optimization theory. Before examining a simple control problem, let us digress a little with some remarks concerning static optimization [6] .
1.2. Problem Formulation
The axiom “A problem well put is a problem half solved” may be a slight exaggeration, but its intent is nonetheless appropriate. In this paper, we shall discuss the important aspects of problem formulation. The formulation of an optimal control problem requires:
1. A mathematical description (model) of the process to be controlled.
2. A statement of the physical constraints.
3. Specification of a performance criterion.
A nontrivial part of any control problem is modeling the process. The objective is to obtain the simplest mathematical description that adequately predicts the response of the physical system to all anticipated inputs. Our discussion will be restricted to systems described by ordinary differential equations (in state variable form). Thus, if
are the state variables ( or simply the states ) of the process at time t, and
are the control inputs to the process at time t, then the system may be described by n first-order differential equations
We can define as the state vector of the system, and as the control vector. The state equations can then be written as:
where [7] .
History of control input values during the interval is denoted by u(t) and is called a control history or simply a control. Also, history of state values in the interval is called a state trajectory and is denoted
by x(t). However, we have to note that the terms “history”, “curve”, “function”, and “trajectory” are used interchangeably [8] .
1.3. A Sketch of the Problem
Let us begin with a rough sketch of the type of economic problems that can be formulated as optimal control problems. In the process we introduce our notation. An economy involving time can be described by n real numbers
(1.3.1)
The amounts of capital goods in n different sectors of the economy might, for example, be suitable state variables. It is often convenient to consider (1.3.1) as defining the coordinates of the vector in. As t varies, this vector occupies different positions in, and we say that the system moves along a curve in, or traces a path in. Let us assume now that the process going on in the economy (causing the’s to vary with t) can be controlled to a certain extent in the sense that there are a number of control functions
(1.3.2)
that influence the process. These control functions, or control variables, also called decision variables or instruments will typically be economic data such as tax rates, interest rates, the allocations of investments to different sectors etc. [9] .
To proceed we have to know the laws governing the behavior of the economy over time, in other words the dynamics of the system. We shall concentrate on the study of systems in which the development is determined by a system of differential equation in the form
(1.3.3)
The functions are given functions describing the dynamics of the economy. The assumption is thus that the rate of change of each state variable, in general, depends on all the state variables, all the control variables, and also explicitly on time t. The explicit dependence of the functions on t is necessary, for example, to allow for the laws underlying (8) to vary over time due to exogenous factors such as technological progress, growth in population, etc.
By using vector notation the system (1.3.3) can be described in a simple form. If we put
then (1.3.3) is equivalent to
(1.3.4)
Suppose that the state of the system is known at time so that, where is a given vector in. If we choose a certain control function defined for and insert it into (1.3.3), we obtain a system of first-order differential equations with n unknown functions.
Since the initial point is given, the system (1.3.3) will, in general, have a unique solution
. Since this solution is “a response” to the control function u(t), it would have been appropriate to denote it by but we usually drop the subscript u.
By suitable choices of the control function u(t) many different evolutions of the economy can be achieved. However, it is unlikely that the possible evolutions will be equally desirable. We assume, then, as is usual in economic analysis, that the different alternative developments give different “utilities” that can be measured. More specifically, we shall associate with each control function and its response the number
(1.3.5)
where is a given function. Here is not necessarily fixed, and might have some terminal condition on it at the end point. The fundamental problem that we study in this chapter can now be formulated:
Among all control functions u(t) that via (1.3.4) bring the system from the initial state to a final state satisfying the terminal conditions, find one (provided there exists any) such that J given by (1.3.5) is as large as possible. Such a control function is called an optimal control and the associated path x(t) is called an optimal path. is often called the criterion functional.
1.4. Motivation for the Study
The study was motivated by the fact that the method used followed sequential order: the Hamiltonian “” was derived from the combination of the state equation (which depends on the admissible control) and the performance measure; the co-state and the state variables could be obtained from the solutions of ordinary differential equations of the co-state and state equations respectively. Also, an admissible control which causes to follow admissible trajectory that minimizes the performance measure could be determined. In addition, the results compared considerably with the computer implementation results.
1.5. Statement of the Problem
The problem statements are stated as follows:
Is it possible to find among all control functions which bring x(t) from the initial point to a point satisfying the given terminal condition, one which makes the integral in (1.3.5) as small as possible (provided such a control function exists)?
Is any correlation exists between Fireman method and computer implementation method in the solutions to this class of problems?
2. Nature of Optimal Control
The task in static optimization is to find a single value for each choice variable, such that a stated objective function will be maximized or minimized as the case maybe. Such a problem is devoid of a time dimension. In contrast, time enters explicitly and prominently in a dynamic optimization problem. In such a problem, we will always have in mind a planning period, say, from an initial time t = 0 to a terminal time t = T, and try to find the best course of action to take during that entire period. Thus the solution for any variable will take the form of not a single value, but a complete time path. Alpha discusses further that, suppose the problem is one of profit maximization over a time period. At any point of time, we have to choose the value of some control variable, u(t), which will then affect the value of some state variables y(t), via a so-called equation of motion. In turn y(t) will determine the profit. Since the objective is to maximize the profit over the entire period, the objective function should take the form of a definite integral of from t = 0 to t = T. To be complete, the problem also specifies the initial value of the state variable, and the terminal value of y, y(T) or alternatively, the range of value that y(T) is allowed to take. Taking into account of preceding, one can state the simplest problem of optimal control as
Maximize
Subject to
, free and for all
The first line of the objective function is an integral whose integral f(t, y, u) stipulates home the choice of the control variable u at time t, along with the resulting y at time g, determines our object of maximization at time t. The second is the equation of the motion for the state variable y. What this equation does is to provide the mechanism whereby our choice of control variable u can be translated into a specific pattern of, movement of the state variable y can be adequately described by a first-order differential equation, then, there is need to transform this equation into a pair of first-order differential equations. In this case an additional state variable will be introduced [10] .
A special case of a general non-linear optimal control problem is the linear quadratic (LQ) optimal control problem. Linear quadratic problem is given as follows;
Minimize the quadratic continuous-time cost functional
(2.1.1)
Subject to the linear first-order dynamic constraints
(2.1.2)
and the initial condition [11] .
A particular LQ problem that arises in many control system problem is that of the linear quadratic regulator (LQR) where all of the matrices (i.e. A, B, Q and R) are constants, the initial time is arbitrarily set to zero, and the terminal time is taken as the (the last assumption is what is known as infinite horizon).
Furthermore, the LQR problem is stated as follows: Minimize the infinite horizon quadratic continuous-time cost function
(2.1.3)
subject to the linear time-invariant first order dynamic constraints, and initial condition.
In the finite-Horizon case, the matrices are restricted in the Q and R are positive semi-definite and positive definite and in the infinite horizon case, however the matrices Q and R are not only semi-definite and positive respectively, but are also constants [12] .
The theory of optimal control is concerned with operating dynamical systems at minimum cost. The case where the system dynamics are described by a set of linear differential equations and the cost is described by a quadratic functional is called the LQ problem. One of the main results in the theory is that the solution is provided by the Linear-Quadratic Regulator (LQR), a feedback controller whose equations are given in Section 2.3. [13] .
General Description of Linear Quadratic Regulator
In layman’s terms this means that the settings of a (regulating) controller governing either a machine or process (like an airplane or chemical reactor) are found by using a mathematical algorithm that minimizes a cost function with weighting factors supplied by a human (engineer). The “cost” (function) is often defined as a sum of the deviations of key measurements from their desired values. In effect this algorithm finds those controller settings that minimize the undesired deviations, like deviations from desired altitude or process temperature. Often the magnitude of the control action itself is included in this sum so as to keep the energy expended by the control action itself limited.
In effect, the LQR algorithm takes care of the tedious work done by the control systems engineer in optimizing the controller. However, the engineer still needs to specify the weighting factors and compare the results with the specified design goals. Often this means that controller synthesis will still be an iterative process where the engineer judges the produced “optimal” controllers through simulation and then adjusts the weighting factors to get a controller more in line with the specified design goals.
The LQR algorithm is, at its core, just an automated way of finding an appropriate state-feedback controller. As such it is not uncommon to find that control engineers prefer alternative methods like full state feedback (also known as pole placement) to find a controller over the use of the LQR algorithm. With these the engineer has a much clearer linkage between adjusted parameters and the resulting changes in controller behavior. Difficulty in finding the right weighting factors limits the application of the LQR based controller synthesis.
2.3. Finite-Horizon, Continuous-Time LQR
For a continuous-time linear system, defined on, described by
(2.3.1)
with a quadratic cost function defined as:
(2.3.2)
The feedback control law that minimizes the value of the cost is
(2.3.3)
where is given by
(2.3.4)
and is found by solving the continuous time Riccati differential equation.
(2.3.5)
The first order conditions for Jmin are
(i). State equation
(2.3.6)
(ii). Co-state equation
(2.3.7)
(iii). Stationary equation
(2.3.8)
(iv). Boundary conditions
(2.3.9)
2.4. Review of Linear Ordinary Differential Equation
Optimal control involves among other things, the ordinary differential equations. Hence, this work deems it necessary to consider the state equation together with the solution of the ordinary differential equation. The solution of an ordinary differential equation (ODE) is given by
Let be the unique solution of the matrix ODE
(2.4.1)
We call, a fundamental solution, and sometimes write
(2.4.2)
the last formula being the definition of the exponential. Observe that:
(2.4.3)
The theorem for solving linear system of ODE as follows.
Theorem 2.4.1
(i) The unique solution of the homogeneous system of ODE
(2.4.4)
is
(ii) The unique solution of the non-homogeneous system
(2.4.5)
is (2.4.6)
And that this expression is the variation of parameters formula [14] .
3. The Algorithmic Frame Work of Continuous Linear Regulator Problems
In this section, we shall now proceed to the steps involved in the algorithmic frame work of linear regulator problems:
Step 1: There is need for noting the coefficient of the state equation (s):
where and are the state and control functions respectively, and A and B are constants. If there is one state equation, we then demand for the values of A1 and B1.
Step2: The initial boundary condition(s). For one state equation, we demand 1x0, and 1t0.
Step3: The coefficient of the performance measure to be minimizes (or maximizes)
where is the integral lower limit, is the integral upper limit, and are constant matrices.
Step 4: The Lagrangian is given by
Step 5: We find the co-state equation. For one state equation, derivative of l, i.e.
Step 6: Derive the optimally condition from the Lagrangian. For one state equation, we find
Step7: We determine the boundary condition with
Step 8: The next is to enter the initial guess value for the optimal control
Step 9: Integrate the state equation, substituting the control initial guess
value i.e. standard integral gives
,
If is a constant. Otherwise, we have
Optimal Control and Hamiltonian
The objective of optimal control theory is to determine the control signals that will cause a process to satisfy the physical constraint and at the same time minimizes (or maximizes) some performance measures. The state of a system is a set of quantities which if known at are determined for by specifying the inputs to the system for. It is worthy to state the various components of maximum principle for problem of the form
Subject to
is free and as follows:
(i)
(ii) (State equation)
(iii) (Co-state equation)
(iv) (Transversality condition)
Condition (i) states that at every time, the value of, the optimal control , must be chosen so as to maximizes the value of the Hamiltonian over all admissible value of. In case the Hamiltonian is differen-
tiable with respect to and yields an interior solution, condition: i) can be replaced by. However, if the control region is a closed set, then boundary solutions are possible and may not apply. In fact the maximum principle does not ever required the Hamiltonian to be differentiable with respect to. Conditions ii) and iii) of the maximum principle and give us two equations of motion referred to as the Hamiltonian system for the given problem. Condition iv), is the transversality condition appropriate
for the free-terminal-state problem only. It is noted that, for optimal control problem, the Lagrange function and the Lagrange multiplier variable are known as the Hamiltonian function and co-state variable. The co-state variable measures the shadow price of the state variable [15] .
4. Fireman Numerical Method for a Class of Existing Optimal Control Problems
This self developed method made use of the application of Geometric Progression and Hamiltonian method to numerical solution of some existing Optimal Control Problems of the form “Find an admissible control which causes to follow admissible trajectory that minimizes:
(4.1)
where”.
Theorem 4.1.
For the initial control and step length, the control update is given as:
(4.2)
Proof
Let be the initial control for the system (4.1), then the successive control updates are given as:
(4.3)
Continuing this way, we have that
Theorem 4.2.
The solution of the state equation in (4.1) is given as:
where
Theorem 4.3.
The solution of the state is given as
as
Theorem 4.4: The algebraic relationship is given as:
4.1. Algorithmic Framework of Fireman Method
1. Find the Hamiltonian: The Hamiltonian is given by
2. Determine the Optimality condition from the Hamiltonian:
3. Obtain and solve the co-state equation
4. Update the control using 0.1 as step length:
5. Find the update of the state:
where
6. Obtain the algebraic relationship:
7. Test for optimality:
8. Test for convergence with:
4.2. Test Problem
In this problem, we consider a non-economic problem. This is finding the shortest path from a given point A to a given straight line. In the diagram below, point A have been plotted on the vertical axis in the t-x plane, and drawn the straight line as vertical one at. Three out of an infinite numbers of admissible paths are also shown, each with a different length. The length of any path is the aggregate of small paths segments, each of which can be considered as the hypotenuse of a triangle (may not be drawn) formed by small movements dt and dx.
By Pythagoras’ theorem, we denote the hypotenuse by
(4.2.1)
(4.2.2)
by simple calculations on division by dt2
Let, so that
(4.2.3)
The total length of the path can then be found by integrating (4.2.3) with respect to from to. If we let be the control variable, then (4.2.3) can be expressed as:
(4.2.4)
The shortest path problem is to find an admissible control, which causes to follow admissible trajectory that minimizes
(4.2.5)
Solution to (4.2.5) is same as that of minimization i.e.
Minimize (using distance problem)
Subject to
(4.2.6)
and
4.3. Analytical Solution to Problem
The problem is same as that of
Max.
Subject to
(4.3.1)
and
The Hamiltonian for the problem is given by
Since ℋ is differentiable in, and is unrestricted, the following first order condition can be used to maximize ℋ.
Making the subject of relation, we have that.
Checking the second order condition we find.
This result verifies that the solution to does maximize the Hamiltonian. Since is a function of l, we need a solution to the co-state variable. From the first order conditions the equation of motion for the co-state variable is
Integrating, we have, where is a constant.
Since is independent of, we have that l is a constant. To definitive this constant we can make use of the transversality condition,. Since can take only a single value now known to be zero, we ac-
tually have . Thus we can write .
It follows that the optimal control is. In addition, if we use the equation of motion for the state variable, we see that or (which is a constant). Thus we may write . Since, then the shortest path from a given point A to a given straight line with zero gradient.
4.4. Results of the Test Problem
Analytic:
Table of Results for Problem
5. Summary
It has been established that the knowledge of the Hamiltonian method and the computer programming method has made the minimization (or maximization) process possible for optimal control problems of the types considered above. The two methods in addition to the solution of differential equations involved made the solutions possible. One of the advantages of Fireman method is that, it compared considerably with the computer implementation results.
5.1. Conclusions
This paper has revealed that mathematics should not be studied in abstraction. There is reality in mathematics. The practical and real life situations should be the main focus of mathematics, like distances, consumption problems etc. The co-state, control variable, associated with optimal control problem coupled with a system of linear dynamic constraints solved by Hamiltonian method, have been found to be a very powerful mathematical tool with numerous applications.
More so, computationally, the Fireman Method was tested on an existing linear-quadratic regulator problem with the result obtained. Our numerical and analytic results for this problem are presented in the table below, with the summary as:
Based on the results, it is obvious that, on determining the optimal controls and trajectories of linear-quadratic regulator problems using iterative numerical technique such as Fireman Method is relevant and recommended for use. It is observed that results obtained from Fireman method are much closed to the numerical result and have higher rate of convergence.
5.2. Recommendation
It is recommended that further researches should be carried out on the application of penalty function method to minimize (or maximize) the tested problems and other problems still in existence. One could consider the use of the steepest descent method.
Cite this paper
J. O. Olademo,A. A. Ganiyu,M. F. Akimuyise, (2015) Fireman Numerical Solution of Some Existing Optimal Control Problems. Open Access Library Journal,02,1-14. doi: 10.4236/oalib.1101352
References
1. 1. Thomas, D. (2001) Optimization Method of Control Processes with Minimis Criteria iii, Automat, (i).Tekmekhem.
2. 2. Mital, K.V. (1976) On Program Pursuit Problems in Linear System. Izvestia Aced, Nauk, Automat, (i).Tekmekhem.
3. 3. Kirk, D.E. (2004) Optimal Control Theory (An Introduction). Dover Publications, INC. Mineola, New York.
4. 4. Seiersted, A. and Sydsaeter, K. (1985) Optimal Control Theory with Economic Application. Amsterdem, Lansame Publisher, New York.
5. 5. Snyman, J.A. (2005) Practical Mathematical Optimization. Springer, USA.
6. 6. Sontag, E.D. (1998) Mathematical Control Theory: Deterministic Finite Dimensional Systems. 2nd Edition, Springer, Berlin.
http://dx.doi.org/10.1007/978-1-4612-0577-7
7. 7. Hull, D.G. (2003) Optimal Control Theory for Applications. Springer-Verlag Inc., New York.
8. 8. Gurman, V.I. and Krotov, V.F. (1973) Methods and Problems of Optimal Control. Nauka, Moskva.
9. 9. Todorov, E. (2006) Optlmal Control Theory. MIT Press, Cambridge.
10. 10. Alpha, C.C. and Kelvin, W. (2005) Fundamental Methods of Mathematical Economics. McGraw-Hill Companies, New York.
11. 11. Rose, M. (2009) Theory and Methods of Projecting Automati Controllers. Automat I, Telemakhan, N.I.
12. 12. Bertsekas, D.P. (2012) Dynamic Programming and Optimal Control. Standard University, Standard.
13. 13. Subbotin, A.I. and Krasovskii, N.N. (1988) Game-Theoretical Control Problems. Springer-Verlag, New York.
14. 14. Evans, L.C. (2001) An Introduction to Mathematical Optimal Control Theory, Versiono. L.
http://math.berkeley.edu/’evans/sde
15. 15. Krotov, V.F. (1996) Global Methods in Optimal Control Theory. Mercel Dekker, Inc., New-York. | 6,436 | 29,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.908828 |
https://dsp.stackexchange.com/questions/24714/distances-in-a-single-image-with-some-real-references/24716 | 1,713,880,872,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818711.23/warc/CC-MAIN-20240423130552-20240423160552-00439.warc.gz | 184,878,479 | 40,695 | Distances In a Single Image With Some Real References
Consider following image:
We have some real references here: Stonehenge measurements (stone dimensions and distances), some strait lines at the back of carts, cart wheels that are almost of the same size for both carts. At the same time we don't know the specification of the camera lens. Question: Is it possible to calculate distance of the cart from stones? Or can we even deduce that the cart belongs to the scene or not?
Well, you can have an approximation. Due to the forms in the image are non regular, completely plain, its hard if not impossible to know the camera distortion. But if you know the size of the wheels and also you have the measures of the stones you can approximate the distance of the cart to the stones by using this equation:
$$D_w = \frac{S_w * f}{P_w},$$
where: $D_w$ is the distance from the camera to the wheel, $S_w$ is the real height of the wheel, $f$ is the distance of the focal to the film when the film is in the camera (You need to search for this parameter to use this approach, also you need to know the proportion of the real impressed picture with the film) and $P_w$ is the height of the wheel in the film. The same thing for the stone:
$$D_r = \frac{S_r * f}{P_r},$$
where: $D_r$ is the distance from the camera to the stone, $S_r$ is the real height of the stone and $P_r$ is the height of the stone in the film. Finally the distance $D_t$ from the wheel to the stone is:
$$D_t= D_r -D_w$$
I recomend you to use the objects that are marked in red in the next image, because are the nearest to the epipolar line.
Hope this could help, Best.
• Thanks. Concise answer. I wonder if you could elaborate on the case that we don't know the focal distance and porportion of the real impressed picture as mentioned in OP. Jul 15, 2015 at 17:42
• Well it can be performed, but in this case I'd rather use the road to try to get how far are the first cart trying to obtain some vanishing point from the shores of the road and assuming that the dimensions of the cart are known try to approximate the distances using these parameters. But I need to emphasize that the approximation could be pretty inaccurate Jul 15, 2015 at 18:37
• What do you think about using two wheels of the front cart which are coplanar circles with the method proposed in: ieeexplore.ieee.org/xpl/… Jul 16, 2015 at 4:12
• Focal length can be calculated using two wheels of the front cart using this method: wakayama-u.ac.jp/~wuhy/Eccv2004Final.pdf Jul 16, 2015 at 19:32
• Wow! this approach is very fascinating, I really wanted to try to answer this question, now I think I've learned a lot of things. Thanks for share your knowledge! Jul 16, 2015 at 21:24 | 682 | 2,730 | {"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} | 3.875 | 4 | CC-MAIN-2024-18 | latest | en | 0.941239 |
https://brainmass.com/biology/cell-biochem/plotting-rsa-versus-protein-percentage-11541 | 1,716,389,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00518.warc.gz | 122,008,319 | 6,928 | Purchase Solution
# Plotting RSA versus Protein Percentage
Not what you're looking for?
Subcellular fractionation of rat liver resulted in the following samples:
Fraction.......Speed (xg)......Volume (ml)......protein (mg/ml)
P-1............2800..................10........................6
P-2............12000..................10......................4
P-3............27000..................8........................4
P-4............75000..................5........................2.5
S............non-sedimentable..25....................6.5
Analysis of the fractions resulted in the following enzyme activities expressed as
nmoles/min/ml of sample fraction.
Fraction......citrate synthase......catalase......phosphatase......aldolase
P-1..................0.20........................0.10............0.80..................1.0
P-2..................8.00........................1.00............9.00..................2.0
P-3..................6.00........................7.00............14.0..................1.8
P-4..................0.50........................17.5............12.0..................0.9
S.....................0.04........................0.40............0.70..................15.0
[RSA
Relative Specific
Activity ]..................P-1............ P-2............ P-3............ P-4............ S
Citrate synthase...0.033..........2.00...........1.50............0.20........ 0.006
Catalase...............0.016..........0.25...........1.75............7.00.........0.061
Phosphatase........0.13............2.25............3.50............4.80........0.107
Aldolase...............0.16............0.50............0.45............0.36.........2.30
Plot the distribution profile for each enzyme as RSA vs. % protein
##### Solution Summary
The answer is given along with an attachment of a diagram to generalize how to plot the graphs.
##### Solution Preview
So you want to plot a distribution profile for each enzyme as RSA vs. protein percentage.
First calculate the percentage of proteins in the given samples. Since the ...
##### Parts of the Brain
This quiz will test your knowledge on different areas of the brain.
##### The Plant Body
This quiz will test your knowledge of the anatomy of a common plant.
##### Cellular Respiration
This quiz is a review for cellular respiration.
##### The Transfer of Energy in an Ecosystem
This quiz will assess your knowledge of how energy is transferred in an ecosystem and the different levels of trophic organization.
##### Human Anatomy- Reproductive System
Do you know your reproductive anatomy? | 599 | 2,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.61288 |
https://www.varsitytutors.com/ged_math-help/numbers-and-operations/numbers?page=2 | 1,642,440,394,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300574.19/warc/CC-MAIN-20220117151834-20220117181834-00461.warc.gz | 1,125,934,651 | 52,392 | # GED Math : Numbers
## Example Questions
### Example Question #11 : Ged Math
How many subsets with three elements does the set have?
Explanation:
The number of ways to select three elements from a set of eight is the number of combinations of three elements chosen from eight:
### Example Question #12 : Ged Math
.
and are integers; they may or may not be distinct.
Which of the following could be equal to ?
Explanation:
20 can be factored as:
I)
II)
III)
The positive difference of the factors can be any of:
Of the four choices, only 8 is possible.
### Example Question #13 : Ged Math
Which of the following numbers is a rational number but not an integer?
Explanation:
is a well-known irrational number and is not the correct choice.
. The square root of an integer is rational only if it is itself an integer; calculation yields a result of 1.414... This is not the correct choice.
, since . This is an integer and is not the correct choice.
, so, by definition, . This is rational and is therefore the correct choice.
### Example Question #14 : Ged Math
To how many of the following sets does the number belong?
I) The set of whole numbers
II) The set of integers
III) The set of rational numbers
None
One
Three
Two
One
Explanation:
, which is not an integer. It is not a whole number either, as the whole numbers consist of all (nonnegative) integers.
. As the quotient of integers, it is rational.
Therefore, belongs to the set of rational numbers but not to the other two sets. The correct response is one.
### Example Question #15 : Ged Math
Which of the following base ten numbers has a base sixteen representation of exactly three digits?
Explanation:
A number in base sixteen has powers of sixteen as its place values; starting at the right, they are .
The lowest base sixteen number with three digits would be
in base ten.
The lowest base sixteen number with four digits would be
in base ten.
Therefore, a number that is expressed as a three-digit number in base sixteen must fall in the range
.
Of the four numbers listed, 1,000 falls in that range.
### Example Question #16 : Ged Math
Which of the following sets does belong to?
(a) Whole numbers
(b) Integers
(b) only
Neither (a) nor (b)
Both (a) and (b)
(a) only
(b) only
Explanation:
The set of whole numbers comprises 0 and the so-called natural, or counting, numbers; that is, it is the set is not one of these numbers.
The set of integers comprises these numbers as well as their (negative) opposites; that is, it is the set is one of these numbers.
### Example Question #17 : Ged Math
Which number is prime?
Explanation:
A prime number is a positive integer which has exactly two factors - 1 and the number itself. 36, 38, and 39 can each be shown to have at least one other factor:
, so 2 and 18 are factors of 36, making 36 not prime.
, so 2 and 19 are factors of 38, making 38 not prime.
, so 3 and 13 are factors of 39, making 39 not prime.
37, however, cannot be evenly divided by any number except for 1 and itself, as can be seen below:
We do not need to go higher, since any higher possible factors have square greater than 37. 37 has been proved a prime number.
### Example Question #18 : Ged Math
How many integers are in this set?
Explanation:
An integer is an element of the set of numbers
that is, the so-called natural, or "counting", numbers, their (negative) opposites, and 0. , 2, , and 1 are elements of this set; and are not. The correct response is 4.
### Example Question #19 : Ged Math
The set of real numbers is divided into several subsets including positive numbers and negative numbers, prime numbers and composite number, rational numbers and irrational numbers, etc. For the following questions, select the function with the specified range.
Which expression is complex? Specifically, which number cannot be written as a real number?
Explanation:
Remember that a complex number is any number that includes or .
is irrational but real.
Similarly, which is irrational but real.
### Example Question #20 : Ged Math
The set of real numbers is divided into several subsets including positive numbers and negative numbers, prime numbers and composite number, rational numbers and irrational numbers, etc. For the following questions, select the answer that is a member of the stated subset.
Which number is prime? | 1,043 | 4,403 | {"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} | 4.0625 | 4 | CC-MAIN-2022-05 | latest | en | 0.853878 |
https://support.office.com/en-us/article/oct2dec-function-87606014-cb98-44b2-8dbb-e48f8ced1554?redirectSourcePath=%252fzh-hk%252farticle%252foct2dec-f0bf4643-dec3-4a2f-a5ba-25154efbaa99&ui=en-US&rs=en-US&ad=US | 1,558,705,578,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257624.9/warc/CC-MAIN-20190524124534-20190524150534-00495.warc.gz | 657,521,115 | 20,599 | # OCT2DEC function
This article describes the formula syntax and usage of the OCT2DEC function in Microsoft Excel.
## Description
Converts an octal number to decimal.
## Syntax
OCT2DEC(number)
The OCT2DEC function syntax has the following arguments:
• Number Required. The octal number you want to convert. Number may not contain more than 10 octal characters (30 bits). The most significant bit of number is the sign bit. The remaining 29 bits are magnitude bits. Negative numbers are represented using two's-complement notation.
## Remark
If number is not a valid octal number, OCT2DEC returns the #NUM! error value.
## Example
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Formula
Description
R esult
=OCT2DEC(54)
Converts octal 54 to decimal form.
44
=OCT2DEC(7777777533)
Converts octal 7777777533 to decimal form.
-165 | 257 | 1,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-22 | latest | en | 0.696475 |
http://www.ask.com/web?qsrc=6&o=102632&oo=102632&l=dir&gc=1&q=1gb+Equal+to+How+Many+MB | 1,464,485,320,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049278244.7/warc/CC-MAIN-20160524002118-00110-ip-10-185-217-139.ec2.internal.warc.gz | 364,223,225 | 18,803 | Web Results
## Gigabyte
en.wikipedia.org/wiki/Gigabyte
The gigabyte is a multiple of the unit byte for digital information. The prefix giga means 10<sup>9</sup> in ... 1024<sup>2</sup>, MiB, mebibyte, MB, megabyte ... engineering,...
## Is 1 GB equal to 1024 MB or 1000 MB? - Quora
www.quora.com/Is-1-GB-equal-to-1024-MB-or-1000-MB
Nov 2, 2014 ... 1 GB = 1000 MB = (1000 x 1000) KB = (1000 x 1000 x 1000) = 1000000000 Bytes ... How many mbs equal to 1 gb? ... A Byte is equal to 8 Bits.
## Convert MB to GB - Conversion of Measurement Units
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## How many MBs are in a GB? - Feature - PC Advisor
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Jun 27, 2011 ... ... are in a gigabyte? Learn the answer in this episode of How Many, by Video Royale. ... 1gb = 1,024mb and 1gb = 8192mib. Read more.
## How much is 1 byte, kilobyte, megabyte, gigabyte, etc.?
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## How many MB in a GB? | Mobot.net
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## GB to MB Conversion Gigabytes to Megabytes
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How Many Megabytes in a Gigabyte? There are 1024 megabytes in a gigabyte. 1 Gigabytes is equal to 1024 Megabytes. 1 GB = 1024 MB ...
## 1 Gigabyte is Not Equal to 1024 Megabytes - Instant Fundas
www.instantfundas.com/2007/08/1-gigabyte-is-not-equal-to-1024_22.html
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## Is MB or KB the larger measurement of data? - Ask.com
In terms of data, one MB is 1,000 times larger than one KB. A kilobyte is 1,000 bytes, while one ... How many MB are in 1 GB? ... How many KB are in a MB? A:.
1 gigabytes is equal to 1,024 megabytes.
Convert to
### how many MB equal 1GB? | Yahoo Answers
Dec 24, 2006 ... The Last Paladin. Depends on which device are you talking about. Normally 1 GB = 1024 MB But Hard drive manufacturers count 1000 MB as ...
### GB to MB Conversion Gigabytes to Megabytes Calculator - GbMb.org
www.gbmb.org
How many Megabytes in a Gigabyte. 1 Gigabyte is equal to 1000 megabytes ( decimal). 1 GB = 10<sup>3</sup> MB in base 10 (SI). 1 Gigabyte is equal to 1024 megabytes ...
### 1GB = 1024MB or 1000MB? | Web Hosting Talk
www.webhostingtalk.com
Nov 19, 2009 ... It is a confusing concept for many people, but it makes perfect sense if you ... it is the smallest value which is equal or greater to 1000 AND a power of 2. ... 1GB will always be 1024MB! ... 1GB = 1 gigabyte = 1024 megabytes | 1,080 | 3,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2016-22 | longest | en | 0.747554 |
http://mathhelpforum.com/calculus/101514-finding-area-region-between-curves.html | 1,480,833,974,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541214.23/warc/CC-MAIN-20161202170901-00508-ip-10-31-129-80.ec2.internal.warc.gz | 177,775,527 | 9,438 | # Thread: Finding the area of the region between the curves
1. ## Finding the area of the region between the curves
Find the area of the region between the curves...
#1.) y= -x^2+6x-5 and y=2x-5
#2.) y=8x^2 and y= (the suare root) of x
#3.) y= 4 divided by x, and y=5-x
2. Hi loutja35
First step : draw the curves and shade the region | 115 | 341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2016-50 | longest | en | 0.839563 |
https://www.programming-idioms.org/impl-create/270 | 1,713,974,114,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819668.74/warc/CC-MAIN-20240424143432-20240424173432-00218.warc.gz | 851,176,990 | 5,402 | # Programming-Idioms
Given a floating point number r1 classify it as follows:
If it is a signaling NaN, print "This is a signaling NaN."
If it is a quiet NaN, print "This s a quiet NaN."
If it is not a NaN, print "This is a number."
New implementation
Be concise.
Be useful.
All contributions dictatorially edited by webmasters to match personal tastes.
Please do not paste any copyright violating material.
Please try to avoid dependencies to third-party libraries and frameworks.
Other implementations
``````if (r1.isNaN) {
print("This is a quiet NaN.");
} else {
print("This is a number.");
}``````
``use, intrinsic:: ieee_arithmetic``
`````` character (len=:), allocatable :: msg
if (ieee_support_nan(r1)) then
if (ieee_class(r1) == ieee_quiet_nan) then
msg = "This s a quiet NaN."
else if (ieee_class(r1) == ieee_signaling_nan) then
msg = "This is a signaling NaN."
else
msg = "This is a number."
end if
else
msg = "NaNs are not supported."
end if
write (*,'(A)') msg``````
``uses sysutils;``
`````` if r1.IsNan then
begin
if (TDoubleRec(r1).Data and \$4000000000000000) = \$4000000000000000 then
writeln('This is a quiet NaN')
else
writeln('This is a signalling NaN');
end
else
writeln('This is a number: ',r1);
``````
``use POSIX qw(:nan_payload nan isnan issignaling setpayload);``
``````my @r = (nan, nan, 1.234);
setpayloadsig \$r[1],'999';
foreach my \$r1 ( @r ) {
if ( isnan \$r1 ) {
printf "This is a %s NaN\n",
issignaling(\$r1) ? 'signaling' : 'quiet';
} else {
printf "This is not a NaN: %s\n", \$r1;
}
}
# Output:
# This is a quiet NaN
# This is a signaling NaN
# This is not a NaN: 1.234
``````
``import math``
``````if math.isnan(r1):
print('This is a quiet NaN.')
else:
print('This is a number.')
`````` | 521 | 1,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-18 | latest | en | 0.636202 |
https://www.chegg.com/homework-help/questions-and-answers/group-students-floated-charged-invisible-tape-anothercharged-tape-determined-approximate-c-q751631 | 1,529,681,230,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864546.30/warc/CC-MAIN-20180622143142-20180622163142-00055.warc.gz | 773,213,361 | 12,965 | A group of students floated a charged invisible tape above anothercharged tape. They determined that the approximate charge on thefloating tape was about 1.3e-08 C. The floating tapewas 15 cm longand 1.25 cm wide.
The ratio of the number of excess electronic charges to the numberof molecules on the surface of the tape is the fraction of themolecules on the surface of the tape have gained (or lost) an extraelectronic charge e = 1.6e-19 C. What is thisfraction? To estimate this, assume that molecules in the tape arearranged in a cubic lattice, as indicated in the accompanyingfigure, and that the diameter of a molecule in the tape is about3e-10 m.
fraction of molecules with an extra charge = (#excess charges per molecule)
The inverse of your previous answer has units of (molecules/excesscharge). This can be interpreted as the ratio of (unchargedmolecules) to (charged molecules). What is this number?
1/fraction = (unchargedmolecules per charged molecule)
Make the assumption that the excess charges are distributeduniformly over the surface, so each excess charge is at the centerof an area containing the number of surface molecules you justcalculated. For example, if there were one charged molecule per 25molecules, each charged molecule would be in the center of a squareof 25 molecules,as shown in the diagram below.
According to your calculations above, how far apart are the excesscharges on these students' tape, measured in atomicdiameters?
moleculardiameters apart
Do your answers suggest that it is a common event or a rare eventfor a molecule to gain (or lose) an electron?
RareCommon
If the electric field at a location in air exceeds about 3e6 N/C,the air will become ionized and a spark will be triggered. InChapter 16 we will see that the electric field in a region veryclose to a uniformly charged disk or plate depends approximatelyonly on the charge per square meter (totalcharge Q divided by total surfacearea A):
Use this model to calculate the magnitude of the electric field ata location in the air very close to your tape (less than 1 mm fromthe surface of the tape), and note how it compares to the electricfield needed to trigger a spark in the air.
E = N/C
This is a significant fraction of the breakdown field for air (3e6N/C). If there were enough charge on a tape to make a field strongenough to trigger a spark, the air would become a conductor, andcharge would leak off the tape. So the amount of excess charge youcan put on a tape is limited by the breakdown strength ofair. | 565 | 2,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-26 | latest | en | 0.914572 |
https://www.itprotoday.com/october-mdx-puzzle-solution-revealed-01-nov-1999 | 1,539,599,659,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509170.2/warc/CC-MAIN-20181015100606-20181015122106-00303.warc.gz | 956,443,800 | 16,316 | October MDX Puzzle Solution Revealed - 01 Nov 1999
How can you write a Multidimensional Expression (MDX) query that returns the Unit Sales for the months of 1997 on columns and the top product brand names that account for 25 percent of Unit Sales for 1997? The most difficult part of this query is figuring out exactly what you're asking for. The first part of the query asks for Unit Sales for the months of 1997. You can use a straightforward query such as:
```SELECT Descendants(\[1997\], \[Month\]) on Columns
FROM Sales
WHERE (\[Unit Sales\])
```
In this query, the Descendants function returns all the time dimension members descending from \[1997\] at the \[Month\] level. In other words, it gets the grandchildren of 1997 because the \[Month\] level is two levels below \[1997\].
Now you need to answer the second part of the question: "How do you get the product brand names that account for 25 percent of Unit Sales for 1997?" This is where the TopPercent MDX function comes in handy. The OLAP Services documentation says the TopPercent sorts a set and returns the top N elements whose cumulative total is at least a specified percentage.
``` TopPercent( Set , Percentage , Numeric
Expression )
```
You pass in all the Brand Name dimension members and find the top 25 percent based on 1997 Unit Sales:
```SELECT Descendants(\[1997\], \[Month\]) on Columns,
TopPercent( \[Product\].\[Brand Name\].Members, 25,
\[1997\]) on Rows
FROM Sales
WHERE (\[Unit Sales\])``` | 375 | 1,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-43 | latest | en | 0.805742 |
https://www.alibabacloud.com/blog/boundary-issues-in-postgis-coordinate-transformation-srid---st-transform_597033 | 1,695,552,463,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506632.31/warc/CC-MAIN-20230924091344-20230924121344-00679.warc.gz | 722,226,606 | 8,551 | # Community
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Community Blog Boundary Issues in PostGIS Coordinate Transformation (SRID) - ST_Transform
# Boundary Issues in PostGIS Coordinate Transformation (SRID) - ST_Transform
By Digoal
## Background
When using PostgreSQL ST_Transform to convert coordinates, a user encountered a boundary issue, causing an inaccurate distance calculation.
Sheyu from Cainiao Network provided a perfect solution for this issue.
It is not about whether to use the 26986 coordinate system. First, understand how the 26986 coordinate system is defined. Why does it function well when calculating a negative longitude (Western Hemisphere)? This is because 26986 is Massachusetts' local coordinate system (regional coordinate system), and its two main definition parameter origins are the longitude and latitude (-71.5, 41). Thus, the central meridian is 71.5°W. It is far away from China (China is centered on 120°E), and therefore the projected coordinate system is inapplicable. Simply put, the farther away from the central meridian, the greater the projection error will be.
To calculate the distance between latitude and longitude on a spherical surface, use functions provided by PostGIS without projecting onto a plane first, unless you want to perform other plane operations such as area calculation. In this case, how to select the projection SRID?
Note that a geographic coordinate system and a projected coordinate system are different. The former is based on a spheroid, and the latter is based on a plane.
Geographic coordinate system:
``````4214 GCS_Beijing_1954
4326 GCS_WGS_1984
4490 GCS_China_Geodetic_Coordinate_System_2000
4555 GCS_New_Beijing
4610 GCS_Xian_1980 ``````
Projected coordinate system:
``````2327 Xian_1980_GK_Zone_13
2328 Xian_1980_GK_Zone_14
2329 Xian_1980_GK_Zone_15
2330 Xian_1980_GK_Zone_16
2331 Xian_1980_GK_Zone_17
2332 Xian_1980_GK_Zone_18
2333 Xian_1980_GK_Zone_19
2334 Xian_1980_GK_Zone_20
2335 Xian_1980_GK_Zone_21
2336 Xian_1980_GK_Zone_22
2337 Xian_1980_GK_Zone_23
2338 Xian_1980_GK_CM_75E
2339 Xian_1980_GK_CM_81E
2340 Xian_1980_GK_CM_87E
2341 Xian_1980_GK_CM_93E
2342 Xian_1980_GK_CM_99E
2343 Xian_1980_GK_CM_105E
2344 Xian_1980_GK_CM_111E
2345 Xian_1980_GK_CM_117E
2346 Xian_1980_GK_CM_123E
2347 Xian_1980_GK_CM_129E
2348 Xian_1980_GK_CM_135E
2349 Xian_1980_3_Degree_GK_Zone_25
2350 Xian_1980_3_Degree_GK_Zone_26
2351 Xian_1980_3_Degree_GK_Zone_27
2352 Xian_1980_3_Degree_GK_Zone_28
2353 Xian_1980_3_Degree_GK_Zone_29
2354 Xian_1980_3_Degree_GK_Zone_30
2355 Xian_1980_3_Degree_GK_Zone_31
2356 Xian_1980_3_Degree_GK_Zone_32
2357 Xian_1980_3_Degree_GK_Zone_33
2358 Xian_1980_3_Degree_GK_Zone_34
2359 Xian_1980_3_Degree_GK_Zone_35
2360 Xian_1980_3_Degree_GK_Zone_36
2361 Xian_1980_3_Degree_GK_Zone_37
2362 Xian_1980_3_Degree_GK_Zone_38
2363 Xian_1980_3_Degree_GK_Zone_39
2364 Xian_1980_3_Degree_GK_Zone_40
2365 Xian_1980_3_Degree_GK_Zone_41
2366 Xian_1980_3_Degree_GK_Zone_42
2367 Xian_1980_3_Degree_GK_Zone_43
2368 Xian_1980_3_Degree_GK_Zone_44
2369 Xian_1980_3_Degree_GK_Zone_45``````
## Example
The following two 4326 coordinates are slightly different. But when converted into 26986 coordinates, their difference becomes significant.
``````postgres=# select ST_AsEWKT(ST_Transform(ST_GeomFromText('POINT(108.50000000001 22.8)', 4326), 26986));
st_asewkt
-----------------------------------------------------
SRID=26986;POINT(8123333.59043839 12671815.6459695)
(1 row)
postgres=# select ST_AsEWKT(ST_Transform(ST_GeomFromText('POINT(108.5000000001 22.8)', 4326), 26986));
st_asewkt
------------------------------------------------------
SRID=26986;POINT(-7723333.59044452 12671815.6459593)
(1 row) ``````
The distance calculated based on the converted coordinates is inaccurate because the central meridian of the SRID is far away from the calculation point. The projection error increases when the distance from the central meridian increases.
``````postgres=# select * from spatial_ref_sys where srid=26986;
-[ RECORD 1 ]-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
srid | 26986
auth_name | EPSG
auth_srid | 26986
proj4text | +proj=lcc +lat_1=42.68333333333333 +lat_2=41.71666666666667 +lat_0=41 +lon_0=-71.5 +x_0=200000 +y_0=750000 +datum=NAD83 +units=m +no_defs
postgres=# select * from spatial_ref_sys where srid=4326;
-[ RECORD 1 ]---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
srid | 4326
auth_name | EPSG
auth_srid | 4326
srtext | GEOGCS["WGS 84",DATUM["WGS_1984",SPHEROID["WGS 84",6378137,298.257223563,AUTHORITY["EPSG","7030"]],AUTHORITY["EPSG","6326"]],PRIMEM["Greenwich",0,AUTHORITY["EPSG","8901"]],UNIT["degree",0.0174532925199433,AUTHORITY["EPSG","9122"]],AUTHORITY["EPSG","4326"]]
proj4text | +proj=longlat +datum=WGS84 +no_defs ``````
PROJCS indicates projected coordinates, and GEOGCS indicates spherical coordinates.
To solve this problem, do not convert the coordinates into 26986 coordinates (planar coordinates). Use the corresponding function of PostGIS to calculate the spherical coordinates. Even if you need to calculate a planar distance, use a coordinate system commonly used in China (whose central meridian is in China, as described earlier).
http://postgis.net/docs/manual-2.0/ST_Distance.html
``````try this:
postgres=# select ST_Distance(ST_GeographyFromText('SRID=4326;POINT(108.51 22.8)'), ST_GeographyFromText('SRID=4326;POINT(108.499999999999999 22.79)'));
-[ RECORD 1 ]--------------------
st_distance | 1510.16913796499989
-- Geography example -- same but note units in meters - use sphere for slightly faster less accurate
-- Geometry example - units in meters (SRID: 26986 Massachusetts state plane meters) (most accurate for Massachusetts) ``````
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# digoal
262 posts | 23 followers
# digoal
262 posts | 23 followers
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Leverage cloud-native database solutions dedicated for FinTech. | 1,955 | 7,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-40 | longest | en | 0.780164 |
http://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.454350.html | 1,369,229,409,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701760529/warc/CC-MAIN-20130516105600-00073-ip-10-60-113-184.ec2.internal.warc.gz | 324,367,358 | 4,808 | # SOLUTION: How do you solve: sin^2(x)+cos(2x)-cos(x)=0 I know that cos(2x) can be equal to cos^2(x)-sin^2(x), or it can be equal to 2cos^2(x)-1, or it can be equal to 1-2sin^2(x) My
Algebra -> Algebra -> Trigonometry-basics -> SOLUTION: How do you solve: sin^2(x)+cos(2x)-cos(x)=0 I know that cos(2x) can be equal to cos^2(x)-sin^2(x), or it can be equal to 2cos^2(x)-1, or it can be equal to 1-2sin^2(x) My Log On
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Algebra: Trigonometry Solvers Lessons Answers archive Quiz In Depth
Question 454350: How do you solve: sin^2(x)+cos(2x)-cos(x)=0 I know that cos(2x) can be equal to cos^2(x)-sin^2(x), or it can be equal to 2cos^2(x)-1, or it can be equal to 1-2sin^2(x) My math teacher told us to use the formula that has the trig-word in it that is in the equation, but since both sin and cos are in the equation, I'm not sure what to use.Answer by richwmiller(9135) (Show Source): You can put this solution on YOUR website!if you use cos^2(x)-sin^2(x) then you will have sin^2(x)+cos^2(x)-sin^2(x) which eliminates sin completely. leaving you cos^(2x)-cos(x) | 410 | 1,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 4 | CC-MAIN-2013-20 | latest | en | 0.793201 |
https://howorigami.com/post/c254971313.html | 1,722,747,772,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640389685.8/warc/CC-MAIN-20240804041019-20240804071019-00038.warc.gz | 247,330,068 | 6,596 | # Bird
Hello, this time I tried folding a flying bird like a hummingbird with origami paper.
It's not that difficult to fold, so please fold it while looking at the explanation.
### How to fold an origami bird
This time, I referred to this video.
Let's fold it now
[ 1 ] Fold it in half so that it becomes a triangle.
[ 2 ] Fold the right corner to the left in half.
[ 3 ] Fold the lower left corner to the upper corner.
[ 4 ] Turn it over and fold the lower right corner to the upper corner.
[ 5 ] Turn around and fold the top corner down to the bottom.
[ 6 ] Turn it over and fold the top corner down to the bottom.
[ 7 ] Open the folded area and fold the red dotted line into a mountain fold and the blue line into a valley fold.
[ 8 ] For the upper part of the paper, fold the red dotted line into a mountain fold and the blue line into a valley fold.
[ 9 ] Align the paper and fold the lower left corner diagonally upward along the dotted line.
[ 10 ] Open the top surface and fold the red dotted line into a mountain fold and the blue line into a valley fold.
[ 11 ] Fold the right corner down along the dotted line.
[ 12 ] Turn it over and fold the right corner up along the dotted line.
[ 13 ] Fold the right corner down along the dotted line, make a crease, and then put it back.
[ 14 ] Turn it over, fold the left corner down along the dotted line, make a crease, and then put it back.
[ 15 ] Open the space between the left corners as shown in the photo, and fold the top and bottom corners inward.
[ 16 ] of the front and back with the magic 2 to draw the eye in place.
[ 17 ] Slightly tilt the front and back wings and shape them to complete.
Thank you for your hard work
How was it?
If you try to make such a shape in 3D, it will be quite complicated. | 445 | 1,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-33 | latest | en | 0.931451 |
https://www.scribd.com/document/260516974/simulacion-tk-skimer-pdf | 1,568,575,815,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572235.63/warc/CC-MAIN-20190915175150-20190915201150-00078.warc.gz | 1,024,466,086 | 66,394 | You are on page 1of 25
# OPTIMIZATION OF THE DESIGN
OF A SKIMMER TANK
Paula Luca Ismirlian
Astra Evangelista S.A.
AESA
Engineering
Manufacturing
Construction
Services
Presence in Argentina, Peru, Bolivia, Brazil, Uruguay.
Problem Description
Location: Santa Cruz, Argentina.
Secondary oil recovery by water injection
Water Treatment Plant for re-injection of water in oil well
Skimmer Tank:
Flotation of oil droplets (discrete phase)
Separation from water (continuous phase)
flow pattern
## Problem Description (cont.)
Skimmer Tank: Simple / Low cost / Low maintenance
Overflow
Inlet
Baffles
Outlet
DESIGN OPTIMIZATION
Configuration of internals and inlet and outlet ducts
Methodology
Initial geometry (Case 1)
Alternative designs (Cases 2 - 3)
## Premise: Mx. ppm of oil in water outlet
Methodology
CFD:
Flow patterns inside the tank
Velocities
## Oil concentration in water outlet
Oil particles and sand particles path
## Quantity of oil and sand escaping from each outlet
Recirculation zones
Analysys
1. Geometry (basic case)
Analysys
2. Meshing
Prisms and
Tetrahedrons
ANSYS Meshing
Analysys
3. Set up
ANSYS CFX
Two simulations:
phase
## Turbulence model: Standard k-epsilon
Stationary regime
Incompressible fluid
Isothermal
Analysys
continuous phase
Transient regime
water)
One-way coupling
## No colision between particles
No coalescence
Indeformable particles
Analysys
Material properties (constant) :
## 3 diameters (Size 1 < Size 2 < Size 3)
Inlet concentrations
Densities
Oil viscosity.
Analysys
Boundary conditions:
Outlet pressure
## Superior wall: overflow condition
Continuous phase
Discrete phase
free slip
absoption coefficient = 1
## Lateral walls/baffles: no roughness
Continuous phase
Discrete phase
no slip
perpendicular and parallel bounce
coefficients
## Results and conclusions
Case 1: Basic geommetry
Water velocities:
INLET
OUTLET
## Results and conclusions
Velocity in a plane:
INLET
OUTLET
## Results and conclusions
Velocities in other planes:
Oil Particles:
size1 BLUE
size2 RED
size3 GREEN
## Results and conclusions
Case 2: Jacketed in/out
Water velocity fields:
INLET
OUTLET
## Results and conclusions
Water Streamlines
(velocities):
## Results and conclusions
Oil concentrations:
Inlet
Outlet
Overflow
Inside
Size 3
Size 2
Size 1
## Results and conclusions
Case 3: Vertical baffles
Horizontal
baffles
Vertical
baffles
Outlet
Inlets
## Results and conclusions
Water streamlines
## Results and conclusions
Size 1 residence time:
## Results and conclusions
Size 3 residence time:
Inlet
Outlet
Overflow
Inside
Mx. ppm
Size 3
Size 2
Size 1
Thank you!
Questions? | 680 | 2,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.715798 |
https://physics.stackexchange.com/questions/31582/compton-scattering-angle | 1,721,901,182,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00855.warc.gz | 399,087,888 | 40,721 | # Compton scattering angle
Say a photon hits a free electron at rest. I understand that there is a formula for the Compton scattering when the photon is scattered with an angle $\theta$, but I don't understand what determines that angle $\theta$. If one photon hits such an electron, won't it be scattered through a unique angle? It seems it is not so, since there is the angle $\theta$ as a parameter.
At low photon energies, the scattering process is sometimes called Thompson scattering. You can ignore relativistic effects, and simply say that the electric field of the photon causes the electron to oscillate in the plane of the electric field. That causes the electron to emit dipole radiation. The probability to scatter in a particular direction is then $\propto \sin^2\theta$, where $\theta$ is the same angle that you're using, and is the angle away from the plane of the electric field of the photon. Jackson derives the angular dependence in chapter 9, section 9.2. The final equation is (9.23). | 218 | 1,009 | {"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} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.918852 |
https://www.hindawi.com/journals/isrn/2014/897834/ | 1,716,500,889,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00111.warc.gz | 701,043,177 | 120,025 | #### Abstract
Let be an odd integer such that is a prime. In this work, we determine all integer solutions of the Diophantine equation and then we deduce the general terms of all -balancing numbers.
#### 1. Introduction
Balancing numbers were first considered by Behera and Panda in [1] when they considered the integer solutions of the Diophantine equation for some positive integers and . In this case is called a balancing number with balancer (or cobalancing number) . For example, and are balancing numbers with cobalancing numbers which are and , respectively.
The th balancing number is denoted by and the th cobalancing number is denoted by . They satisfy the recurrence relations and for , with initial values and .
From (1), one has So is a balancing number if and only if is a perfect square and is a cobalancing number if and only if is a perfect square. Set and . Then is called the th Lucas-balancing number and is called the th Lucas-cobalancing number. Binet formulas for balancing, cobalancing, Lucas-balancing, and Lucas-cobalancing numbers are , , and , respectively, where and (for further details see also [27]).
Recently, there are many studies on balancing numbers. In [8], the authors generalized the theory of balancing numbers to numbers defined as follows. Let such that . Then a positive integer such that is called a -power numerical center for if Later in [9], the authors extended the concept of balancing numbers to the -balancing numbers defined as follows. Let and let be coprime integers. If for some positive integers, and , one has then the number is called an -balancing number and is denoted by .
Like in -balancing numbers, in [10], Dash et al. defined the -balancing number and derived some results on it. Let be an integer. Then a positive integer is called a -balancing number if for some positive integer which is called -balancer (or -cobalancing number). - and -balancing numbers can be given in terms of balancing numbers; indeed, and . So it is assumed that .
The th -balancing number is denoted by and the th -cobalancing number is denoted by . From (5) we see that Hence, is a -balancing number if and only if is a perfect square and is a -cobalancing number if and only if is a perfect square. Set Then is called the th Lucas -balancing number and is called the th Lucas -cobalancing number.
#### 2. Main Results
To determine the general terms of all -balancing numbers, we have to determine all integer (in fact only positive) solutions of some specific Diophantine equations. Let us explain this as follows. We see as above that is a perfect square for a -balancing number . So we set for some integer . Hence, we get the Diophantine equation: If we make the change of variables and for some and , then we get Here we obtain and . Thus (9) becomes which is a Pell equation (see [1114]). Before considering its integer solutions, we need some notations. Now let be a nonsquare discriminant. Then the -order is defined for nonsquare discriminants to be the ring , where if mod 4, or if mod 4). So is a subring of . The unit group is defined to be the group of units of the ring .
For the quadratic form , we can write . So the module of is the -module . Therefore, we get , where So there is a bijection for solving the Pell equation ; that is, . The action of on the set of integral solutions of the equation is most interesting when is a positive nonsquare since is infinite. Therefore, the orbit of each solution will be infinite and so the set is either empty or infinite. Since can be explicitly determined, the set is satisfactorily described by the representation of such a list, called a set of representatives of the orbits. Let be the smallest unit of , that is, greater than , and let if , or if . Then every orbit of integral solutions of contains a solution such that , where if or if . So for finding a set of representatives of the orbits of integral solutions of , we must find, for each integer such that , all integers that satisfy . If , then and so .
For the Pell equation , there are one, two, three, or more sets of representatives (or sets of solutions) depending on . For example, for , the set of representatives is ; for , the set of representatives is ; for , the set of representatives is ; and for , the set of representatives is . So we cannot determine for which values of there are one, two, three, or more sets of representatives. Consequently, we cannot determine all integer solutions of in one way. To determine all integer solutions of , we have to put some restrictions on . From now on, is assumed to be odd such that is a prime.
Thus for the Pell equation , we get and is a square only for in the range . Hence, . Therefore, there is exactly one set of representatives of the orbits and that is a set of representatives; that is, there are two classes of solutions. Here we see that(1) generates the solutions for ,(2) generates the solutions for ,(3) generates the solutions for ,(4) generates the solutions for ,(5) generates the solutions for ,(6) generates the solutions for ,(7) generates the solutions for ,(8) generates the solutions for ,
where by (11). Thus, we can give the following theorem.
Theorem 1. The set of all positive integer solutions of is ,, where
To determine all positive integer solutions of , we have to formulate the th power of . So we can give the following theorem which can be proved by induction on .
Theorem 2. The th power of is where is the th balancing number and is the th Lucas-balancing number.
Consequently from the above two theorems, we can give the following main theorem.
Theorem 3. The set of all positive integer solutions of is ,, where
After formulating all integer solutions of , we can determine all integer solutions of . Notice that and . So applying Theorem 3, we get for and for . Here we note that and are positive integer solutions. By symmetry, and are also integer solutions. Apart from them, we get from generates the solutions for that for and from generates the solutions for that for . So and are also integer solutions, where and . Thus, we proved the following theorem.
Theorem 4. The set of all integer solutions of is ,, where for and for .
Example 5. Let . Then we get So the set of all integer solutions of is
After determining all integer solutions of , we can determine all -balancing numbers.
Theorem 6. The general terms of all -balancing numbers are for and for .
Proof. Since , we get from Theorem 4 that for since and .
Note that and . So we get from (7) that Thus from (6), we obtain and hence since and .
Similarly we deduce that for .
#### Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper. | 1,507 | 6,758 | {"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} | 3.1875 | 3 | CC-MAIN-2024-22 | latest | en | 0.942863 |
https://pastebin.com/qLA90Fjq | 1,582,158,057,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144429.5/warc/CC-MAIN-20200219214816-20200220004816-00104.warc.gz | 536,376,301 | 5,686 | • API
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# Untitled
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1. x=0
2. y=160
3.
4. lagcount = 0
5. table={}
6.
7. -----
8.
9. function Stuff()
10. gui.opacity(0.2)
11. gui.drawbox(148+x,9+y,258+x,42+y, "black")
12. gui.opacity(0.95)
13.
19.
20. gui.text(150+x,9+y," \n \n \n ")
21. gui.text(150+x,9+y," pos speed\n(X):\n(Y):\nlag:", "green")
22. gui.text(172+x,17+y, math.floor(xpos*1000)/1000)
23. gui.text(224+x,17+y, xspeed)
24. gui.text(172+x,25+y, math.floor(ypos*1000)/1000)
25. gui.text(224+x,25+y, yspeed)
26.
27. gui.text(176+x, 33+y, lagcount, "#fd7979")
28. if lag == 1 then gui.text(200+x, 33+y, "*********" , "red") end
29.
30. table=input.get(1)
31. if table.F1 then lagcount = lagcount - lagcount
32. elseif table.F2 then lagcount = lagcount - lagcount
33. elseif table.F3 then lagcount = lagcount - lagcount
34. elseif table.F4 then lagcount = lagcount - lagcount
35. elseif table.F5 then lagcount = lagcount - lagcount
36. elseif table.F6 then lagcount = lagcount - lagcount
37. elseif table.F7 then lagcount = lagcount - lagcount
38. elseif table.F8 then lagcount = lagcount - lagcount
39. elseif table.F9 then lagcount = lagcount - lagcount
40. elseif table.F10 then lagcount = lagcount - lagcount
41. end
42.
43. end
44. gui.register(Stuff)
45.
46. -----
47.
48. while true do
49.
50. if lag == 1 then lagcount = lagcount + 1 end
51. | 534 | 1,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.115503 |
http://mathhelpforum.com/pre-calculus/104581-average-rate-change.html | 1,481,455,400,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544672.33/warc/CC-MAIN-20161202170904-00442-ip-10-31-129-80.ec2.internal.warc.gz | 170,108,277 | 10,042 | # Thread: average rate of change
1. ## average rate of change
Let
f(x)=5 ·x2+2 ·x−3 and let x0= 1.
The average rate of change of f between x= 1 and x= 1.18 equals ?
2. Originally Posted by samtheman17
Let
f(x)=5 ·x2+2 ·x−3 and let x0= 1.
The average rate of change of f between x= 1 and x= 1.18 equals ?
$f(x) = 5x^2 + 2x - 3$.
You have $f(1) = 5(1)^2 + 2(1) - 3$
$= 5 + 2 - 3$
$= 4$
Also $f(1.18) = 5(1.18)^2 + 2(1.18) - 3$
$= 6.962 + 2.36 - 3$
$= 6.322$.
The average rate of change will be given by
$\frac{f(1.18) - f(1)}{1.18 - 1}$
$= \frac{6.322 - 4}{0.18}$
$= \frac{2.322}{0.18}$
$= 12.9$.
3. ok, thank you so much!
that really helped
(ps.anwser is 12.9 as 1.18-1 is .18 not 1) | 333 | 702 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-50 | longest | en | 0.775375 |
https://hackage.haskell.org/package/edenskel-2.0.0.2/candidate/docs/Control-Parallel-Eden-Map.html | 1,660,054,551,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570977.50/warc/CC-MAIN-20220809124724-20220809154724-00295.warc.gz | 288,733,970 | 4,483 | edenskel-2.0.0.2: Semi-explicit parallel programming skeleton library
Copyright (c) Philipps Universitaet Marburg 2009-2014 BSD-style (see the file LICENSE) eden@mathematik.uni-marburg.de beta not portable None Haskell98
Control.Parallel.Eden.Map
Description
This Haskell module defines map-like skeletons for the parallel functional language Eden.
Depends on GHC. Using standard GHC, you will get a threaded simulation of Eden. Use the forked GHC-Eden compiler from http://www.mathematik.uni-marburg.de/~eden for a parallel build.
Eden Group ( http://www.mathematik.uni-marburg.de/~eden )
Synopsis
# Custom map skeletons
These skeletons expose many parameters to the user and thus have varying types
Arguments
:: (Trans a, Trans b) => (a -> b) worker function -> [a] task list -> [b] result list
Basic parMap Skeleton - one process for each list element
Arguments
:: (Trans b, Trans c) => (a -> [b]) input transformation function -> ([c] -> d) result reduction function -> (b -> c) worker function -> a input -> d output
A process ranch is a generalized (or super-) farm. Arbitrary input is transformed into a list of inputs for the worker processes (one worker for each transformed value). The worker inputs are processed by the worker function. The results of the worker processes are then reduced using the reduction function.
Arguments
:: (Trans a, Trans b) => ([a] -> [[a]]) input distribution function -> ([[b]] -> [b]) result combination function -> (a -> b) mapped function -> [a] input -> [b] output
A farm distributes its input to a number of worker processes. The distribution function divides the input list into sublists - each sublist is input to one worker process, the number of worker processes is determined by the number of sublists. The results of the worker processes are then combined using the combination function.
Use `mapFarmS` or `mapFarmB` if you want a simpler interface.
Arguments
:: (Trans a, Trans b) => Int number of processes -> (a -> b) mapped function -> [a] input -> [b] output
Like the `farm`, but uses a fixed round-robin distribution of tasks.
Arguments
:: (Trans a, Trans b) => Int number of processes -> (a -> b) mapped function -> [a] input -> [b] output
Like the `farm`, but uses a fixed block distribution of tasks.
Arguments
:: Trans b => Int number of processes -> ([a] -> [[a]]) input distribution function -> ([[b]] -> [b]) result combination function -> (a -> b) mapped function -> [a] input -> [b] output
Offline farm (alias direct mapping): Like the farm, but tasks are evaluated inside the workers (less communication overhead). Tasks are mapped inside each generated process abstraction avoiding evaluating and sending them. This often reduces the communication overhead because unevaluated data is usually much smaller than evaluated data.
Use `map_offlineFarm` if you want a simpler interface.
Notice: The offline farm receives the number of processes to be created as its first parameter. The task lists structure has to be completely defined before process instantiation takes place.
Arguments
:: (Trans a, Trans b) => Int number of processes -> (a -> b) mapped function -> [a] input -> [b] output
Like the `offlineFarm`, but with fixed round-robin distribution of tasks.
Arguments
:: (Trans a, Trans b) => Int number of processes -> (a -> b) mapped function -> [a] input -> [b] output
Like the `offlineFarm`, but with fixed block distribution of tasks.
# Custom map skeletons with explicit placement
Map skeleton versions with explicit placement
Arguments
:: (Trans a, Trans b) => Places places for instantiation -> (a -> b) worker function -> [a] task list -> [b] result list
Basic parMap Skeleton - one process for each list element. This version takes places for instantiation on particular PEs.
Arguments
:: (Trans b, Trans c) => Places places for instantiation -> (a -> [b]) input transformation function -> ([c] -> d) result reduction function -> (b -> c) worker function -> a input -> d output
A process ranch is a generalized (or super-) farm. This version takes places for instantiation. Arbitrary input is transformed into a list of inputs for the worker processes (one worker for each transformed value). The worker inputs are processed by the worker function. The results of the worker processes are then reduced using the reduction function.
Arguments
:: (Trans a, Trans b) => Places places for instantiation -> ([a] -> [[a]]) input distribution function -> ([[b]] -> [b]) result combination function -> (a -> b) mapped function -> [a] input -> [b] output
A farm distributes its input to a number of worker processes. This version takes places for instantiation. The distribution function divides the input list into sublists - each sublist is input to one worker process, the number of worker processes is determined by the number of sublists. The results of the worker processes are then combined using the combination function.
Use `map_farm` if you want a simpler interface.
Arguments
:: Trans b => Places places for instantiation -> Int number of processes -> ([a] -> [[a]]) input distribution function -> ([[b]] -> [b]) result combination function -> (a -> b) mapped function -> [a] input -> [b] output
Offline farm with explicit placement (alias self-service farm or direct mapping): Like the farm, but tasks are evaluated inside the workers (less communication overhead). Tasks are mapped inside each generated process abstraction, avoiding evaluating and sending them. This often reduces the communication overhead because unevaluated data is usually much smaller than evaluated data.
Use `map_offlineFarm` if you want a simpler interface.
Notice: The task lists structure has to be completely defined before process instantiation takes place.
# Simple map skeleton variants
The map skeletons `farm` and `offlineFarm` can be used to define skeletons with the simpler sequential map interface :: (a -> b) -> [a] -> [b]
Arguments
:: (Trans a, Trans b) => (a -> b) worker function -> [a] task list -> [b] result list
Parallel map variant with map interface using (max (noPe-1) 1) worker processes. Skeletons ending on `S` use round-robin distribution, skeletons ending on `B` use block distribution of tasks.
Arguments
:: (Trans a, Trans b) => (a -> b) worker function -> [a] task list -> [b] result list
Parallel map variant with map interface using (max (noPe-1) 1) worker processes. Skeletons ending on `S` use round-robin distribution, skeletons ending on `B` use block distribution of tasks.
Arguments
:: (Trans a, Trans b) => (a -> b) worker function -> [a] task list -> [b] result list
Parallel map variant with map interface using (max (noPe-1) 1) worker processes. Skeletons ending on `S` use round-robin distribution, skeletons ending on `B` use block distribution of tasks.
Arguments
:: (Trans a, Trans b) => (a -> b) worker function -> [a] task list -> [b] result list
Parallel map variant with map interface using (max (noPe-1) 1) worker processes. Skeletons ending on `S` use round-robin distribution, skeletons ending on `B` use block distribution of tasks.
# Deprecated map skeletons
These skeletons are included to keep old code alive. Use the skeletons above.
Arguments
:: (Trans a, Trans b) => Int number of child processes -> (Int -> [a] -> [[a]]) input distribution function -> ([[b]] -> [b]) result combination function -> (a -> b) mapped function -> [a] input -> [b] output
Deprecated, use the `farm`; `farmClassic` distributes its input to a number of worker processes. This is the Classic version as described in the Eden standard reference "Parallel Functional Programming in Eden". The distribution function is expected to divide the input list into the given number of sublists. In the new farm the number of sublists is determined only by the distribution function.
Use `map_farm` if you want a simpler interface.
Arguments
:: forall a b . Trans b => Int number of child processes -> (Int -> [a] -> [[a]]) input distribution function -> ([[b]] -> [b]) result combination function -> (a -> b) mapped function -> [a] input -> [b] output
Deprecated, use `offlineFarm`; Self service farm. Like the farm, but tasks are evaluated in the workers (less communication overhead). This is the classic version. The distribution function is expected to divide the input list into the given number of sublists. In the new self service farm the number of sublists is determined only by the distribution function.
Use `map_ssf` if you want a simpler interface.
Notice: The task lists structure has to be completely defined before process instantiation takes place.
Arguments
:: Trans b => Int number of processes -> ([a] -> [[a]]) input distribution function -> ([[b]] -> [b]) result combination function -> (a -> b) mapped function -> [a] input -> [b] output
Deprecated: Same as `offlineFarm`.
Arguments
:: (Trans a, Trans b) => (a -> b) worker function -> [a] task list -> [b] result list
Deprecated: Same as `parMap`.
Arguments
:: (Trans a, Trans b) => (a -> b) worker function -> [a] task list -> [b] result list
Deprecated: better use mapFarmS or mapOfflineFarmS instead
Deprecated: Parallel map variants with map interface using noPe worker processes.
Arguments
:: (Trans a, Trans b) => (a -> b) worker function -> [a] task list -> [b] result list
Deprecated: better use mapFarmS or mapOfflineFarmS instead
Deprecated: Parallel map variants with map interface using noPe worker processes.
Arguments
:: (Trans a, Trans b) => (a -> b) worker function -> [a] task list -> [b] result list
Deprecated: better use mapFarmS or mapOfflineFarmS instead
Deprecated: Parallel map variants with map interface using noPe worker processes. | 2,218 | 9,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.716523 |
https://keithbriggs.info/lottery.html | 1,726,782,726,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00536.warc.gz | 298,081,965 | 3,333 | Keith Briggs
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### The lottery - an order statistics paradox?
Suppose 6 balls are drawn uniformly from 49 balls (numbered 1 to 49), without replacement.
Sort the balls by the value of their numeric label and let xi be the value of the ith ranked ball for i from 1 to 6.
Then, for k = 1,2,3,...,49:
(Proof: homework exercise.)
These distributions look like this for i = 1,2,3,4,5,6; and have the mean and mode as in the table. (Proof: another homework exercise.)
imeanmode
150/71
2100/710
3150/720
4200/730
5250/740
6300/749
In other words, the most likely value for the smallest numbered ball is 1, the most likely value for the second smallest numbered ball is 10, and so on. Everything I have claimed so far is correct.
Therefore, you should put your money on balls 1,10,20,30,40, and 49. | 279 | 1,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.822159 |
http://www.windows2universe.org/physical_science/basic_tools/sw_tools.html&dev=1 | 1,454,926,681,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701152987.97/warc/CC-MAIN-20160205193912-00280-ip-10-236-182-209.ec2.internal.warc.gz | 745,455,601 | 12,863 | ## Shop Windows to the Universe
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# Tools for Math and Science
Some ideas are used throughout the sciences. They are "tools" that can help us solve puzzles in different fields of science. These "tools" include units of measurement, mathematical formulas, and graphs.
Scientists use different systems of measurement, like the metric or English systems. Each system has different units, like the gram or pound or meter or foot. Some units, like the meter and mile and pound, are familiar; others, like the ångström or farad or Röntgen, are almost unknown outside of the scientific fields that use them. We need to know how to convert units from one system to another, as when we determine the metric temperature in degrees Celsius when supplied with the English Fahrenheit equivalent.
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## Fundamental Forces
The interactions in the Universe are governed by four forces (strong, weak, electromagnetic and gravitational). Physicists are trying to find one theory that would describe all the forces in nature as...more
## Tools for Math and Science
Some ideas are used throughout the sciences. They are "tools" that can help us solve puzzles in different fields of science. These "tools" include units of measurement, mathematical formulas, and graphs....more
## Starting Points for Science
Some ideas are used in many, many places throughout science. We have grouped these "starting points for science" into three clusters: space, time, and matter. "Space" is the word we use for everything...more
## What is a Supercomputer?
Some scientific problems and processes are so complex that you need SUPERCOMPUTING power to tackle them! Just what is a supercomputer? A supercomputer is a computer that is among the largest, fastest or...more
## Using Computers for Science
In the last decades, computers have become a normal part of life. They are used to send e-mail, write a school report or look up recipes. They are used to keep track of the balance in your bank account....more
## Universal Time
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https://www.usmle-forums.com/threads/hardy-weinberg-equation.1158/ | 1,675,042,644,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499790.41/warc/CC-MAIN-20230130003215-20230130033215-00329.warc.gz | 1,070,049,498 | 25,290 | 1 - 8 of 8 Posts
#### fauzan
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guys help me out with this question.... i ve no clue to how can this be solved.... btw this is a usmlerx qmax question:
Q: A researcher has designed a new test to screen for hereditary spherocytosis. The test has a false-negative rate of 0 and a false-positive rate of 1/16 (6.25%). Suppose the new test is to be used in a population in which the frequency of the disease-causing allele is 0.2.
What is the positive predictive value of the diagnostic test in this setting?
A: 40%
B: 80%
C: 90%
D: 93.75%
E: 100%
#### khanar
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If it is USMLERx question then why didn't you check the answer!
#### rasheed
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Hereditory spherocytosis is autosomal dominant, therefore the frequency of allele is the frequency of the disease itself which is given as 0.2 which means 20/100.
Knowing the false positive and false negative rates we can construct a 2X2 to solve the problem.
2X2 is constructed this way
TP FP
FN TN
let's suppose the total population is 100, then TP must be 20 and FP is one sixteenth of (100-20) which is equal to 5, FN is given as zero.
So our table should look like this
20 5
0 75
Positive predictive value is equal to TP/TP+FP ---> 20/25 which is 80%
http://www.usmle-forums.com/usmle-step-1-bits-pieces/1136-2x2-statistics-epidemiology-table.html
#### fauzan
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according to usmlerx:
hereditary spherocytosis is an autosomal dominant condition. if hardy-weinburgh equatuion is used, the prevalence of disaease is [(0.2 x 0.2) + (2x 0.2 x 0.8)] =0.36. using the information above and creating a 2x2 table or using the formula. ppv= (sensitivity)(prevalence)/[(sensitivity)(prevalence)+(1-specificity)(1- prevalence).
in the latter case, the calculation shows: ppv= (1.0)(0.36)/[(1.0)(0.36)+(1/16)(0.64)] = 0.9
thus ppv = 90%
but i still didnt get it......
#### DrSeddik
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Haardy Weinberg: 1= q2+2qp+p2
Where q2= homozygote for the disease gene
2pq= heterozygote for the gene
In this inappropriately high prevalence of the gene you have to take into consideration p (p doesn't equal 1 here, instead we use 1=p+q and p will be 0.8) and q2 (severly affected and homozygotes for the autosomal dominant gene which are usually discarded due to their insignificantly small number)
So instead of calculating the prevalence of an autosomal domnant disease by 2q, we will use : q2+2pq= 0.2square + 2 x 0.2 x 0.8 = 36
So in a population of 100, 36 will be diseased and 64 will be healthy
-false +ve rate is 1/16, so false positives are 1/16 x 64 = 4
-false -ve rate is 0.0 (FN= 0.0) so true positive is 36
PPV= 36/(36+4) =0.9
First of all this disease is highly prevelant that it affects 36% of the population
It took me 15 minutes to figure it out so I would leave it in the exam...
#### indian
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woe ...
That's the most difficult stat question if seen so far
Nobody can answer it during the exam, even drseddik
#### fauzan
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yep!!! hope not to find such a ques in da real xam
by the way wat abt this formula:
ppv= (sensitivity)(prevalence)/[(sensitivity)(prevalence)+(1-specificity)(1- prevalence).
is this an authentic formula????
i dont remember to come across it any where else!!!!!
#### DrSeddik
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by the way wat abt this formula:
ppv= (sensitivity)(prevalence)/[(sensitivity)(prevalence)+(1-specificity)(1- prevalence).
is this an authentic formula????
i dont remember to come across it any where else!!!!!
If you read it in USMLerX then it's probably correct, it shows the relationship between prevalence and PPV.. I think it's a modulation of the other formulas, you can prove it and learn it, but as long as you got the 2x2 table it's not that necessary
1 - 8 of 8 Posts | 1,179 | 3,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-06 | latest | en | 0.834832 |
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# More efciently notice in the simple approach we use a
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Unformatted text preview: Fall 2010 6 2 5 9 Heapsort • Building the Heap more efficiently // building the heap for(int i = (size 1)/2; i >= 0; i ) HeapRebuild(array, size, i); CPSC 223 Fall 2010 10 5 10/26/10 Heapsort • Sorting from the Heap more efficiently – Notice in the simple approach we use a temporary array – We can avoid this by performing heapsort “in place” – After each deletion, maintain two partitions of the array: • The sorted region (the end of the array) • The Heap region (the beginning of the array Heap array = Sorted … 0 1 2 … last last+1 n 1 CPSC 223 Fall 2010 11 Array-Based Heap Implementation void heapRebuild(int theArray, int size, int index) { int child = 2*index + 1; if(child < size) { int right = child + 1; if(right < size && theArray[right] > theArray[child]) child = right; if(theArray[index] < theArray[child]) { swap(theArray[index], theArray[child]); heapRebuild(theArray, size, child); } } } CPSC 223 Fall 2010 12 6 10/26/10 Heapsort void heapsort(int theArray, int size) { for(int i = (size-1)/2; i >= 0; i--) heapRebuild(theArray, size, i); // build int last = size-1; for(int j = 1; j < size; j++) { swap(theArray[0], theArray[last]); heapRebuild(theArray, last, 0); last--; } // delete } CPSC 223 Fall 2010 13 Comparison of Sorting Algorithms Best Case Average Case Worst Case SelecSon Sort O(n2) O(n2) O(n2) Bubble Sort O(n) O(n2) O(n2) InserSon Sort O(n) O(n2) O(n2) Mergesort O(nlogn) O(nlogn) O(nlogn) Quicksort O(nlogn) O(nlogn) O(n2) Treesort O(nlogn) O(nlogn) O(n2) Heapsort O(nlogn) O(nlogn) O(nlogn) CPSC 223 Fall 2010 14 7...
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## This document was uploaded on 03/18/2014 for the course CPSC 223 at Gonzaga.
Ask a homework question - tutors are online | 697 | 2,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-09 | longest | en | 0.521207 |
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# Compound Arithmetic Function
LabVIEW 2013 Help
Edition Date: June 2013
Part Number: 371361K-01
»View Product Info
Owning Palette: Numeric Functions
Requires: Base Development System
Performs arithmetic on one or more numeric, array, cluster, or Boolean inputs. To select the operation (Add, Multiply, AND, OR, or XOR), right-click the function and select Change Mode from the shortcut menu. When you select this function from the Numeric palette, the default mode is Add. When you select this function from the Boolean palette, the default mode is OR.
The connector pane displays the default data types for this polymorphic function.
value 0..n-1 can be a number or Boolean value, an array of numbers or Boolean values, a cluster, array of clusters, and so on. You can wire a waveform to only one value input. If an input is a waveform, you can have an unlimited number of scalar inputs of varying sizes. If value is an error cluster, only the status parameter of the error cluster passes to the input terminal. result returns the result of the selected operation applied to the value 0..n-1. For AND, OR, or XOR, result returns the bitwise operations on numeric inputs and logical operations on Boolean inputs.
## Compound Arithmetic Details
Add inputs to the node by right-clicking an input and selecting Add Input from the shortcut menu or by resizing the function.
You can invert the inputs or the output of this function by right-clicking the individual terminals and selecting Invert from the shortcut menu. For Add, select Invert to negate an input or the output. For Multiply, select Invert to use the reciprocal of an input or to produce the reciprocal of the output. For AND, OR, or XOR, select Invert to bitwise complement an integer input or output or to logically negate an input or output.
When you use the Compound Arithmetic function to perform an XOR operation on 3 or more values, the Compound Arithmetic function performs an XOR operation on the first pair of inputs, then performs an XOR operation on the result of the first pair of inputs and the next input, and so on until all inputs have been processed.
Note You cannot use this function with fixed-point numbers. If you wire fixed-point numbers to this function, the VI appears with a broken Run button. | 487 | 2,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.642629 |
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# How many Bitcoin addresses are possible
As you can see, there are 633,008 addresses that have at least 1 BTC (1.68 million coins in total), which is 2.18% of all addresses out there. However, one address doesn't necessarily equal one person, so we'll have to dig deeper in order to obtain a realistic range As of September 1, there are 157,000 addresses holding at least 10 BTC—representing the top one percent of all Bitcoin holders—according to data from Coin Metrics Active Addresses last 24h (Number of unique (from or to) addresses per day) 825,606: 100 Largest Transactions: last 24h: 916,788 BTC (\$34,434,715,938 USD) 67.75% Total: First Block (Bitcoin creation date) 2009-01-09: Blockchain Size (Bitcoin database size) 403.25 GB: Reddit subscribers: 2,979,896: Tweets per day #Bitcoin: 114,558: Github. Using Trezor Suite, you can generate any of the three types of address listed above in just a couple of clicks. Add a new account type through the sidebar menu on the Accounts page. Due to how Bitcoin has developed, improving over time, you will find that addresses are backwards-compatible
According to one source, there are 1,461,501,637,330,902,918,203,684,832,716,283,019,655,932,542,976 possible Bitcoin addresses, which means that each person on Earth would be allowed to generate 196,385,600,286,334,710,857,791,565,804,391,698,421 addresses If we place all possible satoshis into a wallet of their own, we would get the maximum number of wallets that could have any balance to them (so the actual number of wallets with bitcoins is obviously less). This is 21x10^6 (BTC) x 10^8 (satoshi/BTC) = 21 x10^14 wallets. It's a huge number, but it's eclipsed by 2^256 possible wallets Bitcoin transactions are pseudo-anonymous, with forensic analysis, any Bitcoin address used in a transaction is likely to be traceable. Why are there only 21 million Bitcoin? There are many theories out there According to the Bitcoin whitepaper, a Bitcoin address is an identifier of 26-35 alphanumeric characters. It begins with the digit 1, 3 or bc1. This describes a potential end for a bitcoin payment. Addresses can be created at zero cost by Bitcoin users
### How Many People Actually Have At Least 1 Bitcoin
1. Bitcoin wallet addresses always start with a 1, 3, or bc1, making them easily distinguishable from public keys and private keys. Bitcoin public keys and addresses are presented in various formats. Here are a few examples. Note: These examples are listed for educational purposes only
2. In this guide, we will look into how many Bitcoins are there, understand what limits are placed on Bitcoins supply, the purpose behind it and the current count of Bitcoins in circulation and of course, some speculations as to what entails beyond the end of Bitcoin's supply. Let's dig in. How many Bitcoins are currently in circulation as of 2020? As of 2020, there are just over 18 Million Bitcoins in existence. However, not all of them are actually usable. Among those 18.
3. ers will probably be supported exclusively by numerous small transaction fees
4. As you can see, over 95% of all bitcoins in circulation are owned by about 4% of the market. In fact, 1% of the addresses control half the entire market. There are a couple limitations in our data. Most importantly, each address can represent more than one individual person
### Number of bitcoin addresses with at least 10 bitcoins
A single entity can own and control multiple Bitcoin addresses holding BTC. Beyond Address Counts In order to obtain a more precise estimate of the actual number of users in the Bitcoin network, and how this number changes over time, more advanced methodologies are required Data from the 2020 article and the current Bitcoin Rich List published on bitinfocharts.com, indicates whale addresses with 10,000 to 100,000 bitcoins dipped from 106 addresses to 85 There are only 21 million bitcoins that can be mined in total. Once bitcoin miners have unlocked all the bitcoins, the planet's supply will essentially be tapped out Unlike a conventional bank account number, these addresses can include both numbers as well as letters and be up to 35 characters in length. In practice though, most addresses tend to be 33 or 34 characters long. Here's what one looks like: A popular misconception among new users is that a Bitcoin address resembles an email address
After all, you can use on-chain analysis to find out how many bitcoin accounts exist and how much is in each account. But you can't see who owns those accounts. So, the actual number of bitcoin millionaires is almost certainly lower than 100,000 if you assume some of these addresses are held by the same individuals The key to keeping your Bitcoin transactions from being traced back to you is preventing others from knowing which addresses are yours.If you're trying to remain anonymous (or more precisely, pseudonymous) with Bitcoin, read on for the most common ways people's true identities are forever associated with their Bitcoin addresses How many of the 18 million Bitcoins accumulated today can enter the circulation channel? How many have become dead coins and should be removed from the 21 million cap forever There are multiple Bitcoin address types, currently P2SH or pay-to-script hash is the default for most wallets. P2PKH was the predecessor and stands for Pay to Public Key Hash
### Bitcoin (BTC) statistics - Price, Blocks Count, Difficulty
• Bitcoin transactions are recorded on a public ledger. Anyone who traces a public address can know the origin and/or destination. There is no protocol-level procedure to anonymize these bitcoins, which is why a Bitcoin mixer is required to hide identity.. Bitcoin mixing is a process that tries to break the linkability or traceability
• With Bitcoin, for example, it is common for a new Bitcoin address to be created for each transaction, and the remaining balance is then transferred to this address. Users can also create multiple addresses for other cryptocurrencies. Probably the best approach is via the user accounts of crypto exchanges
• One of the core features of the Bitcoin currency is its limited supply, which is why it's important to know how many Bitcoins are left. The importance of a limited supply has never been more apparent than now, as central banks around the world ramp up their money printing to unprecedented levels
Bitcoin Address Lookup Search and Alerts. View and research bitcoin ownership, transactions and balance checker by name, bitcoin address, url or keywor Once you have the Bitcoin address, you can then investigate what other transactions that person has made. You can also check how many Bitcoins in total they have! This is where the Wraith Protocol will be different. Check out the example below: John decides to send Bob 100 Verge (XVG) coins. John's address is V123 and Bob's address is V456 With the rapid rise in the number of bitcoin scams, there are easy ways to check if a bitcoin address has been reported as being used by scammers, such as in fake bitcoin giveaways. You can also. With ether, a rival to bitcoin, the top 100 addresses control 40 percent of the supply, and with coins such as Gnosis, Qtum, and Storj, top holders control more than 90 percent. Many large owners. Many online bitcoin services retain their customers private bitcoin keys, which means the accounts are vulnerable to hackers and fraudsters (remember the time Mt. Gox lost 850,000 bitcoins from its customers accounts in 2014?) or governments (like the time BTC-e, a Russian bitcoin exchange, had its domain seized by US District Court for New Jersey in August, freezing the assets of its users)
The Bitcoin reward is divided by 2 every 210,000 blocks, or approximately four years. Some of the Bitcoins in circulation are believed to be lost forever or unspendable, for example because of lost passwords, wrong output addresses or mistakes in the output scripts The crawler maintained by Bitnodes connects from these IP addresses: 88.99.167.175, 88.99.167.186, 2a01:4f8:10a:37ee::2 Bitnodes API v1.0 · Network Snapshot · Charts · Live Map · Network Map · Leaderboard · Bitcoin Client Statu
Cryptographer David Chaum first proposed a blockchain-like protocol in his 1982 dissertation Computer Systems Established, Maintained, and Trusted by Mutually Suspicious Groups. Further work on a cryptographically secured chain of blocks was described in 1991 by Stuart Haber and W. Scott Stornetta. They wanted to implement a system where document timestamps could not be tampered with Bitcoin (₿) is a decentralized digital currency, without a central bank or single administrator, that can be sent from user to user on the peer-to-peer bitcoin network without the need for intermediaries. Transactions are verified by network nodes through cryptography and recorded in a public distributed ledger called a blockchain.The cryptocurrency was invented in 2008 by an unknown person. In 2020, the number of wallets - defined as a set of blockchain addresses controlled by a single entity - holding at least 1,000 bitcoins has increased by 302 (17%) and is now at a record high. A valid Bitcoin address is like a bank account number using which you store your bitcoins and check your balances. For those who are seeing their Bitcoin addresses for the first time, I would say that it won't look like traditional bank account number but instead, it looks like a long alphanumeric string starting usually with '1' or '3
The potential result of this is that Bitcoin transaction fees will become much more important. As the reward gets smaller, the transaction fee may become the main form of compensation for nodes. It is hoped that more people could use layer 2 solutions such as the Lightning Network in the future to reduce transaction costs significantly A Bitcoin private key is simply an integer between one and about 10 77. This may not seem like much of a selection, but for practical purposes it's essentially infinite. If you could process one trillion private keys per second, it would take more than one million times the age of the universe to count them all And (much like the smoke-bear-demon in Lost) Bitcoin doesn't exist in a physical form so you can't drop it in the street or leave it behind somewhere.But you can still lose it, as many people.
An early response from user Coin-1 politely attempts to dissuade anyone from proceeding any further: Let's calculate how much time you need to crack one Bitcoin-address on your machine Also in the software you tell the pool which Bitcoin address payouts should be sent to. If you don't have a Bitcoin wallet or address learn how to get one here. There is mining software available for Mac, Windows, and Linux. Step #6: Is Bitcoin Mining Legal in your Country? Make Sure! This won't be much of an issue in MOST countries
### Bitcoin Addresses and How to Use them by SatoshiLabs
Bitcoin aficionados now believe the cryptocurrency is more useful as a new kind of alternative asset, like gold.Many people on Wall Street have bought into that idea, and the Chicago Mercantile. Step 3: Select Wallet Address and enter the recipient's wallet address in the recipient box. Typically, wallet addresses are easy to copy and paste in or scan when using QR codes. Select the wallet you wish to send from and then input how much you wish to send in USD or in BTC, adding a note if you wish Chapter 4. Keys, Addresses You may have heard that bitcoin is based on cryptography, which is a branch of mathematics used extensively in computer security. Cryptography means secret writing - Selection from Mastering Bitcoin, 2nd Edition [Book
How do Bitcoin Transaction Confirmations Work? To understand what happens to unconfirmed Bitcoin transactions, it is necessary to understand how the whole system works: A transaction is a message sent to the network that includes the public key of a Bitcoin address, a signature corresponding to the private key, an amount of Bitcoin being sent, and a recipient address An address represents the public key of an asymmetric key pair and is a destination for a Bitcoin payment. Each address (public key) has a corresponding private key which is used to move Bitcoins out the associated address (public address or public key) It is a dynamic parameter that controls hash power needed to mine a bitcoin block. New bitcoins are generated roughly every 10 minutes, but your ability to earn those newly created bitcoins is dependent on how much computational power you have relative to how much computational power is on the network So much so, that a singular Bitcoin went from being worth £3,600 in March last year to more than £27,000 now. It is possible to lose your Bitcoin wallet or delete your Bitcoins and lose them. GPU Mining. Aside from using ASIC and CPU, you can use graphic cards (GPU).Yes, if you have a gaming computer with a good dedicated graphics card you can use it to mine Bitcoin. This year, Radeon and Nvidia release cards that can mine Bitcoins at comparable rates. A Nvidia GTX 1070 may cost you from \$699 to \$850.If you use this for gaming for 4 hours, you can allot it to mining for the.
It is also possible to get Bitcoin at specialized ATMs and via P2P exchanges. However, be aware that Bitcoin ATMs were increasingly requiring government-issued IDs as of early 2020 This section shows which address is sending cryptocurrencies associated to it, as well as how much it is sending. You can also click on the address to see its incoming and outgoing transaction history. When you make a Bitcoin transaction, you will automatically send the full amount from your address with the rest sent to your change address I have two bitcoin addresses with private keys (year 2013). I want to create cold storage with Electrum-3.1.3 (as no HD wallet). 1) watch-only wallet: I will create it according to your guide above on the online computer, using Import bitcoin addresses Mining with the latest algorithms allows to make as much Bitcoin as possible. We aim to provide you with the easiest possible way to make money without having to do any of the hard stuff. With data centers around the globe, we aim to keep bills down and mining power high, meaning you can make more in a shorter amount of time than what it would take to mine from your home for instance
Bitcoin is designed so that for every payment you can use a new address that is not tied to any of your previous addresses. When you use a new address for every coin, then it becomes much much more difficult to find out that these coins are yours After you learn how to sell and buy Bitcoin (or any other cryptocurrency), you also need to learn how to transfer Bitcoin between wallets. We have noticed that many people do not even know that something like this is possible. On the other hand, people that do know that this is possible do not know how to do it This shows how Bitcoin is distributed amongst the entire ecosystem and many whale addresses there are. Bear in mind that a single person/entity can hold multiple Bitcoin addresses at any one time! Balance Addresses % Addresses Coins & Value % Coins; No data is available now Bitcoin ATMs are installed in many countries in the world. The current distribution of installations accross the countries can be found on The Chart of Bitcoin ATM number by Continents and Countries. United States (17947 locations) Canada (1495 locations) United Kingdom (199 locations) Austria (159 locations
Now, a Bitcoin wallet can find as many P2PKH addresses as possible, which is ideally a combination of several non-exceptional cryptographic operations. Bitcoin uses ECDSA cryptographic algorithm. Ideally, in blockchain, the wallet address is the result of hashing the public key via cryptographic algorithms and other conversions A brute force attack on a Bitcoin private key is, in theory, much like a brute force attack on any regular password. An exhaustive search of possible combinations is carried out before a private key combination is identified. In reality, brute force attacks on a Bitcoin private key are as close to mathematically impossible as it gets The Luno Learning Portal. You don't have to buy a whole Bitcoin, you can buy half or even a fraction of one. These smaller portions are known as satoshis.. It's never too late to get started with Bitcoin. Learn, buy and use Bitcoin with Luno now Accurate Bitcoin mining calculator trusted by millions of cryptocurrency miners since May 2013 - developed by an OG Bitcoin miner looking to maximize on mining profits and calculate ROI for new ASIC miners. Updated in 2021, the newest version of the Bitcoin mining calculator makes it simple and easy to quickly calculate mining profitability for your Bitcoin mining hardware
### Can You Use the Same Bitcoin Address Twice? - The
How much power does Bitcoin need? Undisputed numbers are hard to come by because of the complex nature of the calculations. Back at the start of 2017, Bitcoin was using 6.6 terawatt-hours of power. Satoshi Nakamoto is the name used by the presumed pseudonymous person or persons who developed bitcoin, authored the bitcoin white paper, and created and deployed bitcoin's original reference implementation. As part of the implementation, Nakamoto also devised the first blockchain database. In the process, Nakamoto was the first to solve the double-spending problem for digital currency using a.
### Can you trace a Bitcoin address? Lun
While those Bitcoin are now worth hundreds of millions of dollars, he lost his passwords many years ago and has put the hard drives containing them in vacuum-sealed bags, out of sight Address: a long string of letters and numbers (in the case of Bitcoin, 26 to 35 characters) that represents a destination for one or more payments. The network ensures all addresses are unique When you receive some bitcoins, say 0.1 BTC, you see them in your bitcoin wallet, listed under a bitcoin address. At the same time, the bitcoins are not actually stored in the wallet, they are just an entry in the public blockchain. What the wallet stores is your secret private key that belongs to that address. Since you control that private.
If you are still planning to buy bitcoins from a bitcoin ATM you might want to use our bitcoin ATM map to find a location or check our tutorial on how to buy bitcoins from bitcoin ATM. Not all bitcoin machines support generating and printing of address on receipts, many don't even support printing receipts at all Many early Bitcoin investors are in a painful predicament. They can't remember the complex security codes they originally created to gain access to their Bitcoin wallet
### Do Bitcoin Addresses Expire? - CryptoTicke
BTC Price Live Data. The live Bitcoin price today is \$38,856.97 USD with a 24-hour trading volume of \$43,381,361,707 USD.. Bitcoin is down 0.06% in the last 24 hours. The current CoinMarketCap ranking is #1, with a live market cap of \$727,402,166,809 USD It is technically possible to mine Bitcoin using a laptop. However, mining Bitcoin these days is primarily done using ASIC miners who are far more advanced than laptops and desktop computers, which makes it unprofitable to do so After you've checked that the addresses match, enter how much Bitcoin you want to send from Coinbase to Exodus. You can enter the amount in USD (or whatever your local currency is) or in BTC (highlighted in orange). You can also enter an optional note (highlighted in purple) to associate with the transaction
### How Many Bitcoins Are There? How Many Left to Mine? (2021
Bitcoin is Secure. Bitcoin miners help keep the Bitcoin network secure by approving transactions. Mining is an important and integral part of Bitcoin that ensures fairness while keeping the Bitcoin network stable, safe and secure. Links. We Use Coins - Learn all about crypto-currency. Bitcoin News - Where the Bitcoin community gets news When you register a new account at NiceHash, you also get access to multiple cryptocurrency addresses (wallets). The following cryptocurrencies are supported: Bitcoin (BTC) Ethereum (ETH) Ripple (XRP) Bitcoin Cash (BCH) Litecoin (LTC) ZCash (ZEC) Dash (DASH) Stellar (XLM) Monero (XMR) EOS (EOS) Tether (USDT) Chainlink (LINK) Basic Attention.
Coinbase is a secure platform that makes it easy to buy, sell, and store cryptocurrency like Bitcoin, Ethereum, and more. Based in the USA, Coinbase is available in over 30 countries worldwide I am sure you know all about Bitcoin and how popular it is, but more and more people are now choosing to mine Litecoin instead! Here's a fun fact for you. When people mine Bitcoin , it takes 10 minutes for the network to confirm each block, but Litecoin is four times faster at just 2.5 minutes
Master The Crypto is a user-first knowledge base featuring everything bitcoin, blockchain and cryptocurrencies. The MTC resource center aims to bridge the gap by featuring easy-to-understand guides that build up and break down the crypto ecosystem for many The '*' means multiple items may be assigned, e.g. an account has one or more users, a user may have multiple calendars and a mailbox may have multiple email aliases (paid feature). By the way, a mailbox also has a main email address Buy Bitcoin online with your credit card, debit card, bank transfer or Apple Pay. Buy Bitcoin Cash (BCH), Bitcoin (BTC) and other cryptocurrencies instantly
### How To Find Out Who Owns A Bitcoin Address - The
In the early days of Lightning, the developers limited how much Bitcoin could be kept inside a Lightning payment channel to 0.1677 BTC; Wumbo channels enable nodes to service larger transactions and higher volumes. Crypto exchange Bitfinex is among those who've announced support for Wumbo channels Can you imagine how you would feel if you somehow lost access to the bitcoin stored in your wallet? It's not as unusual as you might think. In November 2017, Fortune reported on research from the digital forensics analysis firm Chainalysis that suggests as much as 3.78 million, or 23 percent, of all existing bitcoin are already lost. The Best Defense Is a Good Offens Bitcoin transactions are irreversible, once you've sent money to another Bitcoin address it is not possible for you or LocalBitcoins.com to reverse it. To get your Bitcoin back you can try to find the owner of the address and ask them to return the Bitcoin to you
No. Bitcoin isn't a company, so it's not possible to buy shares or Bitcoin stock. However, there are some Bitcoin-related businesses that trade publicly. For example, it's possible to buy shares in Bitcoin mining companies. Historically, the share prices of publicly-traded Bitcoin-related companies rise and fall with Bitcoin prices The Bitcoin world is abuzz with both excitement and curiosity and the opportunity for upside potential to skyrocket. Everyone from everyday Joes to reputable experts are betting on Bitcoin's success One of the current popular bitcoin miners out there is the ANTMINER S9 11.85TH/s bitcoin miner, currently selling for around \$1442 or around R21000 on the manufacturers website and around R45000 locally in South Africa Spoofy on July 23rd at 00:59. These bids were unusual for Spoofy, because this time he left them up for a little bit. It supported the price and he didn't really have to buy very many Bitcoin with his bids
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• Ally Bank IRA ROLLOVER form. | 5,123 | 23,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.920141 |
https://edurev.in/studytube/Geometric-Mean-Measures-of-Central-Tendency--Busin/1265ac3b-231c-49ab-8e99-1b06061fe4f9_v | 1,701,958,747,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100674.56/warc/CC-MAIN-20231207121942-20231207151942-00773.warc.gz | 257,860,772 | 52,971 | Geometric Mean - Measures of Central Tendency, Business Mathematics & Statistics
Geometric Mean - Measures of Central Tendency, Business Mathematics & Statistics Video Lecture - Business Mathematics and Statistics - B Com
115 videos|142 docs
FAQs on Geometric Mean - Measures of Central Tendency, Business Mathematics & Statistics Video Lecture - Business Mathematics and Statistics - B Com
1. What is the geometric mean and how is it calculated?
Ans. The geometric mean is a measure of central tendency used to find the average of a set of values that are exponentially increasing or decreasing. It is calculated by taking the nth root of the product of n values.
2. When is the geometric mean used in business mathematics and statistics?
Ans. The geometric mean is commonly used in business mathematics and statistics when dealing with growth rates, investment returns, interest rates, and other variables that exhibit exponential growth or decay.
3. How does the geometric mean differ from the arithmetic mean?
Ans. The arithmetic mean calculates the average by summing all the values and dividing by the total number of values, while the geometric mean calculates the average by multiplying all the values and taking the nth root of the product. The geometric mean is more suitable for data that has a multiplicative relationship.
4. Can the geometric mean be negative or zero?
Ans. No, the geometric mean cannot be negative or zero. Since it involves taking the nth root, all values must be positive for the calculation to be valid. If there are any negative or zero values, the geometric mean cannot be calculated.
5. What are some practical applications of the geometric mean in business?
Ans. The geometric mean is widely used in finance and investment analysis. It is used to calculate compound annual growth rates (CAGR), average investment returns, and inflation rates. It is also used in industries such as insurance, real estate, and manufacturing to analyze trends and measure performance over time.
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; | 486 | 2,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-50 | latest | en | 0.88641 |
https://www.nevadaappeal.com/news/2012/jan/21/racing-keep-it-slow-and-safe-during-inclimate-weat/ | 1,619,123,341,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039604430.92/warc/CC-MAIN-20210422191215-20210422221215-00591.warc.gz | 1,005,613,518 | 8,847 | # RACING: Keep it slow and safe during inclimate weather
We had our first little taste of winter last week, with more possibly on the way this weekend. So, keeping in mind that very few of us have the car control ability of a Sebastian Vettel, Tony Stewart or Dario Franchitti, here are my annual winter driving tips.
First, a little basic physics. Say you're driving a 4,000 pound vehicle at 30 miles per hour -that's a serious amount of mass and inertia. This collection of steel, aluminum, glass, and plastic is tenuously connected to the road by four small tire contact patches, and any change in speed or direction must be transmitted through those patches. Anything that detracts from the tires' adhesion to the road (snow or ice for instance) tends to limit their response to your control inputs. Tires can transmit only three things from the vehicle to the road - acceleration, deceleration, and change of direction. There is a limit, determined by road surface, temperature, and other factors, of just how much of these inputs the tires can transmit before they lose grip. There is another factor, which is when the tire receives multiple inputs, such as braking and steering simultaneously. This creates a vector of the two forces, but reduces the absolute limit of either force.
OK, we'll let professor Einstein take a break now and put theory into practice. Say your vehicle can stop from 30 mph in 100 feet on dry asphalt pavement. On snow or ice, that distance is going to be doubled, tripled, quadrupled, or may even approach infinity. Remember to slow down in slick conditions, and to increase your distance to the car in front of you. Otherwise, unintended Daytona style bump drafting may occur. Also remember the reference to vectors in our physics lesson. If you try to brake and turn at the same time on a slick surface, both your braking and turning performance will suffer. You will immediately experience what the race driver calls "push" and you will plow straight ahead with your wheels cranked to the right or left. Similarly, if you accelerate too hard while turning, you will be confronted with the condition racers call "loose". You will notice this when the rear of your car passes you.
The thing to remember is to use gentle inputs on all the controls. Brake early and progressively; turn slowly and smoothly, not jerking the steering wheel and accelerate smoothly, feathering the throttle. You also need to stay more alert in slick conditions. Get "up on the wheel" as Darrell Waltrip is fond of saying. Unfortunately, the level of concentration most people bring to their daily driving is just above comatose.
•••
The technical and rule-making folks at NASCAR are busy poring over mountains of data collected at last week's Daytona testing. We may see some tweaks to one or more of the elements of the new Daytona package prior to the Shootout as a result.
Those elements consist of restrictor plate size, cooling system capacity, rear spoiler size, and mandatory spring rates. NASCAR is trying to break up the two-car tandem draft and get back to pack racing at the two restrictor-plate tracks, Daytona and Talladega. So we will probably see more of the old-style pack racing until the last 10 miles or so. Then I'm betting that no matter what NASCAR does, somebody will hook up in a bump draft to go to the front, hopefully taking the checkered flag before the pushing car's engine blows.
•••
Friday night, NASCAR hosted the induction ceremony for the third class to be inducted into the Hall of Fame. The better-known inductees this year are Darrell Waltrip (three Cup championships, 84 race wins), Cale Yarborough (three Cup Championships, 69 race victories), Dale Inman (Richard Petty's cousin and long-time crew chief), and Glenn Wood of the legendary Wood Brothers. Less well known is Richie Evans, who spent his career in NASCAR's Modified division. Evans won an estimated 475 races in approximately 1,300 starts, scoring nine championships before losing his life in a practice accident at Martinsville at age 44. | 859 | 4,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-17 | latest | en | 0.950814 |
http://forums.freshershome.com/search.php?s=c554f5b1c5f87f397ad1727a274e2e5a&searchid=3062333 | 1,618,921,783,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039398307.76/warc/CC-MAIN-20210420122023-20210420152023-00464.warc.gz | 36,975,451 | 7,066 | # Search:
Type: Posts; User: tarun85; Keyword(s):
1. ## Thread: Describe a pattern in the sequence of numbers.Predict the next number.?
by tarun85
Replies
1
Views
5,233
### nt getting the answer for 2 .. 3) 35
nt getting the answer for 2 ..
3) 35
2. ## Thread: Find the next number in all sequences?
by tarun85
Replies
5
Views
7,780
### number one 29..pattern is just adding the two...
number one 29..pattern is just adding the two previous numbers
number two 140...in this the diff. between two number is the square of a number starting from 2 onwards...
3. ## Thread: A SIMPLE MATHEMATICS EQUATION, (for those who think that they are smart)?
by tarun85
Replies
9
Views
7,593
### no need to go tooo long....7*12=84 ...simple...
no need to go tooo long....7*12=84 ...simple table u just know...
4. ## Thread: what is the pattern in this sequence of numbers. predict the next
by tarun85
Replies
9
Views
10,538
### we can also do it like this... 0 1*1+2=3...
we can also do it like this...
0
1*1+2=3
2*2+4=8
3*3+6=15
4*4+8=24
5*5+10=35
square of no starting from 1+the multiple of 2 in sequence.....
5. ## Thread: What's the next number in the sequence?
by tarun85
Replies
16
Views
17,510
### Great man where u find this logic...????
Great man where u find this logic...????
6. ## Thread: What is the 20th number in this sequence?896, 3687, 749, 34448, 836, ...
by tarun85
Replies
5
Views
6,432
### is this sequence given by gracey is right????
is this sequence given by gracey is right????
by tarun85
Replies
1
Views
3,981
B is just the power 4 of a number ...
2 to the power 4= 16.
3 to the power 4= 81
4 to the power 4= 256.
8. ## Thread: solve this math puzzle !!
by tarun85
Replies
2
Views
4,494
### 5101 or 6001 can be the answer...if i m wrong...
5101 or 6001 can be the answer...if i m wrong then plz tell the right answer and the logic...
9. ## Thread: Brain teaser!!! First correct answer gets Best Answer! (I hope you are good at
by tarun85
Replies
7
Views
5,855
### kayla u r very near ...9604 is the no of legs of...
kayla u r very near ...9604 is the no of legs of cats ....and for those 7 girls it is 14...total no of legs are 9618....add 2 more if the driver is included....
Results 1 to 9 of 9 | 725 | 2,237 | {"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} | 3.59375 | 4 | CC-MAIN-2021-17 | latest | en | 0.883926 |
https://whatisconvert.com/788-centimeters-in-kilometers | 1,708,540,548,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00112.warc.gz | 654,993,056 | 7,057 | ## Convert 788 Centimeters to Kilometers
To calculate 788 Centimeters to the corresponding value in Kilometers, multiply the quantity in Centimeters by 1.0E-5 (conversion factor). In this case we should multiply 788 Centimeters by 1.0E-5 to get the equivalent result in Kilometers:
788 Centimeters x 1.0E-5 = 0.00788 Kilometers
788 Centimeters is equivalent to 0.00788 Kilometers.
## How to convert from Centimeters to Kilometers
The conversion factor from Centimeters to Kilometers is 1.0E-5. To find out how many Centimeters in Kilometers, multiply by the conversion factor or use the Length converter above. Seven hundred eighty-eight Centimeters is equivalent to zero point zero zero seven eight eight Kilometers.
## Definition of Centimeter
The centimeter (symbol: cm) is a unit of length in the metric system. It is also the base unit in the centimeter-gram-second system of units. The centimeter practical unit of length for many everyday measurements. A centimeter is equal to 0.01(or 1E-2) meter.
## Definition of Kilometer
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.
## Using the Centimeters to Kilometers converter you can get answers to questions like the following:
• How many Kilometers are in 788 Centimeters?
• 788 Centimeters is equal to how many Kilometers?
• How to convert 788 Centimeters to Kilometers?
• How many is 788 Centimeters in Kilometers?
• What is 788 Centimeters in Kilometers?
• How much is 788 Centimeters in Kilometers?
• How many km are in 788 cm?
• 788 cm is equal to how many km?
• How to convert 788 cm to km?
• How many is 788 cm in km?
• What is 788 cm in km?
• How much is 788 cm in km? | 458 | 1,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-10 | latest | en | 0.878596 |
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