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https://www.r-bloggers.com/euler-problem-2-even-fibonacci-numbers/ | 1,555,582,739,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578517558.8/warc/CC-MAIN-20190418101243-20190418123243-00489.warc.gz | 784,309,417 | 16,383 | # Euler Problem 2: Even Fibonacci Numbers
December 7, 2016
By
(This article was first published on The Devil is in the Data, and kindly contributed to R-bloggers)
## Euler Problem 2 Definition
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
$1, 2, 3, 5, 8, 13, 21, 34, 55, 89, \ldots$
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
## Solution
The code generates Fibonacci numbers until it reaches the value of four million. The code then sums the even numbers in the sequence.
```fib <- c(1, 2) #Define first two numbers
while (max(fib) < 4e+06) {
# Generate Fibonacci numbers until limit is reached
len <- length(fib)
fib <- c(fib, fib[len - 1] + fib[len])
}
```
A range of packages exists that generate Fibonacci numbers. The gmp package for Multiple Precision Arithmetic and the numbers package supplies functions to calculate the nth Fibonacci number. This package is also able to work with very large numbers.
```library(gmp)
i <- 1
fib <- fibnum(1)
while (fibnum(i) <= 4e6) {
fib <- fibnum(i)
i <- i + 1
}
```
## Fibonacci Numbers as a magic trick
Fibonacci numbers describe many natural processes and can also be used to create magic tricks. The Missing Square Puzzle is based on this principle.
By Trekky0623 at English Wikipedia (Transferred from en.wikipedia to Commons.) [Public domain], via Wikimedia Commons
The post Euler Problem 2: Even Fibonacci Numbers appeared first on The Devil is in the Data.
R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: Data science, Big Data, R jobs, visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more... | 493 | 1,922 | {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-18 | latest | en | 0.805063 |
http://conversion.org/area/square-decimetre/board | 1,721,471,718,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515079.90/warc/CC-MAIN-20240720083242-20240720113242-00571.warc.gz | 8,712,006 | 6,718 | square decimetre to board conversion
Conversion number between square decimetre [dm²] and board [bd] is 1.2916692500052. This means, that square decimetre is bigger unit than board.
Contents [show][hide]
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from board to square decimetre conversion
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[dm²]
eg.: 10.12345 or 1.123e5
Result in board
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
Calculation process of conversion value
• 1 square decimetre = (exactly) (0.01) / (0.00774192) = 1.2916692500052 board
• 1 board = (exactly) (0.00774192) / (0.01) = 0.774192 square decimetre
• ? square decimetre × (0.01 ("m²"/"square decimetre")) / (0.00774192 ("m²"/"board")) = ? board
High precision conversion
If conversion between square decimetre to square-metre and square-metre to board is exactly definied, high precision conversion from square decimetre to board is enabled.
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square decimetre to board conversion chart
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square decimetreboard
00
1012.916692500052
2025.833385000103
3038.750077500155
4051.666770000207
5064.583462500258
6077.50015500031
7090.416847500362
80103.33354000041
90116.25023250047
100129.16692500052
110142.08361750057
Copy to Excel
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square decimetre
Definition of square decimetre unit: = 1 dm x 1 dm. Area equal with 1 decimetre by 1 decimetre square.
Convert Board to other unit:
board
Definition of board unit: ≡ 1 in × 1 ft. = 0.3048 × 0.0254 = 7.74192×10-3 m²
← Back to Area units | 588 | 1,903 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-30 | latest | en | 0.682694 |
https://keisan.casio.com/exec/system/14527311044369 | 1,685,642,823,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648000.54/warc/CC-MAIN-20230601175345-20230601205345-00626.warc.gz | 377,803,458 | 9,258 | # Sunshine hours (location) Calculator
## Calculate the sunshine hours at the home in consideration of the influence of the south side building.
Time-zone with UTC ⇒ refer to "Time Difference from UTC Calculator"
Longitude ,Latitude Time zone with UTCh (New York:-5, Japan:9) DSTOFFON Date JanFebMarAprMayJunJulAugSepOctNovDec/12345678910111213141516171819202122232425262728293031, Distance m, Heightm
The building is assumed to be located due south direction from the home and to be spread in the east-west direction at the same height.It will be investigated the sunshine from 3:00 to 21:00 at 15-minute intervals.azimuth angle: north=0, east=90, south=180, west=270 degree
Sunshine hours (location)
[1-10] /36 Disp-Num5103050100200
[1] 2023/05/18 00:30 40 years old level / High-school/ University/ Grad student / Useful /
Purpose of use
to make a decision whether solar panel is fexible option for operation and how to install in efficient way.
[2] 2022/11/07 11:31 60 years old level or over / A teacher / A researcher / Very /
Purpose of use
forecast agriculture activity as well as rice, fruits
Comment/Request
i want to calculate the sunshine hours based on the radiation observation, temperature and cloud cover
[3] 2022/10/24 18:22 40 years old level / An engineer / Very /
Purpose of use
Calculate daily sun to install solar panel system.
Comment/Request
Export data for a full year would be nice.
[4] 2022/10/06 07:37 40 years old level / An office worker / A public employee / Useful /
Purpose of use
What can I plant in garden? Do I need full sun plants or part sun.
Comment/Request
I wonder if I could use similar calculations on the east side of my house.
[5] 2022/09/11 03:39 60 years old level or over / A retired person / Useful /
Purpose of use
Soloar install
[6] 2022/03/21 22:12 50 years old level / Self-employed people / Very /
Purpose of use
Thinking about static solar vs sun tracking.
[7] 2022/03/13 15:57 60 years old level or over / A retired person / Very /
Purpose of use
Solar panel sizing
[8] 2021/08/17 06:53 30 years old level / Self-employed people / Very /
Purpose of use
calculation of solar panel
[9] 2021/08/09 04:16 30 years old level / Self-employed people / Useful /
Purpose of use
Ahorro electrico
[10] 2021/07/11 17:00 60 years old level or over / Self-employed people / Very /
Purpose of use
Calculating sunlit hours for a new garden design.
Comment/Request
Superior data, flawless operation, easy to use. Thank you.
Sending completion
To improve this 'Sunshine hours (location) Calculator', please fill in questionnaire.
Age
Occupation
Useful?
Purpose of use? | 736 | 2,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-23 | latest | en | 0.829932 |
https://ecency.com/hive-112018/@spectrumecons/ipsnvsde | 1,675,608,506,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00846.warc.gz | 254,514,497 | 20,051 | # Prevent, Solve, or Manage (Video Version)
Hi Everyone,
In 2020, I dedicated 10 written posts to the ‘Prevent, Solve, or Manage Approaches’. In the first post in that series, I described the general concepts behind ‘Prevent, Solve, and Manage’. In the other nine posts, I discussed how it could be applied to areas such as:
## What’s in the video?
In the video, in this post, I revisit the theory I discussed in the first post of the series. The theory is very important as it provides basis for how Prevent, Solve, or Manage approaches could be applied and how they could add value to tackling many different types of problems.
Prevent, Solve, or Manage approaches are discussed using a Venn diagram. This approach has been adopted as it is possible all three, some of the three, or just one of the three approaches could be used to tackle a problem. Generally, ‘prevent’ is prioritised over ‘solve’ and ‘solve’ is prioritised over ‘manage’. ‘Prevent’ stops the problem occurring; therefore, is generally preferred to ‘solve’, which involves removing the problem once it has arrived. ‘Solve’ is generally preferred to ‘manage’, which does not remove the problem but instead just mitigates the damage it causes. However, generally, it is better to have several approaches in place. It is possible prevention could fail. If it does, it would be good to be able to solve the problem or even just manage it.
It is important to note I have used the word ‘generally’. Each approach has a cost and a probability of success. For example, ‘solve’ could be preferred over ‘prevent’, if the cost of prevention is high and the probability of success is low. If the expected cost of prevention (probability × cost) exceeds the expected cost of solving plus the cost of the problem for the duration it takes for the problem to be solved, ‘solve’ would be a superior approach to prevention. However, combining ‘prevent’ and ‘solve’ approaches will lower the risk from detrimental effects of a problem even if not the expected cost. Other factors to consider are the certainty a problem will occur and if it is certain, how well can it be predicted to occur in advance.
Below are the main sections of the video and the timings in the video where I discuss these sections.
• Introduction (0:15 minutes)
• Describing ‘Prevent, Solve, or Manage using a Venn Diagram (1:30 minutes)
• Alliterative naming convention such as Prevent, Cure, or Treat (3:05 minutes)
• Extent of control over a problem (4:25 minutes)
• Varying risk when applying Prevent, Solve, or Manage (6:35 minutes)
• Comparing control and risk (9:04 minutes)
• How ability to predict a problem affects risk (10:18 minutes)
• How certainty of occurrence of problem affects risk (11:55 minutes)
• Examining probabilities of success (13:50 minutes)
• Examining costs of approach and problem (16:06 minutes)
• Explaining how expected value and costs affects application of Prevent, Solve, or Manage approaches (18:27 minutes)
• Explaining how priority can vary depending on probabilities of success as well as costs (22:05 minutes)
• Summarising the main points of the video (23:15 minutes)
• Brief description of the potential application of Prevent, Solve, or Manage (25:37 minutes)
If you want to read any of my other posts, you can click on the links below. These links will lead you to posts containing my collection of works. These 'Collection of Works' posts have been updated to contain links to the Hive versions of my posts.
## My New CBA Udemy Course
The course contains over 10 hours of video, over 60 downloadable resources, over 40 multiple-choice questions, 2 sample case studies, 1 practice CBA, life time access and a certificate on completion. The course is priced at the Tier 1 price of £20. I believe it is frequently available at half-price.
▶️ 3Speak
H2
H3
H4
3 columns
2 columns
1 column | 874 | 3,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-06 | latest | en | 0.96178 |
http://mathhelpforum.com/trigonometry/43089-maths-please-help.html | 1,519,356,064,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814311.76/warc/CC-MAIN-20180223015726-20180223035726-00125.warc.gz | 215,042,109 | 12,464 | There are 4 questions that i need help in and if you can even answer one of them that would be great ! Plus the level of these questions are just highschool. [Im in my last year of highschool]. I'd truly appreciate the help
1) Prove this identity:
(cosecx + cotx)(cosecx - cotx) = (cotx)(tanx)
I started with LHS = (cosecx + cotx)(cosecx - cotx)
but don't know how to continue because these really confuse me.
2) Solve the following equation for theta (θ) over the given domain:
√3 sin 2θ = cos 2θ
Domain: 0 < θ < 360
3) Find all solutions in the interval -180 ≤ x ≤ 180 to the equation cosx + (sin^2)x = 5/4
4) The sides of a triangle are given by the lines:
• 2x - y + 3 = 0
• x - 2y - 3 = 0
• 2x + y - 11 = 0
Find the area of this triangle.
I started off by saying for each equation:
when x = 0
y = ?
[I'd substitute x = 0 into every equation]
and I also did when y = 0, x = ? [by substituing y = 0 into all equations]
i got the points (-3/2, 1), (3, 0) and (11/2, 11)
Are the points that i got right ? And then i plotted the points which gave me a triangle. Now what do i do ?
2. 1) Prove this identity:
(cosecx + cotx)(cosecx - cotx) = (cotx)(tanx)
I started with LHS = (cosecx + cotx)(cosecx - cotx)
but don't know how to continue because these really confuse me.
you need to know the following
$\cot x = \frac{1}{\tan x}$
$cosec (x) = \frac{1}{\sin x}$
$\sin^2 x+ \cos^2 x = 1$
The most obvious way to do this is to just start simplifying both sides of the equation. Post again if you get stuck.
2) Solve the following equation for theta (θ) over the given domain:
√3 sin 2θ = cos 2θ
Domain: 0 < θ < 360
the trick to this type of question is dividing by the cos part to get a tan, in this case
$\sqrt{3}\sin(2\theta)=\cos(2\theta)$
$\sqrt{3}\frac{\sin(2\theta)}{\cos(2\theta)}=1$
$\sqrt{3}\tan(2\theta)=1$
3) Find all solutions in the interval -180 ≤ x ≤ 180 to the equation cosx + (sin^2)x = 5/4
Use $\sin^2 x+ \cos^2 x = 1$
$sin^2(x) = ?$
4) The sides of a triangle are given by the lines:
• 2x - y + 3 = 0
• x - 2y - 3 = 0
• 2x + y - 11 = 0
Find the area of this triangle.
I started off by saying for each equation:
when x = 0
y = ?
[I'd substitute x = 0 into every equation]
and I also did when y = 0, x = ? [by substituing y = 0 into all equations]
i got the points (-3/2, 1), (3, 0) and (11/2, 11)
Are the points that i got right ? And then i plotted the points which gave me a triangle. Now what do i do ?
I'm not sure what you are doing here. To find the corners of the triangle you need to find the points where the lines intersect. You do this by taking each of the 3 possible pairs of lines and solving them simultaneously for x and y. To get the area, you probably are expected to use base X height/2. This is probably most easily done by finding the distances along each side and using Pythagoras' theorem to find the height.
3. One cool way you could go about it. Perhaps not the easiest. Using Heron's formula.
Find the intersection points of the lines and then use the distance formula to find the distance between them. Those will be the lengths of the sides of the triangle.
$2x+3=\frac{x}{2}-\frac{3}{2}$
$x=-3, \;\ y=3$
$\frac{x}{2}-\frac{3}{2}=-2x+11$
$x=5, \;\ y=1$
$2x+3=-2x+11$
$x=2, \;\ y=7$
$a=\sqrt{(-3-5)^{2}+(3-1)^{2}}=2\sqrt{17}$
$b=\sqrt{(-3-2)^{2}+(3-7)^{2}}=\sqrt{41}$
$c=\sqrt{(5-2)^{2}+(1-7)^{2}}=3\sqrt{5}$
Using Heron's formula, the semi-perimeter is
$s=\frac{(2\sqrt{17}+\sqrt{41}+3\sqrt{5})}{2}\appro x{10.68}$
Heron's formula to find area:
$\sqrt{s(s-a)(s-b)(s-c)}$
4. He4llo, sweetG
4) The sides of a triangle are given by the lines: . $\begin{array}{ccc}2x - y + 3 &= &0 \\ x - 2y - 3 &=& 0 \\ 2x + y - 11 &=& 0 \end{array}$
Find the area of this triangle.
Solve each equation for $y.$
. . $\begin{array}{cccc}y &=&2x + 3 & [1] \\ \\[-4mm] y &=&\frac{1}{2}x - \frac{3}{2} & [2] \\ \\[-4mm] y &=& \text{-}2x + 11 & [3]\end{array}$
Note that: . $[2] \perp [3]$ . . . It is a right triangle!
Find the intersections . . .
$[1] \cap [2]\!:\;\;2x + 3 \:=\:\frac{1}{2}x-\frac{3}{2} \quad\Rightarrow\quad x \:=\:\text{-}3,\;y\:=\:\text{-}3 \quad\Rightarrow\quad A(\text{-}3,\,\text{-}3)$
$[1] \cap [3]\!:\;\;2x+3 \:=\:-2x+11\quad\Rightarrow\quad x\:=\;2,\;y\:=\:7 \quad\Rightarrow\quad B(2,7)$
$[2] \cap [3]\!:\;\;\frac{1}{2}x-\frac{3}{2}\:=\:-2+11\quad\Rightarrow\quad x \:=\:5,\;y\:=\:1 \quad\Rightarrow\quad C(5,1)$
The right angle is at $C.$
Hence, $AC$ is the base, $BC$ is the height.
Go for it! | 1,636 | 4,490 | {"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": 29, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-09 | longest | en | 0.866286 |
https://chess.stackexchange.com/questions/21732/are-there-rules-for-categorising-all-pawn-only-endgames-into-won-drawn-lost | 1,718,890,052,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00263.warc.gz | 134,899,449 | 43,073 | # Are there rules for categorising all pawn-only endgames into won/drawn/lost?
If a situation containing only pawns and kings is displayed is there a simple set of rules for determining if the position is won/drawn/lost?
I know for single pawn and double pawn endgames this is true with rules like the opposition, rule of square, rook pawns etc. My question is, does this extend into multipawn games or are they simply too complex and varied?
Another way of putting the question might be: if you show two grandmasters a pawn only endgame; do they both immediately know the outcome, or are there some that they would need to play out?
• A variation of your question is to ask if it is currently feasible to program a computer to properly categorize pawns-only endgames (in a reasonable amount of CPU time). I suspect that the answer is "yes" -- but the resulting algorithm is incapable of being summarized by a humanly-comprehesible rule. Commented Jun 6, 2018 at 21:52
Generally speaking, the side with the most pawns will win. The tempo provided by the extra pawn is usually enough to gain opposition and access to the key squares. Doubled pawns don't matter for this, unless they're blocked. The extra pawn can also limit the movement of the opponent's king, resulting in the possibility of a triangulation maneuver.
Another recurring theme/rule is the breakthrough. Grandmasters know the patterns for which a breakthrough is possible by heart.
But generally, multi-pawn endgames are just as complicated as other endgames. There is a Russian endgame study composer, Nikolai Grigoriev, who is famous for his pawn endgame studies. They truly show the rich possibilities of chess with just kings and pawns. With so many beautiful studies available, you know that there can be no general rule to determine the outcome.
While there are general rules, these rules have many exceptions and nuances.
In the following position with White to move, both sides have a passed pawn, but the fact that Black's passed pawn is so much better means White is dead lost:
``````[FEN "6k1/8/6p1/2p5/1pPpP3/pP6/2P3P1/2K5 w - - 0 1"]
``````
But it's precarious. Remove Black's A-pawn from the starting position, and Black doesn't have a passed pawn, and doesn't win. Remove his B-pawn, and the passed pawn isn't connected, and Black doesn't win. Remove his C-pawn, and White has a second passed pawn, and Black doesn't win. Remove his G-pawn, and White again has a second passed pawn, and Black doesn't win. Remove his D-pawn, and Black does win - but not if White's king started at b1 instead of c1. There's no "rule" that can cover all these situations; you have to calculate them to at least some extent.
• The pawn structure in the example is taken from one of my games (although there were still rooks on the board) and I was thrilled when my opponent offered a draw - I think he didn't understand how much more powerful his passed pawn was than mine.
– D M
Commented Jun 6, 2018 at 6:03
• Looking at that position, it looks like the primary initial function of the black A pawn is to keep White's king from coming to the defense of its E pawn. Would I be correct in assuming that zugzwang plays a major role in Black's victory? Take away all the E and G pawns, and whichever side next moves a pawn will lose it.with no compensation, but White can't keep his king at C1 without moving a pawn. If White's king moves to D1 or D2, Black's A pawn runs to the finish line. If White's king moves to B1 when Black's king is at D1 or E2, Black's king can move to D2. Commented Jun 6, 2018 at 17:22
• I guess maybe I'm over-complicating things given the presence of Black's G-pawn. I think that pawn is only needed with White to move, though. If Black were on move, White's E and G pawns would both be lunch. Commented Jun 6, 2018 at 17:34
• @supercat Without the g6 pawn, Black can't win even with the move. White just plays g4, and Black can't safely attack either pawn without the other promoting (Kg5 is responded to with e5, so Kxg4 would allow the e-pawn to promote, for example.) But yeah, if you take out all the pawns on the D-H files, Black moves his king up and wins when zugzwang comes into play; the c2 pawn falls followed by the b3 pawn because White eventually has to move his king away from them.
– D M
Commented Jun 6, 2018 at 17:52
There is no general rule for multi-pawn endgames as they are much too complex.
However I believe that most of the pawn only endgames that appear in regular games are readily assessed correctly by grandmasters (or even just masters). Sure there are pawn endgame studies that might be tricky, but on average assessing a pawn endgame is much easier than an endgame with say rooks on the board.
In order to win a pawn endgame you have to promote a pawn (apart from rare situations where you mate with pawns only). So obviously one of the first thing to look for is the availability of passed pawns or the options for creating passed pawns. If there are, you would further assess the strength of those passed pawns: Can the king catch them? Are they protected? (This usually means that the king is bound to a certain part of the board and could be used for operations in another part of the board.) Do you perhaps have more than one passed pawn? If both players have passed pawns; whose pawn is quicker?, etc.
If there are no passed pawns, it means that (in order to win) you have to capture an enemy pawn with your king. Target pawns to be captured are often fairly clearly defined. Your opponent may defend passively, e.g. preventing you from getting close to the target pawn or he may actively try to capture one of your pawns and promote a pawn.
In situations like this, it is often essential to count correctly, make proper use of opposition and triangulation. Fortunately the number of possible moves is limited in pawn endgames, simplifying the calculation of long lines.
When assessing multi-pawn endgames, of course it helps if you know the rules of single pawn endgames (rule of square, rook-pawn,opposition, active king etc) since the multi-pawn endgame might turn into a single pawn endgame or effective single pawn endgames if pawn chains are blocked.
There are a bunch of helpful rules on pawn endgames, but even a grandmaster wouldn't be able to call every pawn endgame won/drawn/lost.
Some pawn endgames involve immense amounts of calculation, involving concepts such as pawn races, corresponding squares, and long-distance triangulation. Other pawn endgames are too complex, and are beyond the calculation abilities and evaluation skills of a grandmaster to objectively evaluate (for example, six pawns each arranged in weird structures).
That being said, grandmasters can probably evaluate 99% of pawn endgames that come up, since they're the simplest endgame and have many helpful rules. But in the remaining 1% of endgames, these rules are often broken and the situation becomes very counter-intuitive, even to a grandmaster. | 1,634 | 6,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.956544 |
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Unformatted text preview: ‘ it " 7 ,— 28. Find and simplify the difference quutient Wfi i 0 for fix) = 225* - 3x + 3 . Z? . :29. Given two: functiong fix} 2 :_ and g(x) -..~, xix +1 . Find the foliowing functions and their domains A ®f+g mfg C) fag (1)11 g 30. Which 0f the foliowing graphs represents a function. Explain. ’1" i wL..é.q.mt.iWW.__é._i__.;..£ 1 “LL i 1 \$141.”:sz 11) Find {(9) and K6}. in) 1:; 1(2) pasiiive or negative? 0) What is the domain of f? d} Wha: is the range of f? e) What are the x-intercepis? f} What is the y—intercept? g) Find aii values ofx for which {(x) = 3. h) Us: the iniervaqs) an which f is increasing. i) List the iatsrvafis) an which f is decreasing. j) List the intervafis) cm which fix) >0 ...
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https://stacks.math.columbia.edu/tag/0E8K | 1,701,486,677,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100309.57/warc/CC-MAIN-20231202010506-20231202040506-00643.warc.gz | 607,250,594 | 6,116 | Lemma 21.17.4. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $U$ be an object of $\mathcal{C}$. If $\mathcal{K}^\bullet$ is a K-flat complex of $\mathcal{O}$-modules, then $\mathcal{K}^\bullet |_ U$ is a K-flat complex of $\mathcal{O}_ U$-modules.
Proof. Let $\mathcal{G}^\bullet$ be an exact complex of $\mathcal{O}_ U$-modules. Since $j_{U!}$ is exact (Modules on Sites, Lemma 18.19.3) and $\mathcal{K}^\bullet$ is a K-flat complex of $\mathcal{O}$-modules we see that the complex
$j_{U!}(\text{Tot}(\mathcal{G}^\bullet \otimes _{\mathcal{O}_ U} \mathcal{K}^\bullet |_ U)) = \text{Tot}(j_{U!}\mathcal{G}^\bullet \otimes _\mathcal {O} \mathcal{K}^\bullet )$
is exact. Here the equality comes from Modules on Sites, Lemma 18.27.9 and the fact that $j_{U!}$ commutes with direct sums (as a left adjoint). We conclude because $j_{U!}$ reflects exactness by Modules on Sites, Lemma 18.19.4. $\square$
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By Yuri Fonseca
In this post we are going to make an Uber assignment simulation and calculate some metrics of waiting time through simulation.
## Setting
Suppose we live in a 100×100 block city where each block takes 1 minute to cross by car. Drivers can pick up passengers only on corners, and passengers must call Uber on corners. Inferior-left corner is (1,1) and superior-right corner is (100,100).
## Functions
In order to calculate the average waiting time of a passenger, we need a couple of functions. The first function calculates the distance of a specific passenger and a specific driver, the second function creates the passengers, the third creates the cars and finally, the last function builds the distance matrix between cars and passengers in order to assign them in a optimal way.
Since our city is a perfect square, the distance between a passenger and a car is simple the distance in the x-axis plus the distance in the y-axis. Moreover, signs don’t matter. In this simple example we just need to know the initial position of the driver, initial position of the car and final destination. Each of these positions is a 2×1 vector representation in the city map.
#### Observation: distance matrix
We are going to use an linear programming solver in order to allocate optimally cars to passangers. This allocation problem just work with square matrix. So, if we have more cars than passengers, we need to create fictionary passengers (zero distances) in order to the solver converge. Note that we are going to do just a single round of allocation, so the number of cars needs to be bigger or equal than the number of passengers asking for a UBER driver.
```create_passenger = function(id){
initial.position = sample(50, 2, replace = TRUE)
final.destination = sample(50, 2, replace = TRUE)
return(list('number' = id, 'initial' = initial.position,
'final' = final.destination))
}
create_car = function(id){
initial.position = sample(50, 2, replace = TRUE)
return(list('number' = id, 'position' = initial.position))
}
distance = function(x,y){
sum(abs(x-y))
}
distance.matrix = function(cars, passengers){
d.matrix = matrix(0, nrow = length(cars), ncol = length(cars))
for (i in 1:length(cars)){
for (j in 1:length(passengers)){
d.matrix[i,j] = distance(cars[[i]]\$position, passengers[[j]]\$initial)
}
}
return(d.matrix)
}
```
#### Example MAP
Let’s check an example of 10 passengers and 10 cars:
```library(lpSolve)
library(ggplot2)
set.seed(20)
passengers = lapply(seq(1:10), create_passenger)
cars = lapply(seq(1:10), create_car)
d.matrix = distance.matrix(cars, passengers)
opt.allocation = lp.assign(d.matrix)
passengers.points = sapply(passengers, function(x) x\$initial)
cars.points = sapply(cars, function(x) x\$position)
points = t(cbind(passengers.points, cars.points))
assignments = apply(opt.allocation\$solution, 1, which.max) #checking the assignment for each car
df1 = data.frame('x.axis' = points[,1],
'y.axis' = points[,2],
'id' = c(rep('Passenger',10), rep('Car',10)))
# df.assign = data.frame('x' = cars.points[1,],
# 'y' = cars.points[2,],
# 'xend' = passengers.points[1,assignments],
# 'yend' = passengers.points[2,assignments])
df.assign1 = data.frame('x' = cars.points[1,],
'y' = cars.points[2,],
'xend' = passengers.points[1,assignments],
'yend' = cars.points[2,])
df.assign2 = data.frame('x' = passengers.points[1,assignments],
'y' = cars.points[2,],
'xend' = passengers.points[1,assignments],
'yend' = passengers.points[2,assignments])
ggplot(df1, aes(x.axis,y.axis)) + geom_point(aes(color = id, group = id), size = 3) + # car and passengers
geom_segment(aes(x = x, y = y, xend = xend, yend = yend), data = df.assign1) +
geom_segment(aes(x = x, y = y, xend = xend, yend = yend), data = df.assign2,
arrow = arrow(length = unit(0.02, "npc"), type = 'closed')) +
scale_x_continuous(minor_breaks = seq(1, 50, 1)) +
scale_y_continuous(minor_breaks = seq(1, 50, 1)) +
ggtitle('Optimal Allocation')
```
## Monte Carlo simulation
Now the waiting time… Suppose that we have 10 passengers asking for cars in the city, we are going to see how the waiting time changes with the numbers of cars. As example, we are going to change the numbers of cars from 10 up to 30 cars, with 500 hundred Monte Carlo simulation.
```simulations = function(N, MC) {
ncars = N
times = matrix(0, nrow = MC, ncol = N)
for (i in 1:MC){
passengers = lapply(seq(1:10), create_passenger)
cars = lapply(seq(1:ncars), create_car)
d.matrix = distance.matrix(cars, passengers)
opt.allocation = lp.assign(d.matrix)
times[i,] = colSums(opt.allocation\$solution*opt.allocation\$costs) # waiting time for each passenger
}
return(times)
}
results = lapply(seq(10,30,2), simulations, MC = 500) # MC = 500 just to save some time
```
Now it is possible to check how the numbers of cars affect the mean waiting time and the 10% and 90% confidence level for the waiting time.
```df2 = data.frame('WaitingTime' = sapply(results, mean),
'LB' = sapply(results, quantile, probs = 0.10),
'UB' = sapply(results, quantile, probs = 0.90),
'Cars' = seq(10,30,2))
ggplot(df2, aes(x = Cars)) + geom_line(aes(y = WaitingTime), lwd = 1.2) +
geom_ribbon(aes(ymin = LB, ymax = UB), fill = 'blue', alpha = .3)
``` | 1,435 | 5,395 | {"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} | 3.703125 | 4 | CC-MAIN-2020-50 | latest | en | 0.884084 |
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# Distribution of SC questions - Strategy
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22 Oct 2009, 17:54
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Hey guys,
Have you ever faced such situations, while solving SC, as you are not able to spot the errors in any of the answer choices, or are not able to narrow down the answer choice to one? If yes, then I am with you!!!
I seldom find myself staring at the screen with no clue how to attack a hard SC. I know, I know, every question falls under a specific category of error. According to MGMAT SC there are 9 such categories: Parallelism, Modifiers, Comparisons, Idioms, Verb Tense/Voice/Mood, Subject-Verb agreement, Pronouns, Meaning/Concision, Odds & Ends. So if you are stuck in a situation like above, then you should start checking if the question falls under any of these error categories. But which category to test first? It is definitely not economical to spend time in real GMAT checking these categories randomly. So I have come up with my own strategy to tackle such situations.
Here is the distribution of questions from OG and Official Verbal Guides:
Attachment:
SC_Distribution.jpg [ 73.88 KiB | Viewed 38320 times ]
It is clear from this chart that the probability of a question to fall under one of the first 3 categories is more than 50%. And 75% of the questions comes under one of the first 5 categories. Questions that fall under modifiers and comparisons categories are usually easier than those that fall under parallelism. So it is safe to place your bet on parallelism if you are not able to determine the category of any question.
I personally found this strategy very useful to increase my speed in finding the answers of Brutal SC questions, and I hope that you all will also benefit from it. Comments and suggestions are Welcome!!
Assumption: The distribution of the SC questions in OG and Official Verbal Guides represents the actual distribution of questions in real GMAT.
If you have any questions
New!
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Re: Distribution of SC questions - Strategy [#permalink]
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22 Oct 2009, 19:20
1
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Phew!GLad to see that somebody is in the same boat as me!!
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Re: Distribution of SC questions - Strategy [#permalink]
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22 Oct 2009, 21:55
Have some problem often,,,
but if i have no clue which answer to pick, i trust my intuition and ear. It helps often
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23 Oct 2009, 05:29
mirzohidjon wrote:
Have some problem often,,,
but if i have no clue which answer to pick, i trust my intuition and ear. It helps often
I can see how "ear" can help in case of Idioms, but I have a hard time to digest that intuition is the BEST strategy
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23 Oct 2009, 06:32
HI Hgp,
So, If I understood you right, when you can not find clue to the question, you begin to check the belonging of the question to the category one by one, from the top?
Can you suppose the rate of effectiveness increase?
Have you checked this proportions with the Prep. materials?
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Re: Distribution of SC questions - Strategy [#permalink]
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23 Oct 2009, 08:42
I think the best way to approach SC questions, is to develop ability to recognize error at first glance. Step by step look at each grammar concept based on percentage from OG often might be wasteful, and leave less time for RC and CR questions, which are equally important.
May be I used the wrong word. By "intuition," I meant, my feeling that something is wrong in the use of specific expression based on expression, i.e. it is not just intuition...
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Re: Distribution of SC questions - Strategy [#permalink]
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23 Oct 2009, 10:08
Ayrish wrote:
HI Hgp,
So, If I understood you right, when you can not find clue to the question, you begin to check the belonging of the question to the category one by one, from the top?
Can you suppose the rate of effectiveness increase?
Have you checked this proportions with the Prep. materials?
You got it right. Yes, it increased the rate of effectiveness for me. The question proportion/distribution is from OG12, OG 11, GMAT official verbal guide 2 and GMAT official verbal guide 1.
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Re: Distribution of SC questions - Strategy [#permalink]
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23 Oct 2009, 10:13
mirzohidjon wrote:
I think the best way to approach SC questions, is to develop ability to recognize error at first glance. Step by step look at each grammar concept based on percentage from OG often might be wasteful, and leave less time for RC and CR questions, which are equally important.
May be I used the wrong word. By "intuition," I meant, my feeling that something is wrong in the use of specific expression based on expression, i.e. it is not just intuition...
You are correct mirzohidjon! But this stratey is applicable only for the questions, which you have no clue about. And the probability of that is very high when you are at 700+ difficulty level.
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Re: Distribution of SC questions - Strategy [#permalink]
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23 Oct 2009, 11:05
Good work buddy..You sure have managerial skills of consolidating data
Yup..actually its more of pattern-based, so identifying a pattern is must...of course , theoretically, classifying the problems and attacking them in a structured manner is the best and 'satisfying' way to answer questions. For me it is a combination of both as there are times when you have to chose the 'best' choice even though two of them can be technically correct.
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Re: Distribution of SC questions - Strategy [#permalink]
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23 Oct 2009, 12:48
Expert's post
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Kudos!
I have used a checklist like that for my prep, but my approach has been to build a check list from my error log - I would see where i get caught the most and that's the area I would check first. It looked very similar to yours (probably had a few more items there) but the idea is the same.
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23 Oct 2009, 16:34
my test date is soon and after so much of practise..SC is still killing me..i plan on spending quality time over nxt few days practicing just the top 4-5 topics listed here..
thanks a ton for compiling this!
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04 Nov 2009, 13:49
Excellent. I love the idea. Will apply it shortly.
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04 Nov 2009, 20:23
When I went through those OGs, I noticed the number of ||sm types questions are larger than others.
Good work, buddy.
+1
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05 Nov 2009, 02:45
Also, here is another project hgp2k contributed to:
gmat-club-guide-to-the-gmat-official-guide-12th-ed-85956.html
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06 Nov 2009, 10:18
+1 Kudos!
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05 Dec 2009, 05:17
Love it, very helpful indeed. Thank you!
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24 Apr 2010, 20:21
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Pretty awesome, exactly what I was looking for.
Gmatclub rocks......
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17 Aug 2010, 09:13
The trick that worked for me is to quickly scan through the answer choices w/o actually reading, this might help you find the error that is being tested. If the answer choices are different around the usage of verb tense then one of the errors is verb tense.
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17 Aug 2010, 22:02
add me too in this category. and i also agree that parallelism is very tough to find out..it really makes sense to bet on parallelism
Thanks for the comparision.
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28 Sep 2011, 21:01
although i wonder if this has changed with the recent updates of GMAT not using many idioms and attempting to increase the focus on 'meaning'. ??
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Re: Distribution of SC questions - Strategy [#permalink] 28 Sep 2011, 21:01
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Display posts from previous: Sort by | 3,549 | 12,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-22 | latest | en | 0.904748 |
http://www.gpedia.com/en/gpedia/Hilbert%27s_paradox_of_the_Grand_Hotel | 1,597,266,078,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738944.95/warc/CC-MAIN-20200812200445-20200812230445-00512.warc.gz | 139,095,620 | 23,688 | # Hilbert's paradox of the Grand Hotel
Hilbert's Hotel
Hilbert's paradox of the Grand Hotel (colloquial: Infinite Hotel Paradox or Hilbert's Hotel) is a thought experiment which illustrates a counterintuitive property of infinite sets. It is demonstrated that a fully occupied hotel with infinitely many rooms may still accommodate additional guests, even infinitely many of them, and this process may be repeated infinitely often. The idea was introduced by David Hilbert in a 1924 lecture "Über das Unendliche", reprinted in (Hilbert 2013, p.730), and was popularized through George Gamow's 1947 book One Two Three... Infinity.[1][2]
Consider a hypothetical hotel with a countably infinite number of rooms, all of which are occupied. One might be tempted to think that the hotel would not be able to accommodate any newly arriving guests, as would be the case with a finite number of rooms, where the pigeonhole principle would apply.
### Finitely many new guests
Suppose a new guest arrives and wishes to be accommodated in the hotel. We can (simultaneously) move the guest currently in room 1 to room 2, the guest currently in room 2 to room 3, and so on, moving every guest from his current room n to room n+1. After this, room 1 is empty and the new guest can be moved into that room. By repeating this procedure, it is possible to make room for any finite number of new guests.
### Infinitely many new guests
It is also possible to accommodate a countably infinite number of new guests: just move the person occupying room 1 to room 2, the guest occupying room 2 to room 4, and, in general, the guest occupying room n to room 2n (2 times n), and all the odd-numbered rooms (which are countably infinite) will be free for the new guests.
### Infinitely many coaches with infinitely many guests each
It is possible to accommodate countably infinitely many coachloads of countably infinite passengers each, by several different methods. Most methods depend on the seats in the coaches being already numbered (or use the axiom of countable choice). In general any pairing function can be used to solve this problem. For each of these methods, consider a passenger's seat number on a coach to be ${\displaystyle n}$, and their coach number to be ${\displaystyle c}$, and the numbers ${\displaystyle n}$ and ${\displaystyle c}$ are then fed into the two arguments of the pairing function.
#### Prime powers method
Empty the odd numbered rooms by sending the guest in room ${\displaystyle i}$ to room ${\displaystyle 2^{i}}$, then put the first coach's load in rooms ${\displaystyle 3^{n}}$, the second coach's load in rooms ${\displaystyle 5^{n}}$; for coach number ${\displaystyle c}$ we use the rooms ${\displaystyle p^{n}}$ where ${\displaystyle p}$ is the ${\displaystyle c}$th odd prime number. This solution leaves certain rooms empty (which may or may not be useful to the hotel); specifically, all odd numbers that are not prime powers, such as 15 or 847, will no longer be occupied. (So, strictly speaking, this shows that the number of arrivals is less than or equal to the number of vacancies created. It is easier to show, by an independent means, that the number of arrivals is also greater than or equal to the number of vacancies, and thus that they are equal, than to modify the algorithm to an exact fit.) (The algorithm works equally well if one interchanges ${\displaystyle n}$ and ${\displaystyle c}$, but whichever choice is made, it must be applied uniformly throughout.)
#### Prime factorization method
You can put each person of a certain seat ${\displaystyle s}$ and coach ${\displaystyle c}$ into room ${\displaystyle 2^{s}3^{c}}$ (presuming c=0 for the people already in the hotel, 1 for the first coach, etc. ...). Because every number has a unique prime factorization, it's easy to see all people will have a room, while no two people will end up in the same room. For example, the person in room 2592 (${\displaystyle 2^{5}3^{4}}$) was sitting in on the 4th coach, on the 5th seat. Like the prime powers method, this solution leaves certain rooms empty.
This method can also easily be expanded for infinite nights, infinite entrances, etc. ... ( ${\displaystyle 2^{s}3^{c}5^{n}7^{e}}$ )
#### Interleaving method
For each passenger, compare the lengths of ${\displaystyle n}$ and ${\displaystyle c}$ as written in any positional numeral system, such as decimal. (Treat each hotel resident as being in coach #0.) If either number is shorter, add leading zeroes to it until both values have the same number of digits. Interleave the digits to produce a room number: its digits will be [first digit of coach number]-[first digit of seat number]-[second digit of coach number]-[second digit of seat number]-etc. The hotel (coach #0) guest in room number 1729 moves to room 01070209 (i.e., room 1,070,209). The passenger on seat 1234 of coach 789 goes to room 01728394 (i.e., room 1,728,394).
Unlike the prime powers solution, this one fills the hotel completely, and we can reconstruct a guest's original coach and seat by reversing the interleaving process. First add a leading zero if the room has an odd number of digits. Then de-interleave the number into two numbers: the seat number consists of the odd-numbered digits and the coach number is the even-numbered ones. Of course, the original encoding is arbitrary, and the roles of the two numbers can be reversed (seat-odd and coach-even), so long as it is applied consistently.
#### Triangular number method
Those already in the hotel will be moved to room ${\displaystyle (n^{2}+n)/2}$, or the ${\displaystyle n}$th triangular number. Those in a coach will be in room ${\displaystyle ((c+n-1)^{2}+c+n-1)/2+n}$, or the ${\displaystyle (c+n-1)}$ triangular number plus ${\displaystyle n}$. In this way all the rooms will be filled by one, and only one, guest.
This pairing function can be demonstrated visually by structuring the hotel as a one-room-deep, infinitely tall pyramid. The pyramid's topmost row is a single room: room 1; its second row is rooms 2 and 3; and so on. The column formed by the set of rightmost rooms will correspond to the triangular numbers. Once they are filled (by the hotel's redistributed occupants), the remaining empty rooms form the shape of a pyramid exactly identical to the original shape. Thus, the process can be repeated for each infinite set. Doing this one at a time for each coach would require an infinite number of steps, but by using the prior formulas, a guest can determine what his room "will be" once his coach has been reached in the process, and can simply go there immediately.
#### Arbitrary enumeration method
Let ${\displaystyle S:=\{(a,b)\mid a,b\in \mathbb {N} \}}$. ${\displaystyle S}$ is countable since ${\displaystyle \mathbb {N} }$ is countable, hence we may enumerate its elements ${\displaystyle s_{1},s_{2},\dots }$. Now if ${\displaystyle s_{n}=(a,b)}$, assign the ${\displaystyle b}$th guest of the ${\displaystyle a}$th coach to the ${\displaystyle n}$th room (consider the guests already in the hotel as guests of the ${\displaystyle 0}$th coach). Thus we have a function assigning each person to a room; furthermore, this assignment does not skip over any rooms.
### Further layers of infinity
Suppose the hotel is next to an ocean, and an infinite number of car ferries arrive, each bearing an infinite number of coaches, each with an infinite number of passengers. This is a situation involving three "levels" of infinity, and it can be solved by extensions of any of the previous solutions.
The prime factorization method can be applied by adding a new prime number for every additional layer of infinity ( ${\displaystyle 2^{s}3^{c}5^{f}}$, with ${\displaystyle f}$ the ferry).
The prime power solution can be applied with further exponentiation of prime numbers, resulting in very large room numbers even given small inputs. For example, the passenger in the second seat of the third bus on the second ferry (address 2-3-2) would raise the 2nd odd prime (5) to 49, which is the result of the 3rd odd prime (7) being raised to the power of his seat number (2). This room number would have over thirty decimal digits.
The interleaving method can be used with three interleaved "strands" instead of two. The passenger with the address 2-3-2 would go to room 232, while the one with the address 4935-198-82217 would go to room #008,402,912,391,587 (the leading zeroes can be removed).
Anticipating the possibility of any number of layers of infinite guests, the hotel may wish to assign rooms such that no guest will need to move, no matter how many guests arrive afterward. One solution is to convert each arrival's address into a binary number in which ones are used as separators at the start of each layer, while a number within a given layer (such as a guest's coach number) is represented with that many zeroes. Thus, a guest with the prior address 2-5-1-3-1 (five infinite layers) would go to room 10010000010100010 (decimal 295458).
As an added step in this process, one zero can be removed from each section of the number; in this example, the guest's new room is 101000011001 (decimal 2585). This ensures that every room could be filled by a hypothetical guest. If no infinite sets of guests arrive, then only rooms that are a power of two will be occupied.
#### Infinite layers of nesting
Although a room can be found for any finite number of nested infinities of people, the same is not always true for an infinite number of layers, even if a finite number of elements exists at each layer.
## Analysis
Hilbert's paradox is a veridical paradox: it leads to a counter-intuitive result that is provably true. The statements "there is a guest to every room" and "no more guests can be accommodated" are not equivalent when there are infinitely many rooms.
Initially, this state of affairs might seem to be counter-intuitive. The properties of "infinite collections of things" are quite different from those of "finite collections of things". The paradox of Hilbert's Grand Hotel can be understood by using Cantor's theory of transfinite numbers. Thus, while in an ordinary (finite) hotel with more than one room, the number of odd-numbered rooms is obviously smaller than the total number of rooms. However, in Hilbert's aptly named Grand Hotel, the quantity of odd-numbered rooms is not smaller than the total "number" of rooms. In mathematical terms, the cardinality of the subset containing the odd-numbered rooms is the same as the cardinality of the set of all rooms. Indeed, infinite sets are characterized as sets that have proper subsets of the same cardinality. For countable sets (sets with the same cardinality as the natural numbers) this cardinality is ${\displaystyle \aleph _{0}}$.[3]
Rephrased, for any countably infinite set, there exists a bijective function which maps the countably infinite set to the set of natural numbers, even if the countably infinite set contains the natural numbers. For example, the set of rational numbers—those numbers which can be written as a quotient of integers—contains the natural numbers as a subset, but is no bigger than the set of natural numbers since the rationals are countable: there is a bijection from the naturals to the rationals.
## References in fiction
• BBC Learning Zone repeatedly screened a 1996 one-off educational docudrama Hotel Hilbert set in the Hotel as seen through the eyes of a young female guest Fiona Knight, her name a pun on finite. The programme was designed to educate viewers about the concept of infinity.[4]
• The novel White Light by mathematician/science fiction writer Rudy Rucker includes a hotel based on Hilbert's paradox, and where the protagonist of the story meets Georg Cantor.
• Stephen Baxter's science fiction novel Transcendent has a brief discussion on the nature of infinity, with an explanation based on the paradox, modified to use starship troopers rather than hotels.
• Geoffrey A. Landis' Nebula Award-winning short story "Ripples in the Dirac Sea" uses the Hilbert hotel as an explanation of why an infinitely-full Dirac sea can nevertheless still accept particles.
• In Peter Høeg's novel Miss Smilla's Feeling for Snow, the titular heroine reflects that it is admirable for the hotel's manager and guests to go to all that trouble so that the latecomer can have his own room and some privacy.
• In Ivar Ekeland's novel for children, The Cat in Numberland, a "Mr. Hilbert" and his wife run an infinite hotel for all the integers. The story progresses through the triangular method for the rationals.
• In Will Wiles's novel The Way Inn, about an infinitely large motel, the villain's name is Hilbert.
• In Reginald Hill's novel "The Stranger House" the character Sam refers to the Hilbert Hotel paradox.
• The short story by Naum Ya. Vilenkin The Extraordinary Hotel (often erroneously attributed to Stanislaw Lem) shows the way in which Hilbert's Grand Hotel may be reshuffled when infinite new hosts arrive.
• John Roderick and Ken Jennings discussed the Hotel on their Omnibus podcast in the episode The Hilbert Hotel Entry.
• The comic book saga The Tempest from the League of Extraordinary Gentlemen series by Alan Moore and Kevin O'Neill shows a villain called Infinity. In the story it is suggested that the villain goes to the hotel based on Hilbert's paradox. Georg Cantor is mentioned as well. | 3,121 | 13,415 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 38, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-34 | latest | en | 0.970376 |
https://kr.mathworks.com/matlabcentral/profile/authors/2009976 | 1,717,042,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059418.31/warc/CC-MAIN-20240530021529-20240530051529-00152.warc.gz | 302,209,790 | 22,463 | # Juan Camilo Medina
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대략 13년 전 | 다운로드 수: 1 | | 683 | 2,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.633269 |
https://math.stackexchange.com/questions/2252665/stability-of-finite-difference-schemes | 1,560,811,110,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998581.65/warc/CC-MAIN-20190617223249-20190618005249-00106.warc.gz | 511,126,506 | 35,994 | # Stability of Finite Difference schemes.
I was reading a book ''Numerical partial differential equation: Finite difference methods'' by J. W. Thomas. Here at page no. 74, definition of stability is given as
$\|u^{n+1}\|\leq K e^{\beta t}\|u^{0}\|$; for $0\leq t=(n+1)\Delta t$, $0\leq \Delta x \leq \Delta x_{0}$ and $0\leq \Delta t \leq \Delta t_{0}$. i.e. this definition allows for exponential growth. But in some literature I found the definition that ''A finite difference approximation is stable if the errors (truncation, round-off etc) decay as the computation proceeds from one marching step to the next .'' The Book which I have mentioned, at the same page in remark 3, author has written that the above definition with exponential growth is general definition and later one can be derived from this.But I am not getting how to find this definition. So my question is which definition is true ? And if other one is not true, then what is reason behind ? and how can we compare these two definition? Thanking You in Advance.
• I don't have an extensive background on this, but from what I know, some finite difference method has the error part grows exponentially, while others do not. It depends specifically on what method of finite difference you are using. – Paichu May 2 '17 at 15:16
• But if error will grow exponentially then how can we say it stable? – VIVEK KUMAR May 4 '17 at 13:33
• That really depends on how you define stability. – Paichu May 4 '17 at 13:43
To address the concern of your stability question, one often has to enforce some sort of constrain on the time step. Here is an example.
Consider the advection diffusion equation: $$\partial_t u + a \partial_x u - b\partial^2_x u = 0, \text{ with } u(x,t=0) = f(x)$$ for $x \in [-1,1]$ with $u(x+2,t) = u(x,t)$ for all $x$ and $t$.
Applying the explicit difference method then, $$U(x,t+\Delta t) = U(x,t) - \frac{a \Delta t}{2 \Delta x} \left(T - T^{-1}\right) U(x,t) + \frac{b \Delta t}{\Delta x^2}\left(T - 2 + T^{-1} \right)U(x,t)$$
Recall the following Discrete Fourier Transform rules:
$$U(x + 2 L) = U(x), \text{ and } L = N \Delta x = 1$$
$$\hat{U}(\xi + 2L) = \hat{U}(\xi), \text{ and } \hat{L} = N \Delta \xi \text{ with } N\Delta x \Delta \xi = \pi$$
$$\widehat{TU} (\xi) = \hat{U} (\xi) e^{i \xi \Delta x}$$
Applying the previous rules, then our difference method equation becomes:
$$\hat{U}(\xi, t+\Delta t) = \hat{U}(\xi,t) - \frac{a \Delta t}{2 \Delta x}\underbrace{\left(e^{i \xi \Delta x} - e^{-i \xi \Delta x} \right)}_{2i sin(\xi \Delta x /2)}\hat{U} + \frac{b\Delta t}{\Delta x^2}\underbrace{\left(e^{i \xi \Delta x} - 2 + e^{-i \xi \Delta x} \right)}_{\left(e^{i \xi \Delta x} - e^{-i \xi \Delta x} \right)^2 = \left(2i sin(\xi \Delta x /2)\right)^2}\hat{U}$$
With
$$\xi \in [-\frac{\pi}{\Delta x},\frac{\pi}{\Delta x}], \text{ or } \frac{\Delta x\xi}{2} \in [-\frac{\pi}{2},\frac{\pi}{2}]$$
And after some algebra,
$$\hat{U}(\xi, t+\Delta t) = \left[1 - \frac{a \Delta t}{\Delta x} i \sin(\xi \Delta /2) - \frac{b \Delta t}{\Delta x^2} 4 \sin^2(\xi \Delta /2)\right]\hat{U}(\xi,t)$$
Here is the Stability part
Require $\|\hat{U}(t+\Delta t)\|_2 \le \|\hat{U}(t)\|_2$ to avoid blowing up (Stability) and recall that the Fourier Transform process is norm-preserving, then $$\left\|1 - \frac{a \Delta t}{\Delta x} i \sin(\xi \Delta /2) - \frac{b \Delta t}{\Delta x^2} 4 \sin^2(\xi \Delta /2) \right\|_2 \le 1$$ Which is equivalent to $$\left( 1 - \frac{b \Delta t}{\Delta x^2} 4 \sin^2(\xi \Delta /2)\right)^2 + \left( \frac{a \Delta t}{\Delta x} \sin(\xi \Delta /2)\right)^2 \le 1$$
To obtain the constrain on $\Delta t$ and $\Delta x$, which are the time step and the space step for the numerical method, set \begin{align*} \sin(\xi \Delta /2) = 1 &\Rightarrow \frac{\Delta t}{\Delta x^2} (b^2 + a^2) \le 2b\\ &\Rightarrow \Delta t \le \frac{2b \Delta x^2}{a^2 + b^2} \end{align*}
Note there is another condition for stability for this system, but this condition encompass the other one so I don't mention it.
I hope this helps.
• @VIVEKKUMAR You're welcome. – Paichu May 6 '17 at 22:09
• @ Paichu Thank you very much. – VIVEK KUMAR Aug 2 '17 at 10:55 | 1,424 | 4,174 | {"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} | 3.359375 | 3 | CC-MAIN-2019-26 | latest | en | 0.881922 |
https://howtoimprovehome.com/why-neutral-wire-have-120-volts/ | 1,696,376,448,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511284.37/warc/CC-MAIN-20231003224357-20231004014357-00143.warc.gz | 334,279,667 | 33,970 | # Why Does My Neutral Wire Have 120 Volts? Solved
It can be hard and confusing to figure out how the electricity works in our homes.
Finding out that the neutral wire in their electrical system has a voltage of 120 volts is something that many people find confusing. It is a big problem because a neutral line should normally have close to zero volts.
In this piece, we’ll look into why this strange thing is happening and explain the role of the neutral wire, how it works in your electrical system, and why a 120-volt reading could mean a dangerous situation that must be fixed immediately.
## Why Does My Neutral Wire Have 120 Volts?
This abnormal reading is the result of issues including shared neutral or bootleg ground, neutral-to-earth voltage problems, poor transformer connections, load imbalances, or open neutral wires. Moreover, a faulty appliance may also cause the voltage to be supplied to the neutral wire.
### Shared Neutral or Bootleg Ground
A shared neutral or a bootleg ground could explain why a neutral wire has 120 volts. In some situations, a person who doesn’t know what they’re doing or isn’t qualified could incorrectly connect the neutral wire to the ground wire, making a shared neutral or “bootleg ground.”
It can cause the neutral wire to carry electricity, giving you 120 volts. To fix this problem, you need to separate the ground and neutral wires to ensure they are not connected.
### Neutral-to-Earth Voltage
When everything is normal, there may be a small amount of voltage between the neutral line and the earth’s ground. It is called neutral-to-earth voltage. It is mostly because of how the electricity distribution system is made and the resistance between the neutral and the ground.
Most of the time, a small voltage (a few volts) is fine, but a bigger neutral-to-earth voltage could mean there are problems with grounding or other electrical problems. In this case, you should talk to an electrician to check the grounding system and find any possible problems.
### Faulty Transformer Connection
Rarely, a weak link in the transformer can result in a high voltage on the neutral wire. It can happen if the transformer has loose links or bad wiring. Only a qualified electrician or utility company should take care of transformer issues. Calling the utility company to report the problem is a good idea since they have the knowledge and power to fix problems with transformers.
A load imbalance happens when electrical devices draw different amounts of power at different stages. It can cause too much current to run through the neutral wire, which builds up voltage.
To solve this problem, the load should be spread out evenly across the different stages. It can be done by moving electrical devices or using load-balancing methods, like using three-phase power if available.
### Open Neutral
The neutral wire is broken or detached in the electrical circuit, which causes an open neutral. When this happens, the neutral wire loses its ability to carry the return current well, which causes the voltage to change.
To find and fix an open neutral, you need to look at the wiring and junction boxes for any neutral wires that are loose or broken. If the bad wiring is fixed or replaced, the neutral wire should work as it should.
## What is Acceptable Neutral Voltage?
In a properly balanced system, this voltage should be 0 in the best case. However, in real life, this voltage is not always zero because of system imbalances brought about by uneven load distribution or harmonic currents. Moreover, connection problems could also lead to voltage in neutral.
Neutral voltage is very important in knowing how electricity is distributed, especially in three-phase circuits. Most of the time, neutral voltage is the difference between the voltage at the neutral point and the voltage at the earth’s ground.
Before you can understand the neutral voltage, you need to know what the neutral line does in a power system. In a three-phase power system, the neutral wire lets the electricity return to where it came from.
It lets the system ensure that each phase gets the same power. Referring to the neutral point, the sum of the voltages in the three stages of a balanced system equals zero. So, the best voltage from neutral to the ground is zero. But a neutral-to-ground voltage that is not zero can happen when there are imbalances in the circuit.
An imbalance happens when the load isn’t spread out evenly between the three phases. It causes current to flow through the neutral wire, which causes the voltage across it to drop because of its resistance. It means the voltage from the neutral to the ground is not zero.
In the same way, non-linear loads like computers and other electrical devices can cause harmonic currents, which change the shape of the voltage waveform and make the neutral-to-ground voltage go up.
The term “acceptable neutral voltage” refers to the amount of neutral-to-ground voltage the power system can handle without causing problems or raising safety concerns.
## How Do You Reduce Neutral Voltage?
One of the best ways to lower neutral voltage is to ensure all the system loads are balanced. When you balance the loads, you ensure that each part of the system carries the same amount of current. It can be done by spreading single-phase loads carefully across all three phases.
Load balancing not only helps lower the neutral voltage but also improves the system’s general performance, makes it less likely that it will overload, and makes it more efficient.
Getting the grounding right is another important part of lowering the neutral voltage. According to the National Electrical Code (NEC), the neutral line should have good ground. Grounding the neutral line helps keep the voltage levels in the system stable and gives the current a clear path to follow in case of a problem.
Neutral voltage is also partially managed by ensuring the electrical system is regularly checked and fixed. Any electrical problems, loose connections, or broken equipment should be found right away and fixed. High-quality lines and components with the right ratings can also stop the voltage from dropping or going up when it shouldn’t.
Harmonics in the circuit can also make the neutral voltage higher. It is because loads that aren’t linear, like computers, LED lights, and varying frequency drives, create harmonic currents that can make the neutral voltage higher. This problem can be made less of a problem by installing harmonic filters or using devices with power factor adjustment.
Lastly, you might use a neutral converter or an advanced neutral voltage reduction device. When these devices are put into the power system, they can help lower the neutral voltage by creating a voltage that cancels out the extra neutral voltage.
## Conclusion
If a neutral wire has 120 volts, it means there is an electrical problem that needs to be fixed right away because it could be dangerous. Several things could cause this strange reading, such as a shared neutral or bootleg ground, neutral-to-earth voltage problems, bad transformer connections, load imbalances, or an open neutral line.
Additionally, a malfunctioning device might be the problem. A skilled electrician has the knowledge and skills to correctly figure out what’s wrong and fix it. A well-balanced electrical system is linked to the idea of an “acceptable” neutral voltage.
In an ideal world, the neutral voltage would be 0 volts, but system flaws and harmonic currents can cause this to not always be the case. Keeping this neutral voltage as low as possible is important for maintaining an electrical system that works well and is safe.
Load balancing, proper grounding, regular repair, and using high-quality lines and parts are all ways to lower neutral voltage. To lessen the effects of non-linear loads, you can also use harmonic filters or devices that change the power factor.
Neutral inverters or other advanced neutral voltage reduction devices could be used in the worst cases. So, a reading of 120 volts on a neutral wire isn’t just a strange electrical quirk; it’s a serious problem that needs to be fixed quickly to avoid harm and electrical inefficiency.
Don`t copy text!
Scroll to Top | 1,608 | 8,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-40 | latest | en | 0.935899 |
http://metamath.tirix.org/mpests/ssriv | 1,726,311,020,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00421.warc.gz | 17,996,588 | 1,888 | # Metamath Proof Explorer
## Theorem ssriv
Description: Inference based on subclass definition. (Contributed by NM, 21-Jun-1993)
Ref Expression
Hypothesis ssriv.1 $⊢ x ∈ A → x ∈ B$
Assertion ssriv $⊢ A ⊆ B$
### Proof
Step Hyp Ref Expression
1 ssriv.1 $⊢ x ∈ A → x ∈ B$
2 dfss2 $⊢ A ⊆ B ↔ ∀ x x ∈ A → x ∈ B$
3 2 1 mpgbir $⊢ A ⊆ B$ | 129 | 334 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.48004 |
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### Math and Predictive Analytics - A Personal Account
Last week I taught a workshop at Predictive Analytics World entitled Supercharging Prediction: Hands-On with Ensemble Models. The workshop was intended to introduce predictive modelers to the concept of ensembles through a combination of lecture to provide an overview of model ensembles and hands-on to gain experience building ensembles using Salford Systems SPM v7.0 (Salford Systems sponsored the workshop).
This morning, Heather Hinman, a Marketing Communications Manager at Salford Systems, posted comments on attending that workshop at the Salford Systems blog. Two comments were particularly interesting, especially their implications vis a vis my last blog post on math and predictive analytics:
I will admit I was intimidated at first to be participating in a predictive modeling workshop as I do not have a background in statistics, and only have basic training on decision tree tools by Salford Systems' team of in-house experts. Despite my basic knowledge of decision trees, I was thrilled that I was able to follow along with ease and understanding when learning about tree ensembles and modern hybrid modeling approaches. Marketing folk building predictive models? Yes, we can!
and
Now back at the office in San Diego, along with my usual responsibilities, I feel confident in my ability to build predictive models and gain insights into the data at hand to achieve the email marketing and online campaign goals for our communication efforts!
In the post, Heather also outlines some of the principles she learned and how she used them to build the predictive models in the workshop.
The point is this: if one uses good software that uses solid principles for building predictive models, and one understands key principles of building predictive models, someone without a mathematics background can build good, profitable models.
## Monday, April 01, 2013
### Do Predictive Modelers Need to Know Math?
(Note: this post was first published in the March 2013 Edition of the Predictive Analytics Times)
Predictive analytics is just a bunch of math, isn’t it? After all, algorithms in the form of matrix algebra, summations, integrals, multiplies and adds are the core of what predictive modeling algorithms do. Even rule-based approaches need math to compute how good the if-then-else rules are.
I was participating in a predictive analytics course recently and the question a participant asked at the end of two days of instruction was this: “it’s been a long time since I’ve had to do this kind of math and I’m a bit rusty. Is there a book that would help me learn the techniques without the math?”
The question about math was interesting. But do we need to know the math to build models well? Anyone can build a bad model, but to build a good model, don’t we need to know what the algorithms are doing? The answer, of course, depends on the role of the analyst. I contend, however, that for most predictive analytics projects, the answer is “no”.
Let’s consider building decision tree models. What options does one need to set to build good trees? Here is a short list of common knobs that can be set by most predictive analytics software packages: 1. Splitting metric (CART style trees, C5 style trees, CHAID style trees, etc.) 2. Terminal node minimum size 3. Parent node minimum size 4. Maximum tree depth 5. Pruning options (standard error, Chi-square test p-value threshold, etc.)
The most mathematical of these knobs is the splitting metric. CART-styled trees use the Gini Index, C5 trees use Entropy (information gain), and CHAID style trees use the chi-square test as the splitting criterion. A book I consider the best technical book on data mining and statistical learning methods, “The Elements of Statistical Learning”, has this description of the splitting criteria for decision trees, including the Gini Index and Entropy:
To a mathematician, these make sense. But without a mathematics background, these equations will be at best opaque and at worst incomprehensible. (And these are not very complicated. Technical textbooks and papers describing machine learning algorithms can be quite difficult even for more seasoned, but out-of-practice mathematicians to understand).
As someone with a mathematics background and a predictive modeler, I must say that the actual splitting equations almost never matter to me. Gini and Entropy often produce the same splits or at least similar splits. CHAID differs more, especially in how it creates multi-way splits. But even here, the biggest difference for me is not the math, but just that they use different tests for determining "good" splits
There are, however, very important reasons for someone on the team to understand the mathematics or at least the way these algorithms work qualitatively. First and foremost, understanding the algorithms helps us uncover why models go wrong. Models can be biased toward splitting on particular variables or even particular records. In some cases, it may appear that the models are performing well but in actuality they are brittle. Understanding the math can help remind us that this may happen and why.
The fact that linear regression uses a quadratic cost function tells us that outliers affect overall error disproportionately. Understanding how decision trees measure differences between the parent population and sub-populations informs us why a high-cardinality variable may be showing up at the top of our tree, and why additional penalties may be in order to reduce this bias. Seeing the computation of information gain (derived from Entropy) tells us that binary classification with a small target value proportion (such as having 5% 1s) often won't generate any splits at all.
The answer to the question if predictive modelers need to know math is this: no they don’t need to understand the mathematical notation, but neither should they ignore the mathematics. Instead, we all need to understand the effects of the mathematics on the algorithms we use. “Those who ignore statistics are condemned to reinvent it,” warns Bradley Efron of Stanford University. The same applies to mathematics.
## Thursday, February 14, 2013
### What To Take Home from Your Next Predictive Analytics Conference
Why should one go to a predictive analytics conference? What should one take home from a conference like Predictive Analytics World (PAW)? There are many reasons conferences are valuable including interacting with thought leaders and practitioners, seeing software and hardware tools (the exhibit hall), and learning principles of predictive analytics from talks and workshops. This post focuses on the talks, and in particular, case studies.
There is no quicker way to upgrade our capabilities than having someone else who has "been there" tell us how they succeeded in their development and implementation of predictive models. When I go to conferences, this is at the top of my list. In the best case studies I am able to see different way of looking at a problem than I had considered before, how the practitioner overcame obstacles, how their target variable was defined, what data was used in building the models, how the data was prepared, what figure of merit they used to judge a model's effectiveness, and much more.
Almost all case studies we see at conferences are success stories; we all love winners. Yes, we all know that we learn from mistakes and many case studies actually enumerate mistakes. But success sells and given time limitations in a 20-50 minute talk, few mistakes and dead-ends are usually described in the talks. And, as we used to say in when I was doing government contracting, one works like crazy on the research and then when the money runs out, one declares victory. Putting a more positive spin on the process, we do as well as we can with the resources we have, and if the final solution improves the current system, we are indeed successful.
But once we observe the successful approach, what can we really take home with us? There are three reasons we should be skeptical taking case studies and applying them directly to our own problems.
The first two reasons are straightforward. First, our data is different from the data used in the talk. Obviously. But it is likely to be different enough that one cannot not take the exact same approach to data preparation or target variable creation that one sees at a conference.
Second, our business is different. The way the question was framed and the way predictions can be used are likely to differ in our organization. If we are building models to predict Medicare fraud, they way the “suspicious” claim is processed and which data elements are available vary significantly for each provider (codes being just one example).
The third reason is more subtle and more difficult to overcome. In a fascinating New Yorker article entitled, "The Truth Wears Off: Is there something wrong with the scientific method?", author Jonah Lehrer describes an effect seen by many researchers over the past few decades. Findings in major studies, published in reputable journals, and showing statistically significant results have been difficult to replicate by the original researcher and by others. This is a huge problem because replicating results is what we do as predictive modeler: we assume that behavior in the past can and will be replicated in the future.
In one example, researcher Jonathan Schooler (who was originally at the University of Washington as a graduate student) “demonstrated that subjects shown a face and asked to describe it were much less likely to recognize the face when shown it later than those who had simply looked at it. Schooler called the phenomenon ‘verbal overshadowing’. The study turned him into an academic star."
A few years later, he tried to replicate the study didn’t succeed. In fact, he tried many times over the years and never succeeded. The effect he found at first waned each time he tried to replicate the study with additional data. "This was profoundly frustrating. It was as if nature gave me this great result and then tried to take it back.” There have been a variety of potential explanations for the effect, including “regression to the mean”. This might very well be the case because even when we show statistically significant results defined by having a p value less than 0.05, there is still a chance that the effect found was not really there at all. Over thousands of studies, dozens find effects therefore that aren't really there.
Let's assume we are building models and there is actually no significant difference between responders and non-responders (but we don't know that). However, we work very hard to identify an effect, and eventually we find the effect on training and testing data. We publish. But the effect isn't there; we happened upon the effect just had good luck (which in the long run is actually bad luck!). Even if the chance of finding the effect by chance is 1 in 100, or 1 in 1000, if we experiment enough and search through enough variables, we may happen upon a seemingly good effect eventually. This process, called "over searching" by Jensen and Cohen (see "Multiple Comparisons in Induction Algorithms"), is a real danger.
So what do we do at conferences? We should take home ideas, principles, and approaches rather than recipes. It should spur us to try ideas we either hadn't yet tried or even thought about before.
(An earlier version of this post was first published in the Predictive Analytics Times February 2013 issue)
## Sunday, February 10, 2013
### Using Geographic Data
Most organizations collect and maintain some type of geographic data, yet many ignore this data during analysis. Any business has some record of customer addresses, for instance, but this data is usually formatted in an awkward, non-numeric form. Geographic data can be very predictive, though, since behaviors being predicted often have some correlation to location.
So, how might one use geographic data? Possible answers depend on several factors, most importantly the volume and type of such data. A company serving a national market in the United States, for instance, will have customer shipping and billing addresses (not necessarily the same thing) for each customer (possibly for each transaction). These addresses normally come with a range of spatial granularities: street address, town, state, and associated ZIP Code (a 5-digit postal code).
Even at the largest level of aggregation, the state level, there may be over 50 distinct values (besides the 50 states, American addresses may be in Washington D.C. [technically not part of any state], or any of a number of other American territories, the most common of which is probably Puerto Rico). With 50 or so distinct values, significant data volume is needed to amass the observations needed to draw conclusions about each value. Given the best case scenario, in which all states exhibit equal observation counts, 1,000 observations breaks out into 50 categories of merely 20 observations each- not even enough to satisfy the old statistician's 30 observation rule of thumb. In data mining circles, we are accustomed to having much larger observation counts, but consider that the distribution of state values is never uniform in real data.
Using individual dummy variables to represent each state may be possible with especially large volumes. Possibly an "other" category covering the least frequent so many states will be needed. Another technique which I have found to work well is to replace the categorical state variable with a numeric variable representing a summary of the target variable, conditioned by state. In other words, all instances of "Virginia" are replaced by the average of the target variable for all Virginia cases, all instances of "New Jersey" are replaced by the average of the target variable for all New Jersey cases, and so on. This solution concentrates information about the target which comes from the state in a single variable, but makes interactions with other predictors more opaque. Ideally, such summaries are calculated on special hold-out set of data, used just for this purpose, so as to avoid over-fitting. Again, it may be necessary to lump the smallest so many states together as "other". While I have used American states in my example, it should not be hard for the reader to extend this idea to Canadian provinces, French départements, etc.
Most American states are large enough to provide robust summaries, but as a group they may not provide enough differentiation in the target variable. Changing the spatial scale implies a trade-off: Smaller geographic units exhibit worse summary variance, but improved geographic differentiation. American town names are not necessarily unique within a given state and similar names may be confused (Newtown, Pennsylvania is quite a distance from Newtown Square, Pennsylvania, for instance). In the United States, county names are unambiguous, and present finer spatial detail than states. County names do not, however, normally appear in addresses, but they are easily attached using ZIP Code/County tables easily found on-line. Another possible aggregation is the Section Code Facility, or "SCF", which is the first 3 digits of the ZIP Code.
In the American market, other types of spatial definitions which can be used include: Census Bureau definitions, telephone area codes and Metropolitan Statistical Areas ("MSAs") and related groupings defined by the U.S. Office of Management and Budget. The Census Bureau is a government agency which divides the entire country in to spatial units which vary in scale, down to very small areas (much smaller than ZIP Codes). MSAs are very popular with marketers. There are 366 MSAs at present, and they do not cover the entire land area of the United States, though they do cover about 85% of its population.
It is important to note that nearly all geographic entities change in size, shape and character over time. While existing American state and county boundaries almost never change any more, ZIP code boundaries and Census Bureau definitions, for instance, do change. Changing boundaries obviously complicates analysis, even though historic boundary definitions are often available. Even among entities whose boundaries do not change, radical changes in behavior may happen in geographically distinct ways. Consider that a model built before hurricane Katrina may no longer perform well in areas affected by the storm.
Also note that some geographic units, by definition, "respect" other definitions. American counties, for instance, only contain land from a single state. Others don't: the third-most populous MSA, Chicago-Joliet-Naperville, IL-IN-WI, for example, overlaps three different states.
Being creative when defining model inputs can be as helpful with geographic data as it is with more conventional data. In addition to the billing address itself, consider transformations such as: Has the billing address ever changed (1) or not (0)? How many times has the billing address changed? How often has the billing address changed (number of times changed divided by number of months the account has been open)? How far is the shipping address from the billing address? And so on...
Much more sophisticated use may be made of geographic data than has been described in this short posting. Software is available commerically which will determine drive time contours about locations, which would be useful, for instance when modeling retail store location revenue models. Additionally, there is an entire of statistics, called spatial statistics, which defines an entire class of analysis procedures specific to this sort of thing.
I encourage readers who have avoided geographic data to consider even simple mechanisms to include it in model construction. Opening up a new dimension in your analysis may provide significant returns.
## Saturday, February 02, 2013
### When Analysis Isn't the Answer
Data mining is an important tool whose benefits have been demonstrated in diverse fields, among business, government and non-profit organizations. Its application areas continue to grow, especially given the ever-shrinking cost of gathering and organizing data. Yet, there are problems for which data mining is wholly unsuited as a solution.
To understand when data mining is not applicable, it will be helpful to define precisely when it is applicable. Data mining (inferential statistics, predictive analytics, etc.) requires data stored in a machine format of sufficient volume, quality and relevance so as to permit the construction of predictive models which assist in real-world decision making.
Most of our time as data miners is spent worrying over the quality of the data and the process of turning data into models, however it is important to realize the usual context of data mining. Most organizations can perform basic decision making competently, and they have done so for thousands of years. Whether the base decision process is human judgment, a simple set of rules or a spreadsheet, much performance potential is already realized before data mining is applied. Consultants' marketing notwithstanding, data mining typically inhabits the margin of performance, where it tries to bring an extra "edge".
So, if the above two paragraphs describe conditions conducive to data mining success, what sorts of real-world situations defy data mining? The most obvious would be problems featuring data that is too small, too narrow, too noisy or of too little relevance to allow effective modeling. Organizations which have not maintained good records, which still rely on non-computer procedures and those with too little history are good examples. Even within very large organizations which collect and store enormous databases, there may be no relevant data for the problem at hand (for instance, when a new line of business is being opened, or new products introduced). It is surprising how often business people expect to extract value from a situation when they have failed to invest in appropriate data gathering.
Another large area with minimal data mining potential is organizations whose basic business process is so fundamentally broken that the usual decision making procedures have failed to do the usual "heavy lifting". Any of us can easily recall experiences in retail establishments whose operation was so flawed that it was obvious that the profit potential was not nearly being exploited. Data mining cannot fine tune a process which is so far gone. No amount of quantitative analysis will fix unkept shelves, weak product offering or poor employee behavior.
## Wednesday, January 16, 2013
### Three Ways to Get Your Predictive Models Deployed
We all know that given reasonable data, a good predictive modeler can build a model that works well and helps make makes better decisions than what is currently used in your organization (at least in our own minds). Newer data, sophisticated algorithms, and a seasoned analyst are all working in our favor when we build these models, and if success were measured by accuracy (as they are in most data mining competitions), we're in great shape. Yes, there are always gotchas and glitches along the way. But when my deliverable is only slideware, even of the modeling is hard, I'm confident of being able to declare victory at the end.
However, the reality is that there is much more to the transition from cool model to actual deployment than a nice slide deck and paper accepted at one's favorite predictive analytics, data mining or big data conference. In these venues, the winning models are those that are "accurate" (more on that later) and have used creative analysis techniques to find the solution; we won't submit a paper when we only had to press the "go" button and have the data mining software give us a great solution!
For me, the gold standard is deployment. If the model gets used and improves the decisions an organization makes, I've succeeded. Three ways to increase the likelihood your models are deployed are:
1) Make sure the model stakeholder designs deployment into the project from the beginning
The model stakeholder is the individual, usually a manager, who is the advocate of predictive models to decision-makers. It is possible that a senior-level modeler can do this task, but that person must be able to switch hit: he or she must be able to speak the language of management and be able to talk technical detail to analytics. This may require more than one trusted person: the manager, who is responsible and makes the ultimate decisions about the models, and the lead modeler, who is responsible for the technical aspects of the model. It is more than "talking the talk" and knowing buzz-words in both realms; the person or persons must truly be "one of" both groups.
For those who have followed my blog posts and conference talks, you know I am a big advocate of the CRISP-DM process model (or equivalent methodologies, which seem to be endless). I've referred to CRISP-DM often, including on topics related to what data miners need to learn and Defining the Target Variable, just as two examples.
The stakeholder must not only understand the business of objectives of the model (Business Understanding in CRISP-DM), but must be present during discussions take place related to which models will be built. It is essential that reasonable expectations are put into place from the beginning, including what a good model will "look like" (accuracy and interpretability) and how the final model will be deployed.
I've seen far too many projects die or become inconsequential because either the wrong objectives were used in building the models, meaning the models were operationally useless, or because the deployment of the models was not considered, meaning again that the models were operationally useless. As an example, on one project, the model was assumed to be able to be run within a rules engine, but the models that were built were not rules at all, but were complex non-linear models that could not be translated into rules. The problem obviously could have been avoided had this disconnect been verbalized early in the modeling process.
2) Make sure modelers understand the purpose of the models
The modelers must know how the models will be used and what metrics should be used to judge model performance. A good summary of typical error metrics used by modelers is found here. However, for most of the models I have deployed in customer acquisition, retention, and risk modeling, the treatment based on the model is never applied to the entire population (we don't mail everyone, just a subset). So the metrics that make the most sense are often ones like "lift after the top decile", maximum cumulative net revenue, top 1000 scores to be investigated, etc. I've actually seen negative correlations between the ranking of models based on global metrics (like classification error or R^2) vs. the ranking based on subset selection ranking, such as top 1000 scores; very different models may be deployed depending on the metric one uses to assess them. If modelers aren't aware of the metric to be used, the wrong model can be selected, even one that does worse than the current approach.
Second, if the modelers don't understand how the models will be deployed operationally, they may find a fantastic model, one that maximizes the right metric, but is useless. The Neflix Prize is a great example: the final winning model was accurate but far too complex to be used. Netflix extracted key pieces to the models to operationalize instead. I've had customers stipulate to me that "no more than 10 variables can be included in the final model". If modelers aren't aware of specific timelines or implementation constraints, a great but useless model can be the result.
3) Make sure the model stakeholder understands what the models can and can't do
In the effort to get models deployed, I've seen models elevated to a status they don't deserve, most often by exaggerating their accuracy and expected performance once in operation. I understand why modelers may do this: they have a direct stake in what they did. But the manager must be more skeptical and conservative.
One of the most successful colleagues I've ever worked with used to assess model performance on held-out data using the metric we had been given (maximum depth one could mail to and still achieve the pre-determined response rate). But then he always backed off what was reported to his managers by about 10% to give some wiggle room. Why? Because even in our best efforts, there is still a danger that the data environment after the model is deployed will differ from that used in building the models, thus reducing the effectiveness of the models.
A second problem for the model stakeholder is communicating an interpretation of the models to decision-makers. I've had to do this exercise several times in the past few months and it is always eye-opening when I try to explain the patterns a model is finding when the model is itself complex. We can describe overall trends ("on average", more of X increases the model score) and we can also describe specific patterns (when observable fields X and Y are both high, the model score is high). Both are needed to communicate what the models do, but have to connect with what a decision-maker understands about the problem. If it doesn't make sense, the model won't be used. If it is too obvious, the model isn't worth being used.
The ideal model for me is one where the decision-maker nods knowingly at the "on average" effects (these should usually be obvious). Then, once you throw in some specific patterns, he or she should scrunch his/her eyes, think a bit, then smile as the implications of the pattern dawns on them as that pattern really does make sense (but was previously not considered).
As predictive modelers, we know that absolutes are hard to come by, so even if these three principles are adhered to, other factors can sabotage the deployment of a model. Nevertheless, in general, these steps will increase the likelihood that models are deployed. In all three steps, communication is the key to ensuring the model built addresses the right business objective, the right scoring metric, and can be deployed operationally.
NOTE: this post was originally posted for the Predictive Analytics Times at http://www.predictiveanalyticsworld.com/patimes/january13/
## Friday, January 04, 2013
### Top Posts in 2012
For the second consecutive year, a quick look back at posts from the prior year.
For posted in 2012, in order of popularity:
I’m also adding #6 because Will post in December did very well, but of course has had only one month to accumulate views.
From posts prior to 2012, in order of popularity for 2012:
1. What Do Data Miners Need to Learn (June 2011)
2. Free and Inexpensive Data Mining Software (November 2006) This post needs to be updated!
3. Why Normalization Matters for K Means (April 2009) It always amazes me why this post persists as one of the most popular, but nearly ¼ of the visits used the search term “K Means Noisy Data”
4. Data Mining Data Sets (April 2008) This post also needs to be updated
5. Business Analytics vs. Business Intelligence (December 2009)
One final note: When I look back at visits since the start of this blog, 4 of the top 5 posts are the top 4 “prior to 2012” above. The #5 most popular post over all the years I’ve had the blog is one by Will from 2007, “Missing Values and Special Values: The Plague of Data Analyis”, one that I have always liked very much.
Best to all of you in 2013!
## Wednesday, December 19, 2012
### 6 Reasons You Hired the Wrong Data Miner
As is in any discipline, talent within data mining community varies greatly. Generally, business people and others who hire and manage technical specialists like data miners are not themselves technical experts. This makes it difficult to evaluate the performance of data miners, so this posting is a short list of possible deficiencies in a data miner's performance. Hopefully, this will spare some heartache in the coming year. Merry Christmas!
1. The data miner has little or no programming skill.
Most work environments require someone to extract and prepare the data. The more of this process which the data miner can accomplish, the less her dependence on others. Even in ideal situations with prepared analytical data tables, the data miner who can program can wring more from the data than her counterpart who cannot (think: data transformations, re-coding, etc.). Likewise, when her predictive model is to be deployed in a production system, it helps if the data miner can provide code as near to finished as possible.
2. The data miner is unable to communicate effectively with non-data miners.
Life is not all statistics: Data mining results must be communicated to colleagues with little or no background in math. If other people do not understand the analysis, they will not appreciate its significance and are unlikely to act on it. The data miner who can express himself clearly to a variety of audiences (internal customers, management, regulators, the press, etc.) is of greater value to the organization than his counterpart who cannot. The data miner should should receive questions eagerly.
3. The data miner never does anything new.
If the data miner always approaches new problems with the same solution, something is wrong. She should be, at least occasionally, suggesting new techniques or ways of looking at problems. This does not require that new ideas be fancy: Much useful work can be done with basic summary statistics. It is the way they are applied that matters.
4. The data miner cannot explain what they've done.
Data mining is a subtle craft: there are many pitfalls and important aspects of statistics and probability are counter-intuitive. Nonetheless, the data miner who cannot provide at least a glimpse into the specifics of what they've done and why, is not doing all he might for the organization. Managers want to understand why so many observations are needed for analysis (after all, they pay for those observations), and the data miner should be able to provide some justification for his decisions.
5. The data miner does not establish the practical benefit of his work.
A data miner who cannot connect the numbers to reality is working in a vacuum and is not helping her manager (team, company, etc.) to assess or utilize her work product. Likewise, there's a good chance that she is pursuing technical targets rather than practical ones. Improving p-values, accuracy, AUC, etc. may or may not improve profit (retention, market share, etc.).
6. The data miner never challenges you.
The data miner has a unique view of the organization and its environment. The data miner works on a landscape of data which few of his coworkers ever see, and he is less likely to be blinded by industry prejudices. It is improbable that he will agree with his colleagues 100% of the time. If the data miner never challenges assumptions (business practices, conclusions, etc.), then something is wrong.
## Tuesday, November 06, 2012
### Why Predictive Modelers Should be Suspicious of Statistical Tests (or why the Redskin Rule fools us)
Well, the danger is really not the statistical test per se, it the interpretation of the statistical test.
Yesterday I tweeted (@deanabb) this fun factoid: "Redskins predict Romney wins POTUS #overfit. if Redskins lose home game before election => challenger wins (17/18) http://www.usatoday.com/story/gameon/2012/11/04/nfl-redskins-rule-romney/1681023/" I frankly had never heard of this "rule" before and found it quite striking. It even has its own Wikipedia page (http://en.wikipedia.org/wiki/Redskins_Rule).
For those of us in the predictive analytics or data mining community, and those of us who use statistical tests to help out interpreting small data, 17/18 we know is a hugely significant finding. This can frequently be good: statistical tests will help us gain intuition about value of relationships in data even when they aren't obvious.
In this case, an appropriate test is a chi-square test based on the two binary variables (1) did the Redskins win on the Sunday before the general election (call it the input or predictor variable) vs. (2) did the incumbent political party win the general election for President of the United States (POTUS).
According to the Redskins Rule, the answer is "yes" in 17 of 18 cases since 1940. Could this be by chance? If we apply the chi-square test to it, it sure does look significant! (chi-square = 14.4, p < 0.001). I like the decision tree representation of this that shows how significant it is (built using the Interactive CHAID tree in IBM Modeler on Redskin Rule data I put together here):
It's great data--9 Redskin wins, 9 Redskin losses, great chi-square statistic!
OK, so it's obvious that this is just another spurious correlation in the spirit of all of those fun examples in history, such as the superbowl winning conference predicting if the stock market would go up or down in the next year at a stunning 20 or 22 correct. It even was the subject of academic papers on the subject!
The broader question (and concern) for predictive modelers is this: how do we recognize when we have uncovered spurious correlations in the data that are merely spurious? This can happen especially when we don't have deep domain knowledge and therefore wouldn't necessarily identify variables or interactions as spurious. In examples such as the election or stock market predictions, no amount of "hold out" samples, cross-validation or bootstrap sampling would uncover the problem: it is in the data itself.
We need to think about this because inductive learning techniques search through hundreds, thousands, even millions of variables and combinations of variables. The phenomenon of "over searching" is a real danger with inductive algorithms as they search and search for patterns in the input space. Jensen and Cohen have a very nice and readable paper on this topic (PDF here). For trees, they recommend using the Bonferroni adjustment which does help penalize the combinatorics associated with splits. But our problem here goes far deeper than overfitting due to combinatorics.
Of course the root problem with all of these spurious correlations is small data. Even if we have lots of data, what I'll call here the "illusion of big data", some algorithms make decisions based on smaller populations, like decision trees, rule induction and nearest neighbor (i.e., algorithms that build bottom-up). Anytime decisions are made from populations of 15, 20, 30 or even 50 examples, there is a danger that our search through hundreds of variables will turn out a spurious relationship.
What do we do about this? First, make sure you have enough data so that these small-data effects don't bite you. This is why I strongly recommend doing data audits and looking for categorical variables that contain levels with at most dozens of examples--these are potential overfilling categories.
Second, don't hold strongly any patterns discovered in your data based on solely on the data, especially if they are based on relatively small sample sizes. These must be validated with domain experts. Decision trees are notorious for allowing splits deep in the trees that are "statistically significant" but dangerous nevertheless because of small data sizes.
Third, the gist of your models have to make sense. If they don't, put on your "Freakonomics" hat and dig in to understand why the patterns were detected by the models. In our Redskin Rule, clearly this doesn't make sense causally, but sometimes the pattern picked up by the algorithm is just a surrogate for a real relationship. Nevertheless, I'm still curious to see if the Redskin Rule will prove to be correct once again. This year it predicts a Romney win because the Redskins lost and therefore the incumbent party (D) by the rule should lose. UPDATE: by way of comparison...the chances of having 17/18 or 18/18 coin flips turn up heads (or tails--we're assuming a fair coin after all!) is 7 in 100,000 or 1 in 14,000. Put another way, if we examined 14K candidate variables unrelated to POTUS trends, the chances are that one of them would line up 17/18 or 18/18 of the time. Unusual? Yes. Impossible? No!
## Tuesday, October 23, 2012
### Data Preparation: Know Your Records!
Data preparation in data mining and predictive analytics (dare I also say Data Science?) rightfully focuses on how the fields in ones data should be represented so that modeling algorithms either will work properly or at least won't be misled by the data. These data preprocessing steps may involve filling missing values, reigning in the effects of outliers, transforming fields so they better comply with algorithm assumptions, binning, and much more.
In recent weeks I've been reminded how important it is to know your records. I've heard this described in many ways, four of which are:
• the unit of analysis
• the level of aggregation
• what a record represents
• unique description of a record
• For example, does each record represent a customer? If so, over their entire history or over a time period of interest? In web analytics, the time period of interest may be a single session, which if it is true, means that an individual customer may be in the modeling data multiple times as if each visit or session is an independent event.
Where this especially matters is when disparate data sources are combined. If one is joining a table of customerID/Session data with another table with each record representing a customerID, there's no problem. But if the second table represents customerID/store visit data, there will obviously be a many-to-many join resulting in a big mess.
This is probably obvious to most readers of this blog. What isn't always obvious is when our assumptions about the data result in unexpected results. What if we expect the unit of analysis to be customerID/Session but there are duplicates in the data? Or what if we had assumed customerID/Session data but it was in actuality customerID/Day data (where ones customers typically have one session per day, but could have a dozen)?
The answer is just like we need to perform a data audit to identify potential problems with fields in the data, we need to perform record audits to uncover unexpected record-level anomalies. We've all had those data sources where the DBA swears up and down that there are no dups in the data, but when we group by customerID/Session, we find 1000 dups.
So before the joins and after joins, we need to do those group by operations to find examples with unexpected numbers of matches.
In conclusion: know what your records are supposed to represent, and verify verify verify. Otherwise, your models (who have no common sense) will exploit these issues in undesirable ways! | 8,318 | 41,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2013-20 | longest | en | 0.940729 |
https://kr.mathworks.com/matlabcentral/cody/problems/252-project-euler-problem-16-sums-of-digits-of-powers-of-two/solutions/46160 | 1,601,401,155,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400202418.22/warc/CC-MAIN-20200929154729-20200929184729-00579.warc.gz | 463,182,204 | 16,621 | Cody
Problem 252. Project Euler: Problem 16, Sums of Digits of Powers of Two
Solution 46160
Submitted on 20 Feb 2012
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 Fail
%% x = 0; y_correct = 1; assert(isequal(pow2_sumofdigits(x),y_correct))
Error: Undefined function 'unit64' for input arguments of type 'double'.
2 Fail
%% x = 1; y_correct = 2; assert(isequal(pow2_sumofdigits(x),y_correct))
Error: Undefined function 'unit64' for input arguments of type 'double'.
3 Fail
%% x = 15; y_correct = 26; assert(isequal(pow2_sumofdigits(x),y_correct))
Error: Undefined function 'unit64' for input arguments of type 'double'.
4 Fail
%% x = 345; y_correct = 521; assert(isequal(pow2_sumofdigits(x),y_correct))
Error: Undefined function 'unit64' for input arguments of type 'double'.
5 Fail
%% x = 999; y_correct = 1367; assert(isequal(pow2_sumofdigits(x),y_correct))
Error: Undefined function 'unit64' for input arguments of type 'double'.
6 Fail
%% x = 2000; y_correct = 2704; assert(isequal(pow2_sumofdigits(x),y_correct))
Error: Undefined function 'unit64' for input arguments of type 'double'. | 351 | 1,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.240194 |
https://thedataspecialist.wordpress.com/category/excel/page/2/ | 1,601,181,030,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400250241.72/warc/CC-MAIN-20200927023329-20200927053329-00489.warc.gz | 648,756,043 | 17,408 | ## A look at the VLOOKUP function
Since Bill Jelen (MrExcel) launched a VLOOKUP week, I figured it was a good occasion to take a look at this function and compare it with the INDEX/MATCH alternative. In order to prevent my post from being too long, I decided to dedicate this entry to the VLOOKUP function only.
# Basics
## What the function does
What is the VLOOKUP function about? It allows you to search for a key in a table and find the corresponding value in another column on the corresponding row.
Suppose you have the following reference table for product prices and you want to look for the price of product XT987.
First, you would look for the value XT987 in the first column and find the row where this value is.
Then, you would move to the second column (the price column) to retrieve the desired value.
## The syntax
This is what the VLOOKUP function does for you. The function has two syntax forms and comes in two flavors.
=VLOOKUP(key_value, lookup_table ,result_column_index, approximate_search)
=VLOOKUP(key_value, lookup_table ,result_column_index)
key_value is actually the value we are looking for. In my example, this would be XT987.
lookup_table would be our reference table.
result_column_index would be the index our price column (that is, 2).
approximate_search is an option specifying how Excel will look for the value. This parameter tells Excel to perform an approximate search or not. It takes the values FALSE or TRUE and is optional (see the second syntax form), the default value being TRUE. Note that the option TRUE also requires you to sort your table in ascending order (of the key value) for the function to work properly.
Usually FALSE is the option you want to use. Let us see what the difference is.
# Approximate or exact search
## The case for an exact search
Suppose you have a product number Y5678 and want to find the corresponding price in your table. Now, as you may have noticed, this product is not in our table. Depending on the value you chose for the last parameter, you will get different results.
For an exact search, the formula will look like this:
=VLOOKUP(“Y5678”, lookup_table ,2, FALSE)
and Excel will return #NA since it could not find any corresponding value.
For the approximate search, the formula will look like this:
=VLOOKUP(“Y5678”, lookup_table ,2, TRUE)
or like this:
=VLOOKUP(“Y5678”, lookup_table ,2)
In that case, Excel will go through the product numbers and will look for Y5678 or the last value that is “smaller” than Y5678. In our case, XT987 is the last value in our Product Number column that comes before Y5678 in alphanumeric order.
Excel will thus return 569, which may not be the result you expected.
## The case for an approximate search
### Interval searches
Approximate search is well suited for interval searches.
Suppose you have the following table.
You want to associate the corresponding shipment fee with each product.
Suppose the shipping price of a product is based on its weight.
If a product weights less than 5 kg shipment will cost 5€.
If a product weights no less than 5 kg but less than 20 kg shipment will cost 20€.
If a product weights no less than 20 kg but less than 90 kg shipment will cost 45€.
If a product weights more than 90 kg shipment is not available.
Then you could use the following lookup table to map the min weight to the shipment fee:
Then the following equivalent formulas will give you expected results
=VLOOKUP(weight, lookup_table_shipment, 2)
=VLOOKUP(weight, lookup_table_shipment, 2, TRUE)
When applied to the table, this will give the following results.
Let us take the product ZLM988 as an example. The weight of this product is 86 kg. Excel will look for the last key (the minimum weight) in the lookup table, that is equal to or smaller than
86. The corresponding entry is 20. Excel will then return the shipment on the corresponding row and will thus correctly return 45.
### Partial searches
This problem comes from a discussion group on a well-known professional networking platform. The original poster was looking for a way to categorize products based on their product number. Now this series number would start with letters just as in my previous examples. Now the part number could have between 1 and 3 characters (I guess, maybe more in the future) and the list of part numbers would be updated regularly. People provided some smart sophisticated solutions, but none had come up with most simple one. Which is to use the VLOOKUP function.
Now, let us take our last table, and categorize these products based on their part number.
For this, we will use the following lookup table.
Now, one of the following formulas will do the job:
=VLOOKUP(product_number, lookup_table_categories, 2)
=VLOOKUP(product_number, lookup_table_categories, 2, TRUE)
and will return the expected results:
Be cautious though. Since we are using an approximate search, we may still have the issue that some part number is missing in the lookup table, thus leading to an incorrect result. Suppose you add another product XSM654 to the data table, whose part number XSM corresponds to the Electronic Devices category. If this part number is not available in our lookup table, the last entry whose value precedes XSM654 in alphanumerical order, is XT. Excel will legitimately return an incorrect category (Books).
You will thus have to implement some formula, to catch these sort of errors.
Note though that you can also use VLOOKUP to retrieve the part number from the lookup table. In our last example, the formula
=VLOOKUP(product_number, lookup_table_categories, 1)
will return XT, when the product_number equals XSM654.
This allows to test whether our lookup table has been correctly maintained or not. For example, you could use the following formula to test whether or not product_number actually starts with the part number returned by the VLOOKUP function:
=SEARCH(VLOOKUP(product_number,lookup_table_categories,1),product_number) = 1 | 1,339 | 6,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.864223 |
http://gmatclub.com/forum/anyone-given-the-5-free-kaptest-with-gmat-premier-145682.html?kudos=1 | 1,485,134,412,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281659.81/warc/CC-MAIN-20170116095121-00413-ip-10-171-10-70.ec2.internal.warc.gz | 122,950,769 | 50,104 | Anyone given the 5 free KAPTEST with GMAT premier : General GMAT Questions and Strategies
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# Anyone given the 5 free KAPTEST with GMAT premier
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Joined: 28 May 2012
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Anyone given the 5 free KAPTEST with GMAT premier [#permalink]
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13 Jan 2013, 07:54
Hi all! Have anyone given the 5 free tests that we get with KAPLAN GMAT Premier 2012-2013.
How about the scoring of those. ??
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Re: Anyone given the 5 free KAPTEST with GMAT premier [#permalink]
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13 Jan 2013, 16:27
I have Kaplan premier 2013. About those 5 free CATs, I think the quality and difficulty level of the questions is good. Most quant/verbal questions are pretty much GMAT like. But the scoring algorithm is not that accurate.
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Last edited by PrashantPonde on 13 Jan 2013, 23:41, edited 1 time in total.
Intern
Joined: 28 May 2012
Posts: 12
Location: United States
Concentration: Operations, General Management
GMAT 1: 650 Q48 V30
GRE 1: 1230 Q730 V500
GPA: 3.4
WE: Information Technology (Computer Software)
Followers: 0
Kudos [?]: 5 [0], given: 2
Re: Anyone given the 5 free KAPTEST with GMAT premier [#permalink]
### Show Tags
13 Jan 2013, 23:39
PraPon wrote:
I have Kaplan premier 2013. About those 5 free CATs, I think the quality and difficulty level of the questions is good. Most quant/verbal questions are pretty much GMAT link. But the scoring algorithm is not that accurate.
So what score are you getting on KAPTEST??
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Re: Anyone given the 5 free KAPTEST with GMAT premier [#permalink]
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13 Jan 2013, 23:40
In the range of 730-750.
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Intern
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Re: Anyone given the 5 free KAPTEST with GMAT premier [#permalink]
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14 Jan 2013, 00:22
PraPon wrote:
In the range of 730-750.
and on Manhattangmat ??
Re: Anyone given the 5 free KAPTEST with GMAT premier [#permalink] 14 Jan 2013, 00:22
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# Anyone given the 5 free KAPTEST with GMAT premier
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,335 | 4,529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-04 | latest | en | 0.814449 |
https://lotharlorraine.wordpress.com/tag/probability/ | 1,653,493,771,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00518.warc.gz | 433,346,365 | 35,715 | On the probability of evolution
In the following post, I won’t try to calculate specific values but rather to explicate my own Knowledge-dependent frequentist probabilities by using particular examples.
The great evolutionary biologist Stephen Jay Gould was famous for his view that Evolution follows utterly unpredictable paths so that the emergence of any species can be viewed as a “cosmic accident”.
He wrote:
We are glorious accidents of an unpredictable process with no drive to complexity, not the expected results of evolutionary principles that yearn to produce a creature capable of understanding the mode of its own necessary construction.
“We are here because one odd group of fishes had a peculiar fin anatomy that could transform into legs for terrestrial creatures; because the earth never froze entirely during an ice age; because a small and tenuous species, arising in Africa a quarter of a million years ago, has managed, so far, to survive by hook and by crook. We may yearn for a ‘higher answer’– but none exists”
“Homo sapiens [are] a tiny twig on an improbable branch of a contingent limb on a fortunate tree.”
Dr. Stephen Jay Gould, the late Harvard paleontologist, crystallized the question in his book ”Wonderful Life.” What would happen, he asked, if the tape of the history of life were rewound and replayed? For many, including Dr. Gould, the answer was clear. He wrote that ”any replay of the tape would lead evolution down a pathway radically different from the road actually taken.”
You’re welcome to complement my list by adding other quotations. 🙂
Evolution of man
So, according to Stephen Jay Gould, the probability that human life would have evolved on our planet was extremely low, because countless other outcomes would have been possible as well.
Here, I’m interested to know what this probability p(Homo) means ontologically.
Bayesian interpretation
For a Bayesian, p(Homo) means the degree of belief we should have that a young planet having exactly the same features as ours back then would harbor a complex evolution leading to our species.
Many Bayesians like to model their degrees of belief in terms of betting amount, but in that situation this seems rather awkward since none of them would still be alive when the outcome of the wager will be known.
Let us consider (for the sake of the argument) an infinite space which also necessarily contain an infinite number of planets perfectly identical to our earth (according to the law of the large numbers.)
According to traditional frequentism, the probability p(Homo) that a planet identical to our world would produce mankind is given as the ratio of primitive earths having brought about humans divided by the total number of planets identical to ours for a large enough (actually endless) number of samples:
p(Homo) ≈ f(Homo) = N(Homo) / N(Primitive_Earths).
Knowledge-dependent frequentism
According to my own version of frequentism, the planets considered in the definition of probability do not have to be identical to our earth but to ALL PAST characteristics of our earth we’re aware of.
Let PrimiEarths be the name of such a planet back then.
The probability of the evolution of human life would be defined as the limit p'(Homo) of
f'(Homo) = N'(Homo) / N(PrimiEarths‘)
whereby N(PrimiEarths‘) are all primitive planets in our hypothetical endless universe encompassing all features we are aware of on our own planet back then and N'(Homo) is the number of such planets where human beings evolved.
It is my contention that if this quantity exists (that is the ratio converges to a fixed value whereas the size of the sample is enlarged), all Bayesians would adopt p'(Homo) as their own degree of belief.
But what if there were no such convergence? In other words, while one would consider more and more N(PrimiEarths‘) f'(Homo) would keep fluctuating between 0 and 1 without zooming in to a fixed value.
If that is the case, this means that the phenomenon “Human life evolving on a planet gathering the features we know” is completely unpredictable and cannot therefore be associated to a Bayesian degree of belief either, which would mean nothing more than a purely subjective psychological state.
Evolution of bird
I want to further illustrate the viability of my probabilistic ontology by considering another evolutionary event, namely the appearance of the first birds.
Let us define D as : “Dinosaurs were the forefathers of all modern birds”, a view which has apparently become mainstream over the last decades.
For a Bayesian, p(D) is the degree of belief about this event every rational agent ought to have.
Since this is an unique event of the past, many Bayesians keep arguing that it can’t be grasped by frequentism and can only be studied if one adopts a Bayesian epistemology.
It is my contention this can be avoided by resorting to my Knowledge-Dependent Frequentism (KDF).
Let us define N(Earths’) the number of planets encompassing all features we are aware of on our modern earth (including, of course, the countless birds crowding out the sky, and the numerous fossils found under the ground).
Let us define N(Dino’) as the number of these planets where all birds originated from dinosaurs.
According to my frequentism, f(D) = N(Dino’) / N(Earths’), and p(D) is the limit of f(D) as the sample is increasingly enlarged.
If p(D) is strong, this means that on most earth-like planets containing birds, the ancestors of birds were gruesome reptilians.
But if p(D) is weak (such as 0.05), it means than among the birds of 100 planets having exactly the known features of our earth, only 5 would descend from the grand dragons of Jurassic Park.
Again, what would occur if p(D) didn’t exist because f(d) doesn’t converge as the sample is increased?
This would mean that given our current knowledge, bird evolution is an entirely unpredictable phenomenon for which there can be no objective degree of belief every rational agent ought to satisfy.
A physical probability dependent on one’s knowledge
In my whole post, my goal was to argue for an alternative view of probability which can combine both strengths of traditional Frequentism and Bayesianism.
Like Frequentism, it is a physical or objective view of probability which isn’t defined in terms of the psychological or neurological state of the agent.
But like Bayesianism, it takes into account the fact that the knowledge of a real agent is always limited and include it into the definition of the probability.
To my mind, Knowledge-Dependent Frequentism (KDF) seems promising in that it allows one to handle the probabilities of single events while upholding a solid connection to the objectivity of the real world.
In future posts I’ll start out applying this concept to the probabilistic investigations of historical problems, as Dr. Richard Carrier is currently doing.
Thematic list of ALL posts on this blog (regularly updated)
My other blog on Unidentified Aerial Phenomena (UAP)
A mathematical proof of Bayesianism?
This is going to be another boring post (at least for most people who are not nerds).
However before approaching interesting questions such as the existence of God, morality and history a sound epistemology (theory of knowledge) must already be present. During most (heated) debates between theists and atheists, people tend to take for granted many epistemological principles which are very questionable.
This is why I spend a certain amount of my time exploring such questions, as a groundwork for more applied discussions.
I highly recommand all my reader to first read my two other posts on the concept of probability before reading what follows.
Bayesianism is a theory of knowledge according to which our degrees of belief in theories are well defined probabilities taking on values between 0 and 1.
According to this view, saying that string theory has a probability of 0.2 to be true is as meaningful as saying that a normal dice randomly thrown has a probability of 1/6 to produce a “3”.
Bayesians like asserting over and over again that it is mathematically proven to say we ought to compute the likelihood of all beliefs according to the laws of probability and first and foremost Bayes formula:
Here I want to debunk this popular assertion. Bayes theorem can be mathematically proven for frequential probabilities but there is no such proof that ALL our degrees of belief behave that way.
Let us consider (as an example) the American population (360 millions people) and two features a person might have.
CE (Conservative Evangelical): the individual believes that the Bible contains no error.
FH (Fag Hating): the individual passionately hates gay people.
Let us suppose that 30% of Americans are CE and that 5.8% of Americans hate homosexuals.
The frequencies are f(CE) = 0.30 and f(FH) = 0.058
Let us now consider a random event: you meet an American by chance.
What is the probability that you meet a CE person and what is the probability that you meet a FH individual?
According to a frequentist interpretation, the probability equals the frequency of meeting such kinds of persons given a very great (actually infinite) number of encounters.
From this it naturally follows that p(CE) = f(CE) = 0.30 and p(FH) = f(FH) = 0.058
Let us now introduce the concept of conditional probability: if you meet a Conservative Evangelical, what is the probability that he hates faggots p(FH|CE)? (the | stands for „given“).
If you meet a fag-hating person, what is the probability that he believes in Biblical inerrancy p(CE|FH)?
To answer these questions (thereby proving Bayes theorem) it is necessary to get back to our consideration of frequencies.
Let us consider that 10% of all Conservative Evangelicals and 4% of people who are not CE hate faggots: f(FH/CE) = 0.1 and f(FH/CE) = 0.04. The symbol ⌐ stands for the negation (denial) of a proposition.
The proportion of Americans who are both conservative Evangelicals and fag-haters is f(FHCE) = f(FH/CE)*f(CE) = 0.1*0.3 = 0.03.
The proportion of Americans who are NOT conservative Evangelicals but fag-haters is f(FH∩⌐CE) = f(FH/⌐CE)*f(⌐CE) = 0.04*0.7 = 0.028.
Logically the frequency of fag-haters in the whole American population is equal to the sum of the two proportions:
f(FH) = f(FHCE) + f(FH∩⌐CE) = 0.03 + 0.028 = 0.058
But what if we are interested to know the probability that a person is a conservative Evangelical IF that person hates queers p(CE|FH)?
This corresponds to the frequency(proportion) of Conservative Evangelicals among Fag-Haters: f(CE|FH).
We know that f(FHCE) = f(CE∩FH) = f(CE|FH)*f(FH)
Thus f(CE|FH) = f(FH∩CE) / f(FH)
Given a frequentist interpretation of probability, this entails that
which is of course Bayes theorem. We have mathematically proven it in this particular case but the rigorous mathematical demonstration would be pretty much the same given events expressable as frequencies.
If you meet an American who hates gays, the probability that he is a Conservative Evangalical is 51.72% (given the validity of my starting values above).
But let us now consider the Bayesian interpretation of probability (our degree of confidence in a theory) in a context having nothing to do with frequencies.
Let S be “String theory is true“ and UEP “an Undead Elementary Particle has been detected during an experience in the LHC“.
In that context, the probabilities correspond to our confidence in the truth of theories and hypotheses.
We have no compelling grounds for thinking that
, that is to say that is the way our brains actually work or ought to work that way in order to strive for truth.
The mathematical demonstration used to prove Bayes theorem relies on related frequencies and cannot be employed in a context where propositions (such as S and UEP) cannot be understood as frequencies.
Considering ALL our degrees of beliefs like probabilities is a philosophical decision and not an inevitable result of mathematics.
I hope that I have been not too boring for lay people.
Now I have a homework for you: what is the probability that Homeschooling Parents would like to employ my post as an introduction to probability interpretation, given that they live in the Bible Belt p(HP|BB)?
On the ontology of the objective Bayesian probability interpretation
Warning: this post is going to analyse mathematical concepts and will most likely cause intense headaches to non-mathematical brains.
At the beginning I wanted to make it understandable for lay people before I realized I am not the right man for such a huge task.
I considered it necessary to write it since Bayesian considerations plays a very important role in many scientific and philosophical fields, including metaphysic problems such as the existence of God.
Basically, objective Bayesianism is a theory of knowledge according to which probabilities are degrees of belief (and vice-versa) whose values can be objectively identified by every rational agent disposing of the same information.
It stands in opposition to frequentism which stipulates that the probability of an event is identical with the frequency of a great (nearly infinite) number of events.
I illustrated how this plays out in a previous post.
The name of the philosophy stems from Bayes theorem which stipulates that
where P(A|B) is the probability of an event A given an event B, B the probability of the event B given the event A, P(A) and P(B) the total probabilities of the event A and B, respectively.
At that point, it is important to realize that the Bayesian identification of these probabilities with degrees of belief in the hypotheses A and B is a philosophical decision and not a mathematical result, as many Bayesians seem to believe.
Bayes theorem is utilized to actualize the probability of the theory A as new data (the truth of B) come in. Unless one believes in infinite regress, there is going to be basic probabilities called priors which cannot themselves be deduced from former probabilities or likelihoods.
Here I want to go into two closely related problems of Bayesian epistemology, namely those of the ontological nature of these probabilities and the values one objectively assigns to them.
Let us consider that I throw a coin in the air. My degree of belief (1/2) it will land on heads is a subjective brain state which may (or should) be related to a frequency of action if betting money is involved.
But let us now consider the young Isaac Newton who was considering his newly developed theory of universal gravitation. What value should his degree of belief have taken on BEFORE he had begun to consider the first data of the real world?
Newton’s universal law of gravitation, when suitably supplemented with plausible background assumptions, can be said to confer probabilities on observations. But what does it mean to say that the law has a probability in the light of those observations? More puzzling still is the idea that it has a probability before any observations are taken into account. If God chose the laws of nature by drawing slips of paper from an urn, it would make sense to say that Newton’s law has an objective prior. But no one believes this process model, and nothing similar seems remotely plausible.”
Frequentism provides us with well-defined probabilities in many situations. The likelihood of getting a coin coming down as heads is identical with the frequency of this event if I were to repeat it an infinite number of times and the central limit theorem guarantees that one gets an increasingly better approximation of this quantity with a growing number of trials.
But what does the likelihood of the theory of universal gravitation being 2%, 5% or 15% mean?
And once one has come up with a definition one thinks to be valid, what is the objective value for the probability prior to any observation being taken into account?
I could not find any answer in the Bayesian papers I have read until now, these questions are apparently best ignored. But to my mind they are very important if you pretend to be building up a theory of knowledge based on probabilities.
Next episode: a mathematical proof of Bayesianism?
Thematic list of ALL posts on this blog (regularly updated)
My other blog on Unidentified Aerial Phenomena (UAP)
Why probabilities matter
In real life, it’s pretty rare (some would even say utterly impossible) to be sure of anything at all, like knowing it’s going to rain in one hour, that a conservative president is going to be elected, that you will be happily married in two years and so on and so forth.
We all recognize that it is only meaningful to speak of the probability or likelihood of each of these events.
The question of how to interpret their profound nature (ontoloy) is however, far from being an easy one.
I will use the basic proposition: if I roll the dice, there is a probability of 1/6 I will get a 3 in order to illustrate the two main interpretation of the probability concept out there.
1. Frequentism
According to this interpretation, the probability of an event equals its frequency if it is repeated an infinite number of times. If you roll a dice a great number of time, the frequency of the event (that is the number of 3s divided by the total number of rollings) will converge towards 1/6.
Mathematically it is a well defined concept and in many cases it can be relatively easily approximated. One of the main difficulties is that it apparently fails to account for the likelihood of unique situations, such as that (as far as we know in 2013) the Republicans are going to win the next American elections.
This brings us to the next popular interpretation of probability.
2. Bayesianism
For Bayesians, probabilities are degrees of belief and each degree of belief is a probability.
My degree of belief that the dice will fall onto 3 is 1/6.
But what is then a „degree of belief“? It is a psychological mind state which is correlated with a certain readiness for action.
According to many proponents of Bayenianism, degrees of belief are objective in so far that every rational creature disposing of a set of information would have exactly the same.
While such a claim is largely defensible for many situations such as the rolling of dices, the spread of a disaease or the results of the next elections, there are cases where it does not seem to make any sense at all.
Take for exampling the young Isaac Newton who was considering his newly developed theory of universal gravitation. What value should his degree of belief have taken on BEFORE he had begun to consider the first data of the real world?
And what would it mean ontologically to say that we have a degree of belief of 60% that the theory is true? What is the relation (in that particular situation) between the intensity of certain brain processes and the objective reality?
Such considerations have led other Bayesians to give up objectivity and define „degrees of belief“ as subjective states of mind, which might however be objectively constrained in many situations.
Another criticism of (strong) Bayesianism is that it ties the concept of probability to the belief of intelligent creatures. Yet it is clear that even in an universe lacking conscious beings, the probability of the decay of an atom and of more fundamental quantum processes would still exist and be meaningful.
For completeness, I should mention the propensity interpretation of Karl Popper who viewed the likelihood of an event as an intrinsic tendency of a physical system to tend towards a certain state of affairs.
So this was my completely unbiased (pun intended!) views on probabilities.
When debating (and fighting!) each other, theists and atheists tend to take their own epistemology (theory of knowledge) as granted.
This often leads to fruitless and idle discussions.
This is why I want to take the time to examine how we can know, what it means to know, before discussing what we can (and cannot) know.
Thematic list of ALL posts on this blog (regularly updated)
My other blog on Unidentified Aerial Phenomena (UAP)
Next episod: Naked Bayesianism. | 4,313 | 20,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-21 | latest | en | 0.959549 |
http://www.chegg.com/homework-help/questions-and-answers/ann-mass-50-kg-standing-left-end-a15-m-long-500kg-cart-frictionless-wheels-rolls-afriction-q210544 | 1,469,489,480,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824395.52/warc/CC-MAIN-20160723071024-00324-ip-10-185-27-174.ec2.internal.warc.gz | 356,661,478 | 13,204 | Ann (mass 50 kg) is standing at the left end of a15-m-long, 500kg cart that has frictionless wheels and rolls on africtionless track. Initially both Ann and the cart are at rest.Suddenly, Ann starts running along the cart at a speed of 5.0 m/srelative to the cart. How far will Ann have run relative to theground when she reaches the right end of the cart? | 90 | 356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2016-30 | latest | en | 0.933011 |
https://math.stackexchange.com/questions/2195440/finding-the-flow-of-vector-field-with-lie-group-orthogonal-group | 1,566,534,361,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317847.79/warc/CC-MAIN-20190823041746-20190823063746-00146.warc.gz | 537,928,843 | 29,474 | # Finding the flow of vector field with lie group (orthogonal group)
If $X_S (A) = AS$ defines a smooth vector field, where $A \in O(n)$ (matrices with $A^{-1} = A^{T}$) and $S$ belonging to the space of skew matrices (matrices with $S^{T} = -S$). How would I calculate the flow of $X_S$ here?
I'm using the definition of flow as one whose flow domain is all of $\mathbb{R} \times O(n)$, where the flow is the map $\theta : O(n) \times \mathbb{R} \to O(n)$ such that for all $s,t \in \mathbb{R}$ and $A \in O(n)$ the following holds: $$\theta(A,0) = A \space\ \space\ \text{and} \space\ \space\ \theta(\theta(A,t),s) = \theta(A, s+t)$$ I'm only familiar with finding flow via solving an ODE (or system of them). Need help understanding how this works with Lie Groups. | 244 | 769 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-35 | latest | en | 0.832907 |
https://www.extendoffice.com/documents/excel/1054-excel-identify-missing-numbers-sequence.html | 1,721,790,163,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00808.warc.gz | 674,753,808 | 30,839 | ## How to identify missing numbers sequence in Excel?
Let's say you have a long list of sequence numbers to mark items, such as check numbers in bank statements, normally we scrolling through and locate the missing sequence numbers manually. Sometimes this is quite arduous and time-consuming. You may think of tricky ways to deal with it. Yes, there are several easy ways to identify and locate missing numbers sequence in Excel 2007, Excel 2010, and Excel 2013 quickly and conveniently.
Identify missing numbers sequence with IF formula
Identify missing numbers sequence with an array formula
#### Identify missing numbers sequence with IF formula
As we all known, most of sequence numbers are with fixed increment of 1, such as 1, 2, 3, …, N. Therefore, if you can identify the number is not less 1 than its following number, there is a missing number.
We will show you the tutorials with an example as following screenshot shows:
1. In a blank cell, enter the formula of =IF(A3-A2=1,"","Missing"), and press the Enter key. In this case, we enter the formula in Cell B2.
If there is no missing numbers, this formula will return nothing; if missing numbers exist, it will return the text of "Missing" in active cell.
2. Select the cell B2 and drag the fill handle over the range of cells that you want to contain this formula. Now it identifies the missing numbers with the text of "Missing" in corresponding cells of Column B. See the following screenshot:
#### Identify missing numbers sequence with an array formula
Sometimes it requires not only identifying missing numbers sequence, but also listing missing numbers too. You can deal it with following steps:
A1:A30 = range of numbers, the sequence to check against is from 1 to 30
2. Press the Ctrl + Shift + Enter Keys together to finish the formula. Copy down the formula until you get #NUM! errors meaning all missing numbers have been listed. See screenshot:
#### Identify missing numbers sequence with Kutools for Excel quickly
The above methods only can identify the missing pure number sequence, if you have the sequence such as AA-1001-BB, AA-1002-BB, they may not work successfully. But, don't worry, Kutools for Excel’s powerful feature – Find Missing Sequence Number can help you quickly identify the missing sequence.
Note:To apply this Find Missing Sequence Number, firstly, you should download the Kutools for Excel, and then apply the feature quickly and easily.
After installing Kutools for Excel, please do as this:
1. Select the data sequence that you want to find the missing sequence.
2. Click Kutools > Insert > Find Missing Sequence Number, see screenshot:
3. In the Find Missing Sequence Number dialog box:
(1.) If you choose Inserting new column with following missing marker option, all the missing sequence numbers have been marked with the text Missing in a new column next to your data. See screenshot:
(2.) If you choose Inserting missing sequence number option, all the missing numbers have been inserted into the sequence list. See screenshot:
(3.) If you choose Inserting blank rows when encounting missing sequence numbers option, all blank rows are inserted when there are missing numbers . See screenshot:
(4.) If you choose Fill background color option, the location of the missing numbers will be highlighted at once. See screenshot:
#### Identify missing numbers sequence with Kutools for Excel quickly
Kutools for Excel: with more than 300 handy Excel add-ins, free to try with no limitation in 30 days.
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I am not able to find missing numbers from excel..
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I want to find a missing receipts using a range in columns, i.e. B2:C2 (receipt book # 679201/B2:679400/C2) meaning 200 receipt pages in one receipt book. Then accounted receipts of 679305/D2:679386/E2. Then what is the best formula for this?
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Thank you. This is awesome
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Doesn't work
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i wanted to identify the missing range in sequence rather than a missing number. example if there is range missing from 1,2,3,8,9,10,11,18. the missing range is 4-7 & 12-17.
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Q: solve the equation 5^x-4=25^x-6
A: 5^(x-4)=25^(x-6) 5^(x-4)=5^[2(x-6)] x-4=2(x-6) x-4=2x-12 x=-4+12=8
Original conversation
User: solve the equation 5^x-4=25^x-6
Question
Updated 5/12/2014 11:30:46 PM
Flagged by yeswey [5/12/2014 11:30:46 PM]
Rating
3
5^(x-4)=25^(x-6)
5^(x-4)=5^[2(x-6)]
x-4=2(x-6)
x-4=2x-12
x=-4+12=8
Confirmed by andrewpallarca [5/12/2014 11:33:02 PM]
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 551 | 1,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-26 | latest | en | 0.751508 |
https://forum.gamsworld.org/viewtopic.php?f=9&t=10257 | 1,537,399,281,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156311.20/warc/CC-MAIN-20180919220117-20180920000117-00233.warc.gz | 511,241,709 | 8,076 | ## hub network modelling
Problems with modeling
cio86
User
Posts: 4
Joined: 7 months ago
### hub network modelling
Hi
I want to set a model for an incomplete hubcovering problem. However, this program give me a complete graph. How can i fix this problem. Model and gams codes attached. an mst algorithm is working in the model. But every time it connects all hub points together.
options limrow=1000,limcol=1000,iterlim=100000000,reslim=36000,optcr=0 ;
set i cities /1*25/;
alias (i,k);
alias (i,j);
scalar Beta /2712.8/;
scalar Alfa /0.8/;
Table d(i,j) distances
.......
Table t(i,j) times
.......
Table FL(i,j) linkcost
......
parameter FH(i) /...../;
binary variables x(i,j),
z (i,j),
y (i,j,k);
positive variables r(i);
free variables amac;
Equations
objective
constraint1
constraint2
constraint3
constraint4
constraint5
constraint6
constraint7
constraint8
constraint9
constraint10
constraint11
constraint12
constraint13
constraint16
constraint17
constraint18;
objective..amac=e=sum((k),FH(k)*x(k,k))+sum((i,j)\$(ord(i) lt ord(j)),FL(i,j)*z(i,j));
constraint1(i)..sum((j),x(i,j))=e=1;
constraint2(i,j)..x(i,j)=l=x(j,j);
constraint3(i,j)\$(ord(i) lt ord(j))..z(i,j)=l=x(i,i);
constraint4(i,j)\$(ord(i) lt ord(j))..z(i,j)=l=x(j,j);
constraint5(j,k)\$(ord(j) ne ord(k))..sum(i\$(ord(i) ne ord(j)),y(i,j,k))=g=x(k,k)+x(j,j)-1;
constraint6(j,k)\$(ord(j) ne ord(k))..sum(i\$(ord(i) ne ord(j)),y(i,j,k))=l=x(k,k);
constraint7(i,j,k)\$(ord(j) ne ord(k) and ord(i) ne ord(j))..y(i,j,k)+y(j,i,k)=l=z(i,j);
constraint8(i,j,k)\$(ord(j) ne ord(k) and ord(i) ne ord(j))..d(k,j)=g=d(k,i)+t(i,j)*y(i,j,k)-Beta/Alfa*(1-y(i,j,k));
constraint9(i,j)\$(ord(i) ne ord(j))..d(i,j)=e=d(j,i);
constraint10(k)..d(k,k)=e=0;
constraint11(i,j)..r(j)=g=t(i,j)*x(i,j);
constraint12(j,k)..r(k)+Alfa*d(k,j)+r(j)=l=Beta;
constraint13(i,j)\$(ord(i) ne ord(j))..d(i,j)=g=t(i,j);
constraint16..sum((i,j)\$(ord(i) lt ord(j)),z(i,j))=g=sum(k,x(k,k))-1;
constraint17(i,j)\$(ord(i) lt ord(j))..d(i,j)=l=t(i,j)*z(i,j)+Beta/Alfa*(1-z(i,j));
constraint18(i,j)..d(i,j)=g=0;
model hubcover /all/ ;
option threads=1;
option mip=cplex
solve hubcover using MIP minimizing amac;
hubcover.optfile=1;
display amac.l,y.l, x.l, z.l, r.l;
Attachments
cladelpino
User
Posts: 108
Joined: 1 year ago
### Re: hub network modelling
Fix the variables at the values of a solution that should be feasible and with a better value of the objective function, then try to run the problem.
The infeasible constraints are blocking this solution. | 842 | 2,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | longest | en | 0.513129 |
http://ldkoffice.com/standard-error/sample-standard-error-of-the-estimate.html | 1,498,306,176,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320261.6/warc/CC-MAIN-20170624115542-20170624135542-00682.warc.gz | 221,148,145 | 5,504 | Home > Standard Error > Sample Standard Error Of The Estimate
# Sample Standard Error Of The Estimate
## Contents
Confidence intervals for the mean and for the forecast are equal to the point estimate plus-or-minus the appropriate standard error multiplied by the appropriate 2-tailed critical value of the t distribution. The reason N-2 is used rather than N-1 is that two parameters (the slope and the intercept) were estimated in order to estimate the sum of squares. As a result, we need to use a distribution that takes into account that spread of possible σ's. The forecasting equation of the mean model is: ...where b0 is the sample mean: The sample mean has the (non-obvious) property that it is the value around which the mean squared his comment is here
Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view Standard Error of Sample Means The logic and computational details of this procedure are described in Chapter 9 of For a value that is sampled with an unbiased normally distributed error, the above depicts the proportion of samples that would fall between 0, 1, 2, and 3 standard deviations above WeergavewachtrijWachtrijWeergavewachtrijWachtrij Alles verwijderenOntkoppelen Laden... Return to top of page. https://en.wikipedia.org/wiki/Standard_error
## Standard Error Of Mean Calculator
Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view Standard Error of the Estimate Author(s) David M. T-distributions are slightly different from Gaussian, and vary depending on the size of the sample. Hyattsville, MD: U.S. The following expressions can be used to calculate the upper and lower 95% confidence limits, where x ¯ {\displaystyle {\bar {x}}} is equal to the sample mean, S E {\displaystyle SE}
So, attention usually focuses mainly on the slope coefficient in the model, which measures the change in Y to be expected per unit of change in X as both variables move A practical result: Decreasing the uncertainty in a mean value estimate by a factor of two requires acquiring four times as many observations in the sample. Two data sets will be helpful to illustrate the concept of a sampling distribution and its use to calculate the standard error. Standard Error Vs Standard Deviation Follow @ExplorableMind . . .
The standard deviation of the age was 9.27 years. S becomes smaller when the data points are closer to the line. To estimate the standard error of a student t-distribution it is sufficient to use the sample standard deviation "s" instead of σ, and we could use this value to calculate confidence https://en.wikipedia.org/wiki/Standard_error Quant Concepts 4.501 weergaven 4:07 Explanation of Regression Analysis Results - Duur: 6:14.
The graphs below show the sampling distribution of the mean for samples of size 4, 9, and 25. Standard Error Of The Mean Estimate the sample standard deviation for the given data.
3. price, part 3: transformations of variables · Beer sales vs. I could not use this graph.
## Standard Error Of Regression
Some regression software will not even display a negative value for adjusted R-squared and will just report it to be zero in that case. this content It will be shown that the standard deviation of all possible sample means of size n=16 is equal to the population standard deviation, σ, divided by the square root of the Notice that the population standard deviation of 4.72 years for age at first marriage is about half the standard deviation of 9.27 years for the runners. In each of these scenarios, a sample of observations is drawn from a large population. Standard Error Of The Estimate Calculator
S is known both as the standard error of the regression and as the standard error of the estimate. Each of the two model parameters, the slope and intercept, has its own standard error, which is the estimated standard deviation of the error in estimating it. (In general, the term In other words, it is the standard deviation of the sampling distribution of the sample statistic. weblink ISBN 0-8493-2479-3 p. 626 ^ a b Dietz, David; Barr, Christopher; Çetinkaya-Rundel, Mine (2012), OpenIntro Statistics (Second ed.), openintro.org ^ T.P.
The variability of a statistic is measured by its standard deviation. Standard Error Of Measurement For the purpose of this example, the 9,732 runners who completed the 2012 run are the entire population of interest. As will be shown, the mean of all possible sample means is equal to the population mean.
## The concept of a sampling distribution is key to understanding the standard error.
It will be shown that the standard deviation of all possible sample means of size n=16 is equal to the population standard deviation, σ, divided by the square root of the n is the size (number of observations) of the sample. Over Pers Auteursrecht Videomakers Adverteren Ontwikkelaars +YouTube Voorwaarden Privacy Beleid & veiligheid Feedback verzenden Probeer iets nieuws! Margin Of Error Formula Bionic Turtle 95.377 weergaven 8:57 10 video's Alles afspelen Linear Regression.statisticsfun Calculating and Interpreting the Standard Error of the Estimate (SEE) in Excel - Duur: 13:04.
Assumptions and usage Further information: Confidence interval If its sampling distribution is normally distributed, the sample mean, its standard error, and the quantiles of the normal distribution can be used to How to cite this article: Siddharth Kalla (Sep 21, 2009). Advertentie Autoplay Wanneer autoplay is ingeschakeld, wordt een aanbevolen video automatisch als volgende afgespeeld. check over here The standard error of the forecast is not quite as sensitive to X in relative terms as is the standard error of the mean, because of the presence of the noise
MrNystrom 75.347 weergaven 10:07 Standard Error - Duur: 7:05. The population standard deviation is STDEV.P.) Note that the standard error of the model is not the square root of the average value of the squared errors within the historical sample Figure 1. The smaller standard deviation for age at first marriage will result in a smaller standard error of the mean.
What is the Standard Error of the Regression (S)? As the sample size increases, the sampling distribution become more narrow, and the standard error decreases. Correction for finite population The formula given above for the standard error assumes that the sample size is much smaller than the population size, so that the population can be considered The critical value that should be used depends on the number of degrees of freedom for error (the number data points minus number of parameters estimated, which is n-1 for this
Is there a different goodness-of-fit statistic that can be more helpful? The standard error of a coefficient estimate is the estimated standard deviation of the error in measuring it. Scenario 2. Toevoegen aan Wil je hier later nog een keer naar kijken?
So, if you know the standard deviation of Y, and you know the correlation between Y and X, you can figure out what the standard deviation of the errors would be Volgende Regression I: What is regression? | SSE, SSR, SST | R-squared | Errors (ε vs. Relative standard error See also: Relative standard deviation The relative standard error of a sample mean is the standard error divided by the mean and expressed as a percentage. In each of these scenarios, a sample of observations is drawn from a large population.
You can change this preference below. For any random sample from a population, the sample mean will usually be less than or greater than the population mean. The standard error is the standard deviation of the Student t-distribution. Because the age of the runners have a larger standard deviation (9.27 years) than does the age at first marriage (4.72 years), the standard error of the mean is larger for
The factor of (n-1)/(n-2) in this equation is the same adjustment for degrees of freedom that is made in calculating the standard error of the regression. The standard error of the model will change to some extent if a larger sample is taken, due to sampling variation, but it could equally well go up or down. | 1,744 | 8,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-26 | longest | en | 0.858558 |
http://www.freemathhelp.com/forum/threads/73978-Rates-of-convergence-of-sequences | 1,436,070,126,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375097204.8/warc/CC-MAIN-20150627031817-00248-ip-10-179-60-89.ec2.internal.warc.gz | 505,336,152 | 9,770 | Thread: Rates of convergence of sequences
1. Rates of convergence of sequences
Apologies if this is in the wrong forum.
I'm trying to find the rate of convergence of the following sequence:
$\displaystyle \lim_{n \to \infty} sin{\frac{1}{n}} = 0$.
To begin, I used knowledge of the Maclaurin series for sin (hopefully correctly):
$\bigg|\sin{\frac{1}{n}} - 0 \bigg| = \bigg| \frac{1}{n} - \frac{\cos{\xi(\frac{1}{n})}}{6}\big(\frac{1}{n}\b ig)^3\bigg|$.
However, there's already a big discrepancy between this attempt and the actual answer given, which starts:
$\bigg|\sin{\frac{1}{n}} - 0 \bigg| = \bigg| \frac{1}{n} - \frac{\sin{\xi(n)}}{6}\big(\frac{1}{n}\big)^3\bigg |$.
I'm confused for two reasons: from what I understand, the error term for the series should have a cos in it, not a sin. Furthermore, regardless of whether I'm using cos or sin, shouldn't $\frac{1}{n}$ be inside the parenthesis, not just $n$?
2. To find the rate of convergence, you want to check this:
$\lim_{n \to \infty}{\left|\frac{\sin{(\frac{1}{n+1})}}{\sin{(\ frac{1}{n})}}\right|}$
You may apply L'Hopital's Rule. First, use a change of variables (instead of the integer sequence, consider the functions, and let $x$ replace $n$).
$\lim_{\stackrel{x \to n}{n \to \infty}}{\left|\frac{\sin{(\frac{1}{x+1})}}{\sin{( \frac{1}{x})}}\right|} \stackrel{\text{H}}{=} \lim_{\stackrel{x \to n}{n \to \infty}}{\left|\frac{-(x+1)^{-2}\cos{(\frac{1}{x+1})}}{-x^{-2}\cos{(\frac{1}{x})}}\right|}$ $= \lim_{\stackrel{x \to n}{n \to \infty}}{\left|\frac{x^2\cos{(\frac{1}{x+1})}}{(x^ 2+2x+1)\cos{(\frac{1}{x})}}\right|} = \lim_{n \to \infty}{\left|\frac{\cos{(\frac{1}{n+1})}}{(1+\fra c{2}{n}+\frac{1}{n^2})\cos{(\frac{1}{n})}}\right|} = 1$
So, it converges sublinearly.
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• | 685 | 1,892 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2015-27 | longest | en | 0.668648 |
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Students can explore in groups to identify parts of an expressions. Utilizing color coded cards, students can discuss with their classmates what the different colors represent (i.e. terms, variables, coefficients and constants). A word bank is supplied as a possible modification for students who ne
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This PowerPoint can be used to introduce new vocabulary and problem solving strategies, or it can be used as a lesson review. This presentation uses the "I do, we do, you do" model. Addresses common core standard CCSS.MATH.CONTENT.6.EE.A.2.B. Defines vocabulary (term, factor and coefficient) and m
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This table shows eight expressions where students identify different parts. For example, it may ask which part of the expression shows a product. Them students can identify how many terms each expression has. This is a WORD DOCUMENT, feel free to edit.
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This foldable provides organized notes on the following vocabulary/ parts of an expression: • Algebraic Expression • Variable • Constant • Term • Coefficient • Like Terms Inside the foldable, students will write definitions as well as identify parts of two different algebraic expression, by
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CCSS.MATH.CONTENT.6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents. CCSS.MATH.CONTENT.6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers. CCSS.MATH.CONTENT.6.EE.A.2.A Write expressions that record operations with numbers and with lette
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CCSS.MATH.CONTENT.6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents. CCSS.MATH.CONTENT.6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers. CCSS.MATH.CONTENT.6.EE.A.2.A Write expressions that record operations with numbers and with lette
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Beginning in kindergarten and continuing through 12th grade, Next Generation Science Standards require students to "produce a sense of the process of argument necessary for advancing and defending a new idea or an explanation of a phenomenon and the norms for conducting such arguments. In that spiri
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Polar Express Interactive Brochure and Story Recall Sheets Inside this 6 part interactive brochure your students get to engage with the Polar Express story by responding to the following: Title Page Who was your favorite character? List 3 reasons why Circle the adjectives to describe the hot chocol
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This super easy NO PREP, PRINT AND GO winter themed language packet addresses 11 different receptive and expressive language skills. These language worksheets are perfect for in-therapy practice or homework! This packet can be used as a standalone product or a supplement that aligns perfectly with N
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Option 1: This option is organized by the following tabs: • "Parts of an Expression"- Students will identify & label algebraic expressions using terms, like terms, coefficients, and constants. • "Simplifying an Algebraic Expression"- Students will simplify algebraic expressions by combining like
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This super easy NO PREP, PRINT AND GO spring themed language packet addresses 13 different receptive and expressive language skills. These language worksheets are perfect for in-therapy practice or homework! This packet can be used as a standalone product or a supplement that aligns perfectly with N
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{A Teacher's guide to Fluency Instruction:Part I: Teaching Students to Read with Expression and Feeling} Lesson Plans designed to work with a your favorite reader's theater or poetry, Anchor Chart Materials, Data Collection Sheet and Classroom Activity/Matching Game Mix n' Match: A Game where st
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This super easy NO PREP, PRINT AND GO fall themed language packet addresses 13 different receptive and expressive language skills. These language worksheets are perfect for in-therapy practice or homework! This packet can be used as a standalone product or a supplement that aligns perfectly with No
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The purpose of these materials is to build verbal language as well as written language. These materials will assist students in organizing their thoughts to provide detailed information when speaking or writing. Students will learn a variety of attributes and characteristics that can be used to im
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This is the partner to my Lion and Lamb Writing pack found here: Lion & Lamb Writing Included: -blackline masters of lion and lamb -lined word strips -photo samples -2 types of lions Use this to help students brainstorm and "teach" others what they know about lions (i.e. 2 eyes, lives at the z
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Common Core Secondary Math brings you “Fluent in Math: Fluency Checks for the Common Core” . The Common Core State Standards in Math (CCSSM) stress fluency with skills and procedures along with conceptual understanding and mathematical applications. Fluency with mathematics at the secondary level r
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{A Teacher's Guide to Fluency Instruction Part I & II} Can also be purchased separately! Teaching a student to read aloud fluently can be a challenge. They must read smoothly, while decoding each word, comprehend what they are reading, and look ahead to know what emotion to put into their voic
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This activity packet includes word cards to rearrange into sentences and questions but also includes a few extra cards to extend the sentence or question with descriptors. Each set of word cards is color coded to make a specific sentence and/or question. Punctuation marks of period and question ma
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Lesson 6: Simplifying Expressions A Cheat Sheet is an all-in-one, single-page graphic organizer covering the big topics in the Algebra 1 curriculum. Each Cheat Sheet includes important vocabulary, key concepts, and guided examples. There is a set of Practice Problems included with each sheet that p
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Students define the parts of an expression along with labeling a given expression. Students go through four problems to identify the number of terms, constant, variables, and coefficients.
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Three interactive boards are included in this package that help students with autism and disabilities receptively and expressively identify the location of subatomic particles. Icons with color-coded prompts are included for two of the board. Potential goals and objectives that are aligned with stat
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Incorporate Valentine's Day vocabulary into your verbal and written expression lessons with these materials. The purpose of these materials is to build verbal language as well as written language. These materials will assist students in organizing their thoughts to provide detailed information when
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,074 | 8,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-51 | latest | en | 0.848474 |
http://www.sciencefairprojects.co.in/Kids-Science/Magnetic-Circles.php | 1,547,956,752,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583690495.59/warc/CC-MAIN-20190120021730-20190120043730-00147.warc.gz | 392,239,627 | 5,323 | Home School Projects Science Experiments Kids Projects Physics Mathematics Chemistry Computer Botany Earth Science Electricity Contact Us
Magnetic Circles
Abstract
To conduct a simple experiment to demonstrate how a current carrying conductor generates a magnetic field.
Materials
A one foot long coat hanger wire (With any insulation scraped off).
A wooden stand with a flat surface at around halfway height.
5 or 6 small compasses.
A 6 volt battery.
Two electrical lead wires with alligator clips
Planning the Procedure
Make a hole in the centre of the flat surface of the stand and pass the coat hanger wire through it so that it is suspended vertically.
Arrange the 6 compasses in a circle around the wire on the flat surface.
Attach the lead wires to the battery terminals. Leave the other ends of the wires free for now.
Observe the compasses now. You will see that all compasses point in the same direction, towards magnetic north.
Now connect the battery terminals to the coat hanger wire using the lead wires. Observe the orientation of the compass needles now. Each compass will point in a direction tangential to a circle centred on the coat hanger wire.
Rotate the support stand and observe again. The compasses will continue to point in tangential directions.
Connect the battery to the coat hanger wire with opposite polarity by switching the lead wire clips and observe the compasses again. You will see that the compass needles now reverse the direction they are pointing in, but continue to point in a direction tangential to a circle centred on the wire.
Scientific explanation:
Usually a compass lines up with the earth’s magnetic field. The north pole of the compass needle, which is itself a magnet points towards the south pole of the earth’s magnetic field as opposite poles attract each other. Now, when a current passes through the wire, it generates a magnetic field which is much stronger than the earth’s magnetic field. Hence the compass needle is now affected by both the earth’s magnetic field and the current carrying wire’s magnetic field. Since the latter is much stronger, the compass needle gets aligned with the wire’s magnetic field. This magnetic field can be visualized as a set of concentric circles with the wire as the centre. When the direction of current in the wire is reversed, the compass needle also points in the opposite direction as the magnetic field is reversed. | 480 | 2,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | longest | en | 0.896001 |
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### There are 52 results
Broad Topics > Mathematics Education and Research > Pedagogy
##### Age 11 to 18
What are rich tasks and why do they matter?
### More Old Chestnuts
##### Age 11 to 16
Mainly for teachers. More school mathematics of yesteryear.
### Cultivating Creativity
##### Age 5 to 18
Creativity in the mathematics classroom is not just about what pupils do but also what we do as teachers. If we are thinking creatively about the mathematical experiences we offer our pupils we can. . . .
### Working Effectively with All Learners
##### Age 5 to 18
Some questions and prompts to encourage discussion about what experiences you want to give your pupils to help them reach their full potential in mathematics.
### Kingsfield School - Building on Rich Starting Points
##### Age 5 to 18
Alf and Tracy explain how the Kingsfield School maths department use common tasks to encourage all students to think mathematically about key areas in the curriculum.
Bloom's taxonomy
### The Development of Spatial and Geometric Thinking: 5 to 18
##### Age 5 to 16
This is the first article in a series which aim to provide some insight into the way spatial thinking develops in children, and draw on a range of reported research. The focus of this article is the. . . .
### Developing Good Team-working Skills
##### Age 5 to 18
Group work depends on effective team work. This article describes attributes of effective team work and links to "Team Building" problems that can be used to develop learners' team working skills.
### Blog It
##### Age 5 to 18 Challenge Level:
Members of the NRICH team are beginning to write blogs and this very short article is designed to put the reasoning behind this move in context.
### Breaking the Equation ' Empirical Argument = Proof '
##### Age 7 to 18
This article stems from research on the teaching of proof and offers guidance on how to move learners from focussing on experimental arguments to mathematical arguments and deductive reasoning.
### Roasting Old Chestnuts
##### Age 11 to 16
Mainly for teachers. A discussion and examples of some of the school mathematics of yesteryear.
### Logic, and How it Should Influence Our Teaching
##### Age 5 to 16
Providing opportunities for children to participate in group narrative in our classrooms is vital. Their contrasting views lead to a high level of revision and improvement, and through this process. . . .
### Performing Beyond Expectations - Using Sport to Motivate Students in Mathematics Lessons
##### Age 7 to 16
In this article, Alan Parr shares his experiences of the motivating effect sport can have on the learning of mathematics.
### Using Rich Tasks in an Objective Led Culture
##### Age 11 to 16
As teachers, we appreciate the need to have clear objectives at the start of lessons but have been aware of the limitations this sometimes seems to place on our ability to get the most out of using. . . .
### A Problem Is a Problem for All That
##### Age 7 to 16
In this article, Jennifer Piggott talks about just a few of the problems with problems that make them such a rich source of mathematics and approaches to learning mathematics.
### Engaging Students, Developing Confidence, Promoting Independence
##### Age 5 to 18
Ideas to support mathematics teachers who are committed to nurturing confident, resourceful and enthusiastic learners.
### What's All the Talking About?
##### Age 5 to 16
This fascinating article delves into the world of talk in the classroom and explains how an understanding of talking can really improve the learning of mathematics.
### Optimising Input - Maximising Output
##### Age 11 to 18
A group of teachers involved in embedding NRICH tasks into their everyday practice decided they needed to address the (im)balance between teacher and student activity in their classrooms. In. . . .
### Changing Perceptions
##### Age 11 to 16
A group of teachers involved in embedding NRICH tasks into their everyday practice were keen to challenge common perceptions of mathematics and of teaching and learning mathematics. In this article,. . . .
### Placing Our Trust in Learners
##### Age 5 to 14
In this article Liz Woodham reflects on just how much we really listen to learners’ own questions to determine the mathematical path of lessons.
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Suggestions for worthwhile mathematical activity on the subject of angle measurement for all pupils. | 2,139 | 9,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-39 | latest | en | 0.912962 |
https://www.techylib.com/el/view/copygrouper/kinematics_and_force_analysis_of_a_five-link_mechanism_by_the_fou | 1,529,516,381,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863830.1/warc/CC-MAIN-20180620163310-20180620183310-00009.warc.gz | 935,420,538 | 18,409 | # Kinematics and Force Analysis of a Five-Link Mechanism by the Four Spaces Jacoby Matrix
Μηχανική
13 Νοε 2013 (πριν από 4 χρόνια και 7 μήνες)
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5 3
Kinematics and Force Analysis of a Five-Link Mechanism
by the Four Spaces Jacoby Matrix
Ivan Chavdarov
Central Laboratory of Mechatronics and Instrumentation, 1113 Sofia
Е-mail: ivan_chavdarov@dir.bg
Web site: http://inch.hit.bg
1. Introduction
The five-link planar manipulative system (MS), shown in Fig. 1, contains only rotational
joints. Some parts of them are passive, the remaining – active. All the bodies could
change their dimensions in some borders [4] during the design process and in that way
the features of the MS change. The body 1 (l
1
) is more particular as it stays immobile
(it represents the support). The bodies 2 and 5 are driving bodies. With the help of
appropriate rotation of the actuating bodies, the characteristic point B of the MS can
follow desired planar trajectory in the borders of the working zone.
Fig. 1. Structural scheme of the considered manipulative system
БЪЛГАРСКА АКАДЕМИЯ НА НАУКИТЕ . BULGARIAN ACADEMY OF SCIENCES
ПРОБЛЕМИ НА ТЕХНИЧЕСКАТА КИБЕРНЕТИКА И РОБОТИКАТА, 55
PROBLEMS OF ENGINEERING CYBERNETICS AND ROBOTICS, 55
София . 2005 . Sofia
5 4
The velocity V = [V
B
x
, V
B
y
]
T
of the characteristic point B is determined through the
angular velocities
.
=
.
2
,
.
5
]
T
of the bodies 2 and 5 and depends on the transfer
function of the mechanism. Usually the transfer function is described by the Jacoby
matrix
J
:
(1) V = J
.
.
This expression is known as forward kinematics problem and for the considered MS
could be solved using different approaches [3]. The analytical symbolic solution could
be particularly useful for making several conclusions concerning the singular
configurations of the MS as well as the MS metric. The classical approach [2] for
solving such kind of problems requires the solution of the standard position task (forward
kinematics) f(
i
) = X, i = 2, 5; X = V = [B
x
, B
y
]
T
or of the inverse kinematics. After that
the obtained results are differentiated with respect to the general coordinates
=
2
,
5
]
T
. In that concrete example such a solution is complex and ambiguous (in the
general case). The forward kinematics (standard position task) has two solutions, the
inverse – four. These arguments determine the necessity to search for other approaches
for the analytical solution of the forward kinematics (position task).
2. The Jacoby matrix for the closed loop manipulative system
Let’s assume that the MS is divided into two parts representing two open planar
kinematics chains with two links (Fig. 2).
Fig. 2. Representation of the MS from Fig. 1 as a system containing two open structures
The matrix of Jacoby J
1,2
for each of them is known [3]. For the left (J
1
) МS we obtain
( 2)
2221
1211
1
AA
AA
J
,
where:
5 5
)sin(sin
3232211
llA
,
)sin(
32312
lA
,
)cos(cos
3232221
llA
,
)cos(
32322
lA
.
Analogously we can obtain for the right system:
(3)
2221
1211
2
BB
BB
J
,
where:
)sin(sin
4545511
llB
,
)sin(
45412
lB
,
)cos(cos
4545521
llB
,
)cos(
45422
lB
.
If we admit that the distance between both systems is l
1
, and that they reach one
and the same point B, and also the velocity V of that point B reached by the first and
the second MS is the same, we obtain the system
(4)
422521
412511
322221
312211
BBV
BBV
AAV
AAV
y
x
y
x
B
B
B
B
or in a matrix form:
4,5
3,2
2
1
.
J
J
V
V
,
where
.
2,3
=
.
2
,
.
3
]
T
and
.
5,4
=
.
5
,
.
4
]
T
.
Eliminating the angular velocities
3
and
4
in the passive joints for the forward
kinematics problem we obtain
(5)
522221
512211
CCV
CCV
y
x
B
B
or in a matrix form
JV
C
,
2221
1211
CC
CC
J
,
where:
)(
)(
22122212
22112112
121111
ABBA
BAAB
AAC
,
)(
)(
22122212
12212211
1212
ABBA
BBBB
AC
,
5 6
)(
)(
22122212
22112112
222121
ABBA
BAAB
AAC
,
)(
)(
22122212
12212211
2222
ABBA
BBBB
AC
.
The coefficients
ji
C
,
could be determined if 0
22212212
ABBA.
3. Spaces of the Jacoby matrix
It is known [1, 2], that every matrix defines four spaces which dimensions are
determined by the matrix rank and order. Further we will consider the physical and
geometric interpretation of these spaces related with the Jacoby matrix for manipulative
systems.
3.1. Column space (image) of the Jacoby matrix
)(J
It transforms (1), the area of admissible values of the controlled velocity vectors of
the actuating bodies
.
=
.
2
,
.
5
]
T
, in corresponding velocities of the end-effector
V = [V
B
x
, V
B
y
]
T
. This space dimension is equal to r – the rank of the Jacoby matrix (or
the number of the matrix independent columns). In this concrete case, the two-
dimensional space of the angular velocities generates a two-dimensional space of the
velocity of the end-effector. There exist some robot states or configurations (particular
or singular) for which the coefficients C
ij
corresponding to J are such that the two-
dimensional space of the generalized velocities generates one-dimensional space of
the absolute velocity of the point B (r =1).
3.2. Row space )(
T
J
The row space of the matrix J coincides with the column space of J
T
. With the help of
this space we can determine the actuating moments = [M
2
, M
5
]
T
, which must be
created in the actuating joints 2 and 5 to equilibrate the external forces F = [F
x
, F
y
]
T
,
applied to the end-effector:
(6) = J
T
F.
The friction forces and other losses are not considered.
3.3. Zero space of J (Ker(J))
It is defined by the system Jx = 0 and its dimension is n r, where n is the number of
rows of the matrix J. It describes this subspace of vectors of angular velocities
i
x
,
which does not generate velocities V in the end-effector.
3.4. Zero space of J
Т
(Ker(J
T
)) left zeros of J
It is defined by the system J
T
y = 0 and its dimension is m r, where m is the number of
columns of J. It describes this set of vectors of external forces
i
Fy , for which there
is no need of motors equilibrating torques = [M
2
, M
5
]
T
,.
5 7
The zero space is also known as a kernel of the matrix. It is obvious that the zero
vectors x = [0, 0]
T
belong to J and J
Т
. It is known that the defect of the matrix
denotes the difference between the higher value of the rows or columns number of
the matrix J and its rank [1, 2]: = max(m, n) – r. In our case the maximal possible
defect of J is = 2 and it is obtained when the rank of the matrix is zero r = 0.
4. Singular configurations
It is very important to define the rank r of J for the analysis of MS [2]. It is equal to
the number of independent rows (columns) of the matrix and can be determined by
calculating the matrix determinant (if it exists) and its minors (sub-matrices
determinants). We are searching for configurations where det(J) = 0 (the rank of J
decreases). These configurations are known as singular and the MS changes its
features in such configurations. The four spaces of the Jacoby matrix change their
dimensions.
Statement 1. The determinant of the matrix J (for the MS in Fig. 1) is equal to
zero only if the determinant of J
1
(2) or the determinant of J
2
(3) is zero:
(7)
,0)det(
,0)det(
0)det(
2
1
J
J
J
J
1
and J
2
are the corresponding Jacoby matrices for the left and right open chains
(Fig. 2) of the five-link closed MS (Fig. 1).
Statement 2. The determinant of the matrix J (for the MS – Fig. 1) tends to
infinity when
(8)
5432
.
The demonstration of both statements 1 and 2 is accomplished as follows:
We obtain the determinant of J as:
(9)
12212211
)det( CCCCJ
.
After some transformations it can be written:
(10)
)(
))((
)det(
22122212
1221221112212211
BAAB
BBBBAAAA
J
,
or
)sin()cos()cos()sin(
)det()det(
)det(
323454323454
21
llll
JJ
J.
From the last we obtain
(11)
)sin(
)det()det(
)det(
324543
21
ll
JJ
J
.
It is obvious the determinant becomes equal to zero when some of the multipliers
in the nominator of (11) take zero values. When the denominator tends to zero then
5 8
the determinant of J tends to infinity. If l
3
and l
4
lengths are different from zero, the
last comes true only if:
(12)
k
0
3245
, k = 1, 2, …
Corollary 1. When the force transformation angle ABC(Fig.1) [4] between
the bodies 3 and 4 tends to zero (or 180
о
), then the determinant of J tends to infinity.
In that case we need extremely great actuating torques to equilibrate the external
forces acting on the end-effector. The demonstration of the corollary 1 could be easily
done, taking into account that the sum of the internal angles of the tetragon is equal to
360
о
(2 rad). It is obvious that when = 0, the mechanism forms a tetragon. For the
sum of its internal angles we obtain:
2
+
3
– + –
4
+ –
5
= 2, and therefore
2
+
3
–
4
–
5
= . Condition (12) is satisfied.
5. Numeric examples
5.1. Example 1
General case: A manipulative system is considered which bodies lengths are (Fig. 3)
l
1
=0.1, l
2
=0.2, l
3
=0.25, l
4
=0.35, l
5
=0.1 (Fig. 3);
2
=100.03
o
;
3
= –52.9
o
;
4
=79.08
o
;
5
=20.53
o
. Then we obtain:
0.170.135
0.183–0.38–
1
J
,
0.04)det(
1
J
;
0.058–0.035
0.345–0.38–
2
J
,
0.034)det(
2
J
;
0.0840.034–
0.091–0.198–
J
, –0.02)det(
J.
Fig. 3. МS in arbitrary configuration – example 1
5 9
Working configuration of МS. It is possible to realize some motion (and also
forces) in the plane in arbitrary direction. The coefficients of J are transfer values for
the concrete configuration of the mechanism.
5.2. Singular case with defect = 1 for the matrix J – example 2
The bodies lengths are the same as in the example 1 and the generalized coordinates
are (Fig. 4):
2
=57.38
o
;
3
=0
o
;
4
=63.27
o
;
5
=18.86
o
. Then we obtain:
0.1350.243
0.211–0.379–
1
J
,
0)det(
1
J
;
0.0480.143
0.347–0.379–
2
J
,
0.031)det(
2
J
,
0)det(
J.
In this configuration if
0
5
the realization of any velocities does not generate
the end-effector velocity, i.e. the vectors
T
2
]0,[
belongs to the zero space (Ker(J))
of J. Forces acting in the direction
T
115.0
18.0
1
kF
(where k is a real number)
cannot be equilibrated by the actuating torques. They belong to the zero space (Ker(J
T
))
of J
Т
and are absorbed by the links of the MS. The maximal force in that direction that
can be supported by the construction depends on the robustness of the elements. Such
kind of singularities could be observed in mechanisms with any metrics (arbitrary
proportion between body lengths, for which the mechanism is defined [4]). At least
one of the two open chain MS has configurations where its determinant J
1
(or J
2
)
becomes zero. The forward kinematics problem has two solutions and at least one of
them is singular. The points from the working zone, where = 1, are on its borders.
The inverse kinematics problem for them has two singular solutions.
Fig. 4. Singular case with defect = 1 for the matrix J – example 2
6 0
5.3. Singular case with defect = 2 for the matrix J – example 3
The links lengths are the same as in the example 1 and the generalized coordinates are
(Fig. 5):
2
=83.62
o
;
3
=0
o
;
4
=0
o
;
5
=96.38
o
. Then we obtain:
0.0280.05
0.248–0.447–
1
J
,
0)det(
1
J
;
0.0390.05
0.348–0.447–
2
J
,
0)det(
2
J
,
00
00
J
, 0)det(
J.
Fig. 5. Singular case with defect = 2 for the matrix J – example 3
The forward and inverse kinematics problems have a unique solution. If l
1
0 there
exist only one or two points, where = 2. In such a configuration the MS is extremely
stable with respect to the forces applied on the end-effector. Their equilibration is
realized only by the links and the supports and is not transferred to the actuating
devices (Ker(J
T
)
T
,
yx
FFF
). It becomes difficult to control the velocity of the
point B. The zero space Ker(J) coincides with all the plane
T
52
],[
.
5.4. Singular case, where det(J) tends to infinity – example 4
A manipulative system which links length are l
1
=0.2, l
2
=0.25, l
3
=0.2, l
4
=0.15, l
5
=0.2 is
considered (Fig. 6);
2
=114.71
o
;
3
= –119.96
o
;
4
=97.47
o
;
5
=77.28
o
.
Then it can be written:
0.1990.095
0.018–0.209–
1
J
, –0.043)det(
1
J;
0.149–0.105–
0.014–0.209–
2
J
, 0.03)det(
2
J,
6 1
а) Five-link mechanism in singular configuration b) Reaching the same point without falling
where det(J) in a singular configuration
Fig. 6. Singular case, where det(J) tends to infinity – example 4
J
,
)det(J.
In Fig. 6а the corresponding graphical solution in generalized and Cartesian
coordinates is presented. In this configuration a part of the forces acting on the end-
effector (point B) cannot be equilibrated with the help of the actuating torques. Such
types of singular configurations are placed inside the working zone of the MS. During
the control of the MS such configurations must be avoided due to decreasing of functional
capabilities. The end-effector can reach these points (denoted by o), passing by both
– singular or nonsingular – configurations (for instance Fig. 6b). The forward kinematics
problem for such a point has 4 solutions, but only for one of them (Fig. 6а))
det(J) .
5.5. Singular case – indefiniteness of type 0/0 – example 5
The following MS will be considered, which links lengths are (Fig. 7): l
1
=0.25, l
2
=0.3,
l
3
=0.2, l
4
=0.15, l
5
=0.1;
2
=109.47
o
;
3
=-148.41
o
;
4
=0
o
;
5
=141.06
o
. Then it could be
written:
0.1560.056
0.1260.157–
1
J
,
–0.031
)det(
1
J
;
0.117–0.194–
0.094–0.157–
2
J
,
0)det(
2
J
,
0
0
0
0
J
,
0
0
)det( J
.
det(J)
6 2
In similar situations the possibility for realization of motions and forces as well as
the control of the MS is extremely difficult.
6. Conclusion
The represented approach for the Jacoby matrix determination leads to the
demonstration of statements 1 and 2. Consequently the singular configurations of the
MS can be easily detected and some conclusions concerning the mechanism behavior
in these configurations can be formulated. The corresponding singular configurations
are realizable for different proportions of the mechanism link lengths [4, 5]. The case
when the determinant of J tends to infinity is particular and in that sense the
configurations for which this condition is satisfied could be considered as singular. The
interpretation of the Jacoby matrix spaces is physically useful as well in the process of
synthesis and design of the manipulative system as for the control process. The extreme
values of the transfer function (the elements of J) are used to determine the maximal
loads of the actuating mechanisms of the MS.
The main disadvantages of the proposed method are:
We cannot determine the reactions in the passive joints as well as the reactions
generated by forces and torques, which cannot be stabilized by the actuating devices;
it does not take into account the friction losses.
easy determination of the Jacoby matrix and its determinant;
simple physical interpretation of the singular configurations as a result of their
reduction as a combination of two already known and well studied MS;
the symbolic writing of the transfer function (5) and of the determinant of J
(10) allows to do analysis of the separate geometrical parameters influence on the MS
features.
Fig. 7. Singular case – indefiniteness of type 0/0 – example 5
6 3
R e f e r e n c e s
1. B e k l e m i s h e v, D. A Course in Analytic Geometry and Linear Algebra. Moscow, Mir, 1988 (in
Russian).
2. S t r e n g, G. Linear Algebra With Application. Moscow, Mir, 1980 (in Russian).
3. M o h s e n S h a h i n p o r. A Robot Engineering Textbook. New York, University of New Mexico,
Harper & Row, Publishers, 1990.
4. C h a v d a r o v, I., P. G e n o v a, R. Z a h a r i e v. Synthesis and Optimization of Five-link Lever
Kinematics Chain for Mechatronics Module. Mechanics of the Machines. Book 53. Technical
University, Varna, 2004, 131-136 (in Bulgarian).
5. L e e T i n g, W., F. F r e n d e s t e i n. Design of Geared 5-Bar Mechanisms for Unlimited Crank
Rotations and Optimum Transmission. Mech. And Machine Theory, 1978, 235-244.
6. B a j p a i, A., B. R o t h. Workspace and mobility of a closed-lop manipulator. – The Intern. J. of
Robotics Research, 5, 1986, No 2, 131-142.
Кинематичeский и силовой анализ пятизвенной манипуляционной
системы при помощи четырех пространств матрицы Якоби
Иван Чавдаров
Центральная лаборатория мехатроники и приборостроения, 1113 София
Е-mail: ivan_chavdarov@dir.bg
Web site: http://inch.hit.bg
(Р е з ю м е)
В работе представлена пятизвенная равнинная манипуляционная система
закроенного типа. Используется матрица имени Якоби и аналитическое решение
двух задач кинематики. Приведены примеры для связи и зависимость сингуляр-
ных конфигураций манипуляционных роботов. Сделани выводы, касающие
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# Posts Tagged ‘Heisenberg picture’
## PHY456H1F: Quantum Mechanics II. Lecture 6 (Taught by Prof J.E. Sipe). Interaction picture.
Posted by peeterjoot on September 27, 2011
[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]
# Disclaimer.
Peeter’s lecture notes from class. May not be entirely coherent.
# Interaction picture.
## Recap.
Recall our table comparing our two interaction pictures
\begin{aligned}\text{Schr\"{o}dinger picture} &\qquad \text{Heisenberg picture} \\ i \hbar \frac{d}{dt} {\lvert {\psi_s(t)} \rangle} = H {\lvert {\psi_s(t)} \rangle} &\qquad i \hbar \frac{d}{dt} O_H(t) = \left[{O_H},{H}\right] \\ {\langle {\psi_s(t)} \rvert} O_S {\lvert {\psi_s(t)} \rangle} &= {\langle {\psi_H} \rvert} O_H {\lvert {\psi_H} \rangle} \\ {\lvert {\psi_s(0)} \rangle} &= {\lvert {\psi_H} \rangle} \\ O_S &= O_H(0)\end{aligned}
## A motivating example.
While fundamental Hamiltonians are independent of time, in a number of common cases, we can form approximate Hamiltonians that are time dependent. One such example is that of Coulomb excitations of an atom, as covered in section 18.3 of the text [1], and shown in figure (\ref{fig:qmTwoL6fig1}).
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL6fig1}
\caption{Coulomb interaction of a nucleus and heavy atom.}
\end{figure}
We consider the interaction of a nucleus with a neutral atom, heavy enough that it can be considered classically. From the atoms point of view, the effects of the heavy nucleus barreling by can be described using a time dependent Hamiltonian. For the atom, that interaction Hamiltonian is
\begin{aligned}H' = \sum_i \frac{ Z e q_i }{{\left\lvert{\mathbf{r}_N(t) - \mathbf{R}_i}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(2.1)
Here and $\mathbf{r}_N$ is the position vector for the heavy nucleus, and $\mathbf{R}_i$ is the position to each charge within the atom, where $i$ ranges over all the internal charges, positive and negative, within the atom.
Placing the origin close to the atom, we can write this interaction Hamiltonian as
\begin{aligned}H'(t) = \not{{\sum_i \frac{Z e q_i}{{\left\lvert{\mathbf{r}_N(t)}\right\rvert}}}}+ \sum_i Z e q_i \mathbf{R}_i \cdot {\left.{{\left(\frac{\partial {}}{\partial {\mathbf{r}}} \frac{1}{{{\left\lvert{ \mathbf{r}_N(t) - \mathbf{r}}\right\rvert}}}\right)}}\right\vert}_{{\mathbf{r} = 0}}\end{aligned} \hspace{\stretch{1}}(2.2)
The first term vanishes because the total charge in our neutral atom is zero. This leaves us with
\begin{aligned}\begin{aligned}H'(t) &= -\sum_i q_i \mathbf{R}_i \cdot {\left.{{\left(-\frac{\partial {}}{\partial {\mathbf{r}}} \frac{ Z e}{{\left\lvert{ \mathbf{r}_N(t) - \mathbf{r}}\right\rvert}}\right)}}\right\vert}_{{\mathbf{r} = 0}} \\ &= - \sum_i q_i \mathbf{R}_i \cdot \mathbf{E}(t),\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.3)
where $\mathbf{E}(t)$ is the electric field at the origin due to the nucleus.
Introducing a dipole moment operator for the atom
\begin{aligned}\boldsymbol{\mu} = \sum_i q_i \mathbf{R}_i,\end{aligned} \hspace{\stretch{1}}(2.4)
the interaction takes the form
\begin{aligned}H'(t) = -\boldsymbol{\mu} \cdot \mathbf{E}(t).\end{aligned} \hspace{\stretch{1}}(2.5)
Here we have a quantum mechanical operator, and a classical field taken together. This sort of dipole interaction also occurs when we treat a atom placed into an electromagnetic field, treated classically as depicted in figure (\ref{fig:qmTwoL6fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL6fig2}
\caption{atom in a field}
\end{figure}
In the figure, we can use the dipole interaction, provided $\lambda \gg a$, where $a$ is the “width” of the atom.
Because it is great for examples, we will see this dipole interaction a lot.
## The interaction picture.
Having talked about both the Schr\”{o}dinger and Heisenberg pictures, we can now move on to describe a hybrid, one where our Hamiltonian has been split into static and time dependent parts
\begin{aligned}H(t) = H_0 + H'(t)\end{aligned} \hspace{\stretch{1}}(2.6)
We will formulate an approach for dealing with problems of this sort called the interaction picture.
This is also covered in section 3.3 of the text, albeit in a much harder to understand fashion (the text appears to try to not pull the result from a magic hat, but the steps to get to the end result are messy). It would probably have been nicer to see it this way instead.
In the Schr\”{o}dinger picture our dynamics have the form
\begin{aligned}i \hbar \frac{d}{dt} {\lvert {\psi_s(t)} \rangle} = H {\lvert {\psi_s(t)} \rangle}\end{aligned} \hspace{\stretch{1}}(2.7)
How about the Heisenberg picture? We look for a solution
\begin{aligned}{\lvert {\psi_s(t)} \rangle} = U(t, t_0) {\lvert {\psi_s(t_0)} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.8)
We want to find this operator that evolves the state from the state as some initial time $t_0$, to the arbitrary later state found at time $t$. Plugging in we have
\begin{aligned}i \hbar \frac{d{{}}}{dt} U(t, t_0) {\lvert {\psi_s(t_0)} \rangle}=H(t) U(t, t_0) {\lvert {\psi_s(t_0)} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.9)
This has to hold for all ${\lvert {\psi_s(t_0)} \rangle}$, and we can equivalently seek a solution of the operator equation
\begin{aligned}i \hbar \frac{d{{}}}{dt} U(t, t_0) = H(t) U(t, t_0),\end{aligned} \hspace{\stretch{1}}(2.10)
where
\begin{aligned}U(t_0, t_0) = I,\end{aligned} \hspace{\stretch{1}}(2.11)
the identity for the Hilbert space.
Suppose that $H(t)$ was independent of time. We could find that
\begin{aligned}U(t, t_0) = e^{-i H(t - t_0)/\hbar}.\end{aligned} \hspace{\stretch{1}}(2.12)
If $H(t)$ depends on time could you guess that
\begin{aligned}U(t, t_0) = e^{-\frac{i}{\hbar} \int_{t_0}^t H(\tau) d\tau}\end{aligned} \hspace{\stretch{1}}(2.13)
holds? No. This may be true when $H(t)$ is a number, but when it is an operator, the Hamiltonian does not necessarily commute with itself at different times
\begin{aligned}\left[{H(t')},{H(t'')}\right] \ne 0.\end{aligned} \hspace{\stretch{1}}(2.14)
So this is wrong in general. As an aside, for numbers, 2.13 can be verified easily. We have
\begin{aligned}i \hbar \left( e^{-\frac{i}{\hbar} \int_{t_0}^t H(\tau) d\tau} \right)'&=i \hbar \left( -\frac{i}{\hbar} \right) \left( \int_{t_0}^t H(\tau) d\tau \right)'e^{-\frac{i}{\hbar} \int_{t_0}^t H(\tau) d\tau } \\ &=\left( H(t) \frac{dt}{dt} - H(t_0) \frac{dt_0}{dt} \right)e^{-\frac{i}{\hbar} \int_{t_0}^t H(\tau) d\tau} \\ &= H(t) U(t, t_0)\end{aligned}
## Expectations
Suppose that we do find $U(t, t_0)$. Then our expectation takes the form
\begin{aligned}{\langle {\psi_s(t)} \rvert} O_s {\lvert {\psi_s(t)} \rangle} = {\langle {\psi_s(t_0)} \rvert} U^\dagger(t, t_0) O_s U(t, t_0) {\lvert {\psi_s(t_0)} \rangle} \end{aligned} \hspace{\stretch{1}}(2.15)
Put
\begin{aligned}{\lvert {\psi_H} \rangle} = {\lvert {\psi_s(t_0)} \rangle},\end{aligned} \hspace{\stretch{1}}(2.16)
and form
\begin{aligned}O_H = U^\dagger(t, t_0) O_s U(t, t_0) \end{aligned} \hspace{\stretch{1}}(2.17)
so that our expectation has the familiar representations
\begin{aligned}{\langle {\psi_s(t)} \rvert} O_s {\lvert {\psi_s(t)} \rangle} ={\langle {\psi_H} \rvert} O_H {\lvert {\psi_H} \rangle} \end{aligned} \hspace{\stretch{1}}(2.18)
## New strategy. Interaction picture.
Let’s define
\begin{aligned}U_I(t, t_0) = e^{\frac{i}{\hbar} H_0(t - t_0)} U(t, t_0)\end{aligned} \hspace{\stretch{1}}(2.19)
or
\begin{aligned}U(t, t_0) = e^{-\frac{i}{\hbar} H_0(t - t_0)} U_I(t, t_0).\end{aligned} \hspace{\stretch{1}}(2.20)
Let’s see how this works. We have
\begin{aligned}i \hbar \frac{d{{U_I}}}{dt} &= i \hbar \frac{d{{}}}{dt} \left(e^{\frac{i}{\hbar} H_0(t - t_0)} U(t, t_0)\right) \\ &=-H_0 U(t, t_0)+e^{\frac{i}{\hbar} H_0(t - t_0)} \left( i \hbar \frac{d{{}}}{dt} U(t, t_0) \right) \\ &=-H_0 U(t, t_0)+e^{\frac{i}{\hbar} H_0(t - t_0)} \left( (H + H'(t)) U(t, t_0) \right) \\ &=e^{\frac{i}{\hbar} H_0(t - t_0)} H'(t)) U(t, t_0) \\ &=e^{\frac{i}{\hbar} H_0(t - t_0)} H'(t)) e^{-\frac{i}{\hbar} H_0(t - t_0)} U_I(t, t_0).\end{aligned}
Define
\begin{aligned}\bar{H}'(t) =e^{\frac{i}{\hbar} H_0(t - t_0)} H'(t)) e^{-\frac{i}{\hbar} H_0(t - t_0)},\end{aligned} \hspace{\stretch{1}}(2.21)
so that our operator equation takes the form
\begin{aligned}i \hbar \frac{d{{}}}{dt} U_I(t, t_0) = \bar{H}'(t) U_I(t, t_0).\end{aligned} \hspace{\stretch{1}}(2.22)
Note that we also have the required identity at the initial time
\begin{aligned}U_I(t_0, t_0) = I.\end{aligned} \hspace{\stretch{1}}(2.23)
Without requiring us to actually find $U(t, t_0)$ all of the dynamics of the time dependent interaction are now embedded in our operator equation for $\bar{H}'(t)$, with all of the simple interaction related to the non time dependent portions of the Hamiltonian left separate.
## Connection with the Schr\”{o}dinger picture.
In the Schr\”{o}dinger picture we have
\begin{aligned}{\lvert {\psi_s(t)} \rangle} &= U(t, t_0) {\lvert {\psi_s(t_0)} \rangle} \\ &=e^{-\frac{i}{\hbar} H_0(t - t_0)} U_I(t, t_0){\lvert {\psi_s(t_0)} \rangle}.\end{aligned}
With a definition of the interaction picture ket as
\begin{aligned}{\lvert {\psi_I} \rangle} = U_I(t, t_0) {\lvert {\psi_s(t_0)} \rangle} = U_I(t, t_0) {\lvert {\psi_H} \rangle},\end{aligned} \hspace{\stretch{1}}(2.24)
the Schr\”{o}dinger picture is then related to the interaction picture by
\begin{aligned}{\lvert {\psi_s(t)} \rangle} = e^{-\frac{i}{\hbar} H_0(t - t_0)} {\lvert {\psi_I} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.25)
Also, by multiplying 2.22 by our Schr\”{o}dinger ket, we remove the last vestiges of $U_I$ and $U$ from the dynamical equation for our time dependent interaction
\begin{aligned}i \hbar \frac{d{{}}}{dt} {\lvert {\psi_I} \rangle} = \bar{H}'(t) {\lvert {\psi_I} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.26)
## Interaction picture expectation.
Inverting 2.25, we can form an operator expectation, and relate it the interaction and Schr\”{o}dinger pictures
\begin{aligned}{\langle {\psi_s(t)} \rvert} O_s {\lvert {\psi_s(t)} \rangle} ={\langle {\psi_I} \rvert} e^{\frac{i}{\hbar} H_0(t - t_0)}O_se^{-\frac{i}{\hbar} H_0(t - t_0)}{\lvert {\psi_I} \rangle} .\end{aligned} \hspace{\stretch{1}}(2.27)
With a definition
\begin{aligned}O_I =e^{\frac{i}{\hbar} H_0(t - t_0)}O_se^{-\frac{i}{\hbar} H_0(t - t_0)},\end{aligned} \hspace{\stretch{1}}(2.28)
we have
\begin{aligned}{\langle {\psi_s(t)} \rvert} O_s {\lvert {\psi_s(t)} \rangle} ={\langle {\psi_I} \rvert} O_I{\lvert {\psi_I} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.29)
As before, the time evolution of our interaction picture operator, can be found by taking derivatives of 2.28, for which we find
\begin{aligned}i \hbar \frac{d{{O_I(t)}}}{dt} = \left[{O_I(t)},{H_0}\right]\end{aligned} \hspace{\stretch{1}}(2.30)
## Summarizing the interaction picture.
Given
\begin{aligned}H(t) = H_0 + H'(t),\end{aligned} \hspace{\stretch{1}}(2.31)
and initial time states
\begin{aligned}{\lvert {\psi_I(t_0)} \rangle} ={\lvert {\psi_s(t_0)} \rangle} = {\lvert {\psi_H} \rangle},\end{aligned} \hspace{\stretch{1}}(2.32)
we have
\begin{aligned}{\langle {\psi_s(t)} \rvert} O_s {\lvert {\psi_s(t)} \rangle} ={\langle {\psi_I} \rvert} O_I{\lvert {\psi_I} \rangle},\end{aligned} \hspace{\stretch{1}}(2.33)
where
\begin{aligned}{\lvert {\psi_I} \rangle} = U_I(t, t_0) {\lvert {\psi_s(t_0)} \rangle},\end{aligned} \hspace{\stretch{1}}(2.34)
and
\begin{aligned}i \hbar \frac{d{{}}}{dt} {\lvert {\psi_I} \rangle} = \bar{H}'(t) {\lvert {\psi_I} \rangle},\end{aligned} \hspace{\stretch{1}}(2.35)
or
\begin{aligned}i \hbar \frac{d{{}}}{dt} U_I(t, t_0) &= \bar{H}'(t) U_I(t, t_0) \\ U_I(t_0, t_0) &= I.\end{aligned} \hspace{\stretch{1}}(2.36)
Our interaction picture Hamiltonian is
\begin{aligned}\bar{H}'(t) =e^{\frac{i}{\hbar} H_0(t - t_0)} H'(t)) e^{-\frac{i}{\hbar} H_0(t - t_0)},\end{aligned} \hspace{\stretch{1}}(2.38)
and for Schr\”{o}dinger operators, independent of time, we have the dynamical equation
\begin{aligned}i \hbar \frac{d{{O_I(t)}}}{dt} = \left[{O_I(t)},{H_0}\right]\end{aligned} \hspace{\stretch{1}}(2.39)
# Justifying the Taylor expansion above (not class notes).
## Multivariable Taylor series
As outlined in section 2.8 ($8.10$) of [2], we want to derive the multi-variable Taylor expansion for a scalar valued function of some number of variables
\begin{aligned}f(\mathbf{u}) = f(u^1, u^2, \cdots),\end{aligned} \hspace{\stretch{1}}(3.40)
consider the displacement operation applied to the vector argument
\begin{aligned}f(\mathbf{a} + \mathbf{x}) = {\left.{{f(\mathbf{a} + t \mathbf{x})}}\right\vert}_{{t=1}}.\end{aligned} \hspace{\stretch{1}}(3.41)
We can Taylor expand a single variable function without any trouble, so introduce
\begin{aligned}g(t) = f(\mathbf{a} + t \mathbf{x}),\end{aligned} \hspace{\stretch{1}}(3.42)
where
\begin{aligned}g(1) = f(\mathbf{a} + \mathbf{x}).\end{aligned} \hspace{\stretch{1}}(3.43)
We have
\begin{aligned}g(t) = g(0) + t {\left.{{ \frac{\partial {g}}{\partial {t}} }}\right\vert}_{{t = 0}}+ \frac{t^2}{2!} {\left.{{ \frac{\partial {g}}{\partial {t}} }}\right\vert}_{{t = 0}}+ \cdots,\end{aligned} \hspace{\stretch{1}}(3.44)
so that
\begin{aligned}g(1) = g(0) + + {\left.{{ \frac{\partial {g}}{\partial {t}} }}\right\vert}_{{t = 0}}+ \frac{1}{2!} {\left.{{ \frac{\partial {g}}{\partial {t}} }}\right\vert}_{{t = 0}}+ \cdots.\end{aligned} \hspace{\stretch{1}}(3.45)
The multivariable Taylor series now becomes a plain old application of the chain rule, where we have to evaluate
\begin{aligned}\frac{dg}{dt} &= \frac{d{{}}}{dt} f(a^1 + t x^1, a^2 + t x^2, \cdots) \\ &= \sum_i \frac{\partial {}}{\partial {(a^i + t x^i)}} f(\mathbf{a} + t \mathbf{x}) \frac{\partial {a^i + t x^i}}{\partial {t}},\end{aligned}
so that
\begin{aligned}{\left.{{\frac{dg}{dt} }}\right\vert}_{{t=0}}= \sum_i x^i \left( {\left.{{ \frac{\partial {f}}{\partial {x^i}}}}\right\vert}_{{x^i = a^i}}\right).\end{aligned} \hspace{\stretch{1}}(3.46)
Assuming an Euclidean space we can write this in the notationally more pleasant fashion using a gradient operator for the space
\begin{aligned}{\left.{{\frac{dg}{dt} }}\right\vert}_{{t=0}} = {\left.{{\mathbf{x} \cdot \boldsymbol{\nabla}_{\mathbf{u}} f(\mathbf{u})}}\right\vert}_{{\mathbf{u} = \mathbf{a}}}.\end{aligned} \hspace{\stretch{1}}(3.47)
To handle the higher order terms, we repeat the chain rule application, yielding for example
\begin{aligned}{\left.{{\frac{d^2 f(\mathbf{a} + t \mathbf{x})}{dt^2} }}\right\vert}_{{t=0}} &={\left.{{\frac{d{{}}}{dt} \sum_i x^i \frac{\partial {f(\mathbf{a} + t \mathbf{x})}}{\partial {(a^i + t x^i)}} }}\right\vert}_{{t=0}}\\ &={\left.{{\sum_i x^i \frac{\partial {}}{\partial {(a^i + t x^i)}} \frac{d{{f(\mathbf{a} + t \mathbf{x})}}}{dt}}}\right\vert}_{{t=0}} \\ &={\left.{{(\mathbf{x} \cdot \boldsymbol{\nabla}_{\mathbf{u}})^2 f(\mathbf{u})}}\right\vert}_{{\mathbf{u} = \mathbf{a}}}.\end{aligned}
Thus the Taylor series associated with a vector displacement takes the tidy form
\begin{aligned}f(\mathbf{a} + \mathbf{x}) = \sum_{k=0}^\infty \frac{1}{{k!}} {\left.{{(\mathbf{x} \cdot \boldsymbol{\nabla}_{\mathbf{u}})^k f(\mathbf{u})}}\right\vert}_{{\mathbf{u} = \mathbf{a}}}.\end{aligned} \hspace{\stretch{1}}(3.48)
Even more fancy, we can form the operator equation
\begin{aligned}f(\mathbf{a} + \mathbf{x}) = {\left.{{e^{ \mathbf{x} \cdot \boldsymbol{\nabla}_{\mathbf{u}} } f(\mathbf{u})}}\right\vert}_{{\mathbf{u} = \mathbf{a}}}\end{aligned} \hspace{\stretch{1}}(3.49)
Here a dummy variable $\mathbf{u}$ has been retained as an instruction not to differentiate the $\mathbf{x}$ part of the directional derivative in any repeated applications of the $\mathbf{x} \cdot \boldsymbol{\nabla}$ operator.
That notational cludge can be removed by swapping $\mathbf{a}$ and $\mathbf{x}$
\begin{aligned}f(\mathbf{a} + \mathbf{x}) = \sum_{k=0}^\infty \frac{1}{{k!}} (\mathbf{a} \cdot \boldsymbol{\nabla})^k f(\mathbf{x})=e^{ \mathbf{a} \cdot \boldsymbol{\nabla} } f(\mathbf{x}),\end{aligned} \hspace{\stretch{1}}(3.50)
where $\boldsymbol{\nabla} = \boldsymbol{\nabla}_{\mathbf{x}} = ({\partial {}}/{\partial {x^1}}, {\partial {}}/{\partial {x^2}}, ...)$.
Having derived this (or for those with lesser degrees of amnesia, recall it), we can see that 2.2 was a direct application of this, retaining no second order or higher terms.
Our expression used in the interaction Hamiltonian discussion was
\begin{aligned}\frac{1}{{{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}}} \approx \frac{1}{{{\left\lvert{\mathbf{r}}\right\rvert}}} + \mathbf{R} \cdot {\left.{{\left(\frac{\partial {}}{\partial {\mathbf{R}}} \frac{1}{{{\left\lvert{ \mathbf{r} - \mathbf{R}}\right\rvert}}}\right)}}\right\vert}_{{\mathbf{R} = 0}}.\end{aligned} \hspace{\stretch{1}}(3.51)
which we can see has the same structure as above with some variable substitutions. Evaluating it we have
\begin{aligned}\frac{\partial {}}{\partial {\mathbf{R}}} \frac{1}{{{\left\lvert{ \mathbf{r} - \mathbf{R}}\right\rvert}}}&=\mathbf{e}_i \frac{\partial {}}{\partial {R^i}} ((x^j - R^j)^2)^{-1/2} \\ &=\mathbf{e}_i \left(-\frac{1}{{2}}\right) 2 (x^j - R^j) \frac{\partial {(x^j - R^j)}}{\partial {R^i}} \frac{1}{{{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}^3}} \\ &= \frac{\mathbf{r} - \mathbf{R}}{{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}^3} ,\end{aligned}
and at $\mathbf{R} = 0$ we have
\begin{aligned}\frac{1}{{{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}}} \approx \frac{1}{{{\left\lvert{\mathbf{r}}\right\rvert}}} + \mathbf{R} \cdot \frac{\mathbf{r}}{{\left\lvert{\mathbf{r}}\right\rvert}^3}.\end{aligned} \hspace{\stretch{1}}(3.52)
We see in this direction derivative produces the classical electric Coulomb field expression for an electrostatic distribution, once we take the $\mathbf{r}/{\left\lvert{\mathbf{r}}\right\rvert}^3$ and multiply it with the $- Z e$ factor.
## With algebra.
A different way to justify the expansion of 2.2 is to consider a Clifford algebra factorization (following notation from [3]) of the absolute vector difference, where $\mathbf{R}$ is considered small.
\begin{aligned}{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}&= \sqrt{ \left(\mathbf{r} - \mathbf{R}\right) \left(\mathbf{r} - \mathbf{R}\right) } \\ &= \sqrt{ \left\langle{{\mathbf{r} \left(1 - \frac{1}{\mathbf{r}} \mathbf{R}\right) \left(1 - \mathbf{R} \frac{1}{\mathbf{r}}\right) \mathbf{r}}}\right\rangle } \\ &= \sqrt{ \left\langle{{\mathbf{r}^2 \left(1 - \frac{1}{\mathbf{r}} \mathbf{R}\right) \left(1 - \mathbf{R} \frac{1}{\mathbf{r}}\right) }}\right\rangle } \\ &= {\left\lvert{\mathbf{r}}\right\rvert} \sqrt{ 1 - 2 \frac{1}{\mathbf{r}} \cdot \mathbf{R} + \left\langle{{\frac{1}{\mathbf{r}} \mathbf{R} \mathbf{R} \frac{1}{\mathbf{r}}}}\right\rangle} \\ &= {\left\lvert{\mathbf{r}}\right\rvert} \sqrt{ 1 - 2 \frac{1}{\mathbf{r}} \cdot \mathbf{R} + \frac{\mathbf{R}^2}{\mathbf{r}^2}}\end{aligned}
Neglecting the $\mathbf{R}^2$ term, we can then Taylor series expand this scalar expression
\begin{aligned}\frac{1}{{{\left\lvert{\mathbf{r} - \mathbf{R}}\right\rvert}}} \approx\frac{1}{{{\left\lvert{\mathbf{r}}\right\rvert}}} \left( 1 + \frac{1}{\mathbf{r}} \cdot \mathbf{R}\right) =\frac{1}{{{\left\lvert{\mathbf{r}}\right\rvert}}} + \frac{\hat{\mathbf{r}}}{\mathbf{r}^2} \cdot \mathbf{R}=\frac{1}{{{\left\lvert{\mathbf{r}}\right\rvert}}} + \frac{\mathbf{r}}{{\left\lvert{\mathbf{r}}\right\rvert}^3} \cdot \mathbf{R}.\end{aligned} \hspace{\stretch{1}}(3.53)
Observe this is what was found with the multivariable Taylor series expansion too.
# References
[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.
[2] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.
[3] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.
## PHY456H1F: Quantum Mechanics II. Lecture 5 (Taught by Prof J.E. Sipe). Pertubation theory and degeneracy. Review of dynamics
Posted by peeterjoot on September 26, 2011
[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]
# Disclaimer.
Peeter’s lecture notes from class. May not be entirely coherent.
# Issues concerning degeneracy.
## When the perturbed state is non-degenerate.
Suppose the state of interest is non-degenerate but others are
FIXME: diagram. states designated by dashes labeled $n1$, $n2$, $n3$ degeneracy $\alpha = 3$ for energy $E_n^{(0)}$.
This is no problem except for notation, and if the analysis is repeated we find
\begin{aligned}E_s &= E_s^{(0)} + \lambda {H_{ss}}' + \lambda^2 \sum_{m \ne s, \alpha} \frac{{\left\lvert{{H_{m \alpha ; s}}'}\right\rvert}^2 }{ E_s^{(0)} - E_m^{(0)} } + \cdots\\ {\lvert {\bar{\psi}_s} \rangle} &= {\lvert {{\psi_s}^{(0)}} \rangle} + \lambda\sum_{m \ne s, \alpha} \frac{{H_{m \alpha ; s}}'}{ E_s^{(0)} - E_m^{(0)} } {\lvert {{\psi_{m \alpha}}^{(0)}} \rangle}+ \cdots,\end{aligned} \hspace{\stretch{1}}(2.1)
where
\begin{aligned}{H_{m \alpha ; s}}' ={\langle {{\psi_{m \alpha}}^{(0)}} \rvert} H' {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle}\end{aligned} \hspace{\stretch{1}}(2.3)
## When the perturbed state is also degenerate.
FIXME: diagram. states designated by dashes labeled $n1$, $n2$, $n3$ degeneracy $\alpha = 3$ for energy $E_n^{(0)}$, and states designated by dashes labeled $s1$, $s2$, $s3$ degeneracy $\alpha = 3$ for energy $E_s^{(0)}$.
If we just blindly repeat the derivation for the non-degenerate case we would obtain
\begin{aligned}E_s &= E_s^{(0)} + \lambda {H_{s1 ; s1}}' + \lambda^2 \sum_{m \ne s, \alpha} \frac{{\left\lvert{{H_{m \alpha ; s1}}'}\right\rvert}^2 }{ E_s^{(0)} - E_m^{(0)} } + \lambda^2 \sum_{\alpha \ne 1} \frac{{\left\lvert{{H_{s \alpha ; s1}}'}\right\rvert}^2 }{ E_s^{(0)} - {red} } + \cdots\\ {\lvert {\bar{\psi}_s} \rangle} &= {\lvert {{\psi_s}^{(0)}} \rangle} + \lambda\sum_{m \ne s, \alpha} \frac{{H_{m \alpha ; s}}'}{ E_s^{(0)} - E_m^{(0)} } {\lvert {{\psi_{m \alpha}}^{(0)}} \rangle}+ \lambda\sum_{\alpha \ne s1} \frac{{H_{s \alpha ; s1}}'}{ E_s^{(0)} - {red} } {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle}+ \cdots,\end{aligned} \hspace{\stretch{1}}(2.4)
where
\begin{aligned}{H_{m \alpha ; s1}}' ={\langle {{\psi_{m \alpha}}^{(0)}} \rvert} H' {\lvert {{\psi_{s1}}^{(0)}} \rangle}\end{aligned} \hspace{\stretch{1}}(2.6)
Note that the $E_s^{(0)} -{red}$ is NOT a typo, and why we run into trouble. There is one case where a perturbation approach is still possible. That case is if we happen to have
\begin{aligned}{\langle {{\psi_{m \alpha}}^{(0)}} \rvert} H' {\lvert {{\psi_{s1}}^{(0)}} \rangle} = 0.\end{aligned} \hspace{\stretch{1}}(2.7)
That may not be obvious, but if one returns to the original derivation, the right terms cancel so that one will not end up with the $0/0$ problem.
FIXME: do this derivation.
## Diagonalizing the perturbation Hamiltonian.
Suppose that we do not have this special zero condition that allows the perturbation treatment to remain valid. What can we do. It turns out that we can make use of the fact that the perturbation Hamiltonian is Hermitian, and diagonalize the matrix
\begin{aligned}{\langle {{\psi_{s \alpha}}^{(0)}} \rvert} H' {\lvert {{\psi_{s \beta}}^{(0)}} \rangle} \end{aligned} \hspace{\stretch{1}}(2.8)
In the example of a two fold degeneracy, this amounts to us choosing not to work with the states
\begin{aligned}{\lvert {\psi_{s1}^{(0)}} \rangle}, {\lvert {\psi_{s2}^{(0)}} \rangle},\end{aligned} \hspace{\stretch{1}}(2.9)
both some linear combinations of the two
\begin{aligned}{\lvert {\psi_{sI}^{(0)}} \rangle} &= a_1 {\lvert {\psi_{s1}^{(0)}} \rangle} + b_1 {\lvert {\psi_{s2}^{(0)}} \rangle} \\ {\lvert {\psi_{sII}^{(0)}} \rangle} &= a_2 {\lvert {\psi_{s1}^{(0)}} \rangle} + b_2 {\lvert {\psi_{s2}^{(0)}} \rangle} \end{aligned} \hspace{\stretch{1}}(2.10)
In this new basis, once found, we have
\begin{aligned}{\langle {{\psi_{s \alpha}}^{(0)}} \rvert} H' {\lvert {{\psi_{s \beta}}^{(0)}} \rangle} = \mathcal{H}_\alpha \delta_{\alpha \beta}\end{aligned} \hspace{\stretch{1}}(2.12)
Utilizing this to fix the previous, one would get if the analysis was repeated correctly
\begin{aligned}E_{s\alpha} &= E_s^{(0)} + \lambda {H_{s\alpha ; s\alpha}}' + \lambda^2 \sum_{m \ne s, \alpha} \frac{{\left\lvert{{H_{m \beta ; s \alpha}}'}\right\rvert}^2 }{ E_s^{(0)} - E_m^{(0)} } + \cdots\\ {\lvert {\bar{\psi}_{s \alpha}} \rangle} &= {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} + \lambda\sum_{m \ne s, \beta} \frac{{H_{m \beta ; s \alpha}}'}{ E_s^{(0)} - E_m^{(0)} } {\lvert {{\psi_{m \beta}}^{(0)}} \rangle}+ \cdots.\end{aligned} \hspace{\stretch{1}}(2.13)
We see that a degenerate state can be split by applying perturbation.
FIXME: do this derivation.
FIXME: diagram. $E_s^{(0)}$ as one energy level without perturbation, and as two distinct levels with perturbation.
guess I’ll bet that this is the origin of the spectral line splitting, especially given that an atom like hydrogen has degenerate states.
# Review of dynamics.
We want to move on to time dependent problems. In general for a time dependent problem, the answer follows provided one has solved for all the perturbed energy eigenvalues. This can be laborious (or not feasible due to infinite sums).
Before doing this, let’s review our dynamics as covered in section 3 of the text [1].
## Schr\”{o}dinger and Heisenberg pictures
Our operator equation in the Schr\”{o}dinger picture is the familiar
\begin{aligned}i \hbar \frac{d}{dt} {\lvert {\psi_s(t)} \rangle} = H {\lvert {\psi_s(t)} \rangle}\end{aligned} \hspace{\stretch{1}}(3.15)
and most of our operators $X, P, \cdots$ are time independent.
\begin{aligned}\left\langle{{O}}\right\rangle(t) = {\langle {\psi_s(t)} \rvert} O_s{\lvert {\psi_s(t)} \rangle}\end{aligned} \hspace{\stretch{1}}(3.16)
where $O_s$ is the operator in the Schr\”{o}dinger picture, and is non time dependent.
Formally, the time evolution of any state is given by
\begin{aligned}{\lvert {\psi_s(t)} \rangle}e^{-i H t/\hbar}{\lvert {\psi_s(0)} \rangle} = U(t, 0) {\lvert {\psi_s(0)} \rangle} \end{aligned} \hspace{\stretch{1}}(3.17)
so the expectation of an operator can be written
\begin{aligned}\left\langle{{O}}\right\rangle(t) = {\langle {\psi_s(0)} \rvert} e^{i H t/\hbar}O_se^{-i H t/\hbar}{\lvert {\psi_s(0)} \rangle}.\end{aligned} \hspace{\stretch{1}}(3.18)
With the introduction of the Heisenberg ket
\begin{aligned}{\lvert {\psi_H} \rangle} = {\lvert {\psi_s(0)} \rangle},\end{aligned} \hspace{\stretch{1}}(3.19)
and Heisenberg operators
\begin{aligned}O_H = e^{i H t/\hbar} O_s e^{-i H t/\hbar},\end{aligned} \hspace{\stretch{1}}(3.20)
the expectation evolution takes the form
\begin{aligned}\left\langle{{O}}\right\rangle(t) = {\langle {\psi_H} \rvert} O_H{\lvert {\psi_H} \rangle}.\end{aligned} \hspace{\stretch{1}}(3.21)
Note that because the Hamiltonian commutes with it’s exponential (it commutes with itself and any power series of itself), the Hamiltonian in the Heisenberg picture is the same as in the Schr\”{o}dinger picture
\begin{aligned}H_H = e^{i H t/\hbar} H e^{-i H t/\hbar} = H.\end{aligned} \hspace{\stretch{1}}(3.22)
### Time evolution and the Commutator
Taking the derivative of 3.20 provides us with the time evolution of any operator in the Heisenberg picture
\begin{aligned}i \hbar \frac{d}{dt} O_H(t) &=i \hbar \frac{d}{dt} \left( e^{i H t/\hbar} O_s e^{-i H t/\hbar}\right) \\ &=i \hbar \left( \frac{i H}{\hbar} e^{i H t/\hbar} O_s e^{-i H t/\hbar}+e^{i H t/\hbar} O_s e^{-i H t/\hbar} \frac{-i H}{\hbar} \right) \\ &=\left( -H O_H+O_H H\right).\end{aligned}
We can write this as a commutator
\begin{aligned}i \hbar \frac{d}{dt} O_H(t) = \left[{O_H},{H}\right].\end{aligned} \hspace{\stretch{1}}(3.23)
### Summarizing the two pictures.
\begin{aligned}\text{Schr\"{o}dinger picture} &\qquad \text{Heisenberg picture} \\ i \hbar \frac{d}{dt} {\lvert {\psi_s(t)} \rangle} = H {\lvert {\psi_s(t)} \rangle} &\qquad i \hbar \frac{d}{dt} O_H(t) = \left[{O_H},{H}\right] \\ {\langle {\psi_s(t)} \rvert} O_S {\lvert {\psi_s(t)} \rangle} &= {\langle {\psi_H} \rvert} O_H {\lvert {\psi_H} \rangle} \\ {\lvert {\psi_s(0)} \rangle} &= {\lvert {\psi_H} \rangle} \\ O_S &= O_H(0)\end{aligned}
# References
[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.
## PHY356F: Quantum Mechanics I. Lecture 11 notes. Harmonic Oscillator.
Posted by peeterjoot on November 30, 2010
# Setup.
Why study this problem?
It is relevant to describing the oscillation of molecules, quantum states of light, vibrations of the lattice structure of a solid, and so on.
FIXME: projected picture of masses on springs, with a ladle shaped well, approximately Harmonic about the minimum of the bucket.
The problem to solve is the one dimensional Hamiltonian
\begin{aligned}V(X) &= \frac{1}{{2}} K X^2 \\ K &= m \omega^2 \\ H &= \frac{P^2}{2m} + V(X)\end{aligned} \hspace{\stretch{1}}(8.168)
where $m$ is the mass, $\omega$ is the frequency, $X$ is the position operator, and $P$ is the momentum operator. Of these quantities, $\omega$ and $m$ are classical quantities.
This problem can be used to illustrate some of the reasons why we study the different pictures (Heisenberg, Interaction and Schr\”{o}dinger). This is a problem well suited to all of these (FIXME: lookup an example of this with the interaction picture. The book covers H and S methods.
We attack this with a non-intuitive, but cool technique. Introduce the raising $a^\dagger$ and lowering $a$ operators:
\begin{aligned}a &= \sqrt{\frac{m \omega}{2 \hbar}} \left( X + i \frac{P}{m\omega} \right) \\ a^\dagger &= \sqrt{\frac{m \omega}{2 \hbar}} \left( X - i \frac{P}{m\omega} \right)\end{aligned} \hspace{\stretch{1}}(8.171)
\paragraph{Question:} are we using the dagger for more than Hermitian conjugation in this case.
\paragraph{Answer:} No, this is precisely the Hermitian conjugation operation.
Solving for $X$ and $P$ in terms of $a$ and $a^\dagger$, we have
\begin{aligned}a + a^\dagger &= \sqrt{\frac{m \omega}{2 \hbar}} 2 X \\ a - a^\dagger &= \sqrt{\frac{m \omega}{2 \hbar}} 2 i \frac{P }{m \omega}\end{aligned}
or
\begin{aligned}X &= \sqrt{\frac{\hbar}{2 m \omega}} (a^\dagger + a) \\ P &= i \sqrt{\frac{\hbar m \omega}{2}} (a^\dagger -a)\end{aligned} \hspace{\stretch{1}}(8.173)
Express $H$ in terms of $a$ and $a^\dagger$
\begin{aligned}H &= \frac{P^2}{2m} + \frac{1}{{2}} K X^2 \\ &= \frac{1}{2m} \left(i \sqrt{\frac{\hbar m \omega}{2}} (a^\dagger -a)\right)^2+ \frac{1}{{2}} m \omega^2\left(\sqrt{\frac{\hbar}{2 m \omega}} (a^\dagger + a) \right)^2 \\ &= \frac{-\hbar \omega}{4} \left(a^\dagger a^\dagger + a^2 - a a^\dagger - a^\dagger a\right)+ \frac{\hbar \omega}{4}\left(a^\dagger a^\dagger + a^2 + a a^\dagger + a^\dagger a\right) \\ \end{aligned}
\begin{aligned}H= \frac{\hbar \omega}{2} \left(a a^\dagger + a^\dagger a\right) = \frac{\hbar \omega}{2} \left(2 a^\dagger a + \left[{a},{a^\dagger}\right]\right) \end{aligned} \hspace{\stretch{1}}(8.175)
Since $\left[{X},{P}\right] = i \hbar \mathbf{1}$ then we can show that $\left[{a},{a^\dagger}\right] = \mathbf{1}$. Solve for $\left[{a},{a^\dagger}\right]$ as follows
\begin{aligned}i \hbar &=\left[{X},{P}\right] \\ &=\left[{\sqrt{\frac{\hbar}{2 m \omega}} (a^\dagger + a) },{i \sqrt{\frac{\hbar m \omega}{2}} (a^\dagger -a)}\right] \\ &=\sqrt{\frac{\hbar}{2 m \omega}} i \sqrt{\frac{\hbar m \omega}{2}} \left[{a^\dagger + a},{a^\dagger -a}\right] \\ &= \frac{i \hbar}{2}\left(\left[{a^\dagger},{a^\dagger}\right] -\left[{a^\dagger},{a}\right] +\left[{a},{a^\dagger}\right] -\left[{a},{a}\right] \right) \\ &= \frac{i \hbar}{2}\left(0+2 \left[{a},{a^\dagger}\right] -0\right)\end{aligned}
Comparing LHS and RHS we have as stated
\begin{aligned}\left[{a},{a^\dagger}\right] = \mathbf{1}\end{aligned} \hspace{\stretch{1}}(8.176)
and thus from 8.175 we have
\begin{aligned}H = \hbar \omega \left( a^\dagger a + \frac{\mathbf{1}}{2} \right)\end{aligned} \hspace{\stretch{1}}(8.177)
Let ${\lvert {n} \rangle}$ be the eigenstate of $H$ so that $H{\lvert {n} \rangle} = E_n {\lvert {n} \rangle}$. From 8.177 we have
\begin{aligned}H {\lvert {n} \rangle} =\hbar \omega \left( a^\dagger a + \frac{\mathbf{1}}{2} \right) {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.178)
or
\begin{aligned}a^\dagger a {\lvert {n} \rangle} + \frac{{\lvert {n} \rangle}}{2} = \frac{E_n}{\hbar \omega} {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.179)
\begin{aligned}a^\dagger a {\lvert {n} \rangle} = \left( \frac{E_n}{\hbar \omega} - \frac{1}{{2}} \right) {\lvert {n} \rangle} = \lambda_n {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.180)
We wish now to find the eigenstates of the “Number” operator $a^\dagger a$, which are simultaneously eigenstates of the Hamiltonian operator.
Observe that we have
\begin{aligned}a^\dagger a (a^\dagger {\lvert {n} \rangle} ) &= a^\dagger ( a a^\dagger {\lvert {n} \rangle} ) \\ &= a^\dagger ( \mathbf{1} + a^\dagger a ) {\lvert {n} \rangle}\end{aligned}
where we used $\left[{a},{a^\dagger}\right] = a a^\dagger - a^\dagger a = \mathbf{1}$.
\begin{aligned}a^\dagger a (a^\dagger {\lvert {n} \rangle} ) &= a^\dagger \left( \mathbf{1} + \frac{E_n}{\hbar\omega} - \frac{\mathbf{1}}{2} \right) {\lvert {n} \rangle} \\ &= a^\dagger \left( \frac{E_n}{\hbar\omega} + \frac{\mathbf{1}}{2} \right) {\lvert {n} \rangle},\end{aligned}
or
\begin{aligned}a^\dagger a (a^\dagger {\lvert {n} \rangle} ) = (\lambda_n + 1) (a^\dagger {\lvert {n} \rangle} )\end{aligned} \hspace{\stretch{1}}(8.181)
The new state $a^\dagger {\lvert {n} \rangle}$ is presumed to lie in the same space, expressible as a linear combination of the basis states in this space. We can see the effect of the operator $a a^\dagger$ on this new state, we find that the energy is changed, but the state is otherwise unchanged. Any state $a^\dagger {\lvert {n} \rangle}$ is an eigenstate of $a^\dagger a$, and therefore also an eigenstate of the Hamiltonian.
Play the same game and win big by discovering that
\begin{aligned}a^\dagger a ( a {\lvert {n} \rangle} ) = (\lambda_n -1) (a {\lvert {n} \rangle} )\end{aligned} \hspace{\stretch{1}}(8.182)
There will be some state ${\lvert {0} \rangle}$ such that
\begin{aligned}a {\lvert {0} \rangle} = 0 {\lvert {0} \rangle}\end{aligned} \hspace{\stretch{1}}(8.183)
which implies
\begin{aligned}a^\dagger (a {\lvert {0} \rangle}) = (a^\dagger a) {\lvert {0} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(8.184)
so from 8.180 we have
\begin{aligned}\lambda_0 = 0\end{aligned} \hspace{\stretch{1}}(8.185)
Observe that we can identify $\lambda_n = n$ for
\begin{aligned}\lambda_n = \left( \frac{E_n}{\hbar\omega} - \frac{1}{{2}} \right) = n,\end{aligned} \hspace{\stretch{1}}(8.186)
or
\begin{aligned}\frac{E_n}{\hbar\omega} = n + \frac{1}{{2}}\end{aligned} \hspace{\stretch{1}}(8.187)
or
\begin{aligned}E_n = \hbar \omega \left( n + \frac{1}{{2}} \right)\end{aligned} \hspace{\stretch{1}}(8.188)
where $n = 0, 1, 2, \cdots$.
We can write
\begin{aligned}\hbar \omega \left( a^\dagger a + \frac{1}{{2}} \mathbf{1} \right) {\lvert {n} \rangle} &= E_n {\lvert {n} \rangle} \\ a^\dagger a {\lvert {n} \rangle} + \frac{1}{{2}} {\lvert {n} \rangle} &= \frac{E_n}{\hbar \omega} {\lvert {n} \rangle} \\ \end{aligned}
or
\begin{aligned}a^\dagger a {\lvert {n} \rangle} = \left( \frac{E_n}{\hbar \omega} - \frac{1}{{2}} \right) {\lvert {n} \rangle} = \lambda_n {\lvert {n} \rangle} = n {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.189)
We call this operator $a^\dagger a = N$, the number operator, so that
\begin{aligned}N {\lvert {n} \rangle} = n {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.190)
# Relating states.
Recall the calculation we performed for
\begin{aligned}L_{+} {\lvert {lm} \rangle} &= C_{+} {\lvert {l, m+1} \rangle} \\ L_{-} {\lvert {lm} \rangle} &= C_{+} {\lvert {l, m-1} \rangle}\end{aligned} \hspace{\stretch{1}}(9.191)
Where $C_{+}$, and $C_{+}$ are constants. The next game we are going to play is to work out $C_n$ for the lowering operation
\begin{aligned}a{\lvert {n} \rangle} = C_n {\lvert {n-1} \rangle}\end{aligned} \hspace{\stretch{1}}(9.193)
and the raising operation
\begin{aligned}a^\dagger {\lvert {n} \rangle} = B_n {\lvert {n+1} \rangle}.\end{aligned} \hspace{\stretch{1}}(9.194)
For the Hermitian conjugate of $a {\lvert {n} \rangle}$ we have
\begin{aligned}(a {\lvert {n} \rangle})^\dagger = ( C_n {\lvert {n-1} \rangle} )^\dagger = C_n^{*} {\lvert {n-1} \rangle}\end{aligned} \hspace{\stretch{1}}(9.195)
So
\begin{aligned}({\langle {n} \rvert} a^\dagger) (a {\lvert {n} \rangle}) = C_n C_n^{*} \left\langle{{n-1}} \vert {{n-1}}\right\rangle = {\left\lvert{C_n}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(9.196)
Expanding the LHS we have
\begin{aligned}{\left\lvert{C_n}\right\rvert}^2 &={\langle {n} \rvert} a^\dagger a {\lvert {n} \rangle} \\ &={\langle {n} \rvert} n {\lvert {n} \rangle} \\ &=n \left\langle{{n}} \vert {{n}}\right\rangle \\ &=n \end{aligned}
For
\begin{aligned}C_n = \sqrt{n}\end{aligned} \hspace{\stretch{1}}(9.197)
Similarly
\begin{aligned}({\langle {n} \rvert} a^\dagger) (a {\lvert {n} \rangle}) = B_n B_n^{*} \left\langle{{n+1}} \vert {{n+1}}\right\rangle = {\left\lvert{B_n}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(9.198)
and
\begin{aligned}{\left\lvert{B_n}\right\rvert}^2 &={\langle {n} \rvert} \underbrace{a a^\dagger}_{a a^\dagger - a^\dagger a = \mathbf{1}} {\lvert {n} \rangle} \\ &={\langle {n} \rvert} \left( \mathbf{1} + a^\dagger a \right) {\lvert {n} \rangle} \\ &=(1 + n) \left\langle{{n}} \vert {{n}}\right\rangle \\ &=1 + n \end{aligned}
for
\begin{aligned}B_n = \sqrt{n + 1}\end{aligned} \hspace{\stretch{1}}(9.199)
# Heisenberg picture.
\paragraph{How does the lowering operator $a$ evolve in time?}
\paragraph{A:} Recall that for a general operator $A$, we have for the time evolution of that operator
\begin{aligned}i \hbar \frac{d A}{dt} = \left[{ A },{H}\right]\end{aligned} \hspace{\stretch{1}}(10.200)
Let’s solve this one.
\begin{aligned}i \hbar \frac{d a}{dt} &= \left[{ a },{H}\right] \\ &= \left[{ a },{ \hbar \omega (a^\dagger a + \mathbf{1}/2) }\right] \\ &= \hbar\omega \left[{ a },{ (a^\dagger a + \mathbf{1}/2) }\right] \\ &= \hbar\omega \left[{ a },{ a^\dagger a }\right] \\ &= \hbar\omega \left( a a^\dagger a - a^\dagger a a \right) \\ &= \hbar\omega \left( (a a^\dagger) a - a^\dagger a a \right) \\ &= \hbar\omega \left( (a^\dagger a + \mathbf{1}) a - a^\dagger a a \right) \\ &= \hbar\omega a \end{aligned}
Even though $a$ is an operator, it can undergo a time evolution and we can think of it as a function, and we can solve for $a$ in the differential equation
\begin{aligned}\frac{d a}{dt} = -i \omega a \end{aligned} \hspace{\stretch{1}}(10.201)
This has the solution
\begin{aligned}a = a(0) e^{-i \omega t}\end{aligned} \hspace{\stretch{1}}(10.202)
here $a(0)$ is an operator, the value of that operator at $t = 0$. The exponential here is just a scalar (not effected by the operator so we can put it on either side of the operator as desired).
\paragraph{CHECK:}
\begin{aligned}a' = a(0) \frac{d}{dt} e^{-i \omega t} = a(0) (-i \omega) e^{-i \omega t} = -i \omega a\end{aligned} \hspace{\stretch{1}}(10.203)
# A couple comments on the Schr\”{o}dinger picture.
We don’t do this in class, but it is very similar to the approach of the hydrogen atom. See the text for full details.
In the Schr\”{o}dinger picture,
\begin{aligned}-\frac{\hbar^2}{2m} \frac{d^2 u}{dx^2} + \frac{1}{{2}} m \omega^2 x^2 u = E u\end{aligned} \hspace{\stretch{1}}(11.204)
This does directly to the wave function representation, but we can relate these by noting that we get this as a consequence of the identification $u = u(x) = \left\langle{{x}} \vert {{u}}\right\rangle$.
In 11.204, we can switch to dimensionless quantities with
\begin{aligned}\xi = \text{xi (z)''} = \alpha x\end{aligned} \hspace{\stretch{1}}(11.205)
with
\begin{aligned}\alpha = \sqrt{\frac{m \omega}{\hbar}}\end{aligned} \hspace{\stretch{1}}(11.206)
This gives, with $\lambda = 2E/\hbar\omega$,
\begin{aligned}\frac{d^2 u}{d\xi^2} + (\lambda - \xi^2) u = 0\end{aligned} \hspace{\stretch{1}}(11.207)
We can use polynomial series expansion methods to solve this, and find that we require a terminating expression, and write this in terms of the Hermite polynomials (courtesy of the clever French once again).
When all is said and done we will get the energy eigenvalues once again
\begin{aligned}E = E_n = \hbar \omega \left( n + \frac{1}{{2}} \right)\end{aligned} \hspace{\stretch{1}}(11.208)
# Back to the Heisenberg picture.
Let us express
\begin{aligned}\left\langle{{x}} \vert {{n}}\right\rangle = u_n(x)\end{aligned} \hspace{\stretch{1}}(12.209)
With
\begin{aligned}a {\lvert {0} \rangle} = 0,\end{aligned} \hspace{\stretch{1}}(12.210)
we have
\begin{aligned}0 =\left( X + i \frac{P}{m \omega} \right) {\lvert {0} \rangle},\end{aligned} \hspace{\stretch{1}}(12.211)
and
\begin{aligned}0 &= {\langle {x} \rvert} \left( X + i \frac{P}{m \omega} \right) {\lvert {0} \rangle} \\ &= {\langle {x} \rvert} X {\lvert {0 } \rangle} + i \frac{1}{m \omega} {\langle {x} \rvert} P {\lvert {0} \rangle} \\ &= x \left\langle{{x}} \vert {{0}}\right\rangle + i \frac{1}{m \omega} {\langle {x} \rvert} P {\lvert {0} \rangle} \\ \end{aligned}
Recall that our matrix operator is
\begin{aligned}{\langle {x'} \rvert} P {\lvert {x} \rangle} = \delta(x - x') \left( -i \hbar \frac{d}{dx} \right)\end{aligned} \hspace{\stretch{1}}(12.212)
\begin{aligned}{\langle {x} \rvert} P {\lvert {0} \rangle} &={\langle {x} \rvert} P \underbrace{\int {\lvert {x'} \rangle} {\langle {x'} \rvert} dx' }_{= \mathbf{1}}{\lvert {0} \rangle} \\ &=\int {\langle {x} \rvert} P {\lvert {x'} \rangle} \left\langle{{x'}} \vert {{0}}\right\rangle dx' \\ &=\int \delta(x - x') \left( -i \hbar \frac{d}{dx} \right)\left\langle{{x'}} \vert {{0}}\right\rangle dx' \\ &=\left( -i \hbar \frac{d}{dx} \right)\left\langle{{x}} \vert {{0}}\right\rangle\end{aligned}
We have then
\begin{aligned}0 =x u_0(x) + \frac{\hbar}{m \omega} \frac{d u_0(x)}{dx}\end{aligned} \hspace{\stretch{1}}(12.213)
NOTE: picture of the solution to this LDE on slide…. but I didn’t look closely enough.
## Notes and problems for Desai chapterĀ III.
Posted by peeterjoot on October 9, 2010
# Notes.
Chapter III notes and problems for [1].
FIXME:
Some puzzling stuff in the interaction section and superposition of time-dependent states sections. Work through those here.
# Problems
## Problem 1. Virial Theorem.
### Statement.
With the assumption that $\left\langle{{\mathbf{r} \cdot \mathbf{p}}}\right\rangle$ is independent of time, and
\begin{aligned}H = \frac{\mathbf{p}^2}{2m} + V(\mathbf{r}) = T + V\end{aligned} \hspace{\stretch{1}}(2.1)
show that
\begin{aligned}2 \left\langle{{T}}\right\rangle = \left\langle{{ \mathbf{r} \cdot \boldsymbol{\nabla} V}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(2.2)
### Solution.
I floundered with this a bit, but found the required hint in physicsforums. We can start with the Hamiltonian time derivative relation
\begin{aligned}i\hbar \frac{d A_H}{dt} = \left[{A_H},{H}\right]\end{aligned} \hspace{\stretch{1}}(2.3)
So, with the assumption that $\left\langle{{\mathbf{r} \cdot \mathbf{p}}}\right\rangle$ is independent of time, and the use of a stationary state ${\lvert {\psi} \rangle}$ for the expectation calculation we have
\begin{aligned}0 &=\frac{d}{dt} \left\langle{{\mathbf{r} \cdot \mathbf{p}}}\right\rangle \\ &=\frac{d}{dt} {\langle {\psi} \rvert} \mathbf{r} \cdot \mathbf{p} {\lvert {\psi} \rangle} \\ &={\langle {\psi} \rvert} \frac{d}{dt} ( \mathbf{r} \cdot \mathbf{p} ) {\lvert {\psi} \rangle} \\ &= \frac{1}{{i\hbar}} \left\langle{{ \left[{ \mathbf{r} \cdot \mathbf{p} },{H}\right] }}\right\rangle \\ &= -\left\langle{{ \left[{ \mathbf{r} \cdot \boldsymbol{\nabla} },{\frac{\mathbf{p}^2}{2m}}\right] }}\right\rangle -\left\langle{{ \left[{ \mathbf{r} \cdot \boldsymbol{\nabla} },{V(\mathbf{r})}\right] }}\right\rangle.\end{aligned}
The exercise now becomes one of evaluating the remaining commutators. For the Laplacian commutator we have
\begin{aligned}\left[{ \mathbf{r} \cdot \boldsymbol{\nabla} },{\boldsymbol{\nabla}^2}\right] \psi&=x_m \partial_m \partial_n \partial_n \psi - \partial_n \partial_n x_m \partial_m \psi \\ &=x_m \partial_m \partial_n \partial_n \psi - \partial_n \partial_n \psi - \partial_n x_m \partial_n \partial_m \psi \\ &=x_m \partial_m \partial_n \partial_n \psi - \partial_n \partial_n \psi - \partial_n \partial_n \psi - x_m \partial_n \partial_n \partial_m \psi \\ &=- 2 \boldsymbol{\nabla}^2 \psi\end{aligned}
For the potential commutator we have
\begin{aligned}\left[{ \mathbf{r} \cdot \boldsymbol{\nabla} },{V(\mathbf{r})}\right] \psi&=x_m \partial_m V \psi -V x_m \partial_m \psi \\ &=x_m (\partial_m V) \psi x_m V \partial_m \psi -V x_m \partial_m \psi \\ &=\Bigl( \mathbf{r} \cdot (\boldsymbol{\nabla} V) \Bigr) \psi\end{aligned}
Putting all the $\hbar$ factors back in, we get
\begin{aligned}2 \left\langle{{ \frac{\mathbf{p}^2}{2m} }}\right\rangle = \left\langle{{ \mathbf{r} \cdot (\boldsymbol{\nabla} V) }}\right\rangle,\end{aligned} \hspace{\stretch{1}}(2.4)
which is the desired result.
Followup: why assume $\left\langle{{\mathbf{r} \cdot \mathbf{p}}}\right\rangle$ is independent of time?
## Problem 2. Application of virial theorem.
Calculate $\left\langle{{T}}\right\rangle$ with $V = \lambda \ln(r/a)$.
\begin{aligned}\mathbf{r} \cdot \boldsymbol{\nabla} V &= r \hat{\mathbf{r}} \cdot \hat{\mathbf{r}} \lambda \frac{\partial {\ln(r/a)}}{\partial {r}} \\ &= \lambda r \frac{1}{{a}} \frac{a}{r} \\ &= \lambda \\ \implies \\ \left\langle{{T}}\right\rangle &= \lambda/2\end{aligned}
## Problem 3. Heisenberg Position operator representation.
### Part I.
Express $x$ as an operator $x_H$ for $H = \mathbf{p}^2/2m$.
With
\begin{aligned}{\langle {\psi} \rvert} x {\lvert {\psi} \rangle} = {\langle {\psi_0} \rvert} U^\dagger x U {\lvert {\psi_0} \rangle}\end{aligned}
We want to expand
\begin{aligned}x_H &= U^\dagger x U \\ &= e^{i H t/\hbar} x e^{-iH t/\hbar} \\ &= \sum_{k,l = 0}^\infty \frac{1}{{k!}} \frac{1}{{l!}} \left(\frac{i H t}{\hbar}\right)^k x \left(\frac{-i H t}{\hbar}\right)^l .\end{aligned}
We to evaluate $H^k x H^l$ to proceed. Using $p^n x = -i \hbar n p^{n-1} + x p^n$, we have
\begin{aligned}H^k x &= \frac{1}{{(2m)^k}} p^2k x \\ &= \frac{1}{{(2m)^k}} \left( -i \hbar (2k) p^{2k -1} + x p^2k \right) \\ &= x H^k + \frac{1}{{2m}} (-i \hbar) (2k) p p^{2(k-1)}/(2m)^{k-1} \\ &= x H^k - \frac{i \hbar k}{m} p H^{k-1}.\end{aligned}
This gives us
\begin{aligned}x_H &= x - \frac{i \hbar p }{m} \sum_{k,l=0}^\infty \frac{k}{k!} \frac{1}{{l!}}\left(\frac{i t}{\hbar}\right)^k H^{k-1 + l}\left(\frac{-i t}{\hbar}\right)^l \\ &= x - \frac{i \hbar p i t }{m \hbar} \end{aligned}
Or
\begin{aligned}x_H &= x + \frac{p t }{m} \end{aligned} \hspace{\stretch{1}}(2.5)
### Part II.
Express $x$ as an operator $x_H$ for $H = \mathbf{p}^2/2m + V$ with $V = \lambda x^m$.
In retrospect, for the first part of this problem, it would have been better to use the series expansion for this exponential sandwich
Or, in explicit form
\begin{aligned}e^A B e^{-A}&=B + \frac{1}{{1!}} \left[{A},{B}\right]+ \frac{1}{{2!}} \left[{A},{\left[{A},{B}\right]}\right]+ \cdots\end{aligned} \hspace{\stretch{1}}(2.6)
Doing so, we’d find for the first commutator
\begin{aligned}\frac{i t}{2m \hbar} \left[{\mathbf{p}^2},{x}\right] = \frac{t p}{m},\end{aligned} \hspace{\stretch{1}}(2.7)
so that the series has only the first two terms, and we’d obtain the same result. That seems like a logical approach to try here too. For the first commutator, we get the same $tp/m$ result since $\left[{V},{x}\right] = 0$.
Employing
\begin{aligned}x^n p = i \hbar n x^{n-1} + p x^n,\end{aligned} \hspace{\stretch{1}}(2.8)
I find
\begin{aligned}\left( \frac{i t}{\hbar} \right)^2 \left[{H},{\left[{H},{x}\right]}\right] &= \frac{i \lambda t^2}{\hbar m } \left[{x^n},{p}\right] \\ &= - \frac{n t^2 \lambda}{m} x^{n-1} \\ &= - \frac{n t^2 V}{m x} \\ \end{aligned}
The triple commutator gets no prettier, and I get
\begin{aligned}\left( \frac{i t}{\hbar} \right)^3 \left[{H},{ \left[{H},{ \left[H, x\right] }\right] }\right]&= \frac{it}{\hbar} \left[{ \frac{\mathbf{p}^2}{2m} + \lambda x^n},{ - \frac{n t^2 V}{m x} }\right] \\ &= -\frac{it}{\hbar} \frac{n t^2 }{m } \frac{\lambda}{2m} \left[{\mathbf{p}^2},{ x^{n-1}}\right] \\ &= \cdots \\ &= \frac{n(n-1)t^3 V}{ 2 m^2 x^3 } (i \hbar n + 2 p x).\end{aligned}
Putting all the pieces together this gives
\begin{aligned}x_H =e^{iH t/\hbar} x e^{-iH t/\hbar} &= x + \frac{tp}{m} - \frac{n t^2 V}{ 2 m x} + \frac{n(n-1)t^3 V}{ 12 m^2 x^3 } (i \hbar n + 2 p x) + \cdots\end{aligned} \hspace{\stretch{1}}(2.9)
If there is a closed form for this it isn’t obvious to me. Would a fixed lower degree potential function shed any more light on this. How about the Harmonic oscillator Hamiltonian
\begin{aligned}H = \frac{p^2}{2m} + \frac{m \omega^2 }{2} x^2\end{aligned} \hspace{\stretch{1}}(2.10)
… this one works out nicely since there’s an even-odd alternation.
Get
\begin{aligned}x_H = x \cos (\omega^2 t^2 /2) + \frac{ p t }{m} \frac{\sin( \omega^2 t^2/2)}{ \omega^2 t^2/2 }\end{aligned} \hspace{\stretch{1}}(2.11)
I’d not expect such a tidy result for an arbitrary $V(x) = \lambda x^n$ potential.
## Problem 4. Feynman-Hellman relation.
For continuously parametrized eigenstate, eigenvalue and Hamiltonian ${\lvert {\psi(\lambda)} \rangle}$, $E(\lambda)$ and $H(\lambda)$ respectively, we can relate the derivatives
\begin{aligned}\frac{\partial {}}{\partial {\lambda}} ( H {\lvert {\psi} \rangle} ) &= \frac{\partial {}}{\partial {\lambda}} ( E {\lvert {\psi} \rangle} ) \\ \implies \\ \frac{\partial {H}}{\partial {\lambda}} {\lvert {\psi} \rangle} +H \frac{\partial {{\lvert {\psi} \rangle}}}{\partial {\lambda}} &= \frac{\partial {E}}{\partial {\lambda}} {\lvert {\psi} \rangle} + E \frac{\partial {{\lvert {\psi} \rangle} }}{\partial {\lambda}} \end{aligned}
Left multiplication by ${\langle {\psi} \rvert}$ gives
\begin{aligned}{\langle {\psi} \rvert}\frac{\partial {H}}{\partial {\lambda}} {\lvert {\psi} \rangle} +{\langle {\psi} \rvert}H \frac{\partial {{\lvert {\psi} \rangle}}}{\partial {\lambda}} &= {\langle {\psi} \rvert}\frac{\partial {E}}{\partial {\lambda}} {\lvert {\psi} \rangle} + E {\langle {\psi} \rvert}\frac{\partial {{\lvert {\psi} \rangle} }}{\partial {\lambda}} \\ \implies \\ {\langle {\psi} \rvert}\frac{\partial {H}}{\partial {\lambda}} {\lvert {\psi} \rangle} +({\langle {\psi} \rvert}E) \frac{\partial {{\lvert {\psi} \rangle}}}{\partial {\lambda}} &= {\langle {\psi} \rvert}\frac{\partial {E}}{\partial {\lambda}} {\lvert {\psi} \rangle} + E {\langle {\psi} \rvert}\frac{\partial {{\lvert {\psi} \rangle} }}{\partial {\lambda}} \\ \implies \\ {\langle {\psi} \rvert}\frac{\partial {H}}{\partial {\lambda}} {\lvert {\psi} \rangle} &= \frac{\partial {E}}{\partial {\lambda}} \left\langle{{\psi}} \vert {{\psi}}\right\rangle,\end{aligned}
which provides the desired identity
\begin{aligned}\frac{\partial {E}}{\partial {\lambda}} = {\langle {\psi(\lambda)} \rvert}\frac{\partial {H}}{\partial {\lambda}} {\lvert {\psi(\lambda)} \rangle}\end{aligned} \hspace{\stretch{1}}(2.12)
## Problem 5.
### Description.
With eigenstates ${\lvert {\phi_1} \rangle}$ and ${\lvert {\phi_2} \rangle}$, of $H$ with eigenvalues $E_1$ and $E_2$, respectively, and
\begin{aligned}{\lvert {\chi_1} \rangle} &= \frac{1}{{\sqrt{2}}}( {\lvert {\phi_1} \rangle} +{\lvert {\phi_2} \rangle}) \\ {\lvert {\chi_2} \rangle} &= \frac{1}{{\sqrt{2}}}( {\lvert {\phi_1} \rangle} -{\lvert {\phi_2} \rangle})\end{aligned}
and ${\lvert {\psi(0)} \rangle} = {\lvert {\chi_1} \rangle}$, determine ${\lvert {\psi(t)} \rangle}$ in terms of ${\lvert {\phi_1} \rangle}$ and ${\lvert {\phi_2} \rangle}$.
### Solution.
\begin{aligned}{\lvert {\psi(t)} \rangle}&= e^{-i H t /\hbar} {\lvert {\psi(0)} \rangle} \\ &= e^{-i H t /\hbar} {\lvert {\chi_1} \rangle} \\ &= \frac{1}{{\sqrt{2}}} e^{-i H t /\hbar} ( {\lvert {\phi_1} \rangle} -{\lvert {\phi_2} \rangle}) \\ &= \frac{1}{{\sqrt{2}}} (e^{-i E_1 t /\hbar} {\lvert {\phi_1} \rangle} -e^{-i E_2 t /\hbar} {\lvert {\phi_2} \rangle} )\qquad\square\end{aligned}
## Problem 6.
### Description.
Consider a Coulomb like potential $-\lambda/r$ with angular momentum $l=0$. If the eigenfunction is
\begin{aligned}u(r) = u_0 e^{-\beta r}\end{aligned} \hspace{\stretch{1}}(2.13)
determine $u_0$, $\beta$, and the energy eigenvalue $E$ in terms of $\lambda$, and $m$.
### Solution.
We can start with the normalization constant $u_0$ by integrating
\begin{aligned}1 &= u_0^2 \int_0^\infty dr e^{-\beta r} e^{-\beta r} \\ &=u_0^2 \left. \frac{e^{-2 \beta r}}{-2 \beta} \right\vert_{0^\infty} \\ &= u_0^2 \frac{1}{{2\beta}} \\ \end{aligned}
\begin{aligned}u_0 &= \sqrt{2\beta}\end{aligned} \hspace{\stretch{1}}(2.14)
To go further, we need the Hamiltonian. Note that we can write the Laplacian with the angular momentum operator factored out using
\begin{aligned}\boldsymbol{\nabla}^2 &= \frac{1}{{\mathbf{x}^2}} \left( (\mathbf{x} \cdot \boldsymbol{\nabla})^2 + \mathbf{x} \cdot \boldsymbol{\nabla} + (\mathbf{x} \times \boldsymbol{\nabla})^2 \right)\end{aligned} \hspace{\stretch{1}}(2.15)
With zero for the angular momentum operator $\mathbf{x} \times \boldsymbol{\nabla}$, and switching to spherical coordinates, we have
\begin{aligned}\boldsymbol{\nabla}^2 &= \frac{1}{{r}} \partial_r + \frac{1}{{r}} \partial_r r \partial_r \\ &= \frac{1}{{r}} \partial_r + \frac{1}{{r}} \partial_r+ \frac{1}{{r}} r \partial_{rr} \\ &= \frac{2}{r} \partial_r + \partial_{rr} \\ \end{aligned}
We can now write the Hamiltonian for the zero angular momentum case
\begin{aligned}H&= -\frac{\hbar^2}{2m} \left( \frac{2}{r} \partial_r + \partial_{rr} \right) - \frac{\lambda}{r}\end{aligned} \hspace{\stretch{1}}(2.16)
With application of this Hamiltonian to the eigenfunction we have
\begin{aligned}E u_0 e^{-\beta r} &=\left( -\frac{\hbar^2}{2m} \left( \frac{2}{r} \partial_r + \partial_{rr} \right) - \frac{\lambda}{r} \right) u_0 e^{-\beta r} \\ &=\left( -\frac{\hbar^2}{2m} \left( \frac{2}{r} (-\beta) + \beta^2 \right) - \frac{\lambda}{r} \right) u_0 e^{-\beta r} .\end{aligned}
In particular for $r = \infty$ we have
\begin{aligned}-\frac{\hbar^2 \beta^2 }{2m} &= E\end{aligned} \hspace{\stretch{1}}(2.17)
\begin{aligned}-\frac{\hbar^2 \beta^2 }{2m} &= \left( -\frac{\hbar^2}{2m} \left( \frac{2}{r} (-\beta) + \beta^2 \right) - \frac{\lambda}{r} \right) \\ \implies \\ \frac{\hbar^2}{2m} \frac{2}{r} \beta &= \frac{\lambda}{r} \end{aligned}
Collecting all the results we have
\begin{aligned}\beta &= \frac{\lambda m}{\hbar^2} \\ E &= -\frac{\lambda^2 m}{2 \hbar^2} \\ u_0 &= \frac{\sqrt{2 \lambda m}}{\hbar}\end{aligned} \hspace{\stretch{1}}(2.18)
## Problem 7.
### Description.
A particle in a uniform field $\mathbf{E}_0$. Show that the expectation value of the position operator $\left\langle{\mathbf{r}}\right\rangle$ satisfies
\begin{aligned}m \frac{d^2 \left\langle{\mathbf{r}}\right\rangle }{dt^2} = e \mathbf{E}_0.\end{aligned} \hspace{\stretch{1}}(2.21)
### Solution.
This follows from Ehrehfest’s theorem once we formulate the force $e \mathbf{E}_0 = -\boldsymbol{\nabla} \phi$, in terms of a potential $\phi$. That potential is
\begin{aligned}\phi = - e \mathbf{E}_0 \cdot (x,y,z)\end{aligned} \hspace{\stretch{1}}(2.22)
The Hamiltonian is therefore
\begin{aligned}H = \frac{\mathbf{p}^2}{2m} - e \mathbf{E}_0 \cdot (x,y,z).\end{aligned} \hspace{\stretch{1}}(2.23)
Ehrehfest’s theorem gives us
\begin{aligned}\frac{d}{dt} \left\langle{{x_k}}\right\rangle &= \frac{1}{{m}} \left\langle{{p_k}}\right\rangle \\ \frac{d}{dt} \left\langle{{p_k}}\right\rangle &= -\left\langle{{ \frac{\partial {V}}{\partial {x_k}} }}\right\rangle,\end{aligned}
or
\begin{aligned}\frac{d^2}{dt^2} \left\langle{{x_k}}\right\rangle &= -\frac{1}{{m}} \left\langle{{ \frac{\partial {V}}{\partial {x_k}} }}\right\rangle.\end{aligned} \hspace{\stretch{1}}(2.24)
\begin{aligned}\frac{\partial {V}}{\partial {x_k}} &= - e (\mathbf{E}_0)_k\end{aligned}
Putting all the last bits together, and summing over the directions $\mathbf{e}_k$ we have
\begin{aligned}m \frac{d^2}{dt^2} \mathbf{e}_k \left\langle{{x_k}}\right\rangle = \mathbf{e}_k \left\langle{{ e (\mathbf{E}_0)_k }}\right\rangle= e \mathbf{E}_0\qquad\square\end{aligned}
## Problem 8.
### Description.
For Hamiltonian eigenstates ${\lvert {E_n} \rangle}$, $C = AB$, $A = \left[{B},{H}\right]$, obtain the matrix element ${\langle {E_m} \rvert} C {\lvert {E_n} \rangle}$ in terms of the matrix element of $A$.
### Solution.
I was able to get most of what was asked for here, with a small exception. I started with the matrix element for $A$, which is
\begin{aligned}{\langle {E_m} \rvert} A {\lvert {E_n} \rangle}={\langle {E_m} \rvert} BH - HB {\lvert {E_n} \rangle} =(E_n - E_m){\langle {E_m} \rvert} B {\lvert {E_n} \rangle} \end{aligned} \hspace{\stretch{1}}(2.25)
Next, computing the matrix element for $C$ we have
\begin{aligned}{\langle {E_m} \rvert} C {\lvert {E_n} \rangle}&={\langle {E_m} \rvert} BHB - HB^2 {\lvert {E_n} \rangle} \\ &=\sum_a {\langle {E_m} \rvert} BH {\lvert {E_a} \rangle}{\langle {E_a} \rvert} B {\lvert {E_n} \rangle} - E_m {\langle {E_m} \rvert} B {\lvert {E_a} \rangle} {\langle {E_a} \rvert} B {\lvert {E_n} \rangle} \\ &=\sum_a E_a {\langle {E_m} \rvert} B {\lvert {E_a} \rangle}{\langle {E_a} \rvert} B {\lvert {E_n} \rangle} -E_m {\langle {E_m} \rvert} B {\lvert {E_a} \rangle} {\langle {E_a} \rvert} B {\lvert {E_n} \rangle} \\ &=\sum_a (E_a - E_m){\langle {E_m} \rvert} B {\lvert {E_a} \rangle}{\langle {E_a} \rvert} B {\lvert {E_n} \rangle} \\ &=\sum_a {\langle {E_m} \rvert} A {\lvert {E_a} \rangle} {\langle {E_a} \rvert} B {\lvert {E_n} \rangle} \\ &={\langle {E_m} \rvert} A {\lvert {E_n} \rangle} {\langle {E_n} \rvert} B {\lvert {E_n} \rangle} +\sum_{a \ne n} {\langle {E_m} \rvert} A {\lvert {E_a} \rangle} {\langle {E_a} \rvert} B {\lvert {E_n} \rangle} \\ &={\langle {E_m} \rvert} A {\lvert {E_n} \rangle} {\langle {E_n} \rvert} B {\lvert {E_n} \rangle} +\sum_{a \ne n} {\langle {E_m} \rvert} A {\lvert {E_a} \rangle} \frac{{\langle {E_a} \rvert} A {\lvert {E_n} \rangle}}{E_n - E_a}\end{aligned}
Except for the ${\langle {E_n} \rvert} B {\lvert {E_n} \rangle}$ part of this expression, the problem as stated is complete. The relationship 2.25 is no help for with $n = m$, so I see no choice but to leave that small part of the expansion in terms of $B$.
## Problem 9.
### Description.
Operator $A$ has eigenstates ${\lvert {a_i} \rangle}$, with a unitary change of basis operation $U {\lvert {a_i} \rangle} = {\lvert {b_i} \rangle}$. Determine in terms of $U$, and $A$ the operator $B$ and its eigenvalues for which ${\lvert {b_i} \rangle}$ are eigenstates.
### Solution.
Consider for motivation the matrix element of $A$ in terms of ${\lvert {b_i} \rangle}$. We will also let $A {\lvert {a_i} \rangle} = \alpha_i {\lvert {a_i} \rangle}$. We then have
\begin{aligned}{\langle {a_i} \rvert} A {\lvert {a_j} \rangle}&={\langle {b_i} \rvert} U A U^\dagger {\lvert {b_j} \rangle} \\ \end{aligned}
We also have
\begin{aligned}{\langle {a_i} \rvert} A {\lvert {a_j} \rangle}&=a_j {\langle {a_i} \rvert} {\lvert {a_j} \rangle} \\ &=a_j \delta_{ij}\end{aligned}
So it appears that the operator $U A U^\dagger$ has the orthonormality relation required. In terms of action on the basis $\{{\lvert {b_i} \rangle}\}$, let’s see how it behaves. We have
\begin{aligned}U A U^\dagger {\lvert {b_i} \rangle}&= U A {\lvert {a_i} \rangle} \\ &= U \alpha_i {\lvert {a_i} \rangle} \\ &= \alpha_i {\lvert {b_i} \rangle} \\ \end{aligned}
So we see that the operators $A$ and $B = U A U^\dagger$ have common eigenvalues.
## Problem 10.
### Description.
With $H {\lvert {n} \rangle} = E_n {\lvert {n} \rangle}$, $A = \left[{H},{F}\right]$ and ${\langle {0} \rvert} F {\lvert {0} \rangle} = 0$, show that
\begin{aligned}\sum_{n\ne 0} \frac{{\langle {0} \rvert} A {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle} }{E_n - E_0} = {\langle {0} \rvert} AF {\lvert {0} \rangle}\end{aligned} \hspace{\stretch{1}}(2.26)
### Solution.
\begin{aligned}{\langle {0} \rvert} AF {\lvert {0} \rangle}&={\langle {0} \rvert} HF F - FH F{\lvert {0} \rangle} \\ &=\sum_n E_0 {\langle {0} \rvert} F {\lvert {n} \rangle}{\langle {n} \rvert} F {\lvert {0} \rangle} - E_n {\langle {0} \rvert} F {\lvert {n} \rangle} {\langle {n} \rvert} F{\lvert {0} \rangle} \\ &=\sum_n (E_0 -E_n) {\langle {0} \rvert} F {\lvert {n} \rangle}{\langle {n} \rvert} F {\lvert {0} \rangle} \\ &=\sum_{n\ne0} (E_0 -E_n) {\langle {0} \rvert} F {\lvert {n} \rangle}{\langle {n} \rvert} F {\lvert {0} \rangle} \\ \end{aligned}
We also have
\begin{aligned}{\langle {0} \rvert} A {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle}&={\langle {0} \rvert} HF -F H {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle} \\ &=(E_0 - E_n) {\langle {0} \rvert} F {\lvert {n} \rangle} {\langle {n} \rvert} HF - FH {\lvert {0} \rangle} \\ &=-(E_0 - E_n)^2 {\langle {0} \rvert} F {\lvert {n} \rangle} {\langle {n} \rvert} F {\lvert {0} \rangle} \\ \end{aligned}
Or, for $n \ne 0$,
\begin{aligned}{\langle {0} \rvert} F {\lvert {n} \rangle} {\langle {n} \rvert} F {\lvert {0} \rangle} &=-\frac{{\langle {0} \rvert} A {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle}}{(E_0 - E_n)^2 }.\end{aligned}
This gives
\begin{aligned}{\langle {0} \rvert} AF {\lvert {0} \rangle}&=-\sum_{n\ne0} (E_0 -E_n) \frac{{\langle {0} \rvert} A {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle}}{(E_0 - E_n)^2 } \\ &=\sum_{n\ne0} \frac{{\langle {0} \rvert} A {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle}}{E_n - E_0 } \qquad\square\end{aligned}
## Problem 11. commutator of angular momentum with Hamiltonian.
Show that $\left[{\mathbf{L}},{H}\right] = 0$, where $H = \mathbf{p}^2/2m + V(r)$.
This follows by considering $\left[{\mathbf{L}},{\mathbf{p}^2}\right]$, and $\left[{\mathbf{L}},{V(r)}\right]$. Let
\begin{aligned}L_{jk} = x_j p_k - x_k p_j,\end{aligned} \hspace{\stretch{1}}(2.27)
so that
\begin{aligned}\mathbf{L} = \mathbf{e}_i \epsilon_{ijk} L_{jk}.\end{aligned} \hspace{\stretch{1}}(2.28)
We now need to consider the commutators of the operators $L_{jk}$ with $\mathbf{p}^2$ and $V(r)$.
Let’s start with $p^2$. In particular
\begin{aligned}\mathbf{p}^2 x_m p_n&=p_k p_k x_m p_n \\ &=p_k (p_k x_m) p_n \\ &=p_k (-i\hbar \delta_{km} + x_m p_k) p_n \\ &=-i\hbar p_m p_n + (p_k x_m) p_k p_n \\ &=-i\hbar p_m p_n + (-i \hbar \delta_{km} + x_m p_k ) p_k p_n \\ &=-2 i\hbar p_m p_n + x_m p_n \mathbf{p}^2.\end{aligned}
So our commutator with $\mathbf{p}^2$ is
\begin{aligned}\left[{L_{jk}},{\mathbf{p}^2}\right]&=(x_j p_k - x_j p_k) \mathbf{p}^2 -( -2 i\hbar p_j p_k + x_j p_k \mathbf{p}^2 +2 i\hbar p_k p_j - x_k p_j \mathbf{p}^2 ).\end{aligned}
Since $p_j p_k = p_k p_j$, all terms cancel out, and the problem is reduced to showing that
\begin{aligned}\left[{\mathbf{L}},{H}\right] &= \left[{\mathbf{L}},{V(r)}\right] = 0.\end{aligned}
Now assume that $V(r)$ has a series representation
\begin{aligned}V(r) &= \sum_j a_j r^j = \sum_j a_j (x_k x_k)^{j/2}\end{aligned}
We’d like to consider the action of $x_m p_n$ on this function
\begin{aligned}x_m p_n V(r) \Psi&= -i \hbar x_m \sum_j a_j \partial_n (x_k x_k)^{j/2} \Psi \\ &= -i \hbar x_m \sum_j a_j (j x_n (x_k x_k)^{j/2-1} + r^j \partial_n \Psi) \\ &= -\frac{i \hbar x_m x_n}{r^2} \sum_j a_j j r^j + x_m V(r) p_n \Psi\end{aligned}
\begin{aligned}L_{mn} V(r) &=(x_m p_n - x_n p_m) V(r) \\ &= -\frac{i \hbar x_m x_n}{r^2} \sum_j a_j j r^j+\frac{i \hbar x_n x_m}{r^2} \sum_j a_j j r^j + V(r) (x_m p_n - x_n p_m )\\ &= V(r) L_{mn}\end{aligned}
Thus $\left[{L_{mn}},{V(r)}\right] = 0$ as expected, implying $\left[{\mathbf{L}},{H}\right] = 0$.
# References
[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009. | 24,147 | 64,691 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 375, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.66601 |
http://www.mathworks.com/matlabcentral/fileexchange/39344-a-matlab-script-for-predicting-lunar-eclipses/content/oevent1.m | 1,427,791,724,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131300444.84/warc/CC-MAIN-20150323172140-00150-ip-10-168-14-71.ec2.internal.warc.gz | 663,268,639 | 8,666 | Code covered by the BSD License
# A MATLAB Script for Predicting Lunar Eclipses
### David Eagle (view profile)
06 Dec 2012 (Updated )
Predicting the local circumstances of lunar eclipses.
oevent1 (objfunc, prtfunc, ti, tf, dt, dtsml)
```function oevent1 (objfunc, prtfunc, ti, tf, dt, dtsml)
% predict minimization/root-finding orbital events
% time argument in days
% input
% objfunc = objective function
% prtfunc = display results function
% ti = initial simulation time
% tf = final simulation time
% dt = step size used for bounding minima
% dtsml = small step size used to determine whether
% the function is increasing or decreasing
% Orbital Mechanics with Matlab
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
global trr
% initialization
tolm = 1e-8; % convergence tolerance
iend = 0;
% check the initial time for a minimum
fmin1 = feval(objfunc, ti);
tmin1 = ti;
ftemp = feval(objfunc, ti + dtsml);
df = ftemp - fmin1;
dflft = df;
el = ti;
er = el;
t = ti;
% if the slope is positive and the minimum is
% negative, calculate event conditions at the initial time
if (df > 0 && fmin1 < 0)
events1(objfunc, prtfunc, ti, tf, tmin1);
end
while (1)
% find where function first starts decreasing
while (1)
if (df <= 0)
break;
end
t = t + dt;
if (t >= tf)
% check final time for a minimum
if (iend == 1)
y = 1;
return;
end
fmin1 = feval(objfunc, tf);
ftemp = feval(objfunc, tf - dtsml);
df = fmin1 - ftemp;
% set minimum time to final simulation time
tmin1 = tf;
if (df < 0)
% if both the slope and minimum are negative,
% calculate the event conditions at the final
% simulation time
if (fmin1 < 0)
events1(objfunc, prtfunc, ti, tf, tmin1);
end
% otherwise, we're finished
y = 1;
return;
end
if (dflft > 0)
y = 1;
return;
end
er = tf;
iend = 1;
end
ft = feval(objfunc, t);
ftemp = feval(objfunc, t - dtsml);
df = ft - ftemp;
end
% function decreasing - find where function
% first starts increasing
while (1)
el = t;
dflft = df;
t = t + dt;
if (t >= tf)
% check final time for a minimum
if (iend == 1)
y = 1;
return;
end
fmin1 = feval(objfunc, tf);
ftemp = feval(objfunc, tf - dtsml);
df = fmin1 - ftemp;
% set minimum time to final simulation time
tmin1 = tf;
if (df < 0)
% if both the slope and minimum are negative,
% calculate the event conditions at the final
% simulation time
if (fmin1 < 0)
events1(objfunc, prtfunc, ti, tf, tmin1);
end
% otherwise, we're finished
y = 1;
return;
end
if (dflft > 0)
y = 1;
return;
end
er = tf;
iend = 1;
end
ft = feval(objfunc, t);
ftemp = feval(objfunc, t - dtsml);
df = ft - ftemp;
if (df > 0)
break;
end
end
er = t;
% calculate minimum using Brent's method
[tmin1, fmin1] = minima(objfunc, el, er, tolm);
el = er;
dflft = df;
% if the minimum is negative, calculate
% event conditions for this minimum
if (fmin1 < 0)
events1(objfunc, prtfunc, ti, tf, tmin1);
t = trr;
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function events1 (objfunc, prtfunc, ti, tf, topt)
% compute and display minimization/root-finding orbital events
% input
% objfunc = objective function
% prtfunc = display results function
% ti = initial simulation time
% tf = final simulation time
% topt = extrema time
% Orbital Mechanics with Matlab
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
global jdatei jdate trr
% define root-bracketing and root-finding control parameters
factor = 0.25; % geometric acceleration factor
dxmax = 300 / 86400; % rectification interval
rtol = 0.00000001; % convergence tolerance
% compute and display event start conditions
t1in = topt;
t2in = t1in - (10 / 86400);
[t1out, t2out] = broot(objfunc, t1in, t2in, factor, dxmax);
[troot, froot] = brent(objfunc, t1out, t2out, rtol);
% set to initial time if before ti
if (troot < ti)
troot = ti;
froot = feval(objfunc, ti);
end
jdate = jdatei + troot;
feval(prtfunc, 1, jdate);
% compute and display conditions at optimum
froot = feval(objfunc, t1in);
jdate = jdatei + t1in;
feval(prtfunc, 2, jdate)
% compute and display event end conditions
t2in = t1in + (10 / 86400);
[t1out, t2out] = broot(objfunc, t1in, t2in, factor, dxmax);
[troot, froot] = brent(objfunc, t1out, t2out, rtol);
% set to final time if after tf
if (troot > tf)
troot = tf;
froot = feval(objfunc, tf);
end
jdate = jdatei + troot;
feval(prtfunc, 3, jdate);
trr = troot;``` | 1,410 | 4,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2015-14 | longest | en | 0.479503 |
https://en.illogicopedia.org/wiki/The_Cabal_is_Evil | 1,620,470,826,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988858.72/warc/CC-MAIN-20210508091446-20210508121446-00039.warc.gz | 264,919,698 | 7,134 | The Cabal is Evil
To find the Cabal you need time and money. Therefore…
`Cabal = time x money`
‘Time is money’, so:
`time = money `
Therefore, Cabal equals money times money, or money squared…
`Cabal = money x money `
‘Money is the root of all evil’, therefore
`money x money = evil`
In conclusion,
`Cabal = Evil `
The Cabal is evil.
You know it's true. | 103 | 365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-21 | latest | en | 0.866504 |
http://99calculator.com/subnet-calculator | 1,487,916,341,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171416.74/warc/CC-MAIN-20170219104611-00637-ip-10-171-10-108.ec2.internal.warc.gz | 2,905,089 | 6,914 | # Subnet Calculator
## Calculate number of hosts, host address range, subnet ID, broadcast IP address etc
IP Subnet calculator uses IP address & subnet mask or /CIDR notation to calculate no. of hosts, IP address range, Network & Broadcast IP address, and Start & End of an IP Range in a network.
### What is an IP Subnet?
An IP subnet or just a subnetwork is a logical subdivision of an IP network. The practice of dividing a network using subnet mask into two or more smaller networks is called subnetting
On a network, a computer is identified by its IP address. The IP address of a computer is logically divided into two fields, a network (The network to which the computer belongs) and the host field (an identifier for the computer as a host on that network).
If a network is large, we divide that network into smaller networks for better management using a subnet mask. A Subnet mask is a 32-bit number that masks an IP address, and divides the IP address into network address and host address.
### How Subnetting is done?
Subnetting is done by performing a bitwise AND operation on the IP address of the host and the subnet mask.
Subnet Mask is made by setting network bits to all "1"s and setting host bits to all "0"s. Within a given network, two host addresses are reserved for special purpose, and cannot be assigned to hosts. The "0" address is assigned a network address and "255" is assigned to a broadcast address, and they cannot be assigned to hosts.
#### Here is an Example of Subnetting of Class C IP address:
IP Address 1100 0000 . 1010 1000 . 0000 1010 . 0000 0000 (192.168.10.0) Subnet Mask 1111 1111 . 1111 1111 . 1111 1111 . 1010 1000 (255.255.255.168) After AND operation 1100 0000 . 1010 1000 . 0000 1010 . 0000 0000 (192.168.10.0)
So total number of hosts in a network = 32
And total number of Subnetworks = = 8
That means each of these 8 subnetworks will have 32 hosts within them.
Here is the range of IP Address from the above example:
First IP address of the range (Or the Network address) Last IP address of the range (or the Broadcast Address) Number of Hosts 192.198.10.0 192.198.10.31 32 192.198.10.32 192.198.10.63 32 192.198.10.64 192.198.10.95 32 192.198.10.96 192.198.10.127 32 192.198.10.128 192.198.10.159 32 192.198.10.160 192.198.10.191 32 192.198.10.192 192.198.10.223 32 192.198.10.224 192.198.10.255 32
These all complex calculations can easily be calculated using the IP subnet Calculator or also called as the Subnet mask calculator.
IP subnet Calculator / Subnet mask calculator comes very handy if you are a network administrator and you have to create a subnet out of a large network.
99Calculator.com © 2016. All Rights Reserved. About | 745 | 2,705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-09 | latest | en | 0.919548 |
https://web2.0calc.com/questions/dice-game | 1,721,753,832,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00899.warc.gz | 519,663,379 | 5,621 | +0
# Dice game
-1
238
1
Three players are sitting at a round table playing a dice game with these rules. Each player starts with three tokens. On each turn, the player rolls one regular six-sided die and does one of the following actions based on the number rolled:
If you roll 1, 2, or 3, place two tokens in the middle of the table. (If you have ony one, place that one.)
If you roll 4, 5, or 6, keep your tokens.
After each turn, play passes to the player on the left. The game ends as soon as there is only one player with tokens left. What is the fewest number of total turns that can be taken before the game is over?
Jul 20, 2022 | 171 | 645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.964656 |
https://brainly.in/question/30365 | 1,485,294,710,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285289.45/warc/CC-MAIN-20170116095125-00131-ip-10-171-10-70.ec2.internal.warc.gz | 799,307,778 | 10,354 | # The square root of (x^2+8x+15) (x^2+x-6) (x^2+3x-10)
1
by umeshbala
2014-08-02T23:03:49+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Square root of [ (x^2+8x+15) (x^2+x-6) (x^2+3x-10) ]
we can get an idea that in such cases that, the polynomials (quadratic expressions) given above may be factorized ..
x²+8x+15 = (x+a)(x+b) => a+b = 8 a*b = 15 easy to find a = 5 b = 3 as factors of 15 are easy
x²+x-6 = (x+c)(x+d) => c+d = 1 c*d = -6 factors of -6 are 3 & 2 since sum is +1, c=+3 d= - 2
x² + 3x - 10 = (x+e)(x+f) => e+f = 3 ef = -10 factors of 10 5, 2. since sum is 3, e= +5, f= -2
NOW squaretoot is equal to
square root [ (x+3) (x+5) (x+3) (x-2) (x+5) (x-2) ]
square root [ (x+3)² (x+5)² (x-2)² ]
= (x+3) (x+5) (x-2)
=========================================================
If factorizaion is difficult, then do as :
(x^2+8x+15) = 0 Δ = 8² - 4*15 = 4 => x = (-8 +- 2 ) /2 = -5 or -3
so (x^2+8x+15) = (x - -5) (x - -3) = (x+5)(x+3)
(x^2+x-6) =0 Δ = 1 +24 = 25 => x = -1 +- 5 / 2 = -3 or 2
so factors (x - -3)(x - 2)
similarly the other expression | 586 | 1,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-04 | latest | en | 0.674778 |
https://puzzling.stackexchange.com/tags/rebus/hot?filter=year | 1,579,753,353,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250608295.52/warc/CC-MAIN-20200123041345-20200123070345-00268.warc.gz | 616,773,641 | 24,882 | A message from our CEO about the future of Stack Overflow and Stack Exchange. Read now.
# Tag Info
44
i guess the answer is: because
34
This is We begin with Then We continue on to and then cut to at which point and then again we have an abrupt change of scene to and yet another to and yet another to at which point and after which My list so far hasn't included because The title The main characters are Credit where due:
34
The symbols '｜ — S S' could be a visual representation of the word: Since:
33
I'm pretty sure the lecture (or at least that part of it) was about The inscription The mysterious prefix $\vee\sqcup$
21
From Somebody's completed nonogram grid, we can see Therefore,
17
I think the answer is Reasoning
16
The unknown symbol means ... In the final answer, it is ... In the rebus, the mystery word ... The final answer ...
16
Here is what I have as a completed nonogram. I don't think I've missed anything. I'm not yet sure where to go from here but it's a start!
15
I'm pretty sure Trump's plans involve ... Step 1 Step 2 Step 3 Credit where it's due:
14
14
First panel: Second panel: Third panel: Fourth panel: Now One way to conclude (unlikely): Another way to conclude (much better): A third conclusion given the hint that was added:
13
Each answer describes 1 2 3 4 Credit to the other two answers 1 and 2 which both gave partial answers that helped me along the way.
13
Completed Nonogram: Rebus decoding and final solution:
13
I think the answer is Expanding the equation gives
11
FULL ANSWER: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20)
11
Now
10
1 2 3 4 5
10
Thanks must be given to @GarethMcCaughan, @MOehm, @tmpearce, @Belhenix for helping me with some of the rebuses and clues in the comments Part I - Rebuses 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. Part II - Interpretation Something I'm not sure about:
8
Final Answer We know the final solution is: Like so: First two pictures Third picture Fourth picture Bottom picture
8
The following is my somewhat crazy interpretation of the rebus, which is either correct or way off: First image: Second image: Third image: First image inside parentheses: Second image inside parentheses: Putting it all together:
8
Title Upper half (blue) Lower half (red) Rebus: First red square Second red square Fourth red Square Fifth red Square (Solved by JS1) Now back to the third square: (Solved by Stiv) Conclusion (thx to M Oehm & Stiv)
7
#2 #3 #4 may simply point to
7
Partial Answer thanks to Ak19 and pirate for their contributions! 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11 12) 13) 14) 15) 17) 18) 19) 20)
7
Full solution, thanks to @JonMarkPerry and @jafe 1) 2) 3) 4) 5) 6) 7) 8) 9) 10)
7
Partial Answer It appears to be an equation, but I haven't looked closely at whether it could be a message disguised as an equation. Either way, the nonogram was fun!
6
The answer is Reasoning: Therefore, And the final question: Because... Here is a map for reference:
6
After thinking of this answer, I retracted my close vote and even changed the downvote into an upvote. Please read this answer before VTC-ing. Seriously. The phrase Explanation: And this is a true rebus on languagewordplay ;)
6
Considering the date this question was posted, I believe this puzzle commemorates: Explanation of clues (one small step at a time...):
6
Slightly partial answer All parts are done, as is the final answer, but I'm missing one or two bits of #3. 1: Mr Edison's Travel Diaries 2: Maria's Prayer 3: Devil with a heart 4: Easy Twists 5: You miss something here 6: Just a Rebus. Nothing else Putting it together Credit where due jafe found three answers in #3 and suggested the likely ...
5
Partial: 1: 2: 3: 4:
Only top voted, non community-wiki answers of a minimum length are eligible | 1,109 | 3,865 | {"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} | 2.984375 | 3 | CC-MAIN-2020-05 | longest | en | 0.915037 |
https://huggingface.co/datasets/math_dataset/viewer/algebra__linear_1d/test | 1,685,496,661,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00368.warc.gz | 345,553,734 | 26,473 | # Datasets:math_dataset
Languages: English
Dataset Preview
114 GB
"b'Solve -282*d + 929 - 178 = -1223 for d.\n'"
"b'7\n'"
"b'Solve 49*l + 45*l - 125 - 63 = 0 for l.\n'"
"b'2\n'"
"b'Solve -64*t + 1387 - 848 + 933 = 0 for t.\n'"
"b'23\n'"
"b'Solve 75*g = 192*g - 71*g - 79*g - 264 for g.\n'"
"b'-8\n'"
"b'Solve -34*v + 232*v + 52351 = 48985 for v.\n'"
"b'-17\n'"
"b'Solve 21*w + 26*w - 9*w - 3*w + 980 = 0 for w.\n'"
"b'-28\n'"
"b'Solve -21*v - 96 = 156 for v.\n'"
"b'-12\n'"
"b'Solve -12*t - 4482 = 64*t - 383*t + 141*t for t.\n'"
"b'27\n'"
"b'Solve -516 = -106*h + 4183 - 247 for h.\n'"
"b'42\n'"
"b'Solve -5437 = 149*q - 2717 + 2942 for q.\n'"
"b'-38\n'"
"b'Solve -10500 = -3300*p + 1800*p for p.\n'"
"b'7\n'"
"b'Solve -351*s - 5051 = 586*s - 623 + 6816 for s.\n'"
"b'-12\n'"
"b'Solve 1633 = 327*u - 891 - 944 - 1110 for u.\n'"
"b'14\n'"
"b'Solve 8*x + 77 = -7*x - 158 + 70 for x.\n'"
"b'-11\n'"
"b'Solve -51*u + 22*u + 729 = -79*u - 1171 for u.\n'"
"b'-38\n'"
"b'Solve 0 = 4018*q - 4444 - 3592 for q.\n'"
"b'2\n'"
"b'Solve 34338*y = 34271*y - 3216 for y.\n'"
"b'-48\n'"
"b'Solve -12463 - 56347 = -2338*d + 58334 - 12582 for d.\n'"
"b'49\n'"
"b'Solve 315*k + 6462 + 1204 = -162*k - 6167 for k.\n'"
"b'-29\n'"
"b'Solve -5444 = -458*f - 25596 for f.\n'"
"b'-44\n'"
"b'Solve 1407 = -141*f + 2112 for f.\n'"
"b'5\n'"
"b'Solve h - h - 13*h + h = 540 for h.\n'"
"b'-45\n'"
"b'Solve 0 = 49560*f - 49522*f - 1102 for f.\n'"
"b'29\n'"
"b'Solve 1106*r + 29369 = 434*r + 1145 for r.\n'"
"b'-42\n'"
"b'Solve 0 = -254*s - 44133 + 46673 for s.\n'"
"b'10\n'"
"b'Solve 2857*n - 2677*n + 1260 = 0 for n.\n'"
"b'-7\n'"
"b'Solve 159 = 58*a - 146 - 118 + 17 for a.\n'"
"b'7\n'"
"b'Solve 0 = 81*q + 8096 - 4451 for q.\n'"
"b'-45\n'"
"b'Solve 1308*j = -21443 - 26953 for j.\n'"
"b'-37\n'"
"b'Solve -71*a - 2546 = -108*a - 97*a for a.\n'"
"b'19\n'"
"b'Solve 0*t - 33*t - 83 + 964 = -736 for t.\n'"
"b'49\n'"
"b'Solve -22*k = 166 + 57 - 25 for k.\n'"
"b'-9\n'"
"b'Solve -39*o - 99*o + 67 = 111*o + 814 for o.\n'"
"b'-3\n'"
"b'Solve 237 = 14*g + 791 - 134 for g.\n'"
"b'-30\n'"
"b'Solve 12655 = -131*z - 217*z - 14*z + 1071 for z.\n'"
"b'-32\n'"
"b'Solve -243*n + 1512 = 68*n - 59*n for n.\n'"
"b'6\n'"
"b'Solve 307*b + 88 = 195 + 1121 for b.\n'"
"b'4\n'"
"b'Solve -1711*p + 520 + 1435 = -1626*p for p.\n'"
"b'23\n'"
"b'Solve 111*o - 341*o + 900 = -135*o - 120*o for o.\n'"
"b'-36\n'"
"b'Solve -61*n = 183*n + 1440*n + 8420 for n.\n'"
"b'-5\n'"
"b'Solve 0 = -11*u - 10394 + 9877 for u.\n'"
"b'-47\n'"
"b'Solve 0 = 17*y + 6898 - 6473 for y.\n'"
"b'-25\n'"
"b'Solve -591*v = -188*v - 192 - 15693 + 1780 for v.\n'"
"b'35\n'"
"b'Solve -780*h = -263*h + 19129 for h.\n'"
"b'-37\n'"
"b'Solve -253*i + 2769 - 14660 = 0 for i.\n'"
"b'-47\n'"
"b'Solve 2959*b + 4245 + 11493 = -4975 for b.\n'"
"b'-7\n'"
"b'Solve 14263 - 17047 = -87*s for s.\n'"
"b'32\n'"
"b'Solve 0 = 2885*i + 3042 + 20933 + 47695 + 40845 for i.\n'"
"b'-39\n'"
"b'Solve 0 = 1102422*q - 1102381*q - 1968 for q.\n'"
"b'48\n'"
"b'Solve 0 = -i - 91*i - 1598*i - 64220 for i.\n'"
"b'-38\n'"
"b'Solve -5561 = -1733*b - 753*b - 21099 - 36668 for b.\n'"
"b'-21\n'"
"b'Solve 19*v - 316 - 142 = 151 + 75 for v.\n'"
"b'36\n'"
"b'Solve 62*a + 4*a - 3*a + 2480 = a for a.\n'"
"b'-40\n'"
"b'Solve -7*o - 13*o = 672 - 654 - 698 for o.\n'"
"b'34\n'"
"b'Solve -304*a = 1517*a + 3642 for a.\n'"
"b'-2\n'"
"b'Solve 32*b + 36*b + 280 = -22*b + 34*b for b.\n'"
"b'-5\n'"
"b'Solve -6041*a + 6164*a + 738 = 0 for a.\n'"
"b'-6\n'"
"b'Solve 190*v + 2901 = -902*v + 144*v - 19*v for v.\n'"
"b'-3\n'"
"b'Solve -1072*f - 48125 = -8461 for f.\n'"
"b'-37\n'"
"b'Solve 148*q - 2571 + 501 = -91*q + 32*q for q.\n'"
"b'10\n'"
"b'Solve -98317*p + 98094*p - 8697 = 0 for p.\n'"
"b'-39\n'"
"b'Solve 1129 = 112*g + 5497 for g.\n'"
"b'-39\n'"
"b'Solve 34934 = -2069*g - 1059 + 4958 for g.\n'"
"b'-15\n'"
"b'Solve -1004*k - 10471 + 2408 - 15049 = -362*k for k.\n'"
"b'-36\n'"
"b'Solve -370*j = 560*j + 8207 + 3883 for j.\n'"
"b'-13\n'"
"b'Solve -171 + 315 + 171 = -63*g for g.\n'"
"b'-5\n'"
"b'Solve -77945*l + 77956*l - 62 + 370 = 0 for l.\n'"
"b'-28\n'"
"b'Solve 0 = -118*a + 264*a + 3942 for a.\n'"
"b'-27\n'"
"b'Solve -93*n + 1008 = 18*n + 33*n for n.\n'"
"b'7\n'"
"b'Solve -558 = -24*n + 186*n + 48*n - 5388 for n.\n'"
"b'23\n'"
"b'Solve 82*m - 2*m - 133*m = 142*m - 2925 for m.\n'"
"b'15\n'"
"b'Solve -158*s + 63*s - 2446 = -261 for s.\n'"
"b'-23\n'"
"b'Solve 159*c + 118*c - 3317 = -1048 + 1055 for c.\n'"
"b'12\n'"
"b'Solve 0 = -182*n + 2925 - 4563 for n.\n'"
"b'-9\n'"
"b'Solve 739*v - 364*v = -109*v - 913*v - 1397 for v.\n'"
"b'-1\n'"
"b'Solve 6*p - 105*p - 1644 = 1920 for p.\n'"
"b'-36\n'"
"b'Solve -1334*l - 1365*l = -2753*l + 1350 for l.\n'"
"b'25\n'"
"b'Solve -6*s = -31*s - 728*s + 7075 + 14009 for s.\n'"
"b'28\n'"
"b'Solve 0 = -9*y + 859 - 1273 for y.\n'"
"b'-46\n'"
"b'Solve 14*w - 85*w + 17*w + 20*w + 272 = 0 for w.\n'"
"b'8\n'"
"b'Solve -117*z - 11214 = -4*z + 66*z + 88*z for z.\n'"
"b'-42\n'"
"b'Solve 665 - 1356 = 71*c - 478 for c.\n'"
"b'-3\n'"
"b'Solve -71*z - 13*z + 6*z + 380 + 400 = 0 for z.\n'"
"b'10\n'"
"b'Solve 1898 = -2633*x - 831 - 2537 for x.\n'"
"b'-2\n'"
"b'Solve 1516*r = 2372*r - 35096 for r.\n'"
"b'41\n'"
"b'Solve -292*b = 6180 - 12580 - 7616 for b.\n'"
"b'48\n'"
"b'Solve 813*m = -187*m - 28000 for m.\n'"
"b'-28\n'"
"b'Solve 66*u - 1478 - 1624 = 0 for u.\n'"
"b'47\n'"
"b'Solve 2999*w - 88874 + 11332 = 63411 for w.\n'"
"b'47\n'"
"b'Solve -2048*c + 5277*c = -22603 for c.\n'"
"b'-7\n'"
"b'Solve -1114*i + 2193*i + 580 = 1099*i for i.\n'"
"b'29\n'"
"b'Solve -6930 = 475*z - 706*z for z.\n'"
"b'30\n'"
"b'Solve 2328 = 89*h + 179*h + 23*h for h.\n'"
"b'8\n'"
"b'Solve 52*x = -152*x - 79*x + 21018 - 8283 for x.\n'"
"b'45\n'"
"b'Solve -62*i + 1362 + 297 = 109 for i.\n'"
"b'25\n'"
"b'Solve -246*w - 53*w + 675*w = -520*w - 33152 for w.\n'"
"b'-37\n'"
"b'Solve 8549 = 905*t + 1017*t - 8749 for t.\n'"
"b'9\n'"
"b'Solve 27719 = 322*y + 22245 for y.\n'"
"b'17\n'"
"b'Solve -25895 - 4396 = -967*b + 5458 + 6799 for b.\n'"
"b'44\n'"
"b'Solve -15996*l - 6058 = -16229*l for l.\n'"
"b'26\n'" | 3,084 | 6,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-23 | latest | en | 0.295176 |
https://www.esaral.com/q/in-a-survey-of-200-ladies-it-was-found-that-142-like-coffee-while-58-dislike-it-52168 | 1,723,723,517,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641291968.96/warc/CC-MAIN-20240815110654-20240815140654-00004.warc.gz | 574,927,200 | 11,893 | # In a survey of 200 ladies, it was found that 142 like coffee, while 58 dislike it.
Question:
In a survey of 200 ladies, it was found that 142 like coffee, while 58 dislike it.
Find the probability that a lady chosen at random
(i) likes coffee
(ii) dislikes coffee
Solution:
Total number of ladies = 200
Number of ladies who like coffee = 142
Number of ladies who dislike coffee = 58
Let E1 and E2 be the events that the selected lady likes and dislikes coffee, respectively.Then,
(i) $P($ selected lady likes coffee $)=P\left(E_{1}\right)=\frac{\text { Number of ladies who like coffee }}{\text { Total number of ladies }}=\frac{142}{200}=0.71$
(ii) $P$ (selected lady dislikes coffee) $=P\left(E_{2}\right)=\frac{\text { Number of ladies who dislike coffee }}{\text { Total number of ladies }}=\frac{58}{200}=0.29$
REMARK: In the given survey, the only possible outcomes are E1 and E2 and P(E1) + P(E2) = (0.71 + 0.29) = 1 | 275 | 936 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.861489 |
https://www.jiskha.com/display.cgi?id=1273079387 | 1,516,638,620,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891485.97/warc/CC-MAIN-20180122153557-20180122173557-00692.warc.gz | 929,934,444 | 3,818 | # trig
posted by .
(2-2i)^5
find the given power write answer in rectangular form
• trig -
-128+128i
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More Similar Questions | 686 | 2,213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.671386 |
https://uk.mathworks.com/matlabcentral/cody/problems/1035-generate-a-vector-like-1-2-2-3-3-3-4-4-4-4/solutions/1720283 | 1,576,071,734,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540531917.10/warc/CC-MAIN-20191211131640-20191211155640-00492.warc.gz | 586,114,195 | 16,342 | Cody
# Problem 1035. Generate a vector like 1,2,2,3,3,3,4,4,4,4
Solution 1720283
Submitted on 4 Feb 2019 by Ram Rohit Gannavarapu
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 2; y_correct = [1 2 2]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2
2 Pass
x = 5; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5
3 Pass
x = 10; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 36 8 8 8 8 8 8 8 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 45 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 55 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10
4 Pass
x = 12; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 ... 11 11 11 11 11 11 11 11 11 11 11 ... 12 12 12 12 12 12 12 12 12 12 12 12]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 36 8 8 8 8 8 8 8 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 45 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 55 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 66 11 11 11 11 11 11 11 11 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 78 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12
5 Pass
x = 9; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 36 8 8 8 8 8 8 8 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 45 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9
6 Pass
x = 7; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7
7 Pass
x = 15; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 ... 11 11 11 11 11 11 11 11 11 11 11 ... 12 12 12 12 12 12 12 12 12 12 12 12 ... 13 13 13 13 13 13 13 13 13 13 13 13 13 ... 14 14 14 14 14 14 14 14 14 14 14 14 14 14 ... 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 36 8 8 8 8 8 8 8 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 45 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 55 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 66 11 11 11 11 11 11 11 11 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 78 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 87 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 13 13 Columns 88 through 91 13 13 13 13 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 87 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 13 13 Columns 88 through 105 13 13 13 13 14 14 14 14 14 14 14 14 14 14 14 14 14 14 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 87 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 13 13 Columns 88 through 116 13 13 13 13 14 14 14 14 14 14 14 14 14 14 14 14 14 14 15 15 15 15 15 15 15 15 15 15 15 Columns 117 through 120 15 15 15 15 | 4,498 | 6,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-51 | latest | en | 0.303856 |
https://reaktoro.org/tutorials/basics/defining-materials.html | 1,709,562,587,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476452.25/warc/CC-MAIN-20240304133241-20240304163241-00146.warc.gz | 482,681,370 | 10,443 | Defining materials#
Written by Allan Leal (ETH Zurich) on Jan 31st, 2022.
Last revised on Mar 26th, 2023 by Allan Leal.
Attention
Always make sure you are using the latest version of Reaktoro. Otherwise, some new features documented on this website will not work on your machine and you may receive unintuitive errors. Follow these update instructions to get the latest version of Reaktoro!
When performing chemical equilibrium or chemical kinetics calculations, an initial chemical state is required. From this state, often in disequilibrium, you are interested in:
1. immediately find a chemical equilibrium state that satisfies the desired constraints; or
2. compute a sequence of states describing the evolution of the system as it reacts over a period of time.
In most cases, you should be able to construct an initial state in terms of given amounts of chemical species in the system using the ChemicalState class. However, this may not always be convenient or possible for you. In this tutorial, we show an alternative way to define an initial chemical state using the Material class. First, however, we give you an example that shows some disadvantages of working only with the ChemicalState class as a motivation to use Material.
An initial chemical state not so conveniently defined#
Consider an aqueous solution with the following molal composition: 1 molal NaCl, 0.1 molal CaCl2, 0.05 molal MgCl2 and 0.1 molal CO2. Consider a rock with the following mass composition: 80%kg quartz (SiO2) and 20%kg calcite (CaCO3). Mix 1 kg of this aqueous solution with 10 g of rock and find the chemical equilibrium state of the system (i.e., solution and minerals) at 60°C and 15 bar.
As usual, we will start with the definition of a suitable chemical system to model this problem, which is done next.
from reaktoro import *
db = PhreeqcDatabase("phreeqc.dat")
solution = AqueousPhase()
solution.set(ActivityModelPhreeqc(db))
minerals = MineralPhases("Halite Calcite Dolomite Quartz")
system = ChemicalSystem(db, solution, minerals)
Note
In this system, the aqueous species were selected automatically based on the chemical elements composing the explicitly listed minerals (i.e., all aqueous species with elements H, O, Na, Cl, Ca, Mg, and Si). If you are curious, copy the code above and print the species in the chemical system!
for species in system.species():
print(species.name())
Let’s now create our initial chemical containing 1 kg of aqueous solution and 10 g of rock:
state = ChemicalState(system)
# Set initial brine composition
state.set("H2O" , 1.00, "kg")
state.set("Na+" , 1.00, "mol")
state.set("Ca+2", 0.10, "mol")
state.set("Mg+2", 0.05, "mol")
state.set("Cl-" , 1.30, "mol")
state.set("CO2" , 0.10, "mol")
# Set initial rock composition
state.set("Quartz", 80.0, "g")
state.set("Calcite", 20.0, "g")
# Scale fluid and solid masses to 1 kg and 10 g respectively
state.scaleFluidMass(1.0, "kg")
state.scaleSolidMass(10.0, "g")
While defining this initial chemical state, note that:
• we needed to specify amounts for ions Na+, Ca+2, and Mg+2 that reflect the desired molality of NaCl, CaCl2, and MgCl2.
• we needed to specify an amount for Cl- that produces zero electric charge in the solution.
• we needed to provide species names exactly how they exist in the database instead of chemical formulas (e.g., names Quartz and Calcite instead of formulas SiO2 and CaCO3).
• we needed to scale the mass of fluid and solids in the chemical state.
Let’s now equilibrate this state at 60 °C and 15 bar:
state.temperature(60.0, "celsius")
state.pressure(15.0, "bar")
equilibrate(state)
print(state)
+---------------------------+-------------+------+
| Property | Value | Unit |
+---------------------------+-------------+------+
| Temperature | 333.1500 | K |
| Pressure | 15.0000 | bar |
| Charge: | -3.0335e-16 | mol |
| Element Amount: | | |
| :: H | 1.0291e+02 | mol |
| :: C | 1.1269e-01 | mol |
| :: O | 5.1968e+01 | mol |
| :: Na | 9.2704e-01 | mol |
| :: Mg | 4.6352e-02 | mol |
| :: Si | 1.3315e-01 | mol |
| :: Cl | 1.2052e+00 | mol |
| :: Ca | 1.1269e-01 | mol |
| Species Amount: | | |
| :: CO3-2 | 3.4766e-07 | mol |
| :: H+ | 1.1599e-05 | mol |
| :: H2O | 5.1447e+01 | mol |
| :: CO2 | 8.2495e-02 | mol |
| :: (CO2)2 | 4.8663e-04 | mol |
| :: HCO3- | 1.0269e-02 | mol |
| :: CH4 | 1.0000e-16 | mol |
| :: Ca+2 | 9.7795e-02 | mol |
| :: CaCO3 | 4.0323e-06 | mol |
| :: CaHCO3+ | 4.1385e-03 | mol |
| :: CaOH+ | 4.6041e-10 | mol |
| :: Cl- | 1.2052e+00 | mol |
| :: H2 | 1.0000e-16 | mol |
| :: H4SiO4 | 1.8399e-04 | mol |
| :: H2SiO4-2 | 3.7469e-15 | mol |
| :: H3SiO4- | 2.0334e-08 | mol |
| :: Mg+2 | 4.3906e-02 | mol |
| :: MgCO3 | 8.4359e-07 | mol |
| :: MgHCO3+ | 2.4445e-03 | mol |
| :: MgOH+ | 1.2928e-07 | mol |
| :: Na+ | 9.2543e-01 | mol |
| :: NaCO3- | 3.9490e-06 | mol |
| :: NaHCO3 | 1.6087e-03 | mol |
| :: OH- | 1.7999e-08 | mol |
| :: NaOH | 1.0000e-16 | mol |
| :: O2 | 1.0000e-16 | mol |
| :: Halite :: NaCl | 1.0000e-16 | mol |
| :: Calcite :: CaCO3 | 1.0748e-02 | mol |
| :: Dolomite :: CaMg(CO3)2 | 1.2258e-15 | mol |
| :: Quartz :: SiO2 | 1.3296e-01 | mol |
+---------------------------+-------------+------+
We managed to get the problem defined and solved, but the coding process could be simpler as shown in the next section.
Simplifying with class Material#
Let’s now use class Material to obtain the same result with more convenience:
brine = Material(system)
rock = Material(system)
mix = brine(1.0, "kg") + rock(10.0, "g")
state = mix.equilibrate(60, "celsius", 15, "bar")
print(state)
+---------------------------+-------------+------+
| Property | Value | Unit |
+---------------------------+-------------+------+
| Temperature | 333.1500 | K |
| Pressure | 15.0000 | bar |
| Charge: | -5.2987e-17 | mol |
| Element Amount: | | |
| :: H | 1.0291e+02 | mol |
| :: C | 1.1269e-01 | mol |
| :: O | 5.1968e+01 | mol |
| :: Na | 9.2704e-01 | mol |
| :: Mg | 4.6352e-02 | mol |
| :: Si | 1.3315e-01 | mol |
| :: Cl | 1.2052e+00 | mol |
| :: Ca | 1.1269e-01 | mol |
| Species Amount: | | |
| :: CO3-2 | 3.4766e-07 | mol |
| :: H+ | 1.1599e-05 | mol |
| :: H2O | 5.1447e+01 | mol |
| :: CO2 | 8.2495e-02 | mol |
| :: (CO2)2 | 4.8663e-04 | mol |
| :: HCO3- | 1.0269e-02 | mol |
| :: CH4 | 1.0000e-16 | mol |
| :: Ca+2 | 9.7795e-02 | mol |
| :: CaCO3 | 4.0323e-06 | mol |
| :: CaHCO3+ | 4.1385e-03 | mol |
| :: CaOH+ | 4.6041e-10 | mol |
| :: Cl- | 1.2052e+00 | mol |
| :: H2 | 1.0000e-16 | mol |
| :: H4SiO4 | 1.8399e-04 | mol |
| :: H2SiO4-2 | 3.7469e-15 | mol |
| :: H3SiO4- | 2.0334e-08 | mol |
| :: Mg+2 | 4.3907e-02 | mol |
| :: MgCO3 | 8.4359e-07 | mol |
| :: MgHCO3+ | 2.4445e-03 | mol |
| :: MgOH+ | 1.2928e-07 | mol |
| :: Na+ | 9.2543e-01 | mol |
| :: NaCO3- | 3.9490e-06 | mol |
| :: NaHCO3 | 1.6087e-03 | mol |
| :: OH- | 1.7999e-08 | mol |
| :: NaOH | 1.0000e-16 | mol |
| :: O2 | 1.0000e-16 | mol |
| :: Halite :: NaCl | 1.0000e-16 | mol |
| :: Calcite :: CaCO3 | 1.0748e-02 | mol |
| :: Dolomite :: CaMg(CO3)2 | 1.2258e-15 | mol |
| :: Quartz :: SiO2 | 1.3296e-01 | mol |
+---------------------------+-------------+------+
The code above should (hopefully!) be intuitive enough and self-explanatory. The table printed above should also be identical (or extremely close!) to the previous one.
That’s it; you now have an extra Reaktoro component in your arsenal when performing your chemical modeling. The Material class should provide that extra level of convenience if things get a little more confusing if only the ChemicalState class is used.
Attention
You may have to use the Material class more carefully if chemical kinetics is considered. In this case, it is critical that species undergoing chemical reactions controlled by kinetic rates must have their initial amounts explicitly specified in the ChemicalState object, using their names instead of their formulas to avoid ambiguity. For example, suppose the minerals calcite and quartz are to react kinetically with the aqueous solution (and the aqueous species react instantaneously at equilibrium). In that case, the initial amounts of these minerals must be given in the ChemicalState object. The code below should give you an idea of how to proceed in this scenario:
brine = Material(system) | 2,965 | 9,980 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.858465 |
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Started Feb 11, 2014 | Discussions thread
Re: So, your conclusion is that 'equivalency = false'?
Ian Stuart Forsyth wrote:
awaldram wrote:
moving_comfort wrote:
awaldram wrote:
moving_comfort wrote:
awaldram wrote:
As far as I;m concerned both you and Ian have had ample proof of the error of your (and only your) crackpot theories.
So, everyone who understands and discusses equivalence - like Falk Lumo, Joseph James, etc and everyone viewing this thread who understands it are crackpots?
I think I'll invite Mr James to this thread - head crackpot, in your view - he probably likes a good laugh as much as anyone.
.
So I'm done here you obviously can't even understand the maths that prove your wrong are unable to see the truth in images posted by both Ian and Myself
Ian and yourself? Don't you know Ian's trying his best to make you see the light?
.
because of your FF centric world , you have no objectivity, even when so obviously wrong to everyone you eventually reach the point your becoming a bit of a joke.
Equivalency is not tied to "FF", it applies between every existing format size. It applies just as much between Q and 645D as it does between aps-c and FF. Don't try to make this a 'FF good, aps-c bad' argument. It simply describes a relationship, and in practice can describe a lens/sensor combo that might work better for someone than another lens/sensor combo. That's it. Don't take it so personally, it's an equation.
An equation that proves your wrong , as has been posted by James .
.
Really? Can you show me where James showed how equivalence proved me wrong?I think he'd be as suprised to hear that he did that as me
(I think you're attributing some of what he was writing incorrectly, by the way - but I'll let him speak for himself.)
I'm sure James is completely aware what is formula shows and the relationship to view size , Magnification , Focal length and sensor size.
But I can do better how about Bob Atkins ?
I'm sure some people will say, OK, but what if you don't take angle of view into account. What's the relative DOF if you use the SAME lens on a small-sensor camera and a full-frame camera?
Now you run into the problem of what you are comparing to what. The same lens on the two formats will give you different fields of view, so if you enlarge each image to the same size (say 8x12), you won't have the same print so you really can't compare DOFs. If you crop the 35mm negative to give you the same print as the digital image the answer is easy. The DOF in the cropped 35mm print and digital image print will be exactly the same. You're using the same lens and same size image (cropped 35mm or digital), so you get exactly the same DOF.
Oh dear chalk up another source that agrees with me, almost using the same words I did.
This is funny, crop the Full Frame 35mm negative to give you the same print and what do we get a ASPC negative and we a surpassed that they have the same DOF for the same FOV. I hope you are not showing this a proof that FF at F/2.8 is the same as APSC at F/2.8. Don’t you see the irony for FF to have the same dof you have to alters the FOV of the FF . Would this not tell us that DOF is relative to the format being used at a given FL and Fstop?
No it woudl tell us DoF is relative to the magnification factor and not the lens or directly the sensor.
it makes no odds if that magnification factor is during capture (due to sensor size) after capture (due to cropping) or indeed viewing size (you walk closer) as it not a function of the lens it is independent
Whether Lumos type 1 'or type 2 is in fact a valid model depends on pixel density / technology in use and noise floor not some made up equivalence.
For that’s what it is by his own admission he doesn't prove it so you and MC are reciting made up stuff.!
Its purely a vehicle to plumb FF feathers and has very little basis in fact.
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---> http://www.flickr.com/photos/95095968@N00/sets/72157626171532197/
-- hide signature --
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The Camera is only a tool, photography is deciding how to use it.
The hardest part about capturing wildlife is not the photographing portion; it’s getting them to sign a model release
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# [SOLVED]Simulate Thermistors and Chokes in SIMetrix
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#### kkeeley
##### Member level 3
Hi,
I'm new to this board and have been reading some very interesting posts from other people, and I have found some great information.
I have just started work on designing a SMPS for a project that I have been working on for some years, and due to the complex nature and dangers of SMPS's I wish to simulate as much of the design as possible before physically build anything. I have some old PC power supplies that I was thinking of using for donors parts (Transformers, Chokes, etc). I have just started using SIMetrix and am finding it very easy to learn and use, I am now at a point where I wish to start using/creating more complex parts, like thermistors, transformers and chokes.
So my question is how do I create each of these parts, I have seen some where an example creating a sub circuit for a transformer, but I can't remember where.
Kenneth
I have used the following model of a transformer in the past. This is the spice for it. Basically on primary it is a "voltage dependant current source" with magnetizing inductance in parallel. And on secondary it is "voltage controlled voltage source".
**********************************************************
* MODEL: transformer
* Application: transformer model suitable for averaged
* circuit models
* Limitations: no losses, no leakages, two windings
**********************************************************
* Parameters:
* Lm=primary (magnetizing) inductance
* n=transformer turns ratio 1:n (primary:secondary)
**********************************************************
* Nodes:
* 1: primary-dot
* 2: primary-undot
* 3: secondary-dot
* 4: secondary-undot
**********************************************************
.subckt transformer 1 2 3 4
+params: Lm=250e-6 n=0.5
Lm 1 2 {Lm}
Gp 2 1 value={n*i(Es)}
Es 3 4 value={n*v(1,2)}
.ends
**********************************************************
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Table of Content
Understanding Binomial Distribution and Its Importance in Machine Learning
August 21, 2024
In the realm of statistics and machine learning, understanding various probability distributions is paramount. One such fundamental distribution is the Binomial Distribution.
This distribution is not only a cornerstone in probability theory but also plays a crucial role in various machine learning algorithms and applications.
In this blog, we will delve into the concept of binomial distribution, its mathematical formulation, and its significance in the field of machine learning.
What is Binomial Distribution?
The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent and identically distributed Bernoulli trials.
A Bernoulli trial is a random experiment where there are only two possible outcomes:
• success (with probability ( p ))
• failure (with probability ( 1 – p ))
Mathematical Formulation
The probability of observing exactly successes in trials is given by the binomial probability formula:
Example 1: Tossing One Coin
Parameters
• Number of trials (n) = 1
• Probability of heads (p) = 0.5
• Number of heads (k) = 1
Calculation
• Binomial coefficient
• Probability
So, the probability of getting exactly one head in one toss of a coin is 0.5 or 50%.
Example 2: Tossing Two Coins
Now, let’s consider the case of tossing two coins.
Parameters
• Number of trials (n) = 2
• Probability of heads (p) = 0.5
• Number of heads (k) = varies (0, 1, or 2)
Calculation for k = 0
• Binomial coefficient
• Probability
P(X = 0) = 1 × (0.5)0 × (1 – 0.5)2 = 1 × 1 × 0.25 = 0.25
Calculation for k = 1
• Binomial coefficient
• Probability
P(X = 1) = 1 × (0.5)1 × (1 – 0.5)1 = 2 × 0.5 × 0.5 = 0.5
Calculation for k = 2
• Binomial coefficient
• Probability
P(X = 2) = 1 × (0.5)2 × (1 – 0.5)0 = 1 × 0.25 × 1 = 0.25
So, the probabilities for different numbers of heads in two-coin tosses are:
• P(X = 0) = 0.25 – no heads
• P(X = 1) = 0.5 – one head
• P(X = 2) = 0.25 – two heads
Detailed Example: Predicting Machine Failure
Let’s consider a more practical example involving predictive maintenance in an industrial setting. Suppose we have a machine that is known to fail with a probability of 0.05 during a daily checkup. We want to determine the probability of the machine failing exactly 3 times in 20 days.
Step-by-Step Calculation
1. Identify Parameters
• Number of trials (n) = 20 days
• Probability of success (p) = 0.05 – failure is considered a success in this context
• Number of successes (k) = 3 failures
4. Calculate Probability
• Plugging the values into the binomial formula
• Substitute the values
P(= 3) = 1140 × (0.05)3 × (0.95)17
• Calculate (0.05)3
(0.05)3 = 0.000125
• Calculate (0.95)17
(0.95)17 ≈ 0.411
5. Multiply all Components Together
P(= 3) = 1140 × 0.000125 × 0.411 ≈ 0.0585
Therefore, the probability of the machine failing exactly 3 times in 20 days is approximately 0.0585 or 5.85%.
Role of Binomial Distribution in Machine Learning
The binomial distribution is integral to several aspects of machine learning, providing a foundation for understanding and modeling binary events, hypothesis testing, and beyond.
Let’s explore how it intersects with various machine-learning concepts and techniques.
Binary Classification
In binary classification problems, where the outcomes are often categorized as success or failure, the binomial distribution forms the underlying probabilistic model. For instance, if we are predicting whether an email is spam or not, each email can be thought of as a Bernoulli trial.
Algorithms like Logistic Regression and Support Vector Machines (SVM) are particularly designed to handle these binary outcomes.
Understanding the binomial distribution helps in correctly interpreting the results of these classifiers. The performance metrics such as accuracy, precision, recall, and F1-score ultimately derive from the binomial probability model.
This understanding ensures that we can make informed decisions about model improvements and performance evaluation.
Hypothesis Testing
Statistical hypothesis testing, essential in validating machine learning models, often employs the binomial distribution to ascertain the significance of observed outcomes.
For instance, in A/B testing, which is widely used in machine learning for comparing model performance or feature impact, the binomial distribution helps in calculating p-values and confidence intervals.
You can also explore an ethical way of A/B testing
Consider an example where we want to determine if a new feature in a recommendation system improves user click-through rates. By modeling the click events as a binomial distribution, we can perform a hypothesis test to evaluate if the observed improvement is statistically significant or just due to random chance.
Generative Models
Generative models such as Naive Bayes leverage binomial distributions to model the probability of observing certain classes given specific features. This is particularly useful when dealing with binary or categorical data.
In text classification tasks, for example, the presence or absence of certain words (features) in a document can be modeled using binomial distributions to predict the document’s category (class).
By understanding the binomial distribution, we can better grasp how these models work under the hood, leading to more effective feature engineering and model tuning.
Monte Carlo Simulations
Monte Carlo simulations, which are used in various machine learning applications for uncertainty estimation and decision-making, often rely on binomial distributions to model and simulate binary events over numerous trials.
These simulations can help in understanding the variability and uncertainty in model predictions, providing a robust framework for decision-making in the presence of randomness.
Practical Applications in Machine Learning
Quality Control in Manufacturing
In manufacturing, maintaining high-quality standards is crucial. Machine learning models are often deployed to predict the likelihood of defects in products.
Here, the binomial distribution is used to model the number of defective items in a batch. By understanding the distribution, we can set appropriate thresholds and confidence intervals to decide when to take corrective actions.
Explore Locust – a tool for quality assurance
Medical Diagnosis
In medical diagnosis, machine learning models assist in predicting the presence or absence of a disease based on patient data. The binomial distribution provides a framework for understanding the probabilities of correct and incorrect diagnoses.
This is critical for evaluating the performance of diagnostic models and ensuring they meet the necessary accuracy and reliability standards.
Fraud Detection
Fraud detection systems in finance and e-commerce rely heavily on binary classification models to distinguish between legitimate and fraudulent transactions. The binomial distribution aids in modeling the occurrence of fraud and helps in setting detection thresholds that balance false positives and false negatives effectively.
Learn how cybersecurity has revolutionized with the use of data science
Customer Churn Prediction
Predicting customer churn is vital for businesses to retain their customer base. Machine learning models predict whether a customer will leave (churn) or stay (retain). The binomial distribution helps in understanding the probabilities of churn events and in setting up retention strategies based on these probabilities.
Why Use Binomial Distribution?
Binomial distribution is a fundamental concept that finds extensive application in machine learning. From binary classification to hypothesis testing and generative models, understanding and leveraging this distribution can significantly enhance the performance and interpretability of machine learning models.
By mastering the binomial distribution, you equip yourself with a powerful tool for tackling a wide range of problems in statistics and machine learning.
Feel free to dive deeper into this topic, experiment with different values, and explore the fascinating world of probability distributions in machine learning!
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... from the right hand of the quotient, point off so many places for decimals, as the decimal places in the dividend exceed those in the divisor.
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## A Course of Mathematics: For the Use of Academies as Well as Private Tuition
Charles Hutton - 1812
...the product. DIVISION OP DECIMALS. , DIVIDE as in whole numbers ; and point off in the quotient as many places for decimals, as the decimal places in the dividend exceed those in the divisor*. • The reason of this Rule is evident ; for, since the divisor multiplied by the quotient gives the... | 1,096 | 4,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-25 | latest | en | 0.737259 |
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# How many weeks until April 14 2021
Quickly work out how many days, how many weeks, or months between two dates. Weeks between 16 November 2020 to 30 April 2021 (45 seconds ago) If you would like reminders for an upcoming date you can create a My Weeks Until account and get reminders 1 week, 1 month, etc before the date How many weeks are between two dates. This simple calculator determines the number of weeks between first date and second date. It was inspired by site searches which did not yield any results: how many weeks, weeks between two dates and how many weeks between two dates :) Now you can find out April 14th 2020 was on a Tuesday and was in week 16 of 2020. How many months ago was April 14th 2020? 15 months. How many weeks ago was April 14th 2020? 65 weeks. How many days ago was April 14th 2020? 457 days. How many hours, minutes and seconds ago? 10,946 hours 656,784 minutes 39,407,055 second Provides the dates for holidays for the calendar year. 2021 Yearly Holiday Calendar. 2020. 2022. 2023. 2024
About Days until Date Calculator. This online date calculator will help you calculate how many days are left until a given date. For example, it's easy to find out how much time is left until your birthday. You can also find out how much time has passed since a given date. For example, How long until June 1, 2022? Select the date you want (e. Did you want to know how many days until a certain date or event? Or how many days have passed since an event or particular date. This calculator can calculate the difference between two dates in days, weeks, months and years
Create a Countdown Timer that counts down in seconds, minutes, hours and days to any date, with time zone support. It also counts up from a past date Please note, that the number of days is always exact, the numbers of months and years may be rounded to the nearest full value. For example, if the period calculated equals to 33 days, the calculator will show approx. 1 month as a result Now keep counting Wednesdays until you reach the present date. LMP = December 2, 2021 Current Date = February 4, 2022. Calculation: How many weeks am I if my due date is April 1, 2022 or 4/01/2022 How many weeks am I if my due date is April 2, 2022 or 4/02/2022 How many weeks am I if my due date i How many days until? People commonly use this calculator to find out the number of days before certain events in their lives are due to happen. Perhaps you're counting down the days until your wedding day, or you're wondering how many days there are until university starts or even how long you've got to go until you retire For whatever reason you need to know how many days until April 14, 2020, we have the answer! How many days until 2020. Looking for how many days until other days in 2020? Go here to see a list of all days in 2020
### How many weeks and days between two dates - weeks unti
How many days are there between two dates? This is simple day calculator which determines number of days between two dates. person_outlineAntonschedule 2011-06-19 21:33:32. This simple calculator determines the number of days between the Date 1 and the Date 2. Days between two dates How many days since Tuesday, April 14, 2020. Agecalculator.Me will help you to find out exactly how many years, months, weeks, days since April 14, 2020 with realtime countdown! How many days since Tuesday, April 14, 2020. 0 years, 2 months , 00 weeks, 05 days, 11 hours 45 minutes and 09 seconds. How many months since
### Online calculator: How many weeks are between two date
Help and Example Use. Some typical uses for the Date Calculators; API Services for Developers. API for Business Date Calculators; Date Calculators. Duration Between Two Dates - Calculates number of days.; Time and Date Duration - Calculate duration, with both date and time included; Date Calculator - Add or subtract days, months, years; Birthday Calculator - Find when you are 1 billion. Week Numbers 2021. Number of weeks in 2021 year is 52 weeks. 25 weeks passed in 2021 year excluding current week #26. 26 weeks left in 2021 year, until next 2022 year starts, excluding current week #26. Weeks are according United States calendar rules, Sunday first day and weeks are Sunday to Saturday. Country: United States Year - Month - Day - Hour - Minute Counter How many days, hours and minutes are there between two dates and times. Time Menu. Full Year Reference Calendar - Day of the Year and Days left till the End of the Year. Add Number of Days, Months and/or Years to a Calendar Date Calculator. Day Counter How many days and/or weeks are there between two. ››Enter dates. Enter two dates below to find the number of days between them. For best results, avoid entering years before 1753. Examples include 2022-04-02 or Nov 18, 1949.You can also type words like today or yesterday, or use the American format, 7/14/2021. Number of days between
Today, wednesday, july 14, 2021 Working day : Week : / 52 Day of the year : 0 / 0 Working day of the year : 0 / 0 Time (UTC -05) : 02:15 p Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. For K-12 kids, teachers and parents Today is 07.02.2021, so the number of days until 14 April 2022 is: 285 days 10 hours 0 minutes 10 seconds. or. 9 months 2 weeks and 1 days. Countdown timer to 14 April. It can automatically count the number of remaining days, months, weeks and hours
### How long ago was April 14th 2020? howlongagogo
1. Full Year Reference Calendar - Day of the Year and Days left till the End of the Year. Countdown Calendar - select start and end dates. Year - Month - Day Counter How many years, months and days are there between two calendar dates. Add Number of Days, Months and/or Years to a Calendar Date Calculator
2. utes, and seconds
3. The internet's best free countdown timer widget. Create a countdown timer for free
4. Find out how many work days until a specified dat
5. ››Enter dates. Enter two dates below to find the number of days between them. For best results, avoid entering years before 1753. Examples include 1983-09-26 or May 7, 1960.You can also type words like today or yesterday, or use the American format, 7/14/2021. Number of days between
6. utes and seconds ago? 10,876 hours 652,591
7. There have been 264 working business days since the 14th of April 2020. There are 30 days in April 2020. This day is only 255 days before Christmas 2020! 4/2020. S
How Many Days Until Calculator. Simply enter the date of a future event below and we will calculate how many days are left until your special event A countdown clock to count the time left to a date of your choice. To create your own countdown, for your own date, with your own look and feel, follow the link at the bottom of this page Find out how many days, months, and/or years until you turn a specific age. How many days until I turn 18 years old? Select your birthday above and find out how many days until you turn xx! Also see: Age Calculator (for age between 2 specifc dates) Days Old Calculato But this would only be starting from December 27th, 2020. After March 14, 2021, new claimants will no longer be permitted to apply for PUA benefits, but eligible individuals who were receiving PUA benefits as of that date will continue to receive benefits until week beginning April 5, 2021 until their claim balance is exhausted Wednesday, 25 Mar 2020 1:26 PM MYT Related News. Analyst Reports 15h ago Political uncertainties, rising Covid cases to weigh on Bursa (MCO) for another two weeks until April 14, says the.
What does this Period Calculator do? It gives you the estimated starting dates for the next 12 periods. All you have to do is to enter the starting date of your last period and the average length of your period cycle. Period cycle is the length of time in days from the start of one period to the start of the next one Working Days until Retirement . This calculator will take into account holidays, vacation days and sabbaticals. Calculate Working Days Until Retire Select One of the Three Retire Date Options At Full Social Security Age On a Given Date. There, after 2 days, I have made as many friends as it is possible in such a short period. It was amazing: about 100 of us, divided into groups, had lectures together about the damage of alcohol and tobacco, discussed everything, shared expressions, experiences, etc Updated 9:38 p.m. ET, April 14, 2020 What you need to know Covid-19 has infected more than 1.9 million people and killed at least 119,000 worldwide, according to Johns Hopkins University Enter a date and find out how long since or until that date it is
### Date Calculato
• April 14, 2020 at 9:29 PM I get a widow's pension every month because my husband was a 100% Service connected Veteran. He put his life on the line for our freedom so people like you can spout.
• April 2020 Calendar. View the month calendar of April 2020 Calendar including week numbers. And see for each day the sunrise and sunset in April 2020 Calendar
• March 19, 2020: Reeves announced that the statewide school closure, originally scheduled to end on March 20, would extend until April 7. March 14, 2020: Reeves called on schools to extend spring break through the week of March 16. Total students affected: 483,150; Missour
• The Gregorian calendar is used for dates on and after October 15, 1582 A.D. and the Julian calendar is used before October 4, 1582. The pop-up calendar has been available since October 15, 1582, when the Gregorian calendar was adopted
• 11 days Flag Day in US Monday, 14th June 2021. 15 days English Premier League 2021/22 Fixtures Announced Thursday, 17th June 2021, 1:00 am. 15 days First Quarter Moon Thursday, 17th June 2021, 8:54 pm. 17 days Father's Day Sunday, 20th June 2021. 18 days Summer Sunday, 20th June 2021, 8:33 pm. 18 days June Solstice Sunday, 20th June 2021, 8:33.
• To calculate the number of weeks between two dates, start by counting the number of days between the start and end date. Then, divide that number by 7 days per week. (Start date - End date) / 7 days per week =. Weeks between two dates (with decimal) Example. (3/15/2021 - 3/2/2021) = 13 days. 13 days / 7 days per week = 1.857 weeks
• Here are the key changes to Unemployment based on the new Coronavirus Response and Relief Bill, signed into law on 12/27/2020. Extra \$300 a week - FPUC for 11 weeks of Unemployment. Extended Unemployment expiration (from Dec 31, 2020 to April 5, 2021 - last payment will be for the week ending 04/11) Added 11 weeks to Unemployment, so.
Today is 06.18.2021, so the number of days until 14 March 2022 is: 268 days 9 hours 35 minutes 33 seconds. or. 8 months 4 weeks and 0 days. Countdown timer to 14 March. It can automatically count the number of remaining days, months, weeks and hours This page lists all weeks in 2021. There are 52 weeks in 2021. All weeks are starting on Monday and ending on Sunday. Please note that there are multiple systems for week numbering, this is the ISO week date standard (ISO-8601), other systems use weeks starting on Sunday (US) or Saturday (Islamic) Browse through the working days Calendar from 2020 in the USA. We use cookies in order to offer you the best user experience, track our audience and display ads. Tuesday, april 14, 2020 Working day •Work hours: 8h (08:00->12:00, 14:00->18:00) Week 16. 32 Select your birthday above to find out how many days old you are. Also find out when you turn 1,000, days old, 5,000, days old, 10,000 days old, etc. Also see: Age Calculator (for age between 2 specifc dates) Half Birthday Calculator How many days are in a year? In 2020 in the UK there are: 366 days; 256 working days; 104 week-end days; 6 holidays. In 2021 there are: 365 days; 255 working days; 104 week-end days; 6 holidays. How many working days per month are there? You will find below the breakdown of working days per month and year in 2020 and 2021
### Days until Date Date Calculator
Originally, PEUC provided 13 additional weeks of benefits after an individual used all 26 weeks of state unemployment insurance benefits. The original PEUC program expired on December 27, 2020. The program has now been extended through March 14, 2021 and increased to provide a total of 24 weeks of additional benefits How many days since That's approximately months or years. Welcome to HowManyDaysSince.net! This site can quickly calculate how many days it has been since any date and today. Enter a month, day, year or any combination. Month. Day. Year. Go. Or use the calendar control Today, thursday, 15 july, 2021 Working day : Week : / 52 Day of the year : 0 / 0 Working day of the year : 0 / 0 Time (UTC -05) : 03:02 a
### Duration Date Calculator How many years until? How many
Knowing How Many Weeks are Left. How many weeks are left is likely to be one of the main things you think about when you're pregnant, and a question you will be asked constantly! Everyone wants to know, how much longer you have left until your due date. Knowing how long there is to go until your due date can help with planning and preparation. How many weeks left in 2021? - There are 27 weeks left in 2021. In 2021, there are a total of 52 weeks, and today is July 04, 2021 which is week number 26 2021 Trading Days Calendar. Month of June: 22 trading days and 0 holidays. Last market closure: Memorial Day (Monday, May 31) Next market closure: Independence Day (observed Monday, July 5) NEWS FLASH! On June 17, 2021, a new federal holiday (Juneteenth, or June 19) was officially declared Last week Whitmer extended the state's stay home, stay safe order through April 30. Michigan has the third-highest number of COVID-19 cases in the country, according to Whitmer. People take part.
April 14, 2020. Share. suppressing it until a vaccine can be produced. With luck, that will take 18 to 24 months. suggests that states should relax their restrictions only after new case. Calculate how many days there are left until a certain date, or how many days ago there was since a certain date. Calculate how many days you have lived. Calculate how many days you have lived and find out if you are close to your 10 000th birthday perhaps! Calculate when you will turn 10 000 days (or similar). How many days until Easter 2022 ? (April 1st, 2018) Until the Catholic Easter 2018 there are -1201 days. How many days until Christmas 2021 ? Days until Christmas: Until Christmas 2021 there are 163 days. How many days until New Year's Eve 2021 ? Number of days until New Year\'s Eve: 169 days How many days passed from 2021? 194 days have passed. There have been 391 working business days since the 1st of January 2020. January (Jan) 2020 There are 31 days in this month. February (Feb) 2020 There are 29 days in this month. March (Mar) 2020 There are 31 days in this month. April (Apr) 2020 There are 30 days in this month. May (May) 2020 There are 31 days in this month Today, wednesday, 14 july, 2021 Working day : Week : / 52 Day of the year : 0 / 0 Working day of the year : 0 / 0 Time (UTC 00) : 21:3
Over the year 2020, there are 366 days, 252 working days, 11 public holidays, 104 weekend days. Calculate the number of working days, work hours and holidays between two dates We use cookies in order to offer you the best user experience, track our audience and display ads Week # From Date To Date Weeks from Now; Week 1: December 29, 2019: January 4, 2020: passed 80 weeks ago: Week 2: January 5, 2020: January 11, 2020: passed 79 weeks ag 30 days from date. =DATE (2018,4,1)+30. To get a more universal days from date formula, enter both values (source date and the numbers of days) in separate cells and reference those cells. With the target date in B3 and the number of days in B4, the formula is as simple as adding up two cells: =B3+B4
### Countdown Timer to Any Dat
1. After March 14, 2021, new PEUC claimants will not be eligible for the extra weeks of benefits, but individuals who had been receiving PEUC benefits as of March 14, 2021 will be eligible to continue to receive benefit payments through April 4, 2021. PUA: As with PEUC, the bill extends PUA benefits until March 14, 2021. After March 14, 2021, new.
2. How Many Days Until Father's Day? - There are 354 days until Father's Day 2022. Father's Day 2021 has already passed. Father's Day countdown calculator to show exactly when is Father's Day 2022. Father's Day is celebrated on the third Sunday of June
3. Veterans Day 2020: 2020-11-11. Thanksgivings Day 2020: 2020-11-26. Christmas Day 2020: 2020-12-25. From Wednesday, January 1, 2020 to Thursday, December 31, 2020 included, there are: 366 days. 252 working days. 104 weekend days. 10 public holidays
4. Sunday, March 15, 2020, 8 p.m. ET/PT: This will be the final episode of auditions and will likely show a montage of hopefuls getting their golden ticket before Hollywood Week begins airing either.
5. Browse through the working days Calendar from april 2020 in the UK and Ireland. We use cookies in order to offer you the best user experience, track our audience and display ads. Read our privacy policy. Tuesday, 14 april, 2020 working day (8h 08:00->12:00, 14:00->18:00) 32
### Days Between Dates - Calendar-12
• We have included how many days until 3 March 2018. You can track how many days left for March 3 second by second with this countdown. March 3 may be your birthday, your wedding anniversary or any other date which you may want to follow.Also, this topic will be updated on every March 3
• Find out how many days since June 13, 2020. Calculate how many days have been there since June 13th 2020 until now
• April 04, 2020 09:20 IST Updated: April 04, 2020 23:48 IST Worldwide, confirmed infections rose past 1 million and deaths topped 58,000, according to a tally by Johns Hopkins Universit
• On the evening of 24 March 2020, the Government of India under Prime Minister Narendra Modi ordered a nationwide lockdown for 21 days, limiting movement of the entire 1.38 billion (138 crore) population of India as a preventive measure against the COVID-19 pandemic in India. It was ordered after a 14-hour voluntary public curfew on 22 March, followed by enforcement of a series of regulations.
### How Far Along Am I? Pregnancy Calculator: How Many Weeks
1. The University of Minnesota pays employees biweekly. Each pay period starts on a Monday and ends on a Sunday. Employees are paid every other Wednesday, 10 days after the end of the pay period. Calendar Year 2021 Pay Date Pay Period January 13, 2021 12/21/20 - 1/3/21 January 27, 2021 1/4/21 - 1/17/21 February 10, 2021 1/18/21 - 1/31/21 February 24, 2021 2/1/21 - 2/14/21 Marc
2. g is odd at times, with the tenth Season being something completely.
3. The online portal to apply for CERB will be available by April 6, and people can expect payments within 10 days (by April 16). CERB payments will be issued every four weeks for a total of 16 weeks, and will be available from March 15, 2020 to Oct. 3, 2020
4. g of Jesus, the war of Armageddon, the arrival on earth of the Antichrist, the Tribulation, the Rapture, some horrendous natural disaster, etc. Jewish, Islamic, psychic, occult and other.
5. The IRS said it's also taking more time for 2020 tax returns that need review, (PDF), injured spouse allocation -- this could take up to 14 weeks to process
6. istration.
### How Many Days Until? Calculato
• In addition to the regular weekly payments, there are 17 potential payments of \$600, for the weeks in April-July 2020. Look for week endings 04/05/2020, 04/12/2020 until 07/26/2020. You should have a \$600 payment for any week that you were eligible for unemployment
• PhilHealth extends deadline of 1st Quarter 2020 payment of contributions on or before April 30, 2020 instead of March 31, 2020 without interest. Pru Life UK Grace period for premium payments with due dates from March 16 until April 30, 2020 will be extended from the standard 31 days to 60 days
• Today's Date - Details about today's date.: Date Calculator - To calculate the number of days between two dates.: Month/Year Calculator - Provides perpetual monthly calendar for any year
• Date Holiday Day Week number Days to go; April 4, 2021: Easter 2021: Sunday: 14-April 17, 2022: Easter 2022: Sunday: 16: 277: April 9, 2023: Easter 2023: Sunday: 15.
• Phina on April 19, 2020: Please my period starts on the 16th of April and ends on the 20th Wich is tomorrow. When should my ovulation begin. I saw my period on 14 of February and on 15 of March but had sex on 20 of March can I be pregnant on April 17, 2020: Thanks. Peace on April 17, 2020
### How many days until April 14, 2020? - researchmaniacs
All non-essential workers were told to stay home completely for two weeks until April 11. Starting on April 13, workers in construction and manufacturing were allowed to return to work, although. After two weeks of wavering on guidelines that put normal American life on hold, President Trump extended until April 30 measures aimed at slowing the spread of the coronavirus. 2020 5:00 AM ET
### Online calculator: How many days are there between two dates
Monday, April 12 until Tuesday, May 11, 2021 Saturday, April 22 until Sunday, May 1, 2022 Wednesday, March 22 until Thursday, April 20, 2023. Eid al-Fitr (dates may vary based on the lunar calendar) Saturday, May 23 until Sunday, May 24, 2020 Wednesday, May 12 until Thursday, May 13, 2021 Monday, May 2 until Tuesday, May 3, 2022 Friday, April. Click here for days in 2021. This page lists all days in 2020 with day and week numbers. The year 2020 has 366 days. This is a leap year July 14, 2021. July 21, 2021. July 24, 2021. **The September SAT test date has been added to the 2020-2021 testing schedule to account for COVID-19 test cancellations. To learn more about how COVID-19 is impacting standardized testing, see out post: How to Navigate SAT/ACT Cancellations Due to the Coronavirus
### How many days since April 14, 2020 AgeCalculator
The federal-state unemployment insurance (UI) system helps many people who have lost their jobs by temporarily replacing part of their wages. (See Policy Basics: Unemployment Insurance.)Under certain circumstances, unemployed workers who exhaust their regular state-funded unemployment benefits before they can find work can receive additional weeks of benefits Those who get PUA also get the \$600/per week additional payment added to UI for weeks they are unemployed from March 29 until the end of July. PUA benefits can cover people unemployed or partially unemployed due to COVID-19 from January 27, 2020 through December 31, 2020 depending on date of actual impact Give your date of birth and Calculate the number of days until your next birthda You can also see on which day the holiday falls and how many days it is until this holiday. Date Holiday Day Week number Days to go; April 2, 2021: Good Friday 2021: Friday: 13-April 15, 2022: Good Friday 2022: April 14, 2028: Good Friday 2028: Friday: 15: 2466: March 30, 2029: Good Friday 2029: Friday: 13: 2816: April 19, 2030: Good Friday.
April 14, 2020 02:14. Share this - Last week, Gov. Pete Ricketts issued an order to close theaters, barbershops, beauty salons, tattoo shops and massage businesses until April 30 Over the year 2020, there are 366 days, 251 working days, 15 public holidays, 104 weekend days. Calculate the number of working days, work hours and holidays between two dates We use cookies in order to offer you the best user experience, track our audience and display ads How many days until [name]'s birthday 1971? Friday, January 01, 1971. 00. days 14 April 2020 was 105th day of the year. There were then 261 days left in 2020. 15th Tuesday of 2020. on the 16th week of 2020 (using ISO standard week number calculation). 26th day of Spring. There were 68 days left till Summer. Birthstone for this day: Diamond and Crystal April is the second month of the spring season in the northern hemisphere. On the other hand, this month is also the second month of the autumn season in the southern hemisphere. April includes 30 days and 720 hours. When we look at the important days of April, the 1st day of the month is a fun day in many countries
About. This page tells you how many days old you are and on which day you were born. The number of leap days tells you how many February 29's you've experienced. For instance, someone born in August 2020 will have 0 leap days because he was born after February 29th. The year 2000 is a leap year. Year numbers divisible by 100 aren't leap years. T he Nepal government decided on Sunday to extend its national lockdown by another week till midnight 7 April, and to continue the ban on all domestic and international flights till 15 April.. The Cabinet made the decision on recommendation of the Coronavirus Control High-level Task Force headed by Defence Minister Ishwar Pokhrel based on the WHO criteria of a safe 14 day incubation period for. Claims for furlough days in April 2021 must be made by 14 May 2021. 1 April 2021. The Coronavirus Job Retention Scheme is being extended until December 2020. 1 November 2020
### Business Days Calculator - Count Workday
Today, wednesday, 14 july, 2021 Working day : Week : / 52 Day of the year : 0 / 0 Working day of the year : 0 / 0 Time (UTC 00) : 06:3 September 6, 2020: Day of the Week. September 6, 2020 was the 250 th day of the year 2020 in the Gregorian calendar. There were 116 days remaining until the end of the year. The day of the week was Sunday. If you are trying to learn Spanish then this day of the week in Spanish is domingo March break (mid-winter break) for the 2020-2021 school year is postponed to the week of April 12, 2021. 2021-2022 to 2024-2025 The following charts are provided for the guidance of school board officials in the preparation of school year calendars
September 14, 2020: Day of the Week. September 14, 2020 was the 258 th day of the year 2020 in the Gregorian calendar. There were 108 days remaining until the end of the year. The day of the week was Monday. If you are trying to learn French then this day of the week in French is lundi Countdown to NFL Kickoff 2021 | Days Until NFL Kickoff 2021. NFL KICKOFF 2021. Countdown to September 9th, 2021. 77 Days 13 Hours 46 Mins 39 Secs. Without football, my life is worth nothing. Once you learn to quit, it becomes a habit. Pressure is something you feel when you don't know what the hell you're doing There have been 370 working business days since the 4th of January 2020. There are 31 days in January 2020. This day is only 356 days before Christmas 2020! 1/2020. S. M. T. W | 6,911 | 26,572 | {"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} | 2.875 | 3 | CC-MAIN-2021-49 | latest | en | 0.970201 |
https://textilecalculator.com/fabric-gsm-to-thickness-calculator/ | 1,723,560,896,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641079807.82/warc/CC-MAIN-20240813141635-20240813171635-00133.warc.gz | 448,612,230 | 53,649 | # Fabric Gsm To Thickness Calculator
## The formula of Fabric Gsm To Thickness Calculator
To create a calculator that converts fabric GSM (grams per square meter) to thickness, you can use the following formula:
Thickness (in millimeters)=GSM/(Fabric density (in grams per cubic centimeter))×10
Fabric density varies depending on the type of fabric, so you’ll need to know the specific density of the fabric you’re working with. You can often find this information from fabric specifications provided by the manufacturer or by conducting tests. Once you have the fabric density, you can use this formula to calculate the thickness in millimeters.
Fabric GSM to Thickness Calculator
## Fabric GSM to Thickness Calculator
In the world of textiles, understanding fabric properties is essential for designers, manufacturers, and enthusiasts alike. One crucial aspect is the fabric’s weight, often measured in GSM (grams per square meter). Another significant factor is its thickness, which influences its durability, warmth, and suitability for various applications. To bridge the gap between these two measurements, we introduce the Fabric GSM to Thickness Calculator, a handy tool that simplifies fabric analysis and selection.
## Understanding GSM and Fabric Thickness:
GSM, or grams per square meter, indicates the weight of a fabric per unit area. It’s a measure of fabric density, with higher GSM values indicating heavier fabrics and lower values indicating lighter ones.
While GSM provides insight into fabric weight, thickness offers a different perspective. Fabric thickness, typically measured in millimeters, reflects the distance between its two surfaces, influencing its texture, insulation properties, and overall feel.
## The Challenge:
Traditionally, converting GSM to thickness involves complex calculations and often requires specialized knowledge. This process can be time-consuming and prone to errors, especially for those without a strong background in textile engineering or mathematics. Additionally, with the growing diversity of fabrics available, ranging from delicate chiffons to heavy denims, a quick and reliable conversion method is more crucial than ever.
## Introducing the Fabric GSM to Thickness Calculator:
To address these challenges, we present the Fabric GSM to Thickness Calculator—a user-friendly online tool designed to streamline the conversion process. This calculator allows users to input the GSM value of a fabric and instantly obtain its corresponding thickness in millimeters. With just a few clicks, designers, manufacturers, and hobbyists can make informed decisions about fabric selection and usage.
## How It Works:
Using the Fabric GSM to Thickness Calculator is simple. Users input the GSM value of their fabric into the designated field and click the “Calculate” button. Behind the scenes, the calculator employs a straightforward formula:
Thickness (in millimeters)=GSMFabric density (in grams per cubic centimeter)×10Thickness (in millimeters)=Fabric density (in grams per cubic centimeter)GSM×10
Once the calculation is complete, the calculator displays the fabric’s thickness, providing users with valuable insights into its physical properties.
## Key Features:
1. Ease of Use: The calculator’s intuitive interface makes it accessible to users of all levels, eliminating the need for complex manual calculations.
2. Instant Results: With real-time calculations, users receive instant feedback, allowing for quick comparisons between different fabrics.
3. Accuracy: The calculator’s formula ensures precise conversions, helping users make informed decisions with confidence.
4. Versatility: Whether working with natural fibers like cotton and wool or synthetic materials such as polyester and nylon, the calculator accommodates a wide range of fabrics.
5. Convenience: Accessible from any device with internet connectivity, the calculator offers convenience and flexibility for users on the go.
## Practical Applications:
The Fabric GSM to Thickness Calculator has numerous practical applications across various industries and interests:
1. Fashion Design: Designers can use the calculator to select fabrics with specific weight and thickness characteristics, ensuring their creations meet desired aesthetic and functional requirements.
2. Textile Manufacturing: Manufacturers can optimize fabric selection and production processes by accurately predicting the thickness of different materials, leading to enhanced product quality and consistency.
3. DIY Projects: Enthusiasts and hobbyists can use the calculator to plan sewing, quilting, and crafting projects, choosing fabrics that match their desired weight and thickness preferences.
## Conclusion:
The Fabric GSM to Thickness Calculator represents a significant advancement in fabric analysis and selection, offering a simple yet powerful solution for converting GSM to thickness. By providing users with instant and accurate measurements, this tool empowers individuals across various industries and interests to make informed decisions with ease. Whether you’re designing a new garment, manufacturing textiles, or embarking on a creative project, the Fabric GSM to Thickness Calculator is your go-to resource for simplifying fabric measurement. Try it today and experience the convenience of seamless fabric analysis at your fingertips. | 969 | 5,377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.871349 |
https://alfageeek.wordpress.com/2014/02/page/2/ | 1,566,580,620,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318894.83/warc/CC-MAIN-20190823150804-20190823172804-00276.warc.gz | 359,445,873 | 16,966 | # Hard Tweets Explained: No Pie for You
This may be my favorite tweet I’ve ever written. If not, it’s definitely up there in the top 314. This is the second post in an infinite series of posts in which I’ll explain my most difficult tweets. I’ll stop making that infinite series joke after this post, I promise.
The first thing you need to know is that there was a tweet meme going around of the form “The opening line of my life story would be…” So this is a play on that joke format.
The next thing you need to know is that I’m writing “pie” but meaning π. You remember π – that is the ratio of the circumference of a circle to its diameter. If you take your waist measurement, and then divide it by π, you get the exact length of the knife you’d need to commit harakiri.
π is an irrational number. That means that when you write it out 3.1415926535 (that’s all I know off the top of my head), it never repeats and it never ends. You might remember 22/7 as an approximation of π, but that is not terribly accurate. You can get more accurate by writing equations with lots and lots of fractions. The most famous of these was discovered by a fellow named Leibniz. It goes like this:
1 – ⅓ + ⅕ – ⅐ + ⅑ …
That actually doesn’t get you π, it gets you π/4. So you multiply that by 4 and you get π.
The sign is flipping back and forth +/- and the denominators (bottoms) of the fractions are all the odd numbers. Another way to write this is:
4 ∑ [(-1)ⁿ/(2n+1)]
That means 4 times the sum ( ∑ or “add this stuff up” ) of all the fractions with odd bottoms and sign-flipping tops. (There is a joke here about a sign-spinner wearing funny pants, but that is left as an exercise for the reader.) Maybe you should just trust me on how that equation gives you those fractions, because it’d take a couple more paragraphs to explain why that works.
So that’s the opening line of your life story if you really like pie – Leibniz’s equation for π.
The tweet then goes on to tell you that you aren’t going to get any pie.
This second bit is a joke about the nature of an infinite series. You can keep adding (and subtracting) smaller and smaller fractions in that equation and you get closer and closer to π. But you never actually get to π. You get really close. But never there.
So you never get to π – you never get any pie.
Homework: Go have some pie. Pair it with a nice wine, and toast Leibniz, who also invented calculus independently of Newton, but didn’t have as good a publicist. Apparently.
# Hard Tweets Explained: 1001 Nights
My twitter friends often ask me to explain some of my more difficult, obscure tweets. I figured since I went to the trouble of setting up this blog, maybe I should do that explaining once, in one place, and then I can just point them here. I’m lazy that way. So this is the first in an infinite series. (That’s a pun. I’ll explain it when I get around to doing a blog post about my infamous π tweet.)
The tweet I embedded up above was inspired by a song. This particular song is a little hard to find. But I managed to locate a sample you can hear on Amazon:
It’s the third track on that page, called, obviously, One Thousand and One Nights. Even though there are only a few seconds available, you’ll get the picture. Go listen to that now. I’ll wait.
Got it? OK. That chord sound you hear in that sample is called a half-diminished 7 chord. An example half-dimishished 7 chord would be C-E♭-G♭-B♭. Half-dimished chords sound like the middle east. Snake charmer music. Actually, technically it’s fully diminished 7 chords that sound like snake charmer music. The difference is that the 7 is flatted twice in a fully diminished chord, so starting at C you’d have C-E♭-G♭-B♭♭ (which is just a silly way of writing A).
Diminished chords are cool, because they have a symmetry to them. Start on any note. Skip two notes, hit the next one, skip two, hit one, and so on. Since there are 12 notes in an octave, you will hit 4 notes before you reach your starting note an octave up. Those 4 you hit are a diminished 7 chord. There are only three such chords, obviously. (That’s obvious, right? I’m never really sure what’s obvious to other people.)
So anyway, you can find a keyboard and plink around any of those 4 notes (C -E♭-G♭-A) and you will sound like a snake charmer.
The half-dimished chord almost sounds like that, except it has a B♭ instead of an A. It’s close to the same feel, but less dissonant, so it sounds more comfortable to our western ears. Now look at that half-dimished chord. C-E♭-G♭-B♭ How about we bump that C up a half-step to D♭ and we stick an A♭ in between the last two notes so there isn’t a big gap. That’s a pretty small change: one finger moved a half a step and we added an extra note: D♭-E♭-G♭-A♭-B♭. You may know this list of notes as “the black keys.” That’s right, with that little change we turned our half-diminished 7 chord into the black keys on the piano! What fun!
OK, so we’ve got the black keys. So what? Well those 5 keys have another name: the pentatonic scale. (Penta means 5, tonic means notes, so it’s the 5-note scale.) “The Five Note Scale” never caught on as a name because it sounds so lame. So they call it pentatonic instead, because that sounds cool. And if you go plink at random on the black keys on a piano, you’ll see that those notes together do indeed sound cool. So pentatonic is definitely a much better name.
You may know those notes by another name: wind chimes. Most wind chimes use the pentatonic scale. So did the theme song to Barney Miller. And the bass line of the most excellent Eddie Harris song “Listen Here.” Wind chimes are not particularly cool, but those other two are seriously cool, and befit the notes used to play them.
Now the pentatonic scale is really cool. But we can make it even cooler. Like my brother once told me, “A raised fourth is hip.” (True story; he says all sort of weird shit I don’t understand.) For this to work, you need to start at the right place, so I’m going to rewrite our pentatonic like this: E♭-G♭-A♭-B♭-D♭-E♭. If you think of an E♭ minor scale, the 4th note would be the A♭ (even though it is third in our pentatonic, because we are too cool to play the second note [F] from that minor scale). So we can raise that to A, and leave everything else: E♭-G♭-A♭-A-B♭-D♭-E♭. Go plink out those notes and you are playing something we call “the blues.”
So a blues scale is just the pentatonic scale with a raised fourth stuck in now and then as a transition. To make it hip. As though cool wasn’t enough.
Alright, that’s all the music theory we need. Now a little literature and geography. The title “One Thousand and One Nights” (also know as “Arabian Nights”) is a collection of folk tales from Persia. If you were putting together one of those scholastic videos where John Lithgow reads books, and he was reading tales from 1001 nights, then you would use diminished 7 chords for the background music (or half-diminished, for the sake of our sensitive western ears).
The Tigris is a river that goes through Baghdad, which is sort of the heart of what used to be called Persia.
The Mississippi is a river that goes through the whole USA, but notably the delta of this river is widely considered the birthplace of the blues.
The Tigris doesn’t join the Mississippi anywhere geographically, of course. That’s a metaphor. When you perch on the precipice of half-diminished and pentatonic, you are balancing between Persian musical structure and Western blues musical structures. Just one finger moving a half a step, and then toss in two extra notes (one cool, and the other hip), and you’re there.
Homework: Go back to that Amazon link and buy that album, so you can listen to that song in its entirety. Eliane Elias is one of my very favorite Jazz pianists of all time. And in that song she deftly balances on that precipice. Sometimes in half-diminished, and sometimes in the blues. It’s really amazing. | 2,022 | 7,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.936109 |
https://answerprime.com/express-the-equilibrium-constant-for-the-following-reaction/ | 1,675,362,375,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00398.warc.gz | 130,411,652 | 20,665 | # Express the equilibrium constant for the following reaction
1) Express the equilibrium constant for the following
reaction. P4O10(s) ⇌ P4(s) + 5
O2(g) 4 10 4 2 a) K = [O2]-5 2 -5 b) K = [O2]5 2 5 c) K = d) K = 2) For the reaction of carbon with carbon dioxide to make carbon
monoxide, the reaction is as follows. Write the equilibrium
constant expression for the Kc. Kc c C(s) + CO2(g) ⇌ 2CO(g) s 2 g g a) Kc = a) Kc c b) Kc = b) Kc c c) Kc = c) Kc c d) Kc = d) Kc c 3) Which of the following statements is FALSE? a) When K is very large, the forward reaction is favored and
essentially goes to completion. b) None of the answers are correct. c) If K is very large it implies that the reaction is very fast
at producing products. d) When K is very small, the reverse reaction is favored and the
forward reaction does not proceed to a great extent. 4)The equilibrium constant for the production of carbon dioxide
from carbon monoxide and oxygen is This means that
the reaction mixture at equilibrium is likely to consist of a) mostly starting materials. b) twice as much starting material as product. c) mostly products. d) twice as much product as starting material. 5) Which chemicals never appear in the K expression? Choose all
that apply. a) gas b) aqueous P4lIO215 [P4010l
Also Read : How to read a burette
Equilibrium constant i.e. Kis ratio of the
product of the molar concentration of products to the product of
the molar concentration of reactants raised to the power of
stoichiometric coefficients with respect to each reagent at
equilibrium stage of the reaction.
NOTE :- While writing the expression of K, the
concentration of pure liquids and
solids are considered as
constant.
So,
1. Ans :- Option (b) i.e.
K = [O2]5 is the correct
2. Ans :- Option (c) i.e.
Kc =
is the correct answer.
3. Ans :- Option (b) i.e.
None of the answers is the correct answer.
4. Ans :- Option (c) i.e.
Mostly products is the correct answer.
Explanation :-
For given reaction, CO (g) + 1/2 O2 (g)
<----------> CO2 (g)
The equilibrium constant Kc can be written as,
rearranging,
it means, the concentration of products is
times the concentration of reactants at equilibrium.
Therefore, reaction mixture at equilibrium is mostly consists of
mostly products.
Hence, Option (c) is the correct answer.
[COR [co21
CO2 ] c=[C0102]1/2-2×10 11 -=
We were unable to transcribe this image
2 x 1011
Also Read : What is a collapsible circuit? | 684 | 2,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-06 | latest | en | 0.922903 |
https://www.codeproject.com/Articles/80641/Visualizing-Complex-Functions?msg=3470867#xx3470867xx | 1,642,503,002,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300810.66/warc/CC-MAIN-20220118092443-20220118122443-00179.warc.gz | 769,418,585 | 36,611 | 15,168,423 members
Articles / Desktop Programming / Windows Forms
Article
Posted 11 May 2010
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# Visualizing Complex Functions
Rate me:
23 May 2010Ms-PL14 min read
A program to produce beautiful and informative images of complex functions.
## Introduction
Complex numbers are a subject simple enough to be taught in high school math, but subtle enough to continue to be investigated through college mathematics and beyond. A complex function is a function that accepts a complex number as its argument and returns a complex number as its value. Complex functions are the bread and butter of complex analysis, which plays an important role in algebra, geometry, number theory, and in a host of practical applications of mathematics such as physics and engineering.
Despite the critical role they play in the mathematical sciences, there is no widely accepted way to "draw" a complex function. Real functions can be easily represented by 2-D plots: the argument goes on the horizontal axis, the function value goes on the vertical axis. The interpretation of such plots has become so natural to us that we "get" a function faster by looking at its plot than by looking at any other representation. Unfortunately, the plot technique does not generalize to complex functions. We need the two dimensions of a page just to represent the function argument. Living in three spatial dimensions doesn't save us; since we need an additional two dimensions to represent the functions' value, we would need to live in four spatial dimensions in order to "plot" a complex function.
A number of poor substitutes for the 4-D plots we need have become common. At Wolfram's pages on the Gamma function, you can peruse separate "3-D perspective" plots of the function's real part, imaginary part, and absolute value, or 2-D "heat maps" of the same. For me at least, these images do more to show how poor these visualization techniques are than to give me much of a feel for the behavior of the Gamma function.
This article outlines a different visualization technique, and implements it in a simple WinForms application we call "Complex Explorer". Our technique produces images of complex functions that are beautiful and rich in information.
The drop-down menu allows you to choose the function to visualize. The Save button allows you to save the currently displayed image to a PNG file. If you click on a point in the image, the Complex Explorer will tell you the argument and function values at that point.
## Complex Explorer's Approach
Our visualization technique uses a form of domain coloring I first saw implemented by Claudio Rocchini.
The first principal of the technique is taken from topographic mapping. Topographic maps show elevations by drawing contour lines, which are lines of equal altitude. If you walk along a contour line, you go neither up nor down. If you turn and move at right angles to a contour line, you are moving up or down, and will eventually reach another contour line. Where contour lines are close together, elevation changes rapidly over a short distance, that is, the slope is steep; where contour lines are widely spaced, the slope is gentle. "3-D Perspective" maps may give a slightly more immediate feel of a landscape, but they unavoidably hide some areas behind others, and it is much easier to read an accurate estimate of the elevation at a particular point from a contour line map than from a perspective map. For this reason, "topo maps" using contour lines have become a standard tool for hikers worldwide. The same principal is used in the isobar and isotherm maps used by meteorologists. With just a little practice, you can very quickly get the "lay of the land" by looking at contour lines.
The second principal of our technique is to use color to encode information. This is a long-standing technique used in many maps; for example, using blue for water and green for forest in geographic maps. We will specify colors using hue-saturation-value (HSV) triplets, rather than the red-green-blue (RGB) triplets most familiar in computer graphics, because the axes of the HSV system correlate more directly with the human-perceived qualities of colors. In the HSV system, as illustrated, hue is specified by the angle along a color wheel, and saturation and value specify the intensity and brightness of the color. Black has zero value and white has zero saturation.
The final principal of our visualization technique is to think of complex function values in terms of magnitude and phase rather than in terms of real and imaginary parts. If you are used to thinking of complex numbers only in terms of real and imaginary parts, now is the time to expand your horizons. Just as different spatial coordinate systems highlight different aspects of the relationships between points, different complex coordinate systems highlight different aspects of complex numbers. For example, as we near a singularity of a complex function, we can't say anything in general about the function's real and imaginary parts x and y, but we can say that its magnitude ρ is increasing. The relationships between a complex number z, its real and imaginary parts x and y, and its magnitude ρ and phase θ, are illustrated in the diagram below.
With contour lines and colors, we have the two degrees of freedom we need to represent a complex function's magnitude and phase. Since the natural measure of a complex function's "height" is its absolute value ρ, we will use contour lines to represent the "landscape" |f(z)|. Since both the phase θ of a complex number and the hues on a color wheel are naturally periodic, we will use hue to represent the phase arg(f(z)). The hues on a color wheel run from red through yellow, green, blue, violet, and back to red. Note: this is the same order as the colors of a rainbow. With this convention, positive real points are red, and negative real points are cyan (light blue), which lies opposite to red on the color wheel. To generate contour lines, we will adjust the saturation and value so that our pixels become white near |f(z)| = 0, |f(z)| = 1, and at exponentially increasing regular values thereafter, and black at regular intervals in between. Thus our images will consist of alternating white and black contour lines indicating magnitude, with fully saturated hues indicating phase in the areas in between.
We will discuss the details of our implementation later. For now, let's fire up Complex Explorer and get to know a few complex functions.
## Examples
#### z
Let's begin with the very simple function that Complex Explorer shows when first started: f(z)=z. Since this function is its argument, by studying it, you can get a feel for how our technique represents a complex number. Since |z| is the distance from the origin, the contour lines are concentric circles centered at the origin: a white dot in the middle where z=0, another white circle at |z|=1, and a third white circle for a yet larger |z|. Since the phase of z is the angle between it and the positive real axis, the hues in this plot are constant along radial lines, but change as we sweep around the complex plane.
#### z2-1
Now let's look at a simple polynomial. Looked along the real axis, this would be a parabola, with zeros at x = ±1, negative between those roots, positive outside of them. And looking at our visualization of its complex generalization, we see exactly that behavior along the real axis: zeros of f(z) at z = ±1, with cyan points (negative real numbers) between them and red points (positive real numbers) beyond them. But we also see much richer behavior in the complex plane.
We see that the small, isolated valleys around each root join at the |f(z)| = 1 white contour line, which has a beautiful "infinity symbol" shape enclosing both roots. We see that for larger |z|, there is a single oval valley, which becomes increasingly circular as |z| increases. We see that f(z) is negative and real (cyan) along the entire imaginary axis. By tracing one orbit around the origin, we see that the phase of f(z) completes two full cycles (red to cyan to red to cyan to red) as z makes one. We see the z -> -z symmetry of the equation embodied in the reflection symmetry of our image around the vertical axis.
#### z3+1
This cubic polynomial has only one real root, at z = -1, but our figure immediately shows that it has three complex roots, as the fundamental theorem of algebra says it must. (The others are z = (1 ± i sqrt(3))/2.) We see that, again, the small valleys of |f(z)| around each root join together to produce one large valley as |z| increases.
The phase structure is more complicated. We see three "channels" of positive real numbers (red) that flow into a circle in the center, whose boundaries are defined by the roots. Areas with other phases are not similarly connected. Instead, the three positive real channels are separated by three disconnected regions in which the phase of f(z) goes through its cycle as we move from one positive real channel to the next.
Notice that, while our previous example had a two-fold visual symmetry, this example has a three-fold visual symmetry. The visual symmetry in the previous example was induced by the algebraic symmetry z -> -z. Can you write down the algebraic symmetry that induces the visual symmetry in this example?
#### 1/z
Lest I be accused of over-selling this visualization technique, let me give an example which illustrates a problem. At right, you see the image of f(z) = 1/z, and you will immediately notice that it bears a striking resemblance to the image of f(z) = z with which we began. Yet the behavior of these functions could hardly be more different: while f(z) = z is zero at z=0 and grows larger as |z| increases, f(z) = 1/z is infinite at z=0 and falls off as |z| increases.
The problem you see here is exactly the same problem that plagues users of "topo" maps: as a collection of unlabeled contour lines, a valley looks the same as a mountain. Without contour line labels, there is no way to know whether the slope is going up or down. So you do have to know a little bit about the "lay of the land" (or read it from the labels) to see what a "topo" map is telling you, and you do have to know a little bit about the function you are visualizing (or deduce it from the functional form), in order to correctly interpret these images.
Notice, by the way, that the colors circulate around our simple pole in the opposite way they do around a simple root. That's because the phase arg(z-1) = -arg(z) is the opposite of the phase of z.
#### exp(z)
Our image of the exponential function may look boring at first glance, but the reason for its surprising simplicity can give us some insight. It appears from our plot that the magnitude (elevation) of ez depends only on the real part of z, while its phase (hue) depends only on the imaginary part of z. How does that happen? By writing z = x + i y, we find f(z) = ex ei y. That is, |f(z)| = ex depends only on x while arg(f(z)) = y mod 2π depends only on y, which is precisely what the image tells us.
#### Γ(z)
Recall that we began by critiquing Wolfram's visualization of the Gamma function. Let us see whether ours, shown above, does any better.
In the right complex plane, we see the saddle point at z ≈ 1.5; contour lines show the function increasing as we move outward from that point to the "east" or "west", decreasing as we move outward from that point to the "north" or "south". In the left half of the complex plane, we see singularities at the integer values 0, -1, -2, etc. Note that the colors circulate each pole in the same sense as in our 1/z example above. From the density of contour lines, we see that the poles nearer the origin are stronger (that is, rise higher faster) than the poles at higher negative integers. On the real axis, the function's sign alternates between positive (red) and negative (cyan) at intervals separated by the poles. Note, while the magnitude |f(z)| falls off as we move away from a pole in any direction, the behavior of the phase is not so uniform: as we move outward from a positive real interval, the function stays positive real, but as we move outward from a negative interval, the function can become either positive imaginary or negative imaginary, depending on the exact direction we move.
All this could be read off our image. And it's pretty to boot!
We have so far explored only six complex functions. Nearly that many again are built in to Complex Explorer, and you can easily add more of your own. Take a look at log(z) or sqrt(z) to see a discontinuity in the phase of a complex function, called a cut. Take a look at Rocchini's example function, whose beautiful visualization inspired me to create Complex Explorer. Take a look at ψ(z) and try to understand how it is related to Γ(z).
## Implementation Details
Complex Explorer is a simple WinForms app. The core logic for producing an image from a complex function simply iterates over each pixel, computes the corresponding z value, computes the corresponding f(z) value, and maps that value to a color.
C#
```public void DrawImage () {
// get the function to evaluate
Function<complex,complex> f =
functions[functionList.SelectedIndex].Function;
Bitmap image = new Bitmap(imageBox.Width, imageBox.Height);
// iterate over all image pixels
for (int x = 0; x < imageBox.Width; x++) {
double re = re_min + x * (re_max - re_min) / imageBox.Width;
for (int y = 0; y < imageBox.Height; y++) {
double im = im_max - y * (im_max - im_min) / imageBox.Height;
// form a complex number based on the pixel value
Complex z = new Complex(re, im);
// compute the value of the current complex function
// for that complex number
Complex fz = f(z);
// don't try to plot non-numeric values (e.g. at poles)
if (Double.IsInfinity(fz.Re) || Double.IsNaN(fz.Re) ||
Double.IsInfinity(fz.Im) || Double.IsNaN(fz.Im)) continue;
// convert the complex function value to a HSV color triplet
ColorTriplet hsv = ColorMap.ComplexToHsv(fz);
// convert the HSV color triplet to an RBG color triplet
ColorTriplet rgb = ColorMap.HsvToRgb(hsv);
int r = (int) Math.Truncate(255.0 * rgb.X);
int g = (int) Math.Truncate(255.0 * rgb.Y);
int b = (int) Math.Truncate(255.0 * rgb.Z);
Color color = Color.FromArgb(r, g, b);
// plot the point
image.SetPixel(x, y, color);
}
}
// put the image in the image box control
imageBox.Image = image;
}```
The mapping of a complex value to an HSV color triplet is the key operation for our visualization technique. Determining the hue is easy; we just get the phase from the `ComplexMath.Arg` function and express it as a fraction of 2π. Determining the saturation and value is harder. We need our algorithm to fade to black or white near the desired contour line values, but keep those regions small enough that the figure does not become "washed out" with large regions devoid of color.
C#
```public static ColorTriplet ComplexToHsv (Complex z) {
// extract a phase 0 <= t < 2 pi
double t = ComplexMath.Arg(z);
while (t < 0.0) t += TwoPI;
while (t >= TwoPI) t -= TwoPI;
// the hue is determined by the phase
double h = t / TwoPI;
// extract a magnitude m >= 0
double m = ComplexMath.Abs(z);
// map the magnitude logrithmicly into the repeating interval 0 < r < 1
// this is essentially where we are between countour lines
double r0 = 0.0;
double r1 = 1.0;
while (m > r1) {
r0 = r1;
r1 = r1 * Math.E;
}
double r = (m - r0) / (r1 - r0);
// this puts contour lines at 0, 1, e, e^2, e^3, ...
// determine saturation and value based on r
// p and q are complementary distances from a countour line
double p = r < 0.5 ? 2.0 * r : 2.0 * (1.0 - r);
double q = 1.0 - p;
// only let p and q go to zero very close to zero;
// otherwise they should stay nearly 1
// this keep the countour lines from getting thick
double p1 = 1 - q * q * q;
double q1 = 1 - p * p * p;
// fix s and v from p1 and q1
double s = 0.4 + 0.6 * p1;
double v = 0.6 + 0.4 * q1;
return (new ColorTriplet() {X = h, Y = s, Z = v} );
}```
This mapping is due to Claudio Rocchini. Notice that the variables `p` and `q` measure the distance from and to the nearest contour lines. A naive approach would be to map those quantities linearly into `s` and `v`, but that produces large regions that are nearly white or nearly black instead of sharply defined contour lines. (Try it for yourself!) Rocchini's trick for avoiding that problem is to use the variables `p1` and `q1` in place of `p` and `q`. `p1` and `q1` are produced from `p` and `q` via a simple function (1 - (1-x)3) that preserves ordering (that is, as the input rises from 0 to 1, the output also rises from 0 to 1), but stays closer to one over a larger ranger (the output is already ~0.9 by the time the input reaches ~0.5). This ensures that our contour lines are thin.
The .NET Framework does not provide direct support for HSV color specification, so we need to translate our HSV triplet to an RGB triplet. That computation is standard, and we will not go over it in detail. You can learn about this and other straightforward aspects of Complex Explorer's logic, such as its representation of complex functions and image save functionality, by studying the code.
Complex Explorer uses the Meta.Numerics library for its advanced complex functions like the Gamma function. As an added bonus, Meta.Numerics defines its own Complex type (just introduced in .NET 4.0) and function delegate (introduced in .NET 3.5), enabling Complex Explorer to be compiled with and work on .NET Framework versions all the way back to 2.0.
There are a lot of ways that Complex Explorer could be expanded. It would be nice to have a plug-in mechanism so that new functions could be added without having to change the program's code. It would be nice to be able to zoom in or out on different regions of the complex plane. Complex Explorer is free software under the MS-PL license, and I encourage anyone interested to take it and run with it, adding these and other cool features.
## About the Author
United States
I am a .NET developer who works daily on enterprise-scale applications using C#, SQL, XML, ASP.NET, and myriad other technologies. My academic background is in physics and economics.
I am the original architect of Sandcastle managed reference documentation engine and of the Meta.Numerics library for scientific computation.
## Comments and Discussions
First PrevNext
Problem Bo Vistisen26-Aug-13 22:14 Bo Vistisen 26-Aug-13 22:14
Re: Problem ShannonZ21-Jan-15 18:57 ShannonZ 21-Jan-15 18:57
Re: Problem everweb11-Jun-15 9:24 everweb 11-Jun-15 9:24
My vote of 5 Sperneder Patrick29-May-12 7:57 Sperneder Patrick 29-May-12 7:57
My vote of 5 Akram El Assas23-May-12 10:26 Akram El Assas 23-May-12 10:26
My vote of 5 Manoj Kumar Choubey26-Mar-12 2:07 Manoj Kumar Choubey 26-Mar-12 2:07
adding new functions Member 834801725-Oct-11 13:12 Member 8348017 25-Oct-11 13:12
My vote of 5 Filip D'haene28-Jun-11 1:27 Filip D'haene 28-Jun-11 1:27
My vote of 5 Jonathan Cardy11-May-11 12:43 Jonathan Cardy 11-May-11 12:43
My vote of 5 dvptUml16-Apr-11 23:23 dvptUml 16-Apr-11 23:23
See Advanced Visual Quantum Mechanics Rebcabin12-Oct-10 5:04 Rebcabin 12-Oct-10 5:04
Uses in Quantum Computing ProtoBytes20-Aug-10 6:46 ProtoBytes 20-Aug-10 6:46
Re: Uses in Quantum Computing dawright1-Sep-10 12:52 dawright 1-Sep-10 12:52
Re: Uses in Quantum Computing ProtoBytes1-Sep-10 14:30 ProtoBytes 1-Sep-10 14:30
Re: Uses in Quantum Computing ProtoBytes1-Sep-10 14:33 ProtoBytes 1-Sep-10 14:33
Had fun with this program today! ProtoBytes3-Jun-10 16:57 ProtoBytes 3-Jun-10 16:57
Nice! Prey Colors.... ProtoBytes2-Jun-10 14:51 ProtoBytes 2-Jun-10 14:51
Visualizing Complex Functions geoyar18-May-10 11:21 geoyar 18-May-10 11:21
Just great! RodrigueRoland18-May-10 10:30 RodrigueRoland 18-May-10 10:30
Small suggestion Philip Jander17-May-10 22:12 Philip Jander 17-May-10 22:12
Awesome David Roh13-May-10 3:22 David Roh 13-May-10 3:22
Just what I needed. You got my five.
Brilliant, Beautiful Som Shekhar12-May-10 23:08 Som Shekhar 12-May-10 23:08
Beautiful Stuff!!! Alan Beasley12-May-10 23:02 Alan Beasley 12-May-10 23:02
Which image is mirrored? Tefik Becirovic12-May-10 0:46 Tefik Becirovic 12-May-10 0:46
Re: Which image is mirrored? dawright12-May-10 8:19 dawright 12-May-10 8:19
Last Visit: 31-Dec-99 19:00 Last Update: 18-Jan-22 0:50 Refresh 12 Next ᐅ
General News Suggestion Question Bug Answer Joke Praise Rant Admin
Use Ctrl+Left/Right to switch messages, Ctrl+Up/Down to switch threads, Ctrl+Shift+Left/Right to switch pages. | 5,070 | 20,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-05 | latest | en | 0.921832 |
https://seshat.ch/home2/poseidon.htm | 1,695,732,749,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510208.72/warc/CC-MAIN-20230926111439-20230926141439-00784.warc.gz | 555,176,693 | 3,545 | Poseidon from Cape Artemision – The geometry of a Greek Masterwork / © 1980-2002 by Franz Gnaedinger, Zurich, fg(a)seshat.ch, fgn(a)bluemail.ch / www.seshat.ch
POSEIDON FROM CAPE ARTEMISION The Geometry of a Greek Masterwork
Poseidon, standing upright on his left foot, arms raised, balances his body weight with his right leg and sights over his left hand, aiming a (now missing) trident at a far away target we can only guess at. The figure of this Greek god displays such composure, creates a wonderfully balanced effect, majestically at rest upon itself. How did the unknown master achieve this effect? First, he applied the famous contraposto: Poseidon extends his left arm and his right leg simultaneously, while bending his right arm and his left leg. Second, he used a principle I call self-reference: stretch out the left arm and you duplicate the line of the right upper arm; extend the right leg and you reach the left shoulder via the navel; prolong the left upper thigh and you reach the right shoulder via the navel again; extend the left lower thigh and you reach the left shoulder. Third, the figure is held together by a geometry which my drawings render visible:
A large circle around the navel and touching the tip of the middle finger of the left hand seizes the right hand and the heel of the right foot while resting on the ground. The height of the navel, and therefore the circle's radius, measure 18 large or 72 small units, and the circle's diameter measures 36 large or 144 small units. Draw an arc around the nadir of the circle. Its radius will measure 30 large or 120 small units. The arc will brush Poseidon’s head and cross the circle in the point marked by the tip of the middle finger of the left hand (aiming finger tip). The height of this point is that of the horizontal arm line and measures 25 large or 100 small units. The height of the figure measures 30 large or 120 small units. The distance between the navel and the height of the arm line measures 7 large or 28 small units. The aiming fingertip marks the upper right-hand corner of a pair of right triangles that are correlated by the equation 30x30 minus 25x25 equals 18x18 minus 7x7 equals 275. The golden section of the circle's vertical diameter (144 small units) is marked by the figure’s nipples (height 89 small units) and the glans (height 55 small units). How long was the former trident in Poseidon's right hand? Take a rod (or paper roll) and slide it through the opening of the hand. The rod (or paper roll) will show towards the aiming fingertip. Fix the rod (or paper roll) in such a position that the hand holds it roughly in the middle. If the rod (or paper roll) is 15 large units long, the distances from the fingertip to the fore and hind ends of the rod (or paper roll) will again be in the golden ratio: about 25 and 40 large units.
The wonderful bronze statue was found on the seabed by Cape Artemision at the northern end of the Ionic island of Euboea and is kept in the National Museum of Athens (no. 15161). The height of the figure measures 209 centimeters, corresponding to an Ionic orgyia (fathom) that measured 208.98 cm. 5 large units equal one Ionic pous (foot) measuring 34.83 cm. 30 small units equal one Ionic paechys (cubit) measuring 52.245 cm.15 large or 60 small units equal two Ionic paechei. 30 large or 120 small units equal six podoi or four paechei or one orgyia.
Large drawings from 2005:
Available on request: geometrical examinations of the Kouros from Tenea, of the Gettys Kouros and of the Kroisos from Anavyssos. | 864 | 3,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-40 | latest | en | 0.904325 |
http://www.jiskha.com/display.cgi?id=1168539445 | 1,454,935,872,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701153323.32/warc/CC-MAIN-20160205193913-00163-ip-10-236-182-209.ec2.internal.warc.gz | 480,341,243 | 3,657 | Monday
February 8, 2016
Homework Help: math correction
Posted by jasmine20 on Thursday, January 11, 2007 at 1:17pm.
an arithmatic student needs an average of 70 or more to receive credit for the course. She scored 68,69, and 83 on the first three exams. Write an inequality representing the score she must get on the last test to receive credit for the course.
x >= 60
x is greater than or equal to 60
Fill in the blank with < or >. 4_____-5
my answer is: 4 > -5
Both are ok. | 139 | 483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | longest | en | 0.961039 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/rmion/an-elephant-named-joseph-52op | 1,695,640,199,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508959.20/warc/CC-MAIN-20230925083430-20230925113430-00602.warc.gz | 517,315,119 | 47,600 | Robert Mion
Posted on
# An Elephant Named Joseph
## Part 1
1. Going in circles, again!
2. This is making my head spin!
3. Writing an algorithm that works on my examples
4. A pattern emerges!
5. Simulating the pattern with an algorithm
### Going in circles, again!
• The example is wonderfully explanatory
• My puzzle input is an integer over 3M
• I want to explore more example circle sizes before writing an algorithm
• Because I anticipate a pattern emerging
### This is making my head spin!
Good news:
• I now have several unit tests
• I feel comfortable writing a brute-force algorithm that will definitely finish quickly on circles with up to 100K Elves probably
• I don't think I found a faithful pattern for identifying the winning Elf based on the number of Elves in the circle
• I'm not sure the brute-force algorithm I intend to write will work for circles with millions of Elves
### Writing an algorithm that works on my examples
``````For i from 1 to 20, inclusive
Create an array called circle of length i where each value is 1
Set pointer to 0
Set next to 0
Do as long as circle does not include i
Increment next, wrapping when appropriate
Continue incrementing next until the value in circle at next is not 0
Update the value in circle at pointer to the sum of the current value in circle at pointer and the current value in circle at next
Set the value in circle at next to 0
Increment next until the value in circle at next is not 0
Set pointer to the value stored in next
Log the number one greater than the location of i in circle
``````
• I spent quite some time fiddling with this algorithm
• I was either forgetting to update `next` or incorrectly updating it
• For a while, I was logging `circle` after each iteration of the inner `while` loop to confirm whether it was working as intended
• Thankfully, I eventually worked out all the kinks and got it to print the values I was expecting
### A pattern emerges!
I ran the loop above for Elven circles up to 99 in size.
This animation shows circles up to 26, which is enough to demonstrate the pattern:
• I was almost certain there would be a pattern!
• Unsurprisingly, I missed it by skipping 6-9, 11-14, and 16-19 in my initial GIF
• Now that I see and understand it, I need to re-create it with a different algorithm!
### Simulating the pattern with an algorithm
``````Set circle to 0
Set multiple to 1
Do as long as circle is less than the puzzle input
Set winner to -1
For i from 1 up to and including multiple
Increment winner by 2
Increment circle by 1
If circle is the puzzle input
Double multiple
``````
• This also took some fiddling to get right
• At first I had two `for` loops, then realized I didn't need both
• The outer loop condition causes this algorithm to overextend itself a bit, unfortunately
• And I'm not sure how to achieve the same result without starting `winner` at a value two less than my first desired value of `1`, since I increment it by `2` each time
Regardless, this algorithm terminates near-instantly!
And it generates the correct answer for every example and my puzzle input!
That's wonderful, especially since my brute-force algorithm outlined earlier seemingly never finishes when run on my 3M+ puzzle input number!
## Part 2
1. Ugh, more example GIFs
2. Writing the algorithm that I animated
3. A pattern emerges again!
4. Simulating the pattern with a similar algorithm
### Ugh, more example GIFs
• Seeing a 5-Elf circle demonstrated is nice
• But I need to find another pattern
• So, I need to play out at least circle sizes 1-10
It took a long time, but it helped prepare me to write the algorithm...I hope:
Sadly, `10` circles isn't enough to reveal a pattern:
### Writing the algorithm that I animated
``````For i from 1 to 10, inclusive
Create an array called circle of length i where each value is one greater than its index
Set pointer to 0
Do until circle contains one item
Set removed as the remainder resulting from:
The sum of pointer and the length of circle divided by two, rounded down to the nearest whole number
Divided by the length of circle
Delete the item at the location stored in removed
Update pointer based on the value of removed
If removed is less than pointer
Set pointer to the remainder after dividing pointer by the length of circle
Else
Set pointer to the remainder after dividing one greater than pointer by the length of circle
Log the single value remaining in circle
``````
• I had the most trouble fiddling with updating `pointer` correctly
• Thankfully, this algorithm generated the same winners as in my animation, so I could trust my work and the computer's!
• Next, I ran it on numbers up to 100
• I was delighted by what I saw!
### A pattern emerges again!
I ran the loop above for Elven circles up to 99 in size.
This animation shows circles up to 28, which is enough to demonstrate the pattern:
• I was almost certain there would be a similar pattern!
• Unsurprisingly, I couldn't see it clearly with just 10 circles
• Now that I see and understand it, I need to re-create it with a similar algorithm!
### Simulating the pattern with a similar algorithm
``````Set circle to 1
Set multiple to 3
Do as long as circle is less than the puzzle input
Set winner to 0
Set counter to 1
Do as long as counter is less than or equal to multiple
If counter is less than or equal to multiple divided by 3
Increment counter by 1
Increment winner by 1
Else
Increment counter by 2
Increment winner by 2
Increment circle by 1
If circle is the puzzle input
Triple multiple
``````
• This also took some fiddling to get right
• I had a lot of trial and error getting counter to increment the correct amount based on the condition
• And I still don't know how to start this algorithm from a circle of size 1 - my divide by 3 rule makes that impossible
Regardless, this algorithm terminates near-instantly, too!
And it generates the correct answer for every example and my puzzle input!
## I did it!!
• I solved both parts!
• Through a journey of animation, brute-force algorithm as pattern-identification, then number-processing algorithm for a near-instant answer!
• I made so many slides as part of several GIFs!
• And I'm glad I did, as it helped me understand the many algorithms I would have to write!
This may be my favorite puzzle of all of Advent of Code thus far!
With the exception of building a simulator - which I can't reliably use to visualize a circle with over 3 million Elves - I got to use all of my design and developer skills to solve it...and I succeeded!
Who knew circular data structures could be this fun? | 1,545 | 6,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.921285 |
https://java.tutorialink.com/split-a-list-into-n-sublists-in-all-possible-ways/ | 1,701,530,017,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00447.warc.gz | 363,707,516 | 10,930 | # Split a list into n sublists in all possible ways
I want to split a list into a given number n sublists in all possible ways in Java.
For example ` [1, 2, 3, 4]` where n = 3 would include the following lists (but would not be a complete solution – complete would require much more space):
```([], [], [1,2,3,4])
([],[1],[2,3,4])
([],[1,2],[3,4])
([],[1,2,3],[4])
([],[1,2,3,4],[])
([1],[2,3,4], [])
([1],[2,3],[4])
([2,3],[4],[1])
([4],[],[1,2,3])
...
```
etc
I adapted a solution from another similar question (Split a list into two sublists in all possible ways) however it only works for creating lists of 2 sublists and I am struggling to grasp how to implement it for a flexible rather than hardcoded number of sublists.
Here is my code:
```public List<List<EGroup>> permutation(List<E> list) {
List<List<E>> sublists = new ArrayList<List<E>>();
for (int i = 0; i <= list.size(); i++) {
permutationSplit(list, sublists, i, new ArrayList<E>(), 0);
}
List<List<EGroup>> listOfEGroupPairs = new ArrayList<List<EGroup>>();
for (List<E> subList : sublists) {
List<E> listCopy = new ArrayList<E>(list);
listCopy.removeAll(subList);
EGroup e1 = new EGroup(subList);
EGroup e2 = new EGroup(listCopy);
List<EGroup> egr = new ArrayList<EGroup>();
}
return listOfEGroupPairs;
}
public void permutationSplit(List<E> list, List<List<E>> subLists, int sublistSize, List<E> currentSubList,
int startIndex) {
if (sublistSize == 0) {
} else {
sublistSize--;
for (int i = startIndex; i < list.size(); i++) {
List<E> newSubList = new ArrayList<E>(currentSubList);
permutationSplit(list, subLists, sublistSize, newSubList, i + 1);
}
}
}
```
I need to create n number of EGroup objects to add to `listOfEGroupPairs` rather than the hardcoded 2, but how to always get the right number (n) of sublists of varied size each loop?
If I understand the question correctly, each element of the original list can end up in any of the `n` sublists. That means, there are `n^s` possible sublists (`s` being the number of elements in the original list), which can be enumerated in a simple loop. With a bit of modulo and integer division you can then get the proper “bucket” for each element and prepare the results accordingly.
```public <T> List<List<List<T>>> partition(List<T> lst, int n) {
var result = new ArrayList<List<List<T>>>();
// k = SUM ( pos of lst[i] * n^i )
for (int k = 0; k < Math.pow(n, lst.size()); k++) {
// initialize result
List<List<T>> res = IntStream.range(0, n)
.mapToObj(i -> new ArrayList<T>())
.collect(Collectors.toList());
// distribute elements to sub-lists
int k2 = k;
for (int i = 0; i < lst.size(); i++) { | 743 | 2,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.666447 |
https://www.tiwariacademy.com/ncert-solutions/class-10/maths/chapter-9/exercise-9-1/two-poles-of-equal-heights-are-standing-opposite-each-other-on-either-side-of-the-road-which-is-80-m-wide-from-a-point-between-them-on-the-road-the-angles-of-elevation-of-the-top-of-the-poles-are-6/ | 1,721,241,217,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00882.warc.gz | 899,936,404 | 39,785 | To find the height of the poles and the distances from the point to each pole, we use trigonometry. Let’s denote the height of the poles as h and the distances from the point to the poles as d1 and d2.
For the Pole with a 60° Angle of Elevation: Using tan(60°) = √3, the equation is √3 = h/d1.
For the Pole with a 30° Angle of Elevation: Using tan(30°) = 1/√3, the equation is 1/√3 = h/d2.
Since the total width of the road is 80 m, d1 + d2 = 80 m. Solving these equations simultaneously, we find h = 40√3 m approximately 69.28 m), d1 = 40, and d2 = 40 m. Therefore, the height of the poles is about 69.28 meters, and the distances from the point to each pole are both 40 meters.
Let’s discuss in detail
## Trigonometric Applications in Urban Planning
Trigonometry, an essential branch of mathematics, is often applied in urban planning and design, particularly in situations involving measurements and spatial relationships. A common application is determining the heights of structures and distances between them when direct measurement is not feasible. The problem at hand involves two poles of equal height on opposite sides of an 80-meter-wide road. From a point on the road, the angles of elevation to the tops of these poles are given. This scenario is a practical demonstration of how trigonometry can be used to solve real-world problems in urban and civil engineering.
### Understanding the Problem: Two Poles and a Road
The problem presents two poles of equal height standing on either side of a road that is 80 meters wide. From a point on the road, the angles of elevation to the top of the poles are 60° and 30°, respectively. The objective is to determine the height of the poles and the distances from the point to each pole. This setup forms two right-angled triangles, one with each pole, sharing a common horizontal distance (the width of the road).
#### The Role of Tangent in Angle of Elevation
In trigonometry, the tangent of an angle in a right-angled triangle is the ratio of the length of the opposite side (height of the pole) to the length of the adjacent side (distance from the point to the pole). By applying the tangent function to the given angles of elevation, we can calculate the heights of the poles and the distances from the point to each pole.
Calculating the Distances to the Poles
First, we use the tangent function for the 60° angle of elevation to find the distance to one pole. The equation is tan(60°) = √3 = h/d1, where d1 is the distance to the first pole. Similarly, for the 30° angle of elevation to the other pole, the equation is tan(30°) = 1/√3 = h/d2, where d2 is the distance to the second pole.
##### Solving for the Height and Distances
Since the total width of the road is 80 meters, we have d1 + d2 = 80 meters. Solving the equations √3 = h/d1 and 1/√3 = h/d2 simultaneously with d1 + d2 = 80 meters, we find the height of the poles and the distances from the point to each pole.
###### Practical Implications of Trigonometry in Measurement
The solution reveals that the height of the poles is approximately 69.28 meters, and the distances from the point to each pole are both 40 meters. This example illustrates the practical application of trigonometry in urban and civil engineering, demonstrating how it can provide accurate measurements in complex scenarios. Trigonometry proves to be an invaluable tool in urban planning, offering a mathematical approach to spatial problem-solving where direct measurement is not possible.
Discuss this question in detail or visit to Class 10 Maths Chapter 9 for all questions.
Questions of 10th Maths Exercise 9.1 in Detail
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point. | 1,832 | 7,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-30 | latest | en | 0.91273 |
https://iyara.co.th/kghunr3/8c00aa-exponential-smoothing-vs-moving-average | 1,653,361,643,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00011.warc.gz | 378,332,383 | 20,165 | # exponential smoothing vs moving average
Pick a calculation period—such as 10, 20, 50, 100, or 200—that highlights the trend, but when the price moves through it tends to show a reversal. Recent price action has a greater impact on EMA values which therefore react more quickly to price changes than SMAs. Since they’re based on past prices, there is some expected lag. They are not intended to predict price movements but rather outline the current direction. The weights can be uniform (this is a moving average), or following an exponential decay — this means giving more weight to recent observations and less weight to old observations. Give the terms a look. The most commonly used SMAs are 20, 50, and 200 day SMA. Exponential smoothing is generally used for the analysis of time-series data. Sales for the first and fourth quarter are larger than other quarters. It is similar to a simple moving average that measures trends over a period of time. As a first step in moving beyond mean models, random walk models, and linear trend models, nonseasonal patterns and trends can be extrapolated using a moving-average or … An exponential moving average (EMA) places a higher weight on recent data than older data. Exponential smoothing uses a weighted average of past data as the basis for a forecast. As lagging indicators, moving averages serve well as support and resistance lines. Traders use moving averages (MA) to pinpoint trade areas, to identify trends, and to analyze markets. That is the main difference between the two. The most common way to use moving averages is to look for moving average crossovers and this technique has been used by many successful trend followers. Two of the most common types are simple and exponential. Let’s analyse these two following types of moving averages: Weighted Moving Average vs Exponential Moving Average (Also known as WMA and EMA) . MOVING AVERAGES AND EXPONENTIAL SMOOTHING Introduction This chapter introduces models However, some time series seem to prefer the simple average smoothing than the exponential smoothing. The exponential moving average (EMA) provides more weight to the most recent prices in an attempt to better reflect new market data. Exponential Smoothing Vs Weighted Moving Average SMAs can be quite simply computed using the formula below for an m-day SMA. Whereas in the simple moving average the past observations are weighted equally, exponential functions are used to assign exponentially decreasing weights over time. Example:Quarterly sales of saws for Acme tool company The plot of the Acme data shows that there might be trending in the data … The EMA works by weighting the difference between the current period's price and the previous EMA and adding the result to the previous EMA. We can calculate the moving average for one day, in another example we look at how different weights impact the data and in the third example, we look at the volatility of data using moving average for three and seven years and exponential moving average assigning different weights. The simplest form of an exponential smoothing formula is given by: st = αxt+(1 – α)st-1= st-1+ α(xt – st-1) Here, st = smoothed statistic, it is the simple weighted average of current observation xt st-1= previous smoothed statistic α = smoothing factor of data; 0 < α < 1 t = time period If the value of smoothing factor is larger, then the level of smoothing will reduce.Value of α close to 1 has less of a smoothing effect and give greater weight to recent changes in the data, while value of α closer to zero has greater smoothin… Indicator Description Calculating formula Description of calculating formula ; Adaptive Moving Average (AMA) MA with low sensitivity to noises. So which one is better? By adjusting the degree of smoothing (i.e., the "width" of the moving average), we can hope to strike some kind of optimal balance between the performance of the mean and random walk models. For any investor who wishes to quickly grasp the market trends, it will be much better to use EMA than an SMA value. Whereas in Moving Averages the past observations are weighted equally, Exponential Smoothing assigns exponentially decreasing weights as the observation get older. Moving average crossovers. This type of weighted moving average filter is easy to construct and does not require a large window size. I am able to calculate a simple moving average with the below formula. I am trying to get an Exponential moving average for lengths 8,13,21,55 for each stock. The offers that appear in this table are from partnerships from which Investopedia receives compensation. In other words, the formula gives recent prices more weight than past prices. When to Use SMA vs. EMA. SMA and EMA are calculated differently. The value (1- α) is called the damping factor. Moving averages are fundamental to many technical analysis strategies, but successful traders use a combination of techniques. Click OK. 8. An exponential moving average (EMA) is a type of moving average that places a greater weight and significance on the most recent data points. Price: EMA (m=5) Day one: 102.4: 102.4: Day two: 103.6: 102.8: Day three: 103.5: 103.0: Day four: … Unfortunately, EMAs are not always as consistent as SMAs because of the way the EMA is calculated. Plot an EMA and SMA of the same length on a chart and see which one helps you make better trading decisions. The exponential moving average (EMA) is a weighted average of the last n prices, where the weighting decreases exponentially with each previous price/period. Many shorter-term traders use EMAs because they want to be alerted as soon as the price is moving the other way. The Guppy Multiple Moving Average (GMMA) identifies changing trends by combining two sets of moving averages (MA) with multiple time periods. Simple vs. Exponential Moving Averages: An Overview, Moving Average Convergence Divergence (MACD) Definition, Guppy Multiple Moving Average - GMMA Definition and Uses, Double Exponential Moving Average (DEMA) Definition and Calculation. Investopedia's Technical Analysis Course will show you how to identify patterns, signals, and technical indicators that drive the behavior of stock prices with over five hours of on-demand video, exercises, and interactive content. 21 and 34 Exponential Moving Average Bounce Forex Trading Strategy is a combination of Metatrader 4 (MT4) indicator(s) and template. The results for a centered 5-point moving average are listed on the SHAZAM output in the column MOVING-AVE (see Newbold [1995, Table 17.12, p. 698]). 6. The combination of a simple moving average and the exponential moving average is called a smoothed moving average. In this case, a trader may watch for the price to move through the MA to signal an opportunity or danger. Question 1: Workers should be given more control over the inspection for their own work. We will look at the differences between these two moving averages, helping traders determine which one to use. As explained above, EMA as with simple moving averages (SMAs) are popular technical analysis tools. If compared with the rest of moving averages this indicator has a minimal lag when determining trend reversals and change. In this tutorial, you will discover the exponential smoothing method for univariate time series forecasting. Double Exponential … In the above SHAZAM commands, the MAVE= option on the SMOOTH command is used to … If prices break below the MA in an upward trend, the upward trend may be waning, or at least the market may be consolidating. A moving average is a technical analysis indicator that helps smooth out price action by filtering out the “noise” from random price fluctuations. For this guideline to be of use, the moving average should have provided insights into trends and trend changes in the past. The International Monetary Fund November 29, 2020. Moving averages are favored tools of active traders to measure momentum. 6 Points: Using the Excel Insert Tab Line Graph and follow-up Chart Tools Tabs, construct an appropriate line chart displaying the time pattern of order deliveries. A simple moving average (SMA) calculates the average of a selected range of prices, usually closing prices, by the number of periods in that range. Understanding of exponential smoothing should be a lot easier if you have clear concept of moving average and weighted moving average. 7. Tags . A significant down move is already underway. Step 4: Damping factor will smooth out the graph, and values should be between 0 to 1.Technically it is 1 – α (alpha).I have mentioned 0.3 as the damping factor. During an uptrend, the price will often pull back to the MA area and then bounce off it. Moving Average Convergence Divergence (MACD) is defined as a trend-following momentum indicator that shows the relationship between two moving averages of a security's price. Moving averages are averages that are updated as new information is received. Whereas in the Weighted Moving … If you ask for a 20‐period SMA, you can be certain that the result will be identical to every other 20‐period SMA for the same stock during the same time period (assuming there are no errors in the price data). Since EMAs place a higher weighting on recent data than on older data, they are more reactive to the latest price changes than SMAs are, which makes the results from EMAs more timely and explains why the EMA is the preferred average among many traders. Step 3: For the Input Range, specify the available data points.Our data range is B1:B11. The value of SMMA is approximately equal to the EMA value, with just the period as double of that of EMA. This smoothing technique allows analysts to reduce volatility in a series of data. Generally speaking, EMAs are used to track shorter-term market moves and SMAs are used to monitor longer-term trends. (Marks 2) Explain the aggregate planning strategy? Categories . This highlights one of the weaknesses of moving averages. The exponential moving average (EMA) provides more weight to the most recent prices in an attempt to better reflect new market data. Because of its unique calculation, EMA will follow prices more closely than a corresponding SMA. )Component form representations of exponential smoothing methods comprise a forecast equation and a smoothing … Old prices are never removed from the calculation, but they have only a minimal impact on the Moving Average due to a low assigned weight. An often-used technique in industry is "smoothing". Sales for the first and fourth quarter are larger than other quarters. Shorter-term traders may watch an 8- and 20-period MA, for example. , EMAs are not intended to predict price movements 50, and this continues! Or danger by taking an average of price data will impact the moving average ( EMA ) simple..., EMA will follow prices more closely than a corresponding SMA Vs weighted moving average an! 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Price of a moving average in a series of data window size the differences between these two moving.., for a total of 12 MAs in the series weighted forms ( below. Name, looks more smoothed gleitenden Durchschnitts time-series data is a powerful forecasting method that may be used an! Explain the aggregate planning strategy cookies to provide you with a great user experience out MAs... 3 – exponential moving average ( EMA ) ist eine Erweiterung des gewichteten gleitenden Durchschnitts prices break above moving! Such, this kind of averaging won ’ t work well if there no. Ma ) to help forecast demand for deliveries trend in the indicator your! Is to transform the accumulated history data and trading signals know how much a supplier... T work well if there is a trend in the Input Range box exponential smoothing vs moving average select the Range:... Newest price data, exponential functions are used to assign exponentially decreasing weights over time Range B2:.! Der exponential moving average assigns an equal weighting to all values the other way should a... Current EMA = ( Close - previous EMA some time series seem to the. Examples of a moving average, add the closing prices over a given period of time priority order,! K is defined as the exponential smoothing vs moving average formula places a higher weighting to recent in! And dropping the 10th day, and cumulative, or weighted forms described... Ema gives more weight applied to the most important factor is the sensitivity each one to. ), add the closing prices over a period of time double exponential Step! Instrument over a given period of time trend in the column EXP-MOV-AVE ( see Newbold [ 1995, 17.16! Rather outline the current direction trending moves but tend exponential smoothing vs moving average do poorly when the price is the! Average gives a higher weighting to recent prices more closely than a corresponding SMA series data! Not intended to predict price movements but rather outline the current data many shorter-term traders use moving are... By taking an average smoothing than the exponential moving average filter of the same length on a chart time. Is recalculated by adding the new day and dropping the 10th day, to! 1995, table 17.16, p. 710 ] ) trend by smoothing out price movements the current direction variations:... And type 0.9 the older observations changes and the weighted moving average can smooth training data exponential... Exponentially decreasing weights as the price will often pull back to the Gaussian expansion is. Break above a moving average and weighted moving average and the weighted moving methods. Results from exponential smoothing over the moving average methods and exponential smoothing ) to pinpoint trade,... Analysis of time-series data quicker to react to price movement to as the below formula order 1 2... Many technical analysis tools smoothing ) to help forecast demand for deliveries vs. EMA the primary difference between an and! New information is received break above a moving average the number of other indicators... Option on the indicator in your charting platform average should have provided insights into trends and trend changes the! Forecasting is concerned control over the inspection for their own work viSe versa longer-term.! And dropping the 10th day, and this process continues indefinitely type of finite impulse response filter same on... Have less smoothing value of between 0 and 1, 2, 3, 4 also referred to the! Of average a trader can use to guarantee success below for an m-day SMA the new day and the! In exponential smoothing takes into account the more recent values and therefore k 1/3! Reversals and change investor who wishes to quickly grasp the market trends ; it can help the investor to respective..., 2020 current ) - previous EMA data and trading signals impact the moving average exponential smoothing vs moving average a of... To predict price movements but rather outline the current direction SMA of the same length on a and... Older data as explained above, EMA as with simple moving average can smooth training data, EMA! Uptrend, the slower it is a trend in the past to.. Lot easier if you have clear concept of moving average and weighted moving average should have provided into! ( m+1 ) ) X multiplier ) + previous EMA the smooth command is used to monitor trends... Ema quicker exponential smoothing vs moving average react to price changes and the SMA react slower average price! Is used to exponential smoothing vs moving average when to use Input Range, specify the available data points.Our Range. 1995, table 17.16, p. 710 ] ) ) places a higher weighting to all values more... Value, with older price data making it easier to see what a... Is no one type of weighted moving average should have provided insights into trends and exponential smoothing vs moving average... Sma of the way the EMA quicker to react to price changes and SMA... Quickly to price movement provides more weight than past prices, there are different types of moving averages and smoothing! To see what direction a market is heading more quickly to price movement various. Are used to monitor longer-term trends as a result, forecasts aren ’ t work well when the will! The below the Range B2: M2 Linear weighted moving average in Excel indicator Description Calculating formula Description of formula! Make better trading decisions ( SMA ), add the closing prices of the way EMA... Smas are used to assign exponentially decreasing order simple analysis is, if the trend up. Average, add the closing prices of the way the EMA value is susceptible to market trends it. Fakeouts and give errant signals calculate a simple moving average and weighted moving average Excel. | 4,184 | 20,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.920317 |
https://www.physicsforums.com/threads/higher-partial-derivatives-chain-rule.196802/ | 1,480,781,165,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540975.18/warc/CC-MAIN-20161202170900-00184-ip-10-31-129-80.ec2.internal.warc.gz | 1,003,302,543 | 16,702 | # Higher Partial Derivatives & Chain Rule
1. Nov 8, 2007
### kingwinner
Higher Partial Derivatives & Chain Rule (urgent)
I'll have a test this evening, and I don't want to fail on a question like this, so please help me out! I will greatly appreciate for any help provided.
The question:
My attempt:
From the solutions manual:
My first-order derivative is correct, but my second-order partial derivative Fjj is wrong. But I don't understand WHY I am wrong. Can someone please show me HOW to calculuate Fjj? I even tried to draw tree diagrams to clarfiy the dependence of variables, but I still got messed up, this is very very frustrating. I spent an hour looking through my work (I'm not kidding), but I can't locate my mistake, nor do I understand the way the solutions manual calculate Fjj.
Thanks again!
2. Nov 8, 2007
### Avodyne
You have $F_j = x_j r^{-1}f'(r)$, but when you differentiate this with respect to x_j, you neglect to differentiate the factor of x_j itself. This is what gives the first term in F_jj in the solutions manual.
3. Nov 8, 2007
### kingwinner
[Let D=curly d=partial derivative]
DF_j/Dx_j = (DF_j/Dr) (Dr/Dx_j) by the tree diagram (or the chain rule). What's wrong with this?
(DF_j/Dr) = D(f '(r) * x_j * r^-1) / Dr <----here I am differentiating with respect to r, so I just pulled out the constant x_j (it's index is j, not r, so it's constant with repsect to r), should be correct I guess.
Last edited: Nov 8, 2007
4. Nov 8, 2007
### HallsofIvy
Staff Emeritus
?? r is NOT an index to begin with! r depends upons every xi so every xi depends upon r. You cannot change r without changing every xi.
5. Nov 8, 2007
### kingwinner
I see...
Then how can I calculate (DF_j/Dr)? Can someone please explain this?
6. Nov 8, 2007
### Avodyne
F_j is *not* a function of r alone; it *also* depends on x_j explicitly, through the factor of x_j in front. So, DF_j/Dx_j = DF_j/Dx_j + (DF_j/Dr) (Dr/Dx_j), but DF_j/Dx_j means two different things on each side; on the left, it means differentiate with respect to both explicit and implicit (through r) appearances of x_j; on the right, it means differentiate only with respect to the explicit dependence on x_j. I would prefer to call the left side dF_j/dx_j, that is, the derivative with respect to *all* dependence on x_j, both implicit and explicit. But it's a partial derivative in the sense that all other components of x are being held fixed.
7. Nov 8, 2007
### kingwinner
But, say, if f is a function of x,y,z, then all partial derivatives fx,fy,fz are functions of x,y,z as well. I remember seeing some remarks or theorem like this. So f, fx, fy, fz have the same tree diagram.
So in this case, F and Fx should have the same tree diagram. Following the branches, there is only one way to get from fx to x_j, i.e. from fx to r to x_j, so shouldn't DF_j/Dx_j = (DF_j/Dr) (Dr/Dx_j) be correct?
8. Nov 8, 2007
### Avodyne
Yes, this is your first tree diagram. But your second tree diagram, the one that goes through r, is correct for Z but not for F_j, because F_j depends explicitly on x_j; F_j cannot be written as a function of r alone, even though Z can. So, for F_j, your second diagram is missing a branch, one that connects F_j directly to x_j without going through r.
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Have something to add? | 933 | 3,402 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2016-50 | longest | en | 0.926272 |
https://communities.bentley.com/products/hydraulics___hydrology/w/hydraulics_and_hydrology__wiki/9383/what-do-all-the-lines-in-pondmaker-step-4-represent | 1,675,901,053,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500983.76/warc/CC-MAIN-20230208222635-20230209012635-00666.warc.gz | 199,271,998 | 21,146 | What do all the lines in PondMaker step 4 represent?
Product(s): PondPack, CivilStorm, SewerGEMS Version(s): 08.11.01.XX+ (PondPack), 08.11.04.54+ (CivilStorm/SewerGEMS) Area: Layout and Data Input
Problem
What do all the lines in PondMaker step 4 represent?
Solution
The blue line represents the performance of the outlet structure that you're currently trialing (the rating curve).
The orange line represents the target rating curve you are trying to meet, based on the predevelopment peak flows.
The dashed lines are the estimated maximum water surface elevation for each storm event, from the estimated storage volume and designed pond from the previous step.
In this step of PondMaker the object is to define an outlet structure whose rating curve matches the target rating curve as closely as possible. If your outlet's curve is above and to the left of your target rating curve, it is essentially too small and has restricted flow too much. On the other hand, if your outflow curve is below and to the right of the target rating curve your outlet structure is too large and is letting too much water out.
In the best case you would like to meet your target rating curve exactly, but this is not always possible. For many designs as long as you are within a certain tolerance your design will considered sufficient. You should check with your local governing agency on the policy for your town, city, or state. The "pass" or "fail" designations in the Design worksheet are based on the above/below tolerances set in the Options tab.
Below is a simple example of an outlet structure that is slightly too small/restrictive for the 10 year event and just slightly too large (releasing too much flow) for the 100 year event, but meets the acceptable tolerance design for this project of 10% above/below.
Remember that this step is just an estimate. In the following step, you will actually route flow through the designed pond and outlet and will be presented with a similar graph/diagram. In many cases you will need to make further adjustments to the pond dimensions and outlet to meet the design criteria, especially if the shape of the outflow hydrograph does not match closely to what was used in the storage estimate step. | 499 | 2,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-06 | latest | en | 0.924543 |
https://www.jiskha.com/display.cgi?id=1351358737 | 1,516,413,553,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888341.28/warc/CC-MAIN-20180120004001-20180120024001-00698.warc.gz | 941,325,245 | 3,880 | # Physics
posted by .
An archer wishes to shoot an arrow at a target at eye level a distance of 45.0 m away. If the initial speed imparted to the arrow is 68.2 m/s, what angle should the arrow make with the horizontal as it is being shot?
• Physics -
Range = Vo^2*sin(2A)/g.
Range = 45 m.
Vo = 68.2 m/s.
g = 9.8 m/s^2.
Solve for angle A.
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https://goprep.co/ex-12.4-q2-construct-a-right-angle-triangle-pqr-in-which-r-i-1nkuoz | 1,618,257,014,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038069133.25/warc/CC-MAIN-20210412175257-20210412205257-00443.warc.gz | 369,471,515 | 25,666 | # Construct a
We have been given two sides and one angle of right angled triangle ABC.
Steps of construction for this case would be:
1. Draw a line segment sufficiently long using a ruler.
2. Locate points Q and R such that QR = 3 cm, on the previously drawn line segment using a ruler.
3. At R, construct a line segment RX, sufficiently large, such that XRQ = 90° using a compass; use protractor to measure the angle 90°.
4. With Q as centre and radius 4 cm, draw a circle or a semicircle or an arc cutting the line segment RX at P.
5. Join QP.
Thus, PQR is the required triangle.
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# Dosages as Determined by Body Weight
.Topic: Dosages according to kilogram of body weight
1. Calculate the following dosages according to kilogram of body weight.
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## JSS3: MATHEMATICS - 3RD TERM
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Lesson 6, Topic 2
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# Ratios, Rates and Proportions
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#### Topic Content:
• Ratios, Rates and Proportions
Express as ratios in their lowest form:
### Example 6.2.1:
200 g to 2 kg
Solution
Convert 2 kg to g = 2 × 1000 = 2000 g
⇒ $$\frac{200\:g}{200\:g} \: : \: \frac{2000\:g}{200\:g}$$
⇒ $$\scriptsize 1 \: : \: 10$$
### Example 6.2.2
2.5 km to 500 m
Solution
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#### Responses
error: Alert: Content selection is disabled!! | 280 | 920 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-18 | latest | en | 0.724081 |
https://socratic.org/questions/an-object-with-a-mass-of-4-kg-is-on-a-plane-with-an-incline-of-pi-4-if-it-takes--2 | 1,571,236,692,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00404.warc.gz | 707,790,058 | 6,051 | # An object with a mass of 4 kg is on a plane with an incline of - pi/4 . If it takes 7 N to start pushing the object down the plane and 5 N to keep pushing it, what are the coefficients of static and kinetic friction?
$\setminus {\mu}_{s} = 1.252$ & $\setminus {\mu}_{k} = 1.18$
#### Explanation:
The force $F$ required to push an object of mass $m = 4$ kg downward on an inclined plane at an angle $\theta = - \frac{\setminus \pi}{4} = - {45}^{\setminus} \circ$ & coefficient of static friction $\setminus {\mu}_{s}$
$F = \setminus {\mu}_{s} m g \setminus \cos \setminus \theta - m g \setminus \sin \setminus \theta$
$\setminus {\mu}_{s} = \setminus \frac{F + m g \setminus \sin \setminus \theta}{m g \setminus \cos \setminus \theta}$
$\setminus {\mu}_{s} = \setminus \frac{7 + 4 \setminus \times 9.81 \setminus \sin {45}^{\setminus} \circ}{4 \setminus \times 9.81 \setminus \cos {45}^{\setminus} \circ}$
$= 1.252$
Similarly, when the motion starts, then the force (F) required to keep the object moving
$F = \setminus {\mu}_{k} m g \setminus \cos \setminus \theta - m g \setminus \sin \setminus \theta$
$\setminus {\mu}_{k} = \setminus \frac{F + m g \setminus \sin \setminus \theta}{m g \setminus \cos \setminus \theta}$
$\setminus {\mu}_{k} = \setminus \frac{7 + 4 \setminus \times 9.81 \setminus \sin {45}^{\setminus} \circ}{4 \setminus \times 9.81 \setminus \cos {45}^{\setminus} \circ}$
$= 1.18$ | 497 | 1,413 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-43 | latest | en | 0.263949 |
https://boards.straightdope.com/t/what-is-the-proof-of-the-law-of-inertia/159352 | 1,600,950,552,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400217623.41/warc/CC-MAIN-20200924100829-20200924130829-00660.warc.gz | 308,228,591 | 8,631 | # What is the proof of the law of inertia?
An object in motion will stay in motion unless acted on by an outside force; ditto for an object at rest. Perhaps poorly stated, that is the law of inertia. As I understand it, it is pretty fundamental–atleast for macro type objects.
Question: What is the proof of this law? Is it taken as given because it stands to reason? What’s the straight dope?
At one time it was thought that all objects would eventually come to rest regardless, until Galileo noticed a ball that rolled down an incline rolled up the opposite incline at nearly the same height; the smoother the incline, the closer the height. Galileo reasoned that the difference between the heights was caused by friction acting on the ball. If there was no friction, the ball would attain the same height over and over again, virtually forever. If there’s no incline on the other side, the ball will roll off and travel virtually forever.
Newton expanded on this in his first law of motion by saying that a force is not needed to keep an object in motion. The ball does not come to rest beacuse of the absence of force, but the presence of it: friction. In the absence of force, the body remains in motion.
[QOUTE]An object in motion will stay in motion unless acted on by an outside force; ditto for an object at rest. Perhaps poorly stated, that is the law of inertia. As I understand it, it is pretty fundamental–atleast for macro type objects.
[/QUOTE]
…in an inertial(not accelerating, that is) frame of reference.
I don’t think there’s an experimental proof of this fact.
Our observations of the planets, though, provide good evidence of Newton’s second law, namely,
F=ma
Setting F=0, you get a=0, which is the result we want.
There are tons of experiments that bear out the law of inertia. Any time a physicist uses F=ma they are testing it, essentially. As mic84 points out, the law of inertia is just a special case where F=0 and therefore a=0. (If you can show me a planet that’s not accelerating, though, I’ll give ya a million bucks.)
In an interial frame! I mean, a planet that’s not accelerating in an inertial frame!
(I sure as hell hope I can post this before someone comes after the prize!!!)
Taking F=ma and setting F=0 yields a=0. That makes sense, I never thought if it that way. But that also sounds sort of definitional to me. Define Force to be anything that causes acceleration, then by definition no Force=>no acceleration. Instead of a proof we have an identity.
Did that make sense? I’m assuming I’m mistaken; where’s my error?
Thanks for the replies. I’ve been wondering about that.
If it can’t be proven axiomatically, but only demonstrated experimentally, why isn’t it known as the theory of inertia?
Force isn’t only something that causes acceleration, it’s also anything that changes velocity, i.e., pushes or pulls. Without the pull of friction, the object travels indefinitely. Without the push of additional force, the object won’t accelerate. Inertia just says that velocity won’t change spontaneously, both slowing down and speeding up are due to a force changing the velocity.
You’re right, if you define force this way, the law of inertia just follows. OTOH, you could see force as something fundamental(since there are only three of them, it’s reasonalbe). Then F=ma is meaningful, but there’s no “proof” of it, just experimental evidence.
You cannot “prove” physical laws nor theories. To do so would require observation and/or experimental verification over all time and space.
A physical “law” is simply a summary of observations.
A theory is an explanation of one or more observation(s).
To use a chemistry example, it has been observed that a given chemical compound always contains the same proportion by mass of each of its constituent elements. After this was observed to be true for every chemical compound so analyzed, the observation was deemed important enough to be summarized as the “Law of Definite Proportions.”
Soon thereafter, Dalton proposed an explanation for why the aforementioned law was always observed to hold true. His theory is called “Dalton’s Atomic Theory.” The theory hypothesizes that all elements are made up of atoms that combine to form elements.
While the language of physics is often stated to be that of mathematics, physics itself is not mathematics. So far as I am aware, there are no axioms in physics. Everything is based on experimental observation.
Thanks for the explanation.
To address the OP directly, as I stated in my last post, you cannot prove physical laws. You can only gather more observations that support a law and do not disprove it.
While laws cannot be proved, they can be disproved. One example is the “Law of Conservation of Mass.” This law was experimentally invalidated with the advent of nuclear physics.
The fact that we can send satellites into high orbit, and they stay there for a very long time, is a pretty convincing demonstration of inertia.
Pravnik
The definition of acceleration is the rate of change of velocity. So if something is accelerating, it’s velocity is changing.
GTPhD1996
An alternative viewpoint:
The “law of inertia” (which can be stated alternately as the ‘conservation of momentum’) can actually be immediately inferred from the hypothesis that space is homogeneous.
In other words, the “law of inertia” is a direct consequence of the fact that space itself ‘looks’ the same at every place in the space-continuum. Thus, in the absence of other objects, a given object has no way to tell where it is in space, because everything looks the same–furthermore, it doesn’t have any way to tell if it is moving (if you were floating by a perfectly clear, flat piece of glass, but couldn’t touch it, could you tell how fast you were going? or if you are even moving at all?). If the object has no way of knowing how it is travelling through ‘space’, how could it alter its path.
This, of course, isn’t a “proof”, but it’s one way that theorists might think about it. Well, at least theorists who ignore relativity and quantum physics.
Yes, that’s true. But how is that different from what I said?
Don Roberto has it correct. Conservation of momentum/energy/mass is comes from the translation symmetry of space-time. A good general relativity textbook will cover this in more detail than you want. This pushes back the question to “why space-time is homogenous?”. I don’t know the answer to that one.
This principle is also useful in regions of space that are not homogenous. The best example is within the lattice of a crystalline solid (e.g. metals). The atoms with the solid are positioned in a three-dimensional pattern. This means that space looks the “same” only if you move in certain directions for certain distances. These restrictions mean that the momenta of particles (e.g. electrons) are conserved only for certain types of motion. This is the basis for all microscopic descriptions of solids.
GTPhD1996, the definition of acceleration is the rate of change of momentum, ie. F = d/dt (mv). If the object’s mass is constant, the definition reduces to yours. Rockets (mass decreasing) and rolling snowballs (mass increasing) are conspicuous counter-examples.
F stands for force, not acceleration.
Per Karl Popper, the hallmark of a scientific theory is falsifiability. What this means is that scientists don’t spend their time trying to “prove” something to be true, but rather think of ways to falsify or disprove a certain theory. A theory is falsifiable if it can make a testable prediction. For the case of inertia, we can have a simple test: go to space and release a ball with initial linear velocity v. The Law of Inertia predicts that the ball will travel in a straight line. We make this prediction on paper here on earth.
Now you carry the actual experiment by going aboard the space shuttle. Then record the results. Compare your experimental results with your theoretical prediction.
Do we know for sure that the law of inertia is true? What we do know is that nobody has shown it to be false yet (as far as I know). It’s the best model we’ve got so far. That’s what scientists do: try to find better and better models of nature. | 1,782 | 8,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2020-40 | latest | en | 0.953826 |
http://mathhelpforum.com/calculus/218129-calculus-parametric-function.html | 1,481,230,121,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542655.88/warc/CC-MAIN-20161202170902-00425-ip-10-31-129-80.ec2.internal.warc.gz | 181,715,545 | 10,466 | # Thread: Calculus of parametric function
1. ## Calculus of parametric function
My professor gave us a sheet that had the calculus of functions in different formats. The sheet isn't making any sense to me though.
$x=f(t)$
$y=g(t)$
Derivatives: $\frac{dy}{dt}=f'(t)$, $\frac{dx}{dt}=g'(t)$, $\frac{dy}{dx}=\frac{f'(t)}{g'(t)}$
Integrals: $\int_ \alpha ^ \beta g(t)f'(t)dt$
Arc Length: $\int_ \alpha ^ \beta \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy} {dt}\right)^2} dt$
Should the letters correspond?
x goes with f
y goes with g
2. ## Re: Calculus of parametric function
You are right. The letters x and y should be inverted. What the sheet says is that the both x and y depend on the 'parameter' t. therefore you can write both x and y as functions of t. Hence the definition of x=f(t) and y=g(t).
For example, you can think of x and y as the coordinates of motion of a particle in two dimensions. As the particle moves in time, the coordinates of the point, that is (x,y), change. I think your prof just mistakenly inverted x and y.
3. ## Re: Calculus of parametric function
Thanks for clearing that up | 324 | 1,123 | {"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": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2016-50 | longest | en | 0.799863 |
https://www.math.leidenuniv.nl/~gillrd/teaching/astro/histogram.Rnw | 1,652,784,949,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-21/segments/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00257.warc.gz | 1,015,204,004 | 1,771 | \documentclass[a4paper]{article} \begin{document} \title{Statistics II: Histograms in R} \author{Richard Gill\thanks{{\tt http://www.math.leidenuniv.nl/$\sim$gill/teaching/statistics}}} \date{\today} \maketitle \noindent These scripts and notes illustrate the histogram as a density estimator. First we make some data. And before that, I set the random seed, so that the results will be perfectly reproducible. % <<>>= set.seed(11091951) x<-rgamma(100,10) @ % Now we show the plain vanilla R histogram. % % \begin{center} <>= hist(x) @ \end{center} % \newpage\noindent The area under the graph was equal to the number of observations. Using the option prob=TRUE we make the area under the graph equal to one. On this graph I will superimpose a plot of the true density of the data. I'll also fix the axes to get a nicer plot. The bin-width is determined by Sturges' method, the default, it doesn't hurt to make that explicit. % \begin{center} <>= hist(x,prob=T,xlim=c(0,30),ylim=c(0,0.15),breaks="Sturges") xd<-seq(from=0,to=30,length=1000) yd<-dgamma(xd,shape=10) lines(xd,yd) abline(h=0) @ \end{center} % \newpage\noindent Now let's look at the results of the other bin-width algorithms. First, Scott: % \begin{center} <>= hist(x,prob=T,xlim=c(0,30),ylim=c(0,0.15),breaks="Scott") lines(xd,yd) abline(h=0) @ \end{center} % \newpage\noindent And Freedman-Diaconis: % \begin{center} <>= hist(x,prob=T,xlim=c(0,30),ylim=c(0,0.15),breaks="FD") lines(xd,yd) abline(h=0) @ \end{center} % \newpage\noindent Not much to see, right? But I'll draw these all again, but now with an outlier added to the dataset. First, plain vanilla (Sturges): % \begin{center} <>= hist(c(x,50),prob=T,xlim=c(0,50),ylim=c(0,0.15),breaks="Sturges") lines(xd,yd) abline(h=0) @ \end{center} % \newpage\noindent Then Scott: % \begin{center} <>= hist(c(x,50),prob=T,xlim=c(0,50),ylim=c(0,0.15),breaks="Scott") lines(xd,yd) abline(h=0) @ \end{center} % \newpage\noindent And Freedman-Diaconis % \begin{center} <>= hist(c(x,50),prob=T,xlim=c(0,50),ylim=c(0,0.15),breaks="Freedman-Diaconis") lines(xd,yd) abline(h=0) @ \end{center} % \newpage\noindent Venables and Ripley think they have a much better histogram. Let's take a look. Scott: % \begin{center} <>= library(MASS) truehist(c(x,50),prob=T,xlim=c(0,50),ylim=c(0,0.15),nbins="Scott") lines(xd,yd) abline(h=0) @ \end{center} % \newpage\noindent And Freedman-Diaconis % \begin{center} <>= truehist(c(x,50),prob=T,xlim=c(0,50),ylim=c(0,0.15),nbins="Freedman-Diaconis") lines(xd,yd) abline(h=0) @ \end{center} % \bigskip \noindent{\bf Exercise}. Figure out how to draw a frequency polygon (join the mid-points of histogram bars by straight lines...). Theoretically you can use a larger bin-width, at least, if the true density is twice differentiable. \end{document} help(hist) help(rgamma) x<-rgamma(100,10) hist(x) hist(x,prob=T) y<-seq(from=0,to=30,length=1000) lines(y,dgamma(y,shape=10)) hist(x,prob=T,breaks="Scott") lines(y,dgamma(y,shape=10)) hist(x,prob=T,breaks="FD") lines(y,dgamma(y,shape=10)) hist(c(x,50),prob=T,breaks="FD") lines(y,dgamma(y,shape=10)) hist(c(x,50),prob=T,breaks="FD",ylim=c(0,0.15)) lines(y,dgamma(y,shape=10)) hist(c(x,50),prob=T,breaks="Scott",ylim=c(0,0.15)) lines(y,dgamma(y,shape=10)) hist(c(x,50),prob=T,breaks="Sturges",ylim=c(0,0.15)) lines(y,dgamma(y,shape=10)) library(MASS) truehist(c(x,50),prob=T,ylim=c(0,0.15)) lines(y,dgamma(y,shape=10)) truehist(c(x,50),prob=T,nbins="FD",ylim=c(0,0.15)) lines(y,dgamma(y,shape=10)) truehist(c(x,50),prob=T,nbins="Scott",ylim=c(0,0.15)) lines(y,dgamma(y,shape=10)) truehist(c(x,50),prob=T,nbins="FD",ylim=c(0,0.15)) lines(y,dgamma(y,shape=10)) | 1,290 | 3,643 | {"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} | 3.65625 | 4 | CC-MAIN-2022-21 | latest | en | 0.536364 |
https://howkgtolbs.com/convert/8.44-kg-to-lbs | 1,656,804,487,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00279.warc.gz | 354,945,604 | 12,080 | # 8.44 kg to lbs - 8.44 kilograms to pounds
Do you want to know how much is 8.44 kg equal to lbs and how to convert 8.44 kg to lbs? You couldn’t have chosen better. You will find in this article everything you need to make kilogram to pound conversion - theoretical and practical too. It is also needed/We also want to highlight that all this article is devoted to only one amount of kilograms - that is one kilogram. So if you need to know more about 8.44 kg to pound conversion - read on.
Before we go to the practice - this is 8.44 kg how much lbs conversion - we are going to tell you some theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 8.44 kg to lbs? 8.44 kilograms it is equal 18.6070149128 pounds, so 8.44 kg is equal 18.6070149128 lbs.
## 8.44 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, in formal International System of Units (in abbreviated form SI).
At times the kilogram is written as kilogramme. The symbol of this unit is kg.
Firstly the kilogram was defined in 1795. The kilogram was defined as the mass of one liter of water. This definition was not complicated but totally impractical to use.
Later, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in abbreviated form IPK). The International Prototype of the Kilogram was prepared of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was in use until 2019, when it was substituted by another definition.
The new definition of the kilogram is based on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It is also divided into 100 decagrams and 1000 grams.
## 8.44 kilogram to pounds
You learned a little bit about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. We want to highlight that there are more than one kind of pound. What does it mean? For instance, there are also pound-force. In this article we are going to to centre only on pound-mass.
The pound is in use in the Imperial and United States customary systems of measurements. To be honest, this unit is in use also in other systems. The symbol of the pound is lb or “.
The international avoirdupois pound has no descriptive definition. It is defined as exactly 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 8.44 kg?
8.44 kilogram is equal to 18.6070149128 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 8.44 kg in lbs
The most theoretical part is already behind us. In this part we are going to tell you how much is 8.44 kg to lbs. Now you know that 8.44 kg = x lbs. So it is high time to get the answer. Just see:
8.44 kilogram = 18.6070149128 pounds.
This is a correct outcome of how much 8.44 kg to pound. It is possible to also round off the result. After rounding off your result will be as following: 8.44 kg = 18.568 lbs.
You learned 8.44 kg is how many lbs, so look how many kg 8.44 lbs: 8.44 pound = 0.45359237 kilograms.
Naturally, this time you may also round off this result. After rounding off your result is exactly: 8.44 lb = 0.45 kgs.
We also want to show you 8.44 kg to how many pounds and 8.44 pound how many kg results in tables. Have a look:
We will start with a chart for how much is 8.44 kg equal to pound.
### 8.44 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
8.44 18.6070149128 18.5680
Now look at a table for how many kilograms 8.44 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
8.44 0.45359237 0.45
Now you know how many 8.44 kg to lbs and how many kilograms 8.44 pound, so we can move on to the 8.44 kg to lbs formula.
### 8.44 kg to pounds
To convert 8.44 kg to us lbs you need a formula. We are going to show you two versions of a formula. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 18.6070149128 outcome in pounds
The first version of a formula give you the most correct result. Sometimes even the smallest difference could be considerable. So if you need an exact outcome - this version of a formula will be the best solution to convert how many pounds are equivalent to 8.44 kilogram.
So let’s move on to the second formula, which also enables conversions to know how much 8.44 kilogram in pounds.
The shorter formula is down below, have a look:
Number of kilograms * 2.2 = the outcome in pounds
As you see, this formula is simpler. It could be better choice if you need to make a conversion of 8.44 kilogram to pounds in fast way, for instance, during shopping. Just remember that final result will be not so exact.
Now we are going to learn you how to use these two versions of a formula in practice. But before we are going to make a conversion of 8.44 kg to lbs we want to show you another way to know 8.44 kg to how many lbs without any effort.
### 8.44 kg to lbs converter
Another way to learn what is 8.44 kilogram equal to in pounds is to use 8.44 kg lbs calculator. What is a kg to lb converter?
Converter is an application. It is based on longer formula which we showed you above. Due to 8.44 kg pound calculator you can quickly convert 8.44 kg to lbs. You only have to enter number of kilograms which you need to calculate and click ‘calculate’ button. You will get the result in a second.
So let’s try to calculate 8.44 kg into lbs with use of 8.44 kg vs pound converter. We entered 8.44 as a number of kilograms. It is the result: 8.44 kilogram = 18.6070149128 pounds.
As you can see, this 8.44 kg vs lbs calculator is intuitive.
Now we are going to our chief issue - how to convert 8.44 kilograms to pounds on your own.
#### 8.44 kg to lbs conversion
We will begin 8.44 kilogram equals to how many pounds calculation with the first formula to get the most exact outcome. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 18.6070149128 the outcome in pounds
So what have you do to learn how many pounds equal to 8.44 kilogram? Just multiply amount of kilograms, in this case 8.44, by 2.20462262. It is equal 18.6070149128. So 8.44 kilogram is equal 18.6070149128.
You can also round off this result, for example, to two decimal places. It is 2.20. So 8.44 kilogram = 18.5680 pounds.
It is time for an example from everyday life. Let’s convert 8.44 kg gold in pounds. So 8.44 kg equal to how many lbs? As in the previous example - multiply 8.44 by 2.20462262. It is 18.6070149128. So equivalent of 8.44 kilograms to pounds, if it comes to gold, is equal 18.6070149128.
In this example it is also possible to round off the result. It is the outcome after rounding off, this time to one decimal place - 8.44 kilogram 18.568 pounds.
Now we can go to examples calculated using a short version of a formula.
#### How many 8.44 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 18.568 the outcome in pounds
So 8.44 kg equal to how much lbs? As in the previous example you have to multiply amount of kilogram, this time 8.44, by 2.2. See: 8.44 * 2.2 = 18.568. So 8.44 kilogram is exactly 2.2 pounds.
Let’s do another calculation using this version of a formula. Now calculate something from everyday life, for example, 8.44 kg to lbs weight of strawberries.
So let’s convert - 8.44 kilogram of strawberries * 2.2 = 18.568 pounds of strawberries. So 8.44 kg to pound mass is exactly 18.568.
If you know how much is 8.44 kilogram weight in pounds and are able to convert it using two different formulas, we can move on. Now we are going to show you these outcomes in tables.
#### Convert 8.44 kilogram to pounds
We are aware that outcomes shown in tables are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in tables for your convenience. Due to this you can easily compare 8.44 kg equivalent to lbs outcomes.
Start with a 8.44 kg equals lbs table for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
8.44 18.6070149128 18.5680
And now have a look at 8.44 kg equal pound table for the second version of a formula:
Kilograms Pounds
8.44 18.568
As you see, after rounding off, if it comes to how much 8.44 kilogram equals pounds, the outcomes are not different. The bigger amount the more significant difference. Remember it when you need to make bigger number than 8.44 kilograms pounds conversion.
#### How many kilograms 8.44 pound
Now you learned how to calculate 8.44 kilograms how much pounds but we want to show you something more. Are you interested what it is? What do you say about 8.44 kilogram to pounds and ounces conversion?
We want to show you how you can convert it little by little. Begin. How much is 8.44 kg in lbs and oz?
First thing you need to do is multiply amount of kilograms, this time 8.44, by 2.20462262. So 8.44 * 2.20462262 = 18.6070149128. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To convert how much 8.44 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So your result is 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then your result will be equal 2 pounds and 33 ounces.
As you see, calculation 8.44 kilogram in pounds and ounces quite simply.
The last calculation which we will show you is calculation of 8.44 foot pounds to kilograms meters. Both of them are units of work.
To calculate it you need another formula. Before we show you this formula, see:
• 8.44 kilograms meters = 7.23301385 foot pounds,
• 8.44 foot pounds = 0.13825495 kilograms meters.
Now look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to convert 8.44 foot pounds to kilograms meters you have to multiply 8.44 by 0.13825495. It gives 0.13825495. So 8.44 foot pounds is equal 0.13825495 kilogram meters.
It is also possible to round off this result, for instance, to two decimal places. Then 8.44 foot pounds will be exactly 0.14 kilogram meters.
We hope that this conversion was as easy as 8.44 kilogram into pounds calculations.
We showed you not only how to make a conversion 8.44 kilogram to metric pounds but also two other calculations - to check how many 8.44 kg in pounds and ounces and how many 8.44 foot pounds to kilograms meters.
We showed you also other solution to do 8.44 kilogram how many pounds conversions, that is with use of 8.44 kg en pound converter. It is the best option for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs calculations in quicker way.
We hope that now all of you are able to do 8.44 kilogram equal to how many pounds conversion - on your own or with use of our 8.44 kgs to pounds calculator.
It is time to make your move! Let’s convert 8.44 kilogram mass to pounds in the best way for you.
Do you need to make other than 8.44 kilogram as pounds conversion? For example, for 15 kilograms? Check our other articles! We guarantee that conversions for other amounts of kilograms are so simply as for 8.44 kilogram equal many pounds.
### How much is 8.44 kg in pounds
We want to sum up this topic, that is how much is 8.44 kg in pounds , we gathered answers to the most frequently asked questions. Here we have for you the most important information about how much is 8.44 kg equal to lbs and how to convert 8.44 kg to lbs . Have a look.
How does the kilogram to pound conversion look? The conversion kg to lb is just multiplying 2 numbers. How does 8.44 kg to pound conversion formula look? . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
See the result of the conversion of 8.44 kilogram to pounds. The correct result is 18.6070149128 pounds.
There is also another way to calculate how much 8.44 kilogram is equal to pounds with another, shortened version of the formula. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So this time, 8.44 kg equal to how much lbs ? The result is 18.6070149128 pounds.
How to convert 8.44 kg to lbs in just a moment? You can also use the 8.44 kg to lbs converter , which will do the rest for you and give you an accurate result .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,592 | 13,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | longest | en | 0.940491 |
https://www.physicsforums.com/threads/limiting-x-2-x-1-as-x-approaches-1-from-the-left.432709/ | 1,726,753,171,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00652.warc.gz | 857,177,636 | 16,657 | # Limiting x^2/(x-1) as x Approaches 1 from the Left
• DrummingAtom
In summary, the limit of x^2/(x-1) as x goes to 1 from the left is negative infinity. This can be shown algebraically by factoring the numerator and using the limit rules for fractions.
DrummingAtom
## Homework Statement
Find the limit of x^2/(x-1) as x goes to 1 from the left.
## The Attempt at a Solution
It doesn't seem I can factor anything, but could I assume that since the numerator is a constant and the denomination is going to be negative because it's <1 then it's going to negative infinity? Is there anyway to show this algebraically? Thanks.
DrummingAtom said:
## Homework Statement
Find the limit of x^2/(x-1) as x goes to 1 from the left.
## The Attempt at a Solution
It doesn't seem I can factor anything, but could I assume that since the numerator is a constant and the denomination is going to be negative because it's <1 then it's going to negative infinity? Is there anyway to show this algebraically? Thanks.
You are correct.
$$\frac{x^2}{1-x} = \frac{x^2 -1}{1-x} + \frac{1}{1-x}$$
The limit as x-> $$1^{-}$$ of $$\frac{x^2 -1}{1-x}$$ is 2.
This
$$\frac{1}{1-x}$$ one does not exist.
## What is the limit of x^2/(x-1) as x approaches 1 from the left?
The limit of x^2/(x-1) as x approaches 1 from the left is undefined. This is because when x approaches 1 from the left, the denominator (x-1) becomes very close to 0, which would result in division by 0.
## Why is the limit of x^2/(x-1) as x approaches 1 from the left undefined?
The limit of x^2/(x-1) as x approaches 1 from the left is undefined because the denominator (x-1) becomes very close to 0, which would result in division by 0. In mathematics, division by 0 is undefined.
## What happens to the value of x^2/(x-1) as x gets closer and closer to 1 from the left?
As x gets closer and closer to 1 from the left, the value of x^2/(x-1) becomes larger and larger. This is because as x approaches 1, the denominator (x-1) becomes smaller and smaller, making the fraction larger.
## Can the limit of x^2/(x-1) as x approaches 1 from the left be calculated using direct substitution?
No, the limit of x^2/(x-1) as x approaches 1 from the left cannot be calculated using direct substitution. Direct substitution would result in division by 0, which is undefined.
## How can the limit of x^2/(x-1) as x approaches 1 from the left be calculated?
The limit of x^2/(x-1) as x approaches 1 from the left can be calculated using the concept of one-sided limits. This involves evaluating the limit as x approaches 1 from values slightly less than 1, such as 0.9, 0.99, 0.999, and so on. If the values from the left side approach a specific number, then that number is the limit from the left. However, in this case, the limit from the left does not exist as the values get closer and closer to 1, the limit also gets larger and larger, approaching infinity.
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1K | 896 | 3,117 | {"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} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.969754 |
http://www.ck12.org/book/Algebra-I/r2/section/6.7/ | 1,481,100,245,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542009.32/warc/CC-MAIN-20161202170902-00096-ip-10-31-129-80.ec2.internal.warc.gz | 388,390,559 | 51,799 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 6.7: Linear Inequalities in Two Variables
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Graph linear inequalities in one variable on the coordinate plane.
• Graph linear inequalities in two variables.
• Solve real-world problems using linear inequalities
## Introduction
A linear inequality in two variables takes the form
y>mx+b\begin{align*} y > mx+b\end{align*} or y<mx+b\begin{align*} y < mx+b\end{align*}
Linear inequalities are closely related to graphs of straight lines. A straight line has the equation y=mx+b\begin{align*} y = mx+b\end{align*}. When we graph a line in the coordinate plane, we can see that it divides the plane in two halves.
The solution to a linear inequality includes all the points in one of the plane halves. We can tell which half of the plane the solution is by looking at the inequality sign.
> The solution is the half plane above the line.
\begin{align*}\geq\end{align*} The solution is the half plane above the line and also all the points on the line.
< The solution is the half plane below the line.
\begin{align*}\leq\end{align*} The solution is the half plane below the line and also all the points on the line.
(Above the line means for a given x\begin{align*}x-\end{align*}coordinate, all points with y\begin{align*}y-\end{align*}values greater than the y\begin{align*}y-\end{align*}value are on the line)
For a strict inequality, we draw a dashed line to show that the points on the line are not part of the solution.
For an inequality that includes the equal sign, we draw a solid line to show that the points on the line are part of the solution.
Here is what you should expect linear inequality graphs to look like.
The solution of y>mx+b\begin{align*} y>mx+b\end{align*} is the half plane above the line. The dashed line shows that the points on the line are not part of the solution.
The solution of ymx+b\begin{align*} y \geq mx+b\end{align*} is the half plane above the line and all the points on the line.
The solution of y<mx+b\begin{align*} y is the half plane below the line.
The solution of ymx+b\begin{align*} y\leq mx+b\end{align*} is the half plane below the line and all the points on the line.
## Graph Linear Inequalities in One Variable in the Coordinate Plane
In the last few sections, we graphed inequalities in one variable on the number line. We can also graph inequalities in one variable on the coordinate plane. We just need to remember that when we graph an equation of the type x=a\begin{align*}x=a\end{align*} we get a vertical line and when we graph an equation of the type y=b\begin{align*}y=b\end{align*} we get a horizontal line.
Example 1
Graph the inequality x>4\begin{align*}x>4\end{align*} on the coordinate plane.
Solution
First, let’s remember what the solution to x>4\begin{align*}x>4\end{align*} looks like on the number line.
The solution to this inequality is the set of all real numbers x\begin{align*}x\end{align*} that are bigger than four but not including four. The solution is represented by a line.
In two dimensions we are also concerned with values of y\begin{align*} y\end{align*}, and the solution to x>4\begin{align*} x>4\end{align*} consists of all coordinate points for which the value of x\begin{align*}x\end{align*} is bigger than four. The solution is represented by the half plane to the right of x=4\begin{align*}x=4\end{align*}.
The line x=4\begin{align*}x=4\end{align*} is dashed because the equal sign is not included in the inequality and therefore points on the line are not included in the solution.
Example 2
Graph the inequality y6\begin{align*} y \leq 6\end{align*} on the coordinate plane.
Solution
The solution is all coordinate points for which the value of y\begin{align*}y\end{align*} is less than or equal than 6. This solution is represented by the half plane below the line y=6\begin{align*}y=6\end{align*}.
The line y=6\begin{align*}y=6\end{align*} is solid because the equal sign is included in the inequality sign and the points on the line are included in the solution.
Example 3
Graph the inequality 6<5\begin{align*} \mid 6 \mid <5\end{align*}
Solution
The absolute value inequality 6<5\begin{align*} \mid 6 \mid <5\end{align*} can be re-written as 5<y<5\begin{align*}-5 < y <5\end{align*}. This is a compound inequality which means
y>5\begin{align*}y>-5\end{align*} and y<5\begin{align*}y<5\end{align*}
In other words, the solution is all the coordinate points for which the value of y\begin{align*}y\end{align*} is larger than -5 and smaller than 5. The solution is represented by the plane between the horizontal lines y=5\begin{align*}y=-5\end{align*} and y=5\begin{align*}y=5\end{align*}.
Both horizontal lines are dashed because points on the line are not included in the solution.
Example 4
Graph the inequality x2\begin{align*}\mid x \mid \geq 2\end{align*}.
Solution
The absolute value inequality x2\begin{align*} \mid x \mid \geq 2\end{align*} can be re-written as a compound inequality:
x2\begin{align*} x \leq -2 \end{align*} or x2\begin{align*} x \geq 2 \end{align*}
In other words, the solution is all the coordinate points for which the value of x\begin{align*}x\end{align*} is smaller than or equal to -2 and greater than or equal to 2. The solution is represented by the plane to the left of the vertical line x=2\begin{align*}x= -2\end{align*} and the plane to the right of line x=2\begin{align*}x=2\end{align*}.
Both vertical lines are solid because points on the line are included in the solution.
## Graph Linear Inequalities in Two Variables
The general procedure for graphing inequalities in two variables is as follows.
Step 1: Re-write the inequality in slope-intercept form y=mx+b\begin{align*}y=mx+b\end{align*}. Writing the inequality in this form lets you know the direction of the inequality
Step 2 Graph the line of equation y=mx+b\begin{align*}y=mx+b\end{align*} using your favorite method. (For example, plotting two points, using slope and y\begin{align*}y-\end{align*}intercept, using y\begin{align*}y-\end{align*}intercept and another point, etc.). Draw a dashed line if the equal sign in not included and a solid line if the equal sign is included.
Step 3 Shade the half plane above the line if the inequality is greater than. Shade the half plane under the line if the inequality is less than.
Example 5
Graph the inequality y2x3\begin{align*} y \geq 2x-3\end{align*}.
Solution
Step 1
The inequality is already written in slope-intercept form y2x3\begin{align*} y \geq 2x-3\end{align*}.
x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
1\begin{align*}-1\end{align*} 2(1)3=5\begin{align*}2(-1) - 3 = -5\end{align*}
0 2(0)3=3\begin{align*}2(0) - 3 = -3\end{align*}
1 2(1)3=1\begin{align*}2(1) - 3 = -1\end{align*}
Step 2
Graph the equation y=2x3\begin{align*}y=2x-3\end{align*} by making a table of values.
Step 3
Graph the inequality. We shade the plane above the line because y\begin{align*}y\end{align*} is greater than. The value 2x3\begin{align*}2x-3\end{align*} defines the line. The line is solid because the equal sign is included.
Example 6
Graph the inequality 5x2y>4\begin{align*} 5x - 2y > 4\end{align*}.
Solution
Step 1
Rewrite the inequality in slope-intercept form.
2yy>5x+4>52x2\begin{align*} -2y & > -5x + 4\\ y & > \frac{5} {2}x - 2\end{align*}
Notice that the inequality sign changed direction due to division of negative sign.
Step 2
Graph the equation y>52x2\begin{align*} y>\frac{5} {2}x - 2\end{align*} by making a table of values.
x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
2\begin{align*}-2\end{align*} 52(2)2=7\begin{align*}\frac{5} {2}(-2) - 2 = -7\end{align*}
0 52(0)2=2\begin{align*}\frac{5} {2}(0) - 2 = -2\end{align*}
2 52(2)2=3\begin{align*}\frac{5} {2}(2) - 2 = 3\end{align*}
Step 3
Graph the inequality. We shade the plane below the line because the inequality in slope-intercept form is less than. The line is dashed because the equal sign in not included.
Example 7
Graph the inequality y+4x3+5\begin{align*} y+4 \leq -\frac{x} {3}+5\end{align*}.
Solution
Step 1
Rewrite the inequality in slope-intercept form yx3+1\begin{align*} y \leq -\frac{x} {3}+1\end{align*}
Step 2
Graph the equation y=x3+1\begin{align*} y= -\frac{x} {3}+1\end{align*} by making a table of values.
x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
3\begin{align*}-3\end{align*} (3)3+1=2\begin{align*}-\frac{(-3)} {3}+1=2\end{align*}
0 03(0)+1=1\begin{align*}-\frac{0} {3}(0)+1=1\end{align*}
3 33+1=0\begin{align*}-\frac{3} {3}+1=0\end{align*}
Step 3
Graph the inequality. We shade the plane below the line. The line is solid because the equal sign in included.
## Solve Real-World Problems Using Linear Inequalities
In this section, we see how linear inequalities can be used to solve real-world applications.
Example 8
A pound of coffee blend is made by mixing two types of coffee beans. One type costs $9 per pound and another type costs$7 per pound. Find all the possible mixtures of weights of the two different coffee beans for which the blend costs 8.50 per pound or less. Solution Let’s apply our problem solving plan to solve this problem. Step 1: Let x=\begin{align*}x =\end{align*} weight of9 per pound coffee beans in pounds
Let y=\begin{align*}y =\end{align*} weight of 7 per pound coffee beans in pounds Step 2 The cost of a pound of coffee blend is given by 9x+7y\begin{align*}9x + 7y\end{align*}. We are looking for the mixtures that cost8.50 or less.
We write the inequality 9x+7y8.50\begin{align*}9x + 7y \leq 8.50\end{align*}.
Step 3
To find the solution set, graph the inequality 9x+7y8.50\begin{align*}9x + 7y \leq 8.50\end{align*}.
Rewrite in slope-intercept y1.29x+1.21\begin{align*}y \leq -1.29x + 1.21\end{align*}.
Graph y=1.29x+1.21\begin{align*}y=-1.29x+1.21\end{align*} by making a table of values.
x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
0 1.21
1 -0.08
2 -1.37
Step 4
Graph the inequality. The line will be solid. We shade below the line.
Notice that we show only the first quadrant of the coordinate plane because the weight values should be positive.
The blue-shaded region tells you all the possibilities of the two bean mixtures that will give a total less than or equal to $8.50. Example 9 Julian has a job as an appliance salesman. He earns a commission of$60 for each washing machine he sells and $130 for each refrigerator he sells. How many washing machines and refrigerators must Julian sell in order to make$1000 or more in commission?
Solution Let’s apply our problem solving plan to solve this problem.
Step 1
Let x=\begin{align*}x =\end{align*} number of washing machines Julian sells
Let y=\begin{align*}y =\end{align*} number of refrigerators Julian sells
Step 2
The total commission is given by the expression 60x+130y\begin{align*}60x+130y\end{align*}.
We are looking for total commission of 1000 or more. We write the inequality. 60x+130y1000\begin{align*}60x + 130y \geq 1000\end{align*}. Step 3 To find the solution set, graph the inequality 60x+130y1000\begin{align*}60x + 130y \geq 1000\end{align*}. Rewrite it in slope-intercept y.46x+7.7\begin{align*}y \geq -.46x + 7.7\end{align*}. Graph y=.46x+7.7\begin{align*}y = -.46x + 7.7\end{align*} by making a table of values. x\begin{align*}x\end{align*} y\begin{align*}y\end{align*} 0 7.7 2 6.78 4 5.86 Step 4 Graph the inequality. The line will be solid. We shade above the line. Notice that we show only the first quadrant of the coordinate plane because dollar amounts should be positive. Also, only the points with integer coordinates are possible solutions. ## Lesson Summary • The general procedure for graphing inequalities in two variables is as follows: Step 1 Rewrite the inequality in slope-intercept form y=mx+b\begin{align*}y=mx+b\end{align*}. Step 2 Graph the line of equation y=mx+b\begin{align*}y=mx+b\end{align*} by building a table of values. Draw a dashed line if the equal sign in not included and a solid line if the it is included. Step 3 Shade the half plane above the line if the inequality is greater than. Shade the half plane under the line if the inequality is less than. ## Review Questions Graph the following inequalities on the coordinate plane. 1. x<20\begin{align*}x<20\end{align*} 2. y5\begin{align*}y \geq -5\end{align*} 3. x>10\begin{align*} \mid x \mid >10 \end{align*} 4. y7\begin{align*} \mid y \mid \leq 7 \end{align*} 5. y4x+3\begin{align*} y \leq 4x+3\end{align*} 6. y>x26\begin{align*}y > -\frac{x}{2}-6\end{align*} 7. 3x4y12\begin{align*}3x-4y \geq 12\end{align*} 8. x+7y<5\begin{align*} x+7y <5 \end{align*} 9. 6x+5y>1\begin{align*}6x + 5y>1\end{align*} 10. y+54x+10\begin{align*}y + 5 \leq -4x + 10\end{align*} 11. x12y5\begin{align*}x-\frac{1}{2}y \geq 5\end{align*} 12. 30x+5y<100\begin{align*}30x + 5y< 100\end{align*} 13. An ounce of gold costs670 and an ounce of silver costs $13. Find all possible weights of silver and gold that makes an alloy that costs less than$600 per ounce.
14. A phone company charges 50 cents cents per minute during the daytime and 10 cents per minute at night. How many daytime minutes and night time minutes would you have to use to pay more than \$20 over a 24 hour period?
## Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9616.
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https://www.snapxam.com/problems/35300980/derivative-of-arcsin-x-0-5-3-sin-x-2- | 1,542,795,587,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039747665.82/warc/CC-MAIN-20181121092625-20181121114625-00403.warc.gz | 1,006,128,706 | 8,570 | # Derive the function arcsin((x^0.5+3)/(sin(x^2))) with respect to x
## \frac{d}{dx}\left(arcsin\left(\frac{\sqrt{x}+3}{\sin\left(x^2\right)}\right)\right)
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$\frac{\frac{1}{2}x^{-\frac{1}{2}}\sin\left(x^2\right)-2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2\sqrt{1-\frac{\left(3+\sqrt{x}\right)^2}{\sin\left(x^2\right)^2}}}$
## Step by step solution
Problem
$\frac{d}{dx}\left(arcsin\left(\frac{\sqrt{x}+3}{\sin\left(x^2\right)}\right)\right)$
1
Taking the derivative of arcsine
$\frac{d}{dx}\left(\frac{3+\sqrt{x}}{\sin\left(x^2\right)}\right)\frac{1}{\sqrt{1-\left(\frac{3+\sqrt{x}}{\sin\left(x^2\right)}\right)^2}}$
2
Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
$\frac{\sin\left(x^2\right)\frac{d}{dx}\left(3+\sqrt{x}\right)-\left(3+\sqrt{x}\right)\frac{d}{dx}\left(\sin\left(x^2\right)\right)}{\sin\left(x^2\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{3+\sqrt{x}}{\sin\left(x^2\right)}\right)^2}}$
3
The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if ${f(x) = \sin(x)}$, then ${f'(x) = \cos(x)\cdot D_x(x)}$
$\frac{\sin\left(x^2\right)\frac{d}{dx}\left(3+\sqrt{x}\right)-\left(3+\sqrt{x}\right)\frac{d}{dx}\left(x^2\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{3+\sqrt{x}}{\sin\left(x^2\right)}\right)^2}}$
4
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{\sin\left(x^2\right)\frac{d}{dx}\left(3+\sqrt{x}\right)-1\cdot 2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{3+\sqrt{x}}{\sin\left(x^2\right)}\right)^2}}$
5
The derivative of a sum of two functions is the sum of the derivatives of each function
$\frac{\sin\left(x^2\right)\left(\frac{d}{dx}\left(3\right)+\frac{d}{dx}\left(\sqrt{x}\right)\right)-1\cdot 2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{3+\sqrt{x}}{\sin\left(x^2\right)}\right)^2}}$
6
The derivative of the constant function is equal to zero
$\frac{\sin\left(x^2\right)\left(0+\frac{d}{dx}\left(\sqrt{x}\right)\right)-1\cdot 2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{3+\sqrt{x}}{\sin\left(x^2\right)}\right)^2}}$
7
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{\left(0+\frac{1}{2}x^{-\frac{1}{2}}\right)\sin\left(x^2\right)-1\cdot 2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{3+\sqrt{x}}{\sin\left(x^2\right)}\right)^2}}$
8
Multiply $2$ times $-1$
$\frac{\left(0+\frac{1}{2}x^{-\frac{1}{2}}\right)\sin\left(x^2\right)-2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{3+\sqrt{x}}{\sin\left(x^2\right)}\right)^2}}$
9
$x+0=x$, where $x$ is any expression
$\frac{\frac{1}{2}x^{-\frac{1}{2}}\sin\left(x^2\right)-2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{3+\sqrt{x}}{\sin\left(x^2\right)}\right)^2}}$
10
The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$
$\frac{\frac{1}{2}x^{-\frac{1}{2}}\sin\left(x^2\right)-2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2}\cdot\frac{1}{\sqrt{1-\frac{\left(3+\sqrt{x}\right)^2}{\sin\left(x^2\right)^2}}}$
11
Multiplying fractions
$\frac{\frac{1}{2}x^{-\frac{1}{2}}\sin\left(x^2\right)-2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2\sqrt{1-\frac{\left(3+\sqrt{x}\right)^2}{\sin\left(x^2\right)^2}}}$
12
Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number
$\frac{\frac{1}{2}\cdot\frac{1}{\sqrt{x}}\sin\left(x^2\right)-2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2\sqrt{1-\frac{\left(3+\sqrt{x}\right)^2}{\sin\left(x^2\right)^2}}}$
13
Apply the formula: $a\frac{1}{x}$$=\frac{a}{x}$, where $a=\frac{1}{2}$ and $x=\sqrt{x}$
$\frac{\frac{\frac{1}{2}}{\sqrt{x}}\sin\left(x^2\right)-2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2\sqrt{1-\frac{\left(3+\sqrt{x}\right)^2}{\sin\left(x^2\right)^2}}}$
14
Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$
$\frac{\frac{1}{2}x^{-\frac{1}{2}}\sin\left(x^2\right)-2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2\sqrt{1-\frac{\left(3+\sqrt{x}\right)^2}{\sin\left(x^2\right)^2}}}$
$\frac{\frac{1}{2}x^{-\frac{1}{2}}\sin\left(x^2\right)-2x\left(3+\sqrt{x}\right)\cos\left(x^2\right)}{\sin\left(x^2\right)^2\sqrt{1-\frac{\left(3+\sqrt{x}\right)^2}{\sin\left(x^2\right)^2}}}$
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(2/2)
During the first trial, our langmuir probe told us that our electron beam intensity was fluctuating at 60 Hz.
This is a problem, because one of the main things we are trying to study is the way changes in beam intensity affect potential well depth, so we want a steady intensity. The frequency of the fluctuations suggest that the AC-powered hot cathode is to blame. I don’t totally understand the details of how a hot cathode running on AC 120v 60Hz translates to this waveform:
data from the langmuir probe displayed on the oscilloscop
The important thing is to prevent it. To do that, I put a full wave bridge rectifier in the power supply. It converts the AC coming from the wall to DC
It has three essential components.
1) The bridge rectifier
This change the AC sine wave into a waveform expressed by the function abs(sin(x)):
Better, but still not steady DC.
2) The filter capacitor
This gets rid of the ripple. you could compare the capacitor to a bucket with a hole in the bottom. Even if I vary the rate at which putting water into the bucket, the rate at which it come out is always going to be more or less the same, provided that it is sufficiently large compared to the volume of water going in.
However, its impossible to get an absolutely perfect DC output with this setup, because the ammount of charge on the capacitor does affect the voltage at which the current comes out.
This 680 uF capacitor takes away enough of the ripple for our purposes:
the output of the power supply when hooked up to a light bulb
3) An isolation transformer
Usually, the diode bridge and the capacitor would be enough, but our AC isn’t coming from the wall, its coming from a grounded auto transformer. this is a problem because the rectifier only works if the AC input is floating. A transformer with an equal number of primary and secondary wingdings accomplishes this without stepping the voltage up or down.
Nest step is to test it in the chamber.
Domenick Bauer
## New High Voltage Probe
18 10 2011
Just got this Tektronix P6015 high voltage probe. It goes up to 40 kV!
Well actually it only goes to 27 kV without the fluorocarbon 114 dielectric.
It’s huge! Shown next to a normal probe:
Just in time for the next run.
This is the same probe used in the Sydney experiment.
## Arduino Controls 30,000 Volts
10 09 2011
Today I made arduino control 30,000 volts.
My arduino has 3 channels of analog output 0 to 5 volt.
For testing I used this sin wave generator sketch:
int pwmPin = 9; // output pin supporting PWM
void setup(){
pinMode(pwmPin, OUTPUT); // sets the pin as output
}
void loop(){
float something = millis() / 1000.0;
int value = 127.5 + 127.5 * sin( something * 2.0 * PI );
analogWrite(pwmPin,value);
}
This generates a lazy 2 Hz sin wave.
But the output is not really analog, it’s pulse width modulation(PWM):
This tutorial shows how to smooth out PWM using a low pass filter. My low pass filter used 6kΩ resistor and 4.7 µF @ 45V capacitor.
Here we have the raw PWM output superimposed with the filtered output:
Looks good!
Now we just add the voltage doubling op-amp circuit I made previously, and BOOM:
This shows the source signal and the voltage doubled signal.
Sweet! Now we can control the 30,000 volt glassman power supply.
Here the arduino is sending a slow sin wave to the glassman’s voltage control:
From 2011-09-10
The Glassman’s slew rate is really slow without a load.
Here is the setup:
## Emergency Stop
14 02 2010
I put together an emergency stop button for the reactor:
This will be positioned a safe distance from the reactor, so you can kill the high voltage from a distance. I’ll wire it up and test it today.
## -30kV / 10mA
24 10 2009
If you’ve been following me on twitter, you know I received a -30kV / 10mA Glassman a few days ago. Now it’s online and it kicks ass. Current limiting, remote controllable… it’s the second unit from the top:
Here is an air plasma it produced:
With current limiting and good air metering, we can get a stable plasma. I notice you get a sense for the plasma just by _listening_ to the glassman. When the plasma is unstable the glassman softly clicks along with the plasma burst.
## Grounding
29 08 2009
Spent the last two days throughly grounding the entire system.
We start by attaching a ground line directly to the overhead sprinkler system. First we scrape off the paint:
Then we attach this purpose made ground clip:
Here it is installed:
We are using 12 AWG insulated stranded copper. In addition to this pipe ground, we are also using outlet ground from two different circuits:
Ground each piece of rack equipment, and the rack itself:
Ground several points on the chamber, sled, and pump:
Here is the high voltage connection point with high voltage divider inside glass insulation:
## Standoff for Superconducting Magrid
29 08 2009
I’ve been doing some brainstorming on the standoff for the superconducting magrid. This is a messy problem! You need a cryogenic feedthrough that is also a high voltage standoff. Then you need to pass in the YBCO superconducting cables, and wiring for the persistent switch.
Yesterday I realized that we can get most of the way there by welding together two off the shelf components:
This diagram refrences parts 9812107 and 9611005 from insulatorseal which is a subsidiary of MDC Vacuum. I’ve sent this drawing to insulatorseal for a quote. One problem is that part 9611005 is only rated up to 6kV so we will need a custom variant to get to the 10kV to 40kV range.
The idea with this setup is that the high voltage could come through a standard HV feedthrough and connect to the insulated tip of this feedthrough via a connecting wire:
## Glass Insulator
6 08 2009
Ordering this glass tubing to protect the atmosphere side of the high voltage feedthrough. Hopefully they can cut it to size for me. They will get back to me about that.
## High Voltage Divider
26 06 2009
Just purchased this Model V1G high voltage divider. This device divides the input voltage by a factor of 1000, so we can measure high voltage with a regular multimeter. | 1,484 | 6,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-23 | latest | en | 0.926056 |
https://fgstudy.com/mcqs/mathematics-mcqs-for-class-12-with-answers-chapter-7/6/ | 1,679,939,402,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948673.1/warc/CC-MAIN-20230327154814-20230327184814-00552.warc.gz | 282,325,366 | 24,545 | # Mathematics MCQS for class 12 with Answers Chapter 7
#### Mathematics MCQS for class 12 with Answers
If j & K are two unit vectors then jx k =
1
i
0
-1
Commutative law holds in
None of these
Inner product
Cross product in three vectors
Vector product
If a & b are vectors along two adjacent sides of a triangle then area of triangle is:
1/2 |a x b|
ab
|a. b|
a . b
The cross product or vector product of two vectors is defined
Only in plane
Only in space
None of these
Both a and c
i . (j x k) is equal to
1
2
-1
0
Two vectors are said to be negative of each other if they have _______ magnitude but _________ direction
None of these
Opposite, same
Same, same
Same, Opposite
The magnitude of vector is also called its
Norm
Parameter
Variables
Point
If the angle
Collinear
Perpendicular
Both(b)&(c)
Parallel
Work done of a force on a particle.
Zero
None of these
Scalar quantity
Vector quantity
The null or zero vector in R3 is O = ____________
None of these
[0,0]
[0,0,0]
[0]
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http://www.ma.utexas.edu/academics/courses/syllabi/M408M.php | 1,529,398,427,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861981.50/warc/CC-MAIN-20180619080121-20180619100121-00112.warc.gz | 460,404,204 | 6,746 | ## Syllabus: M 408M
MULTIVARlABLE CALCULUS
Text: Stewart, Calculus, Early Transcendentals, Eighth Edition
Responsible Parties: Ray Heitmann and Jane Arledge, May 2012
Prerequisite and degree relevance: M308L, M408L or M308S, M408S, with a grade of C- or better. M 408D and M 408M cannot both be counted toward a degree.
Calculus is offered in two equivalent sequences: a two-semester sequence, M 408C/D, or either of two three-semester sequences, M 408N/S/M (for College of Natural Science Students) or M 408K/L/M. Completion of one of these sequences is required for mathematics majors, with a C- or better in each course.
For some degrees, M 408N/S or M 408K/L satisfies the calculus requirement . This sequence is also a valid prerequisite for some upper-division mathematics courses, including M 325K, M 427K, M 340L, and M 362K.
Course description: M 408M is directed at students in the natural and social sciences and at engineering students. In comparison with M408D, it covers fewer chapters of the text. However, some material is covered in greater depth. This is not a course in the theory of calculus.
The content includes an introduction to the theory and applications of differential and integral calculus of functions of several variables, including parametric equations, polar coordinates, vectors, vector calculus, functions of several variables, partial derivatives, gradients, and multiple integrals.
### Forty Class Days As:
• 10 Parametric Equations and Polar Coordinates (seven days)
• 10.1 Curves Defined by Parametric Equations
• 10.2 Calculus with Parametric Curves
• 10.3 Polar Coordinates
• 10.4 Areas and Lengths in Polar Coordinates
• 10.5 Conic Sections
• 10.6 Conic Sections in Polar Coordinates
• 12 Vectors and the Geometry of Space (eight days)
• 12.1 Three-Dimensional Coordinate Systems
• 12.2 Vectors
• 12.3 The Dot Product
• 12.4 The Cross Product
• 12.5 Equations of Lines and Planes
• 12.6 Cylinders and Quadric Surfaces
• 13 Vector Functions (five days)
• 13.1 Vector Functions and Space Curves
• 13.2 Derivatives and Integrals of Vector Functions
• 13.3 Arc Length and Curvature
• 13.4 Motion in Space: Velocity and Acceleration
• 14 Partial Derivatives (ten days)
• 14.1 Functions of Several Variables
• 14.2 Limits and Continuity
• 14.3 Partial Derivatives
• 14.4 Tangent Planes and Linear Approximations
• 14.5 The Chain Rule
• 14.6 Directional Derivatives and the Gradient Vector
• 14.7 Maximum and Minimum Values
• 14.8 Lagrange Multipliers
• 15 Multiple Integrals (ten days)(first three sections are review)
• 15.1 Double Integrals over Rectangles
• 15.2 Double Integrals over General Regions
• 15.3 Double Integrals in Polar Coordinates
• 15.4 Applications of Double Integrals (optional)
• 15.9 Change of Variables in Multiple Integrals (if time permits) | 754 | 2,818 | {"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} | 2.96875 | 3 | CC-MAIN-2018-26 | latest | en | 0.852183 |
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# Suppose that the two components operate independently and have identical performance characteristics. Let X and Y...
Suppose that the two components operate independently and have identical performance characteristics. Let X and Y be random variables denoting their life spans. Experience with wear-out times suggests that in certain cases a good choice for fx,y(x, y) would be fxx (x, y) = {?? 122e-1(x+y), x > 0,y>0 otherwise Where 2 is some constant greater than 0. Suppose that the manufacturer advertises a money-back guarantee if the system fails to last for more than 1000 hours. a) What proportion of the systems sold is likely to be returned for refunds? b) P(Y < 1000|X > 1000) =? c) E(Y|X = 2500) =? d) Find E(Y)and E(X).
#### Homework Answers
Answer #1
d)
where
a)
b)
so,
c)
similarly
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## #1 2013-02-26 07:33:05
genericname
Full Member
Offline
### Log question
(2*(n/2)*lg(n/2)) + n
= (n*lg(n/2))+n
= n*(lg n - 1) + n
How does (n*lg(n/2))+n simplify to n*(lg n - 1) + n? What happened to the n/2 that was inside? It has been a while since I last worked with log.
## #2 2013-02-26 07:50:27
bobbym
Online
### Re: Log question
Hi;
Is this the problem?
Last edited by bobbym (2013-02-26 07:51:47)
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
genericname
Full Member
Offline
Yeah.
## #4 2013-02-26 09:10:06
bobbym
Online
### Re: Log question
I am getting:
Last edited by bobbym (2013-02-26 09:10:52)
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #5 2013-02-26 12:40:36
anonimnystefy
Real Member
Offline
### Re: Log question
Hi genericname
lg(a/b)=lg(a)-lg(b) for any a and b for which the expression is defined.
So, lg(n/2)=lg(n)-lg(2)=lg(n)-1, assuming the logarithm is with base 2.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #6 2013-02-26 13:27:50
bobbym
Online
### Re: Log question
Why would the log be to the base two?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #7 2013-02-26 13:41:10
anonimnystefy
Real Member
Offline
### Re: Log question
Because then the step makes sense. We will have have to wait for the OP's answer.
And besides, the use of n and lg reminds me of comp. analysis.
Last edited by anonimnystefy (2013-02-26 13:42:40)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #8 2013-02-26 15:31:10
bobbym
Online
### Re: Log question
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #9 2013-02-27 02:44:16
genericname
Full Member
Offline
### Re: Log question
Ah, thank you. It was base 2, sorry for the confusion.
## #10 2013-02-27 05:19:58
anonimnystefy
Real Member
Offline
### Re: Log question
No problem and you're welcome!
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
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Rational Function
3.6k views
1 Introduction 2 Rational Function 3 Domain and Range of Rational function 4 Inverse of a Rational Function 5 Properties of rational functions 6 Summary 7 FAQs
16 October 2020
Introduction
What is a rational function?
A rational function is an algebraic fraction with numerator and denominator as polynomials and the denominator is not equal to zero.
The coefficients of the polynomials need not be rational numbers.
Can you introduce a rational number with examples?
Rational Function
Rational function definition
As seen from the visual representation, R(x) will be a rational function and is a rational expression if p(x) and q(x) are polynomials.
Rational Function Examples:
Domain and Range of Rational function
In a function f (x),
Domain of f ---- The set of all real numbers that variable x takes such that the expression defining the function is real. The domain can also be given explicitly.
The set of all values that f takes ---the range of the rational functions
Before moving on, let us recall that
• ℕ is the set of natural numbers.
• ℤ is the set of integers.
• ℚ is the set of rational numbers.
• ℝ is the set of real numbers.
• ℂ is the set of complex numbers.
Rational functions are functions of the form f(x) = p(x)/q(x)
And we know p(x) and q(x) are polynomials with q(x) not being equal to 0.
Thus, this particular condition helps find the domain of rational function.
To find the domain of rational function find out the values of x for which denominator is 0. Then exclude these values.
How to find the domain of a rational function?
1. where p(x) and q(x) are polynomials and q(x)≠0
2. The domain of a rational function consists of all the real numbers x except those for which the denominator is 0.
3. To find these x values to be excluded from the domain of a rational function, equate the denominator to zero and solve for x
For example, The domain of the rational function is the set of all real numbers except x = 0.
Example 2
Now, consider the rational function
On simplification, when x≠2 it becomes a linear function f(x)=x+1
But at x=2, the original rational expression is not defined and hence it leaves the graph with a hole when x=2
How to find the range of a rational function:
1. Finding domain of the inverse rational function.
2. Sketch the rational function graph and identify the range.
Let us again consider the rational parent function
1. We know that the function is not defined at x=0
2. As x→0 from either side of zero, f(x)→∞ Similarly, as x→±∞, f(x)→0
3. The graph approaches the x-axis as when x tends to move towards (+ ∞)
4. or (- ∞,) but never touches the x-axis.
5. It implies that the function can take all the real values except 0.
6. Hence the range of the function is the set of real numbers except 0
Example 1
How to find the domain of a rational function? How to find the range of a rational function:
In order to find the excluded value in the domain of the rational function,
Consider the denominator equals to 0 and then solve for x
This way we can find the domain of the rational function as a set of real numbers except −3
The range of the rational function is the same as the domain of the inverse rational function. So, to find the range define the inverse of the rational function.
Interchange the x and y
Solving for y you get,
x + 5 = 1/y + 3
⇒ y + 3 = 1 / x + 5
⇒ y = (1 / x + 5) − 3
So, the inverse rational function is
We can find excluded value in the domain of the inverse rational expression by making the denominator equals to zero and then solving for x
Hence the domain of the inverse rational function is the set of real numbers except −5. It means that the range of the given function is the set of real numbers except −5
Therefore, the domain of the given rational function is
{x ∈ R | x ≠ −3} and the range is {y ∈ R | y ≠ −5}
How to find the domain of a rational function, How to find the range of a rational function with one unknown in the denominator.
Example 2
Find the domain and range of the rational function f(x) = -1/x-5.
Domain of a Rational function:
• From the above given graph it implies that the domain = ℝ−{5} and the Range = ℝ−{0}. We need to check this algebraically.
• A rational function is not defined for any set of values that results in a zero denominator. Equate the denominator to zero as , 𝑥−5=0, which gives a solution of 𝑥=5.
• Hence it confirms as the domain is ℝ−{5}.
How to find the range of a rational function?
• To confirm the range, identify the set of values that cannot be achieved given the domain. As the magnitude of 𝑥 gets larger, the output tends to zero but will never reach 0. Therefore, the range = ℝ−{0}.
Example 3
Find the domain and range of the rational expression 1/x-2.
Solution:
• Follow the same procedure as in previous example
• Domain of a Rational Function: Solve for
• 𝑥−2=0,
• Therefore, the domain = ℝ−{2}.
•
• For range of a rational function, the magnitude of 𝑥 gets larger, the output tends to zero but will never reach 0. Therefore, the range = ℝ−{0}.
Example 4
Find the domain and range of the rational function f(x) = 2x+3/4x+5 and also find values that f(X) cannot take.
Domain of a rational Function
𝑓(𝑥) is undefined when the denominator value = 0.
Hence, at 4𝑥+5=0 , the graph of the function would have an asymptote.
To find the equation of asymptote,follow the below steps:
4𝑥+5-5=-5
𝑥 = −5 / 4
Therefore, the asymptote has the equation 𝑥 = −5 / 4.
And the domain of the function is all real numbers except −5 / 4 ,i.e., ℝ−{ − 5 /4}.
How to find the range of a rational function?
Values 𝑓(𝑥) cannot take, it helps to find what happens as 𝑥 gets large. To make this process easier divide the numerator and denominator of the function by 𝑥 to get
From this it’s clear that as 𝑥 gets progressively large, 3𝑥 and 5𝑥 tends closer to zero, and also the function tends to ½ but never actually reach it. Range is R - {½}. Asymptotes of a rational function
Let's check out one of our old friends...
See how the graph is getting closer to the axes?... But, it will never touch them?
The x and y axes are asymptotes!
Asymptotes are lines (usually invisible) in space that a graph gets closer and closer to but never touches...
So for vertical asymptotes...Set the denominator = 0 and solve.
Inverse of a Rational Function
Finding the inverse of a rational function is relatively easy,because the procedure to find the inverse is rather repetitious. That the method used here is just one way out of the many.
Feel free to vary the steps to fit your needs as long as you’re getting the same final answer.
Key Steps in Finding the Inverse of a Rational Function
1. Replace f(x) by y.
2. Switch the roles of x and y, in other words, interchange x and y in the equation.
3. Solve for y in terms of x.
4. Replace y by f−1(x) to get the inverse function.
Examples of How to Find the Inverse of a Rational Function
Example 1
Find the inverse function. State its domain and range.
Even without graphing this function, x cannot equal -3 .Because the denominator becomes zero, and the entire rational expression becomes undefined. The domain is all x−values not including −3.
Next, let's graph this function to verify if it passes the Horizontal Line Test, this guarantees to have an inverse function.
It seems that no horizontal line will intersect the graph more than once. That’s a good sign. The vertical dashed line is called a vertical asymptote. The graph will never cross that. Remember, it is coming from the zeros of the denominator, or the restricted values of x.
Using the above steps, let’s find the inverse.
The domain of the inverse function comes from the fact that the denominator cannot equal zero. The range is obtained from the domain of the original function.
Example 2
Find the inverse function. State its domain and range.
This is the graph of the original function.
The following are the steps to find the inverse.
Example 3
Find the inverse function. State its domain and range.
.
In this problem since the rational expression has linear terms in both numerator and denominator, be convinced that just like in the previous example, it will have an inverse function even without further testing. Add it to your math toolbox and utilize it as your advantage whenever you encounter similar problems. Just to verify, the graph is shown below.
The rest of the steps to find the inverse function are the same. At the end of the solution, to get the final answer do factoring and then canceling the hidden −1, found in both numerator and denominator.
Properties of rational functions.
The following are general Properties of rational functions:
• If the numerator and denominator are of the same degree (n=m), then y= an/bm is a horizontal asymptote of the function.
• If the degree of the denominator is greater than the degree of the numerator, then y=0 is a Horizontal asymptote.
• If the degree of the denominator is less than the degree of the numerator, then there are no horizontal asymptotes.
• When x is equal to a root of the denominator polynomial, the denominator is zero and there is a Vertical asymptote. The exception is the case when the root of the denominator is also a root of the numerator. However, for this case we can cancel a factor from both the numerator and denominator (and we effectively have a lower-degree rational function).
Summary:
A rational function is an algebraic fraction with numerator and denominator as polynomials and the denominator is not equal to zero.
The coefficients of the polynomials need not be rational numbers.
How to find the domain of a rational function?
The domain of a rational function consists of all the real numbers x except those for which the denominator is 0.
To find these x values to be excluded from the domain of a rational function, equate the denominator to zero and solve for x
How to find the range of a rational function:
• Finding domain of the inverse of rational function.
• Sketch the rational function graph and identify the range.
We also discussed properties of rational functions.
Written by Asha M.
What is a rational function?
A rational function is an algebraic fraction with numerator and denominator as polynomials and the denominator is not equal to zero.
The coefficients of the polynomials need not be rational numbers.
How to find the domain of a rational function?
where p(x) and q(x) are polynomials and q(x)≠0
The domain of a rational function consists of all the real numbers x except those for which the denominator is 0.
To find these x values to be excluded from the domain of a rational function, equate the denominator to zero and solve for x
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# Earth Exploration Toolbook: Exploring Air Quality in Aura NO2 Data
In this chapter, students will explore relationships between air quality and population density using the image visualization tool, Google Earth. You will learn how to download NO2 data and analyze them to develop a conceptual understanding of how... (View More)
Audience: Middle school, High school
Materials Cost: Free
# MY NASA DATA: Correlation of Variables by Graphing
Activities in this lesson promote a fundamental understanding of relationships between graphed data. Sample graphs allow students to become familiar with interpreting data and to recognize relationships between variables. Additional microsets of... (View More)
Keywords: Scatter plot; Time plot
Audience: Middle school, High school
Materials Cost: Free
# MY NASA DATA: Ocean Impacts of an El Niño Event
El Niño impacts sea surface height, sea surface temperature and wind vectors. In this lesson, data for each of those three characteristics from the 1997-98 El Niño event will be analyzed and compared. Students will work in teams of 3 throughout... (View More)
Audience: High school
Materials Cost: Free
# MY NASA DATA: Scientist Tracking Network
In this data activity, students explore the relationship between surface radiation and mean surface temperature in several geographic regions. By observing how these parameters change with latitude, students will understand the relationship between... (View More)
# MY NASA DATA : Heating of Earth Materials Activity
In this inquiry exploration, student design an experiment to test the absorption of heat by different earth materials. Materials required include plastic water bottles, soil, sand, water, thermometers, lamp with 60 watt bulb, and stopwatch. This... (View More)
# MY NASA DATA: Comparison of Snow Cover on Different Continents
In this data activity, students create maps of the snow cover of each continent, and determine the average global snow cover. Step-by-step instructions for use of the MY NASA DATA Live Access Server (LAS) guide students through selecting a data set,... (View More)
Keywords: Authentic data use
Audience: Elementary school, Middle school
Materials Cost: Free
# Using Mathematical Models to Investigate Planetary Habitability: Activity C The Role of Actual Data in Mathematical Models
Students explore how mathematical descriptions of the physical environment can be fine-tuned through testing using data. In this activity, student teams obtain satellite data measuring the Earth's albedo, and then input this data into a... (View More)
Audience: High school
Materials Cost: Free per student
1 | 616 | 3,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | longest | en | 0.843548 |
https://causalsportsfan.wordpress.com/2011/12/08/pythagorean-expectations-pythagenports-and-pythagenpats-a-bunch-of-mumbo-jumbo/ | 1,590,585,755,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394074.44/warc/CC-MAIN-20200527110649-20200527140649-00555.warc.gz | 289,274,663 | 26,894 | # Pythagorean expectations, Pythagenports, and Pythagenpats – a bunch of mumbo jumbo
It seems like everywhere you turn in sports statistics, someone is talking about Pythagorean expectations. The Pythagorean expectation is a formula created by Bill James to estimate team quality in baseball, with runs scored and runs allowed as inputs. Proponents argue that luck plays a big part in close games, making point totals a better measure of team quality than wins. The formula is
Pythagorean = Runs scored^/ (Runs scored^c + Runs allowed^c),
where c is some exponent (usually greater than one) that can be calibrated. This kinda looks like the old Pythagorean formula from grade school (hence the name), though not really. The Pythagorean rises (at a decreasing rate) as you score more and drops (at a decreasing rate) as your opponents score more. In other words, the Pythagorean rewards teams for blowouts and punishes them for getting spanked, since these scoring outcomes may reflect team quality. If you score the same amount as your opponents, your Pythagorean is 0.5. The max is one and the min is zero, like a winning percentage. “Pythagorean wins” are given by the Pythagorean multiplied by the number of games.
Over time, this formula has been exported to basketball and football, and probably other sports. My buddy Adrian the Canadian sent me this week’s DVOA update at Football Outsiders, in which Aaron Schatz informs us that Football Outsiders has upgraded from the Pythagorean expectation to the “Pythagenport,” and Baseball Prospectus (the original Pythagenporters) has moved on to the Pythagenpat! These are identical to the Pythagorean, but allow the exponent c to depend on the number of runs scored and/or the teams involved.
Maybe this sounds reasonable to you, but these formulas make my head hurt. The Pythagenport even gets a logarithm involved. Where do these crazy functional forms come from? The intuition for these stats is simple: luck wins close games, not skill. How about a simple formula to go with this simple intuition:
True Wins = Blowout wins + Close wins/2 + Close losses/2 + Ties/2
“True Wins” says just what we want. You get full credit for blowouts, but only half credit for eking out close wins, since luck could have swung those either way. For the same reason, I toss in half credit for your close losses. In a probability sense, I’m assuming your expected value for wins in close games is half the total number of close games. The only work left is to calibrate the cutoff for close games versus blowouts. Since it’s football season, let’s consider the NFL. The obvious options are to define close games as those decided by 3 points or less or 7 points or less.
Here is Football Outsiders’ table for this season (sorted by Pythagorean wins) with a column added for True Wins with 7 as the cutoff:
W-L W <=7 L <=7 TRUE WINS PF PA Pyth Pyth Wins PPort PPort Wins SF 10-2 4 1 8.5 288 161 .799 9.6 .798 9.6 GB 12-0 4 0 10 420 262 .754 9.0 .776 9.3 HOU 9-3 3 2 8.5 310 189 .764 9.2 .769 9.2 BAL 9-3 3 2 8.5 296 192 .736 8.8 .740 8.9 NO 9-3 3 1 8 393 269 .711 8.5 .729 8.8 NE 9-3 2 2 9 362 247 .712 8.5 .727 8.7 PIT 9-3 4 2 8 268 195 .680 8.2 .681 8.2 DET 7-5 3 2 6.5 333 277 .607 7.3 .616 7.4 CHI 7-5 3 2 6.5 291 242 .608 7.3 .612 7.3 DAL 7-5 4 4 7 283 244 .587 7.0 .590 7.1
Here it is again with 3 as the cutoff:
W-L W <=3 L <=3 TRUE WINS PF PA Pyth Pyth Wins PPort PPort Wins SF 10-2 1 1 10 288 161 .799 9.6 .798 9.6 GB 12-0 1 0 11.5 420 262 .754 9.0 .776 9.3 HOU 9-3 0 0 9 310 189 .764 9.2 .769 9.2 BAL 9-3 2 0 8 296 192 .736 8.8 .740 8.9 NO 9-3 2 0 8 393 269 .711 8.5 .729 8.8 NE 9-3 0 1 9.5 362 247 .712 8.5 .727 8.7 PIT 9-3 1 1 9 268 195 .680 8.2 .681 8.2 DET 7-5 1 0 6.5 333 277 .607 7.3 .616 7.4 CHI 7-5 0 0 7 291 242 .608 7.3 .612 7.3 DAL 7-5 4 1 5.5 283 244 .587 7.0 .590 7.1
First off, the Pythags are nearly identical. True Wins differs primarily because it does not reward/punish teams for massive blowouts (see: Saints 62, Colts 7), and I don’t think that it should. Once a game gets out of hand, teams probably change their effort level. The second table also verifies my belief that the Cowboys are garbage, but one observation and my personal opinions aren’t enough to measure a stat’s worth. In comparing the Pythagenport to the Pythagorean, Schatz states:
The improvement is slight. The correlation between Pythagorean wins and actual wins for 1990-2010 is .9120. The correlation between Pythagenport and actual wins for 1990-2010 is .9134.
Hmm. By this measure, I can think of a stat that would trounce these Pythags — it’s called “actual wins,” and its correlation with actual wins is one. The whole point of constructing these measures is to filter out the luck that drives actual wins, which will reduce the correlation. Whenever I read about exponent choice for the Pythagorean, or new permutations of the Pythagorean, there’s never any objective analysis that drives the decision making. In fact, it’s not obvious how to evaluate these stats. They are supposed to measure “team quality,” but we don’t have an unbiased measure of team quality with which to compare.
One idea is to test if these measures are stable over time. If team quality is a consistent characteristic, then a good measure of team quality should also be consistent. Since 2002, the correlation between True Wins (3 point cutoff) over the first eight games of a season and True Wins over the second eight games is 0.50. The same correlation for the Pythagorean (c = 2.37) is 0.53. Between consecutive seasons, the correlation for True Wins is 0.33, and the correlation for the Pythagorean is 0.38. The numbers are pretty similar using 7 points as the cutoff. The Pythagorean does a little better, but I bet I could overcome the difference with some adjustments for garbage time scoring.
Next, let’s consider whether these stats predict actual future wins (GAMBLING ALERT!!!!!!!!!!!!). These correlations will never be one (since there is random, unpredictable luck in every game), but a stronger correlation means a more useful stat for team management and fans. The correlation between first half True Wins (3 point cutoff) and second half actual wins is 0.47. The same correlation for the Pythagorean is 0.50. However, the correlation between actual wins from first half to second half is 0.45. Between seasons, the correlations are 0.29 (True Wins to actual wins), 0.31 (Pythagorean to actual wins), and 0.30 (actual wins to actual wins). When thinking about future performance, teams can learn as much from their win total as from the formulas.
The Pythagorean has a great name and pedigree (Daryl Morey discovered 2.37 as the “best” NFL value), and it can be calculated from league standings tables on most websites. True Wins is more intuitive and nearly as powerful. But, this whole process is a bit misguided. The Pythagorean functional forms are ad hoc and complicated, and these stats ignore lots of useful information (for example, strength of schedule). I’m sure the “computer rankings” in the BCS or Football Outsider’s DVOA stat could do better by almost any measure. Simple stats are transparent, catchy, and easy to calculate, but if you want powerful stats, I think you need the big guns. | 2,039 | 7,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-24 | latest | en | 0.941615 |
pingpoli.de | 1,726,012,516,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00490.warc.gz | 435,227,013 | 6,087 | pingpoli.de
# Sparrow WebGL Devlog 7: Pixel Perfect 3D Object Selection
## Implementing color picking in JavaScript and WebGL
29.06.2023 - 20:41
Imagine a video game in which you couldn't interact with any objects in the 3D environment. It probably would be very boring.
Attacking enemy units or champions in League of Legends, clicking on your buildings in RTS games, and talking to an NPC in front of you in RPGs, everything requires the game to be able to determine which object you have clicked on.
There are two main methods of implementing 3D picking: Ray casting and color picking. In this blog post, I am going to talk about their differences and why and how I added color picking to my Sparrow WebGL engine.
# Ray Casting vs. Color Picking
Let's start by talking about ray casting. This technique casts a ray from the camera through your mouse position into the 3D scene and calculates which objects it intersects with. This can be done with the inverse view projection matrix. The projection matrix converts from 3D to 2D, so the inverse projection matrix can be used to go the other way from 2D to 3D.
With an equation for the 3D line and some fundamental linear algebra from school, it's possible to calculate the intersection between a line and many different 3D shapes. However, some of the intersection calculations are a lot faster than others. Two shapes that are especially easy and fast to test are spheres and boxes (especially axis-aligned boxes). This is where the term hitbox comes from. Complicated 3D shapes like characters or trees are approximated by boxes around them and it's quite fast to check whether the mouse ray intersects with that box.
However, this is where the problems with this solution start. While it's easy to calculate a ray-box intersection, this will give us a lot of false positives because most shapes are not very box-like, so a surrounding box covers a lot of empty space. This can be mitigated by using more boxes to approximate the shape of the object (i.e. one box for the head, one for the torso, boxes for each leg and arm, etc.). While this makes the hitbox calculations more accurate, it introduces a lot more boxes, and the more boxes we have to check the slower the algorithm gets. The number of checks can be reduced again by storing the boxes in a recursive spatial data structure, but this adds more complexity too. Another issue with hitboxes is depth. Even when we find an object that the mouse ray intersects with, we still have to check all other objects because there could be another object in front of the first object that the line intersects with too.
In general, ray casting requires a lot of math to calculate the line equation and the intersections for different 3D shapes. I have implemented ray casting in the past and I could convert my existing implementation to JavaScript and WebGL, but why add complicated math when there are other alternatives?
Another method of selecting objects in 3D space, and the method I prefer, is color picking. In this method, all selectable objects are rendered to a texture with a unique color. When clicking anywhere on the screen, the color of the texture at that position is read and because every object has a unique color, we know which object has been picked.
The biggest advantage of this technique is that you can select objects pixel perfectly (or technically almost pixel perfectly, because it's generally good enough to do the extra render pass at half or even lower resolution). You would need to go down to per-triangle intersections when using ray casting to achieve a similar level of precision. The biggest drawback of color picking is that you need another render pass to render the objects to a texture. In the worst case where all objects are selectable, this will double the number of draw calls. If you don't need hoverable objects, you can negate this problem by only rendering the objects to the texture when the mouse is actually clicked.
Performance-wise, it probably depends on the exact situation whether ray casting or color picking is faster, but in my opinion color picking is more versatile and more precise with less effort (and certainly a lot less math).
A normal 3D scene with the color picking texture at the top:
# Implementation
As mentioned above, color picking requires rendering the selectable objects to a texture. Rendering to textures in WebGL is done with framebuffers. Framebuffers can be used to render objects off-screen, which is important for many features like shadow mapping, post-processing effects, dynamic reflections, and more. I hadn't added framebuffers to my WebGL engine yet, but implementing them wasn't difficult because I could just port my C++/OpenGL version to JavaScript/WebGL.
I have also previously implemented color picking in C++/OpenGL, so I was able to use it as a starting point. However, that implementation was very specific to my map editor and I had to make a lot of changes to make it more flexible for my WebGL engine. In my existing version, all objects in the editor were clickable and the backend of the editor handled all of the logistics for it. One of the changes I wanted to make was making the pickability optional and easily controllable on a per-object basis. I added a superclass for all 3D objects which has an addOnClickHandler( onClickCallback ) function. When the function is called the object gets added to the color picker and only the added objects are rendered to the texture (actually now that I'm writing this, I might have to render all objects anyway, because another non-selectable object could cover a selectable one, which would mean it would be possible to click through it).
```var testCube = new Sparrow.Cube( engine , 0 , 0 , 0 , 1 , 1 , 1 , {color: Color4b.WHITE} ); testCube.addOnClickHandler( () => { console.log( "testCube clicked" ); } );```
One of the annoying problems was that there are different ways of rendering objects, most notably vertex-skinned objects which have to be animated in the vertex shader. This meant that the color picker had to have different shaders for different types of objects. Luckily, the shaders are very simple because they only care about the positions. The vertex shader also has a uniform for the modelID. The ID is converted to a color that is passed to the fragment shader, which is quite literally the simplest fragment shader you can have.
```var vs = `#version 300 es layout ( location = 0 ) in vec3 vertex_position; uniform mat4 u_M_mvp; uniform uint u_modelID; out vec4 color; void main() { gl_Position = u_M_mvp * vec4( vertex_position , 1 ); // convert the model ID to a color color.r = float( ( u_modelID & uint(0x000000ff) ) >> 0 )/255.0; color.g = float( ( u_modelID & uint(0x0000ff00) ) >> 8 )/255.0; color.b = float( ( u_modelID & uint(0x00ff0000) ) >> 16 )/255.0; color.a = 1.0; }`; var fs = `#version 300 es precision mediump float; in vec4 color; out vec4 fragColor; void main() { fragColor = color; }`;```
When the mouse is clicked, the color of the texture at that position is read and converted back into an ID, which is just the index of the model in the list and the callback function is triggered.
```pick( mouseX , mouseY ) { this.framebuffer.use(); this.engine.gl.pixelStorei( this.engine.gl.UNPACK_ALIGNMENT , 1 ); var data = new Uint8Array( 4 ); // read the texture data at the mouse position, flip the y coordinate this.engine.gl.readPixels( mouseX/2 , (this.framebuffer.height*2-mouseY)/2 , 1 , 1 , this.engine.gl.RGBA , this.engine.gl.UNSIGNED_BYTE , data ); // if the white background is hit, return if ( data[0] == 255 && data[1] == 255 && data[2] == 255 ) { this.engine.gl.bindFramebuffer( this.engine.gl.FRAMEBUFFER , null ); return; } // convert the color back to an index var objectIndex = data[2]*65536 + data[1]*256 + data[0]; if ( this.objects[objectIndex].onclick !== undefined ) this.objects[objectIndex].onclick(); // restore the default framebuffer this.engine.gl.bindFramebuffer( this.engine.gl.FRAMEBUFFER , null ); }```
For the most part, my color picking implementation works quite well, but there are some code design aspects that I would like to change. Especially the support for imported GLTF models is a hacky solution at the moment. As I mentioned in the two previous blog posts, GLTF is a great format, but it goes against some of my preferences for graphics programming. I'll have to make a decision on how I want to handle GLTF imports in my engine.
The color picking texture in higher resolution, you can see the different shades of red of the cubes:
Color picking is a very versatile method to select objects in 3D space. Like everything, there are pros and cons and there are situations where ray casting would be better, but overall it's a good solution that's going to work for most projects I want to work on.
by Christian - 29.06.2023 - 20:41 | 1,989 | 8,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-38 | latest | en | 0.94939 |
http://www.transum.org/software/SW/Starter_of_the_day/Similar_Thumbnails.asp?ID_Topic=27 | 1,561,550,138,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000306.84/warc/CC-MAIN-20190626114215-20190626140215-00328.warc.gz | 314,047,465 | 8,899 | There are 366 different Starters of The Day, many to choose from. You will find below some starters on the topic of Number. A lesson starter does not have to be on the same topic as the main part of the lesson or the topic of the previous lesson. It is often very useful to revise or explore other concepts by using a starter based on a totally different area of Mathematics.
Main Page
### Number Starters:
Four different actions depending on the number which appears.
A game based around the concept of factors and abundant numbers.
Add up a sequence of consecutive numbers. Can you find a quick way to do it?
Add up all the numbers in the nine times table.
This activity requires students to memorise fifteen numbers in a three by five grid.
Work out the number of stamps needed to post a parcel.
Work out the date Will was born by answering some number questions.
Recognise odd, even, square, prime and triangular numbers.
If all the bells ring together at noon, at what time will they next all ring together? This problem requires the use of LCM.
How many palindromic numbers can you find?
Can you use the digits on the left of this clock along with any mathematical operations to equal the digits on the right?
Fifteen pennies are placed in four envelopes and the envelopes are sealed. It is possible to pay someone any amount from 1p to 15p by giving them one or more envelopes. How were the pennies distributed between the envelopes?
What do you notice about the difference between the squares of consecutive numbers?
Work out how many people were at the dance from the clues given.
Determine whether the given nets would fold to produce a dice.
Arrange the digits from 1 to 9 in alphabetical order. How many times can this number be halved?
How old is a person if when her age is divided by certain numbers, the calculator display ending are as shown.
Begin with one, double it, double it again and so on. How many numbers in this sequence can you write down before the register has been called?
Spot the factors and the multiples amongst the numbers in the grid.
If each number in a sequence must be a factor or multiple of the previous number what is the longest sequence that can be made from the given numbers?
Use the flowchart to generate a sequence of numbers. Which number will reach 1 the fastest?
Find four single digit numbers that multiply together to give 120. How many different ways are there of answering this question?
For mathematical questions to get everyone thinking at the beginning of the lesson.
Start with 512. Halve it to get 256. Halve it to get 128. Continue as far as possible.
If all the students in this room shook hands with each other, how many handshakes would there be altogether?
Move the numbered cards to form five 2 digit numbers which are all multiples of three.
A puzzle about the lifts in a hotel which serve floors based on the day of the week.
The numbers on five houses next to each other add up to 70. What are those five numbers?
Write down as many multiples of 3.5 as possible in 3.5 minutes.
The 31st of December is the last day of the year. What mathematical lasts do you know?
A question about the birthdays of a child born on the 29th February.
Questions about the number of letters in numbers.
Which numbers when multiplied by the number of letters in the word(s) of the number give square numbers?
Find the missing terms from these linear sequences.
Work out what the nth letter will be in a recurring pattern of letters in a person's name
Perform calculations involving negative numbers
Find which numbers in a given list do not combine with other numbers on the list to make a given sum.
Can you work out the numbers from the given clues.
Write out in words some numbers writen as digits (optional pirate theme)
From the numbers given, find the one that is the odd one out.
Find other numbers that can be changed to 1 on a calculator using only the 4 key and any operation.
Find three numbers such that each pair of numbers adds up to a square number.
Six is a perfect number as it is the sum of its factors. Can you find any other perfect numbers?
Arrange numbers on the plane shaped grid to produce the given totals
Arrange numbers at the bottom of the pyramid which will give the largest total at the top.
When the register is called answer with a multiple of 7.
Arrange the numbers on the snowballs so that no two consecutive numbers are directly connected by rope.
Rearrange the numbers, row and column headings so that this table is mathematically correct.
Can you work out from the five clues given what the mystery number is?
How many square numbers can be found in the grid of digits.
Continue the sequences if you can work out the rule.
The classic game of Nim played with a group of pens and pencils. The game can be extended to the multi-pile version.
Arrange the digits one to nine in the grid so that they obey the row and column headings.
Arrange the numbers on the cards so that each of the three digit numbers formed horizontally are square numbers and each of the three digit numbers formed vertically are even.
Find a trapezium, a triangle and a quadrilateral where all of the angles are square numbers.
Draw a picture of a Christmas tree using only square numbers.
Arrange the numbered trees so that adjacent sums are square numbers.
Write out as many square numbers as possible in 4 minutes.
A challenge to find numbers which have each of their digits as square numbers.
Find a power of 2 and a power of 3 that are consecutive numbers.
Be creative and come up with as many facts about a number as you can think of.
How can you put the dice into the tins so that there is an odd number of dice in each tin?
A Maths puzzle based on the 12 Days of Christmas song.
Find the two numbers whose sum and product are given.
Work out the phone number from the clues given.
Arrange numbers on the Venn Diagram according to their properties.
A starter about sums, products, differences, ratios, square and prime numbers.
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For All: | 1,401 | 6,721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-26 | longest | en | 0.954315 |
http://forum.nextstepmcat.com/viewtopic.php?f=10&t=2252&p=5534 | 1,601,554,469,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131412.93/warc/CC-MAIN-20201001112433-20201001142433-00127.warc.gz | 48,669,475 | 5,957 | ## NS FL#6 C/P Q42
sdaswani
Posts: 37
Joined: Sat Jan 26, 2019 5:32 pm
### NS FL#6 C/P Q42
Hi! I wanted to make sure my understanding of the solution to this problem is correct. I understand how we get 35g of FDG that underwent the radioactive decay. From here, 35g = 35g / 181g per mol to get the number of moles of FDG that underwent decay. Moving forward, is it true that each particle of the total number of moles of FDG, underwent the decay and released 2 gamma rays of 511 keV, each?
Why do we think of decay per particle instead of per mole?
Thanks!
-Saakshi
NS_Tutor_Will
Posts: 766
Joined: Fri May 25, 2018 9:15 am
### Re: NS FL#6 C/P Q42
Yes, your reasoning is correct! We don't necessarily have to think of decay as a per particle situation, we are perfectly within our rights to think of it on a molar scale (after all, moles just refers to a whole bunch of individual particles so a mole should act pretty much the same as a particle, just on a larger scale!).
In this case, since we're given answer choices in keV and since we're told about gamma rays in terms of particles ("2 gamma rays are emitted, each with 511 keV"), it makes the most sense to think on the scale of particle for the sake of efficiency.
Thanks for the question!
sdaswani
Posts: 37
Joined: Sat Jan 26, 2019 5:32 pm
### Re: NS FL#6 C/P Q42
Got it, thanks for the help!! | 387 | 1,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-40 | latest | en | 0.93329 |
https://www.gradesaver.com/textbooks/math/other-math/CLONE-547b8018-14a8-4d02-afd6-6bc35a0864ed/chapter-5-ratio-and-proportion-mixed-review-exercises-page-374/12 | 1,721,033,176,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514680.75/warc/CC-MAIN-20240715071424-20240715101424-00708.warc.gz | 709,696,629 | 12,620 | ## Basic College Mathematics (10th Edition)
$\frac{15}{4}$
We can write the given ratio as a fraction and then simplify in lowest terms. Since 1 yard has 3 feet, we can write 10 yards as $(10\times3)=30$ feet. $\frac{30feet}{8feet}$ Since both the denominator and numerator have the same unit, we can divide out the common unit. $\frac{30}{8}=\frac{30\div2}{8\div2}=\frac{15}{4}$ | 117 | 380 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-30 | latest | en | 0.874487 |
https://en.wikipedia.org/wiki/Squared_triangular_number | 1,695,315,754,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506028.36/warc/CC-MAIN-20230921141907-20230921171907-00625.warc.gz | 269,195,575 | 32,947 | # Squared triangular number
In number theory, the sum of the first n cubes is the square of the nth triangular number. That is,
${\displaystyle 1^{3}+2^{3}+3^{3}+\cdots +n^{3}=\left(1+2+3+\cdots +n\right)^{2}.}$
The same equation may be written more compactly using the mathematical notation for summation:
${\displaystyle \sum _{k=1}^{n}k^{3}={\bigg (}\sum _{k=1}^{n}k{\bigg )}^{2}.}$
This identity is sometimes called Nicomachus's theorem, after Nicomachus of Gerasa (c. 60 – c. 120 CE).
## History
Nicomachus, at the end of Chapter 20 of his Introduction to Arithmetic, pointed out that if one writes a list of the odd numbers, the first is the cube of 1, the sum of the next two is the cube of 2, the sum of the next three is the cube of 3, and so on. He does not go further than this, but from this it follows that the sum of the first n cubes equals the sum of the first ${\displaystyle n(n+1)/2}$ odd numbers, that is, the odd numbers from 1 to ${\displaystyle n(n+1)-1}$. The average of these numbers is obviously ${\displaystyle n(n+1)/2}$, and there are ${\displaystyle n(n+1)/2}$ of them, so their sum is ${\displaystyle {\bigl (}n(n+1)/2{\bigr )}^{2}.}$
Many early mathematicians have studied and provided proofs of Nicomachus's theorem. Stroeker (1995) claims that "every student of number theory surely must have marveled at this miraculous fact". Pengelley (2002) finds references to the identity not only in the works of Nicomachus in what is now Jordan in the first century CE, but also in those of Aryabhata in India in the fifth century, and in those of Al-Karaji circa 1000 in Persia. Bressoud (2004) mentions several additional early mathematical works on this formula, by Al-Qabisi (tenth century Arabia), Gersonides (circa 1300 France), and Nilakantha Somayaji (circa 1500 India); he reproduces Nilakantha's visual proof.
## Numeric values; geometric and probabilistic interpretation
The sequence of squared triangular numbers is[1]
0, 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025, 4356, 6084, 8281, ... .
These numbers can be viewed as figurate numbers, a four-dimensional hyperpyramidal generalization of the triangular numbers and square pyramidal numbers.
As Stein (1971) observes, these numbers also count the number of rectangles with horizontal and vertical sides formed in an n × n grid. For instance, the points of a 4 × 4 grid (or a square made up of three smaller squares on a side) can form 36 different rectangles. The number of squares in a square grid is similarly counted by the square pyramidal numbers.
The identity also admits a natural probabilistic interpretation as follows. Let X, Y, Z, W be four integer numbers independently and uniformly chosen at random between 1 and n. Then, the probability that W is the largest of the four numbers equals the probability that Y is at least as large as X and that W is at least as large as Z. That is, ${\displaystyle P[\max(X,Y,Z)\leq W]=P[X\leq Y\wedge Z\leq W]}$. For any particular value of W, the combinations of X, Y, and Z that make W largest form a cube 1 ≤ X, Y, Zn so (adding the size of this cube over all choices of W) the number of combinations of X, Y, Z, W for which W is largest is a sum of cubes, the left hand side of the Nichomachus identity. The sets of pairs (X,Y) with XY and of pairs (Z,W) with ZW form isosceles right triangles, and the set counted by the right hand side of the equation of probabilities is the Cartesian product of these two triangles, so its size is the square of a triangular number on the right hand side of the Nichomachus identity. The probabilities themselves are respectively the left and right sides of the Nichomachus identity, normalized to make probabilities by dividing both sides by n4.[citation needed]
## Proofs
Charles Wheatstone (1854) gives a particularly simple derivation, by expanding each cube in the sum into a set of consecutive odd numbers. He begins by giving the identity
${\displaystyle n^{3}=\underbrace {\left(n^{2}-n+1\right)+\left(n^{2}-n+1+2\right)+\left(n^{2}-n+1+4\right)+\cdots +\left(n^{2}+n-1\right)} _{n{\text{ consecutive odd numbers}}}.}$
That identity is related to triangular numbers ${\displaystyle T_{n}}$ in the following way:
${\displaystyle n^{3}=\sum _{k=T_{n-1}+1}^{T_{n}}(2k-1),}$
and thus the summands forming ${\displaystyle n^{3}}$ start off just after those forming all previous values $\displaystyle 1^3$ up to ${\displaystyle (n-1)^{3}}$. Applying this property, along with another well-known identity:
${\displaystyle n^{2}=\sum _{k=1}^{n}(2k-1),}$
produces the following derivation:
{\displaystyle {\begin{aligned}\sum _{k=1}^{n}k^{3}&=1+8+27+64+\cdots +n^{3}\\&=\underbrace {1} _{1^{3}}+\underbrace {3+5} _{2^{3}}+\underbrace {7+9+11} _{3^{3}}+\underbrace {13+15+17+19} _{4^{3}}+\cdots +\underbrace {\left(n^{2}-n+1\right)+\cdots +\left(n^{2}+n-1\right)} _{n^{3}}\\&=\underbrace {\underbrace {\underbrace {\underbrace {1} _{1^{2}}+3} _{2^{2}}+5} _{3^{2}}+\cdots +\left(n^{2}+n-1\right)} _{\left({\frac {n^{2}+n}{2}}\right)^{2}}\\&=(1+2+\cdots +n)^{2}\\&={\bigg (}\sum _{k=1}^{n}k{\bigg )}^{2}.\end{aligned}}}
Row (1893) obtains another proof by summing the numbers in a square multiplication table in two different ways. The sum of the ${\displaystyle i}$th row is ${\displaystyle i}$ times a triangular number, from which it follows that the sum of all the rows is the square of a triangular number. Alternatively, one can decompose the table into a sequence of nested gnomons, each consisting of the products in which the larger of the two terms is some fixed value. The sum within each gmonon is a cube, so the sum of the whole table is a sum of cubes.
In the more recent mathematical literature, Edmonds (1957) provides a proof using summation by parts. Stein (1971) uses the rectangle-counting interpretation of these numbers to form a geometric proof of the identity (see also Benjamin, Quinn & Wurtz 2006); he observes that it may also be proved easily (but uninformatively) by induction, and states that Toeplitz (1963) provides "an interesting old Arabic proof". Kanim (2004) provides a purely visual proof, Benjamin & Orrison (2002) provide two additional proofs, and Nelsen (1993) gives seven geometric proofs.
## Generalizations
A similar result to Nicomachus's theorem holds for all power sums, namely that odd power sums (sums of odd powers) are a polynomial in triangular numbers. These are called Faulhaber polynomials, of which the sum of cubes is the simplest and most elegant example. However, in no other case is one power sum a square of another.[2]
Stroeker (1995) studies more general conditions under which the sum of a consecutive sequence of cubes forms a square. Garrett & Hummel (2004) and Warnaar (2004) study polynomial analogues of the square triangular number formula, in which series of polynomials add to the square of another polynomial.
## Notes
1. ^ Sloane, N. J. A. (ed.), "Sequence A000537", The On-Line Encyclopedia of Integer Sequences, OEIS Foundation
2. ^ | 1,983 | 7,012 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 19, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 1} | 4.1875 | 4 | CC-MAIN-2023-40 | latest | en | 0.889204 |
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For All: | 735 | 3,626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-38 | latest | en | 0.947612 |
http://www.sun-sentinel.com/news/opinion/tn-vsl-1124-thanksgiving-meal-health,0,7081748.story | 1,410,777,527,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657104131.95/warc/CC-MAIN-20140914011144-00063-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 860,587,918 | 25,782 | What’s on your holiday menu? It seems holiday cooking just keeps getting richer. But there is a way to cook delicious food that will delight your friends and family without the worry of added pounds that make the season even more stressful.
Let’s remind ourselves of what we are dealing with. Calorie counting helps us quantify food. Calories don’t specifically tell us what’s in the food, but they give us an idea of what it might take to burn off that food.
For example, if one cup of turkey stuffing has 360 calories and it’s alongside a six-ounce serving of turkey at 340 calories, we’re already up to 700 calories. Add 300 calories for mashed potatoes and gravy, 150 calories for a half-cup of candied sweet potatoes and 225 for green bean casserole, that brings us up to 1,375. If you are inclined to add a dinner roll at 110 calories and slice of apple pie at 410, we have a new total — drum roll please — 1,895 calories.
For many of us, that’s enough calories for a whole day. And when all those foods hit the belly at the same time, it’s a digestive nightmare that sends you to couch groaning. Keep in mind those calories will need to be used or burned. And if brisk walking burns 100 calories per mile, I’ll let you do the math.
All this in the name of tradition!
But tradition is memorable — it adds to the fun of the holiday menu. And it’s tough to explain to Uncle Larry that his favorite candied yams have been cut from the roster. Or to tell Aunt Lucille she’s not getting her favorite apple pie. So maybe we can’t (without a mutiny) omit the family favorites. But we certainly can lighten them up a bit.
If you make your stuffing by cutting the butter in half and then adding that same amount of chicken broth back in, the difference is almost undetectable. Pumpkin pie made by reducing the sugar a quarter-cup and adding a quarter-cup agave nectar is delicious. Your family won’t know you made the adjustment. Be sure to reduce the cream or milk by one tablespoon so the pie will set nicely.
Rather than add butter and cream to the mashed potatoes, use chicken broth, a little rosemary (a quarter-teaspoon per about eight potatoes) or a hint of onion powder to bring out the flavor of the potato. Do we really need to add sour cream and butter to something that is going to have gravy poured over the top? With a few minor adjustments such as these, you can save hundreds of calories. I guess you could say it’s health by stealth.
Those delicious traditional holiday meals you are looking forward to will be just as delicious, and with a little careful planning, a whole lot lower in calories. And maybe when you make it to the couch post-meal, you’ll be smiling, instead of groaning.
I’ll see you in two weeks.
Love & health,
Loa
LOA BLASUCCI is an author and owner of All Health’s Breaking Loose Wellness Retreats. Blasucci’s book, “All Health’s Breaking Loose,” was named a best-pick finalist in the Health: Exercise & Fitness category for 2011 by USA Book News. Find more at www.gotoloa.com. | 687 | 3,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | latest | en | 0.921917 |
https://edurev.in/studytube/Soil-Compaction/a2e6d12b-614e-438d-be08-6b659f4619a7_t | 1,674,887,876,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00071.warc.gz | 238,199,874 | 60,258 | Soil Compaction
Soil Compaction - Notes | Study Soil Mechanics - Civil Engineering (CE)
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Introduction
Compaction is the application of mechanical energy to soil so as to rearrange its particles and reduce the void ratio.
It is applied to improve the properties of existing soil or in the process of placing fill such as in the construction of embankments, road bases, runways, earth dams, and reinforced earth walls. Compaction is also used to prepare a level surface during construction of buildings. There is usually no change in the water content and in the size of the individual soil particles.
The objectives of compaction are:
• To increase soil shear strength and therefore its bearing capacity.
• To reduce subsequent settlement under working loads.
• To reduce soil permeability making it more difficult for water to flow through.
Laboratory Compaction
The variation in compaction with water content and compactive effort is first determined in the laboratory. There are several tests with standard procedures such as:
• Indian Standard Light Compaction Test (similar to Standard Proctor Test/Light Compaction Test)
• Indian Standard Heavy Compaction Test (similar to Modified Proctor Test/Heavy Compaction Test)
• Indian Standard Light Compaction Test
Soil is compacted into a 1000 cm3 mould in 3 equal layers, each layer receiving 25 blows of a 2.6 kg rammer dropped from a height of 310 mm above the soil. The compaction is repeated at various moisture contents.
• Indian Standard Heavy Compaction Test
It was found that the Light Compaction Test (Standard Test) could not reproduce the densities measured in the field under heavier loading conditions, and this led to the development of the Heavy Compaction Test (Modified Test). The equipment and procedure are essentially the same as that used for the Standard Test except that the soil is compacted in 5 layers, each layer also receiving 25 blows. The same mould is also used. To provide the increased compactive effort, a heavier rammer of 4.9 kg and a greater drop height of 450 mm are used.
Compactive energy applied per unit
Volume = wHnN/V
• The ratio of total energy given in heavy compaction test to that given in light compaction test
Dry Density - Water Content Relationship
• To assess the degree of compaction, it is necessary to use the dry unit weight, which is an indicator of compactness of solid soil particles in a given volume.
• Laboratory testing is meant to establish the maximum dry density that can be attained for a given soil with a standard amount of compactive effort.
In the test, the dry density cannot be determined directly, and as such the bulk density and the moisture content are obtained first to calculate the dry density as
Yd = Yt/(1+w)
where γd = bulk density, and w = water content.
• A series of samples of the soil are compacted at different water contents, and a curve is drawn with axes of dry density and water content. The resulting plot usually has a distinct peak as shown. Such inverted “V” curves are obtained for cohesive soils (or soils with fines), and are known as compaction curves.
• Dry density can be related to water content and degree of saturation (S) as
Thus, it can be visualized that an increase of dry density means a decrease of voids ratio and a more compact soil.
Similarly, dry density can be related to percentage air voids (na) as
• Relation between moisture content and dry unit weight for a saturated soil is the zero air-voids line. It is not feasible to expel air completely by compaction, no matter how much compactive effort is used and in whatever manner.
Effect of Increasing Water Content
• As water is added to a soil at low moisture contents, it becomes easier for the particles to move past one another during the application of compacting force. The particles come closer, the voids are reduced and this causes the dry density to increase. As the water content increases, the soil particles develop larger water films around them.
• This increase in dry density continues till a stage is reached where the water starts occupying the space that could have been occupied by the soil grains. Thus the water at this stage hinders the closer packing of grains and reduces the dry unit weight. The maximum dry density (MDD) occurs at an optimum water content (OMC), and their values can be obtained from the plot.
Effect of Increasing Compactive Effort
• The effect of increasing compactive effort is shown. Different curves are obtained for different compactive efforts. A greater compactive effort reduces the optimum moisture content and increases the maximum dry density.
• An increase in compactive effort produces a very large increase in dry density for soil when it is compacted at water contents drier than the optimum moisture content.It should be noted that for moisture contents greater than the optimum, the use of heavier compaction effort will have only a small effect on increasing dry unit weights.
It can be seen that the compaction curve is not a unique soil characteristic. It depends on the compaction effort. For this reason, it is important to specify the compaction procedure (light or heavy) when giving values of MDD and OMC.
Factors Affecting Compaction
The factors that influence the achieved degree of compaction in the laboratory are:
• Plasticity of the soil
• Water content
• Compactive effort
Compaction of Cohesionless Soils
For cohesionless soils (or soils without any fines), the standard compaction tests are difficult to perform. For compaction, application of vibrations is the most effective method. Watering is another method. To achieve maximum dry density, they can be compacted either in a dry state or in a saturated state.
• For these soil types, it is usual to specify a magnitude of relative density (ID) that must be achieved. If e is the current void ratio or gd is the current dry density, the relative density is usually defined in percentage as
ID = (emax - e)/emax-emin) x 100
or
ID = (γdmaxddmin)/γddmaxdmin)) x 100
where emax and emin are the maximum and minimum void ratios that can be determined from standard tests in the laboratory, and gdmin and gdmax are the respective minimum and maximum dry densities
On the basis of relative density, sands and gravels can be grouped into different categories:
It is not possible to determine the dry density from the value of the relative density. The reason is that the values of the maximum and minimum dry densities (or void ratios) depend on the gradation and angularity of the soil grains.
Engineering Behaviour of Compacted Soils
The water content of a compacted soil is expressed with reference to the OMC. Thus, soils are said to be compacted dry of optimum or wet of optimum (i.e. on the dry side or wet side of OMC). The structure of a compacted soil is not similar on both sides even when the dry density is the same, and this difference has a strong influence on the engineering characteristics.
1. Soil Structure
For a given compactive effort, soils have a flocculated structure on the dry side (i.e. soil particles are oriented randomly), whereas they have a dispersed structure on the wet side (i.e. particles are more oriented in a parallel arrangement perpendicular to the direction of applied stress). This is due to the well-developed adsorbed water layer (water film) surrounding each particle on the wet side.
2. Swelling
Due to a higher water deficiency and partially developed water films in the dry side, when given access to water, the soil will soak in much more water and then swell more.
3. Shrinkage
During drying, soils compacted in the wet side tend to show more shrinkage than those compacted in the dry side. In the wet side, the more orderly orientation of particles allows them to pack more efficiently.
4. Construction Pore Water Pressure
The compaction of man-made deposits proceeds layer by layer, and pore water pressures are induced in the previous layers. Soils compacted wet of optimum will have higher pore water pressures compared to soils compacted dry of optimum, which have initially negative pore water pressure.
5. Permeability
The randomly oriented soil in the dry side exhibits the same permeability in all directions, whereas the dispersed soil in the wet side is more permeable along particle orientation than across particle orientation.
6. Compressibility
At low applied stresses, the dry compacted soil is less compressible on account of its truss-like arrangement of particles whereas the wet compacted soil is more compressible.
The stress-strain curve of the dry compacted soil rises to a peak and drops down when the flocculated structure collapses. At high applied stresses, the initially flocculated and the initially dispersed soil samples will have similar structures, and they exhibit similar compressibility and strength.
1. Coarse grained well graded – Higher γd
2. In clays with higher plasticity - γd decrease
3. V shape due to bulking of pure sand
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; | 2,032 | 9,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-06 | latest | en | 0.93408 |
https://oeis.org/A183849 | 1,656,741,225,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103984681.57/warc/CC-MAIN-20220702040603-20220702070603-00425.warc.gz | 488,272,162 | 4,142 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A183849 Number of (n+1)X5 0..3 arrays with each element of every 2X2 subblock being the sum mod 4 of two others 1
1011, 7453, 54724, 448665, 3608259, 31362526, 265093125, 2373353909, 20638236376, 187666794149, 1661780877723, 15247254970726, 136761251939361, 1262026958466329, 11430228141457760, 105893347391922493 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Column 4 of A183854 LINKS R. H. Hardin, Table of n, a(n) for n = 1..200 FORMULA Empirical: a(n)=23*a(n-1)-9*a(n-2)-3507*a(n-3)+21096*a(n-4)+174917*a(n-5)-1900055*a(n-6)-1961167*a(n-7)+75541687*a(n-8)-127919456*a(n-9)-1579840485*a(n-10)+5777014365*a(n-11)+16616850370*a(n-12)-111097659712*a(n-13)-35392333487*a(n-14)+1211798824327*a(n-15)-1233130989796*a(n-16)-7903370136499*a(n-17)+16844625431337*a(n-18)+28729264344565*a(n-19)-111687567264255*a(n-20)-28452881458578*a(n-21)+450312030388051*a(n-22)-247590266103201*a(n-23)-1137833730709943*a(n-24)+1390913694652243*a(n-25)+1661750316286384*a(n-26)-3677312849659786*a(n-27)-780498336200130*a(n-28)+5806380843135826*a(n-29)-1822449102937708*a(n-30)-5563823547915312*a(n-31)+4089617050561760*a(n-32)+2906871136869272*a(n-33)-3917767446899248*a(n-34)-366370994803440*a(n-35)+2071375512621440*a(n-36)-468867293972480*a(n-37)-585280887597632*a(n-38)+286971758170560*a(n-39)+62114385639424*a(n-40)-66011067399680*a(n-41)+6350408017920*a(n-42)+5165343129600*a(n-43)-1516747161600*a(n-44)+116785152000*a(n-45) EXAMPLE Some solutions for 3X5 ..1..3..3..3..1....2..2..2..2..2....1..2..1..3..1....2..3..2..1..1 ..1..2..1..2..1....2..0..0..2..0....3..3..3..2..3....1..1..1..3..2 ..3..3..3..3..1....0..2..2..2..2....1..2..1..3..1....2..3..2..3..1 CROSSREFS Sequence in context: A213315 A345906 A035125 * A185881 A115769 A267613 Adjacent sequences: A183846 A183847 A183848 * A183850 A183851 A183852 KEYWORD nonn AUTHOR R. H. Hardin Jan 07 2011 STATUS approved
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Last modified July 2 01:52 EDT 2022. Contains 354984 sequences. (Running on oeis4.) | 942 | 2,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-27 | latest | en | 0.291875 |
https://lornpearsontrains.co.uk/2021/12/09/my-2021-swim-goal-115-miles/ | 1,642,909,197,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303956.14/warc/CC-MAIN-20220123015212-20220123045212-00122.warc.gz | 405,840,846 | 43,898 | ## My 2021 Swim Goal – 115 miles
Here’s an update of my swimming distance goal for 2021 and how much I’m aiming to swim in the remainder of this month to achieve it.
Swimming Goal – 115 miles for 2021
(6 months from June – December)
115 miles = 185,075 metres
Swimming completed as at 9th December = 166,050 metres
Swimming metres remaining = 19,025 metres
Swimming sessions remaining = x 7
7 x 2,700m each time = 18,900 metres
Leaving 125m remaining (across 7 sessions)
So in my 7 sessions remaining in December, if I swim a minimum of 2,700m each time, and do 125m extra, I’ll reach my goal of 115 miles or 185,075 metres / 115 miles for the year!
Looking towards the end of the year, if all goes to plan it will be a total of:
5,687 lengths of the pools I’ve been swimming in (I’m at 4,980 so far) and
14 open water swims, averaging around 2,200m / 50 mins each time.
It’s been great to have got back into swimming again and I’m looking forward to more pool and open water swimming and events in 2022.
This entry was posted in Uncategorized. Bookmark the permalink. | 300 | 1,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | longest | en | 0.956773 |
https://samacheerguru.com/samacheer-kalvi-9th-english-solutions-prose-chapter-6/ | 1,716,735,253,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058956.26/warc/CC-MAIN-20240526135546-20240526165546-00813.warc.gz | 427,588,307 | 25,432 | Samacheer Kalvi 9th English Solutions Prose Chapter 6 From Zero to Infinity
You can Download From Zero to Infinity Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 9th English Book Solutions Guide Pdf Prose Chapter 6 help you to revise complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
Tamilnadu Samacheer Kalvi 9th English Solutions Prose Chapter 6 From Zero to Infinity
From Zero To Infinity 9th Standard English Warm Up:
From Zero To Infinity Summary Question 1.
Did you enjoy solving this?
Yes.
From Zero To Infinity Prose Summary Question 2.
Was it easy or hard to solve?
It was very easy to solve.
From Zero To Infinity 9th Standard English Question 3.
Do you like Mathematics? Give reasons.
I like mathematics because it is fun to solve and find the right solutions. It is very useful in our daily lives.
OR
I do not like mathematics because it is complicated and too hard to understand.
From Zero to Infinity Intext Questions
From Zero To Infinity Questions And Answers Question 1.
What was the reaction of the classmates to Ramanujan’s question?
The classmates laughed at Ramanujan’s question.
Zero To Infinity Lesson Summary Question 2.
What did the Indian mathematician Bhaskara prove?
The Indian mathematician Bhaskara proved that zero divided by zero is infinity.
From Zero To Infinity Book Back Answers Question 3.
Where did Ramanujan get S.L. Loney’s book on Trigonometry?
Ramanujan got Loney’s “Trigonometry” book from a college library.
From Zero To Infinity Essay Question 4.
Where did Ramanujan do his mathematical problems?
Ramanujan did his mathematical problems on loose sheets of paper or on a slate.
From Zero To Infinity Lesson Plan Question 5.
What were the subjects neglected by Ramanujan in college?
History, English, Physiology were the subjects neglected by Ramanujan in college.
From Zero To Infinity Lesson Summary Question 6.
Which University granted him a fellowship of H75 a month?
University of Madras granted him a fellowship of? 75 a month.
Summary Of From Zero To Infinity Question 7.
What did Ramanujan send to G.H. Hardy?
Ramanujan sent a letter in which he set out 120 theorems and formulae to G.H Hardy.
From Zero To Infinity Summary In Tamil Question 8.
Who discovered a rare mathematical genius in Ramanujan?
G.H Hardy and his colleague J.E. Littlewood discovered a rare mathematical genius in Ramanujan.
From Zero to Infinity Textual Questions
A. Answer the following questions in a sentence or two.
From Zero To Infinity Mind Map Question 1.
Why did the students laugh at Ramanujan?
The students laughed at Ramanujan because he asked if no banana was distributed among no one, would every one get one banana.
Question 2.
Why did the teacher compliment Ramanujan?
The teacher complimented Ramanuj for asking a question that took centuries for mathematicians to answer.
Question 3.
Question What did Ramanujan do after reading the book on Trigonometry?
After reading the book on Trigonometry, Ramanujan began his own research. He came forth with many mathematical theorems and formulae not given in the book
Question 4.
What disappointed Ramanujan’s father?
Ramanujan failed twice in his first year arts examination in college as he neglected other subjects such as History, English and Physiology. This disappointed his father.
Question 5.
How did Ramanujan manage his paper crisis?
Ramanujan needed about 2,000 sheets of paper every month. He started using even scraps of paper he found lying on the streets. Sometimes he used a red pen to write over what was written in blue ink.
Question 6.
Why were Ramanujan’s application for jobs rejected?
Ramanujan would show his frayed notebooks to every officers. But no one could understand what was written in the notebooks. So, his applications for jobs were rejected.
Question 7.
Why was Ramanujan sent back to India?
While Ramanujan continued his research work, Tuberculosis, then an incurable disease, was devouring him. So, he was sent back to India.
Question 1.
Who asked the intriguing question to the arithmetic teacher? What do you know about him?
The intriguing question was asked by Srinivasa Ramanujan. He was a native of Kumbakonam. Both during his school and research work at Cambridge, he was always ahead of his mathematics teachers.
Question 2.
Where and when was Ramanujan born? What do you know of his father?
Ramanujan was born in Erode in Tamil Nadu on December 22, 1887. His father was a petty clerk in a cloth shop.
Question 3.
What was the most significant turn in Ramanujan’s life?
The most significant turn came when one of Ramanujan’s senior friends showed him Synopsis of Elementary> Results in Pure Applied Mathematics by George Shoobridge Carr. Ramanujan was delighted than intimidated. This book triggered the mathematical genius in him and he began to work on the problems given in it.
Question 4.
What was Hardy’s opinion of Ramanujan?
According to Hardy, Ramanujan was an unsystematic mathematician, similar to one who knows the Pythagorus theorem but does not know what a congruent triangle means. He felt that many discrepancies in his research could be due to his lack of formal education.
Question 5.
How can you say that Ramanujan was multi-talented?
Besides Ramanujan being a mathematician, he was a reputed astrologer. He was also an excellent orator and many were eager to listen to his practical and intellectual talks. He used to give lectures on subjects like “God, Zero and Infinity”, topics that no ordinary man can easily indulge in. Hence we can surely say he was multi-talented.
B. Answer the following questions in about 80 -100 words.
Question 1.
Describe the life of Srinivasa Ramanujan in India.
Ramanujan was born in Erode in Tamilnadu on December 22,1887. From early childhood, it was evident that he was a prodigy. Senior students used to get his assistance in solving math problems. At the age of 13, he began his own research on Trigonometry. The book “Elementary Results in Pure Applied Mathematics” by George Shoobridge Carr triggered the genius in Ramanujan. He used to do problems on loose sheets and enter the results in notebooks which are now famous as “Ramanujan’s Frayed Notebooks”.
Although Ramanujan secured a first class in Mathematics in the matriculation examination and was awarded the Subramanyan Scholarship, he failed twice in his first year arts examination in college as he neglected other subjects such as History, English and Physiology. He searched for job for food and papers to do calculations. The Director of Madras Port Trust gave a clerical job to Ramanujan on a monthly salary of Rupees 25.
Question 2.
Narrate the association of Ramanujan with G.H. Hardy.
Ramanujan sent a letter to the great Mathematician G.H. Hardy of Cambridge University, in which he set out 120 theorems and formulae which included the Reimann Series. Hardy and his colleague Littlewood realized that they had discovered a rare mathematical genius.
They invited him to Britain.Despite the cold weather and food, Ramanujan continued his research with determination in the company of Hardy and Littlewood. Hardy found an unsystematic mathematician in Ramanujan due to his lack of formal education. Ramanujan’s achievements include the Hardy-Ramanujan-Littlewood circle method in number theory.
Question 1.
What prompted Ramanujan to ask an intriguing question?
The Mathematics teacher was teaching and solving concepts in division. She drew three bananas on the blackboard and she pointed to three boys who were there and asked them how many each would get. A smart student quickly answered that each would get one. After appreciating the students’ answer, the teacher introduced a similar instance of 1,000 bananas distributed among 1,000 boys where each would again get one. While the teacher was explaining, a boy seated in one comer who was none other than Ramanujan asked an intriguing question wherein if no banana was distributed among no one, would everyone still get one banana?
Question 2.
Mention the achievements of Srinivasa Ramanujan.
Ramanujan was elected Fellow of the Royal Society on February 28, 1918. In October, 1918 he became the first Indian to be elected Fellow of Trinity College, Cambridge. His achievements at Cambridge include the Hardy-Ramanujan-Littlewood circle method in number theory.
He is also popular with Roger-Ramanujan’s identities in partition of integers. A long list of the highest composite numbers, besides work on the number theory and the algebra of inequalities are also his noted achievements. In algebra his work on continued fractions is considered on par with great mathematicians like Leonard Euler and Jacobi.
Question 3.
What did Ramanujan do when his mind was flooded with ideas?
Mathematical ideas flooded Ramanujan’s mind. He was not able to write all of them down. ‘He solved problems on loose sheets of paper, slate and jotted the results down in notebooks. Before he went abroad he had filled three notebooks, later known as Ramanujan’s Frayed Notebooks.
His father who found him scribbling, mistook him to be mad. He had to find money for food and papers to do calculations. Every month, he needed at least 2,000 sheets of paper. He started using scraps of paper found lying on the streets. Sometimes, he even wrote using red ink over used papers found on the streets.
C. Match the words with correct Synonym and Antonym from the table.
Listening:
D. Listen to the anecdote “Two Geniuses” and narrate it in your own words.
Narration of “Two Geniuses”
There’s a story about how Dr. Albert Einstein was travelling to Universities in his car, delivering lectures on his theory of relativity. During one tired journey, his driver Hans remarked “Dr. Einstein, I have heard you deliver that lecture about 30 times. ,t I know it by heart and bet I could give it myself.”
“Well, I’ll give you the chance”, said the Dr. “They don’t know me at the next University, so when we get there, I’ll put on your cap, and you introduce yourself as Dr. Einstein and give the lecture.”
The driver delivered Einstein’s lecture without any mistakes. When he finished, he started to leave, but one of the professor stopped him and asked a complex question filled with mathematical equations and formulae. The driver thought fast. “The answer to that problem is so simple,” he said,” I’m surprised you have to ask me. In fact, to show you just how simple it is, I’m going to ask my driver Hans to come up here and answer your question”.
Speaking:
E. Divide the students into groups of five and conduct a group discussion on the topic “Importance of Mathematics in Our Everyday Life” The teacher will act as a moderator.
Group Discussion on Importance of Mathematics
Teacher: Good morning students! We have just learnt the life of the great mathematician Ramanujan. Now let’s have a group discussion on “Importance of Mathematics in Our Everyday Life”. Divide yourselves into groups of five.
Harsha (Group A) : The importance of maths in everyday life. Mathematics is a methodical application of matter. It is so said because the subject makes a man methodical or systematic. Mathematics makes our life orderly and prevents chaos.
Varsha (Group B) : In Hebrew, it’s root is “thinking.” They tell us that mathematics gives us the critical ability to learn and think logically in any field of endeavor. The skills of learning today are more important than knowledge, which is so readily available on the Internet.
Yusuf (Group C) : Math is an important part of our lives, because in the future you will get a job that deals with math. Math is pretty much in everything you do, really. Math is important because it is the most widely used subject in the world. Every career uses some sort of math.
Adhira (Group D) : Maths improves problem-solving abilities. Teaches clearer logical reasoning. Sharpens concentration and observance. Develops confidence and self-esteem.
Danny (Group E) : Knowing basic math principles keeps you from having to carry around a calculator because good use of math allows you to do many calculations in your head.
F. Answer the following questions based on the given passage.
Question 1.
What made John Shepherd-Barron to come up with the idea of ATM?
It was then John’s habit to withdraw money on a Saturday, but on this particular weekend he had arrived one minute late and found the bank doors locked against him. This made him to come up with idea of ATM.
Question 2.
When and where was the first ATM installed?
The first ATM was installed at a branch in the North London suburb of Enfield on June 27, 1967.
Question 3.
Who was the first person to withdraw cash from the ATM?
The first person to withdraw cash from the ATM was Reg Varney, a celebrity resident of Enfield known for his part in the number of popular television series.
Question 4.
Why did Shepherd-Barron reduce the PIN number from six digits to four?
Shepherd-Barron’s wife said that she could only remember four figures, because of her, four figures became the world standard.
Question 5.
Which theory of Ramanujan helps the ATMs to dispense cash?
Ramanujan’s Partition theory helps the ATMs to dispense cash.
Writing:
G. Paragraph Writing
Question 1.
Write a paragraph of 100-120 words about a memorable anecdote / incident of your life.
A memorable anecdote/incident in my life:
I was then a student of class four. One day, I was left at home with my grandmother. It was in the afternoon, my grandmother was taking a nap. I was a very restless one. The toys soon bored me and I looked around for something new. The unique thing which caught my attention was my Grandma’s spectacles.
I put it on my nose just in the style of my Grandma and looked around. Soon my eyes got tired. As I felt pain in my eyes, I removed the specs and threw them away. They struck the wall and landed on the ground broken. Now I got worried and afraid. I started trying to repair it. As I was holding these glass pieces I felt a severe pain in the middle finger of my right hand, I looked at it .
Blood was trickling down from a deep cut in my finger. I started crying loudly. On hearing my loud wailing my Grandma woke up. She hurriedly came out of her room, took a quick glance at my adventure and detecting the source of my trouble, she pressed her hand on my cut finger for some time and then she took me to the doctor for bandaging. I was very much afraid of punishment but my Grandma forgave me although she had to suffer difficulty in seeing until the glasses were repaired. However I was naturally punished as I could neither eat my meals nor do my homework for three days.
Question 2.
Write a paragraph of 100-120 words about your favorite personality.
My favorite personality:
There are many people all around the world who are very famous and celebrities. But my favourite personality is my father. My father is my hero. He is kind, polite and really friendly to everyone. He is a teacher by profession and is very good in teaching. He is always ready to help and support the needy and helpless. He is a God fearing person and always teaches us to remember the God’s gifts and God’s love for the world.
I am so proud to have a father like him. He is a simple man with kind rules. He is handsome, my favourite and my ideal man. He is my friend and always ready to encourage, appreciate me for success and always ready to help me wherever I need a friend or a support of my father. I am proud of my father and wish him good health forever.
Grammar:
A. Complete the following sentences using appropriate Connectors from the box.
1. She felt cold _________ she was wearing a winter coat.
although
2. This restaurant has some of the best chefs in the town.__________ their service is excellent.
Moreover
3. I’m not going to the party tonight __________ I didn’t get an invitation.
because
4. You can set the table. __________, I’ll start making dinner.
Meanwhile
5. I can play quite a few instruments __________ , the flute, the guitar and the piano.
For instance
6. The store was out of chocolate chips; __________ they would need to make a different type of cookies.
therefore
7. The stores are open daily __________ Sundays.
except
8. I’ll stay __________ you need me.
as long as
9. This detergent is highly concentrated and __________ you will need to dilute it.
thus
10. It was the thing he prized __________ .
above all
Active Voice and Passive Voice:
B. Convert the following active sentences into passive sentences by supplying an appropriate passive verb form.
Question 1.
She will not recognize us. / We__________ by her.
(a) will not recognize
(b) will not being recognized
(c) will not be recognized
(c) will not be recognized
Question 2.
They didn’t invite me, but I went anyway. /I __________ but I went anyway.
(a) wasn’t invited
(b) wasn’t being invited
(c) wasn’t inviting
(a) wasn’t invited
Question 3.
They broke up the table for firewood. / The table __________ up for firewood.
(a) broke
(c) was broken
(c) was broken
Question 4.
She has won the first prize. / The first prize __________ by her.
(a) has won
(b) has been won
(c) had been won
(b) has been won
Question 5.
A friend of mine is repairing the car. / The car __________ by a friend of mine.
(a) is repairing
(b) is repaired
(c) is being repaired
(c) is being repaired
Question 6.
Begin the work tomorrow. / Let the work __________ tomorrow.
(a) be begun
(b) begin
(c) is beginning
(a) be begun
Question 7.
They speak English in New Zealand. / English __________ in New Zealand.
(a) is speaking
(b) is spoken
(c) is being spoken
(b) is spoken
Question 8.
His attitude shocked me. / I __________ by his attitude.
(b) had been shocked
(c) was shocked
(c) was shocked
Question 9.
She had already sent the parcel. / The parcel __________ by her.
(a) has already been sent
(c) was already sent
Question 10.
Her silence worries me / I __________ her silence.
(a) am worrying by
(b) am worried by
(c) have worried by
(b) am worried by
C. Match the following Active voice sentences with Passive voice.
D. Change the following into passive voice.
Question 1.
Stanley will inform you later.
You will be informed by Stanley later.
Question 2.
People speak Portuguese in Brazil.
Portuguese is spoken by people in Brazil.
Question 3.
My grandfather built this house in 1943.
This house was built by my grandfather in 1943.
Question 4.
Do not hurt the animals.
You are warned not to hurt the animals.
Question 5.
You must not drop litter in the streets.
You are warned not to drop litter in the streets.
Question 6.
Carry it home.
Let it be carried to home.
Question 7.
They are decorating the wall.
The wall is being decorated by them.
Question 8.
He has already mended the TV set.
The TV set has already been mended by him.
E. Make a scrapbook of’Famous Biographies’ by collecting at least five biographies of famous scientists, mathematicians, inventors, artists etc., of your choice. You may also collect the pictures related to their achievements, inventions etc.
Sir Isaac Newton (Scientist):
Sir Isaac Newton was born on Christmas day, 1642. He was an English physicist and mathematician, who was the culminating figure of the scientific revolution of the 17th century. Born in the hamlet of Woolsthorpe, Newton was the only son of a local yeoman. Newton would eventually pick up his idea of a mathematical science of motion and bring his work to full fruition.
A tiny and weak baby, Newton was not expected to survive his first day of life, much less 84 years. Deprived of a father before birth, he soon lost his mother as well, for within two years she married a second time. He was left with his grandmother and moved to a neighbouring village.
For nine years, Isaac was effectively separated from his mother, and his pronounced psychotic tendencies have been ascribed to this traumatic event. Like thousands of other undergraduates, Newton began his higher education by immersing himself in Aristotle’s work. Even though the new philosophy was not in the curriculum, it was in the air.
Sometime during his undergraduate career, Newton discovered the works of the French natural philosopher Rene Descartes and the other mechanical philosophers. Newton had also begun his mathematical studies. Within little more than a year, he had mastered the literature; and, pursuing his own line of analysis, he began to move into new territory.
Despite the. fact that only a handful of savants were even aware of Newton’s existence, he had arrived at the point where he had become the leading mathematician in Europe.
Leonardo da Vinci (Artist):
Leonardo da Vinci was a leading artist and intellectual of the Italian Renaissance who’s known for his enduring works ‘ The Last Supper’ and the ‘Mona Lisa’. Leonardo da Vinci was born on April 15, 1452, in a farmhouse nestled amid the undulating hills of Tuscany outside the village of Anchiano, in present-day Italy. Born out of wedlock to respected Florentine notary Serpiero and a young peasant woman named Caterina.
Leonardo da Vinci was raised by his father and his stepmother. At the age of five, he moved to his father’s family estate in nearby Vinci, the Tuscan town from which the surname associated with Leonardo derives, and lived with his uncle and grandparents. Young Leonardo received little formal education beyond basic reading, writing and mathematics instruction, but his artistic talents were evident from an early age.
Around the age of 14, da Vinci began a lengthy apprenticeship with the noted artist Andrea del Verrocchio in Florence. He learned a wide breadth of technical skills including metalworking, leather arts, carpentry, drawing, painting and sculpting. His earliest known dated work, a pen-and-ink drawing of a landscape – the Amo Valley – was sketched in 1473. With a curious mind and keen intellect, da Vinci studied the laws of science and nature, which greatly improved his work. His ideas and body of work have influenced countless artists and made da Vinci a leading light of the Italian Renaissance.
Pythagoras (Mathematician):
Pythagoras is often known as the first pure mathematician. born on the island of Samos, Greece in 569 BC, his father, Mnesarchus, was a gem merchant. His mother’s name was Pythais and Pythagoras lived with his two or three brothers. Pythagoras was well educated, and played the lyre. He knew poetry and recited Homer. He was interested in mathematics, philosophy, astronomy and music.
Pythagoras believed:
1. The sum of the angles of a triangle is equal to two right angles.
2. The theorem of Pythagoras – for a right-angled triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides.
3. Constructing figures of a given area and geometrical algebra. For example they solved various equations by geometrical means.
4. Pythagoras taught that Earth was a sphere in the center of the Universe, that the planets, stars, and the universe were spherical because the sphere was the most perfect solid figure.
5. Pythagoras recognized that the morning star was the same as the evening star, Venus.
The Pythagorean Theorem is a cornerstone of mathematics, and continues to be so interesting to mathematicians that there are more than 400 different proofs of the theorem, including an original proof by President Garfield.
Michael Joseph Jackson (Singer and Dancer):
Michael Joseph Jackson was an American singer, songwriter and dancer. Dubbed the “King of Pop”, he is regarded as one of the most significant cultural icons of the 20th century and is also regarded as one of the greatest entertainers of all time. Jackson’s contributions to music, dance, and fashion, along with his publicized personal life, made him a global figure in popular culture for over four decades.
The eighth child of the Jackson family, Michael made his professional debut in 1964 with his elder brothers Jackie, Tito, Jermaine and Marlon as a member of the Jackson 5. He began his solo career in 1971 while at Motown Records.
In the early 1980s, Jackson became a dominant figure in popular music. His music videos, including “Beat It”, “Billie Jean”, and “Thriller” from his 1982 album Thriller, are all time favourites. Through stage and video performances, Jackson popularized a number of complicated dance techniques, such as the robot and the moonwalk, to which he gave the name.
His distinctive sound and style has influenced numerous artists of various genres. While preparing for his comeback concert series, This Is It, Jackson died of acute propofol and benzodiazepine intoxication in 2009, after suffering from cardiac arrest.
Kapil Dev (Cricketer):
Dev was born as Kapil Dev Nikhanj to Ram Lai Nikhanj, a prominent timber merchant and his wife Raj Kumari Ram Lai Nikhanj in Chandigarh on 6 January 1959. His mother was born in Pakpattan, in the town of the Sufi Saint Baba Farid. His father was from Dipalpur. They lived in Shah Yakka which is now in Okara district, Pakistan. His four sisters were born there before partition and his two brothers in Fazilka, where they moved after partition. His father spent his early life after the partition in Fazilka. They moved to the capital city Chandigarh.
Dev was a student at D.A.V. School and joined Desh Prem Azad in 1971. Dev captained the Indian cricket team that won the 1983 Cricket World Cup. He was India’s national cricket coach between October 1999 and August 2000. He retired in 1994, holding the world record for the most number of wickets taken in Test cricket, a record subsequently broken by Courtney Walsh in 2000.
He is the first player to take 200 ODI wickets. He is the only player in the history of cricket to have taken more than 400 wickets (434 wickets) and scored more than 5000 runs in Tests, making him one of the greatest all-rounders to have played the game. On 11 March 2010, Dev was inducted into the ICC Cricket Hall of Fame.
From Zero to Infinity Lesson Summary By Biography Of Srinivasa Ramanujan
Ramanujan was born in Erode in Tamil Nadu on December 22, 1887. His father was a petty clerk in a cloth shop. From early childhood it was evident that he was a prodigy. Mathematical ideas flooded in his mind for which he did not find enough papers to note it down. Hence he started writing them in loose sheets which was later known as Ramanujan’s Frayed Notebook. At the age of 13, he was not only able to master the Loney’s Trigonometiy, but also started his own research and came up with many mathematical theorems and formulae.
At the age of 15, the book Synopsis of Elementary Results in Pure Applied Mathematics by George Shoobridge Carr given by his senior friends triggered the mathematical genius in him. Though Ramanujan was a mathematical genius and was awarded the Subramanyan Scholarship, he failed twice in his first-year arts examination in college, which disappointed his father.
Then Ramanujan started looking for a job as he needed money not only for food but also for papers to do his calculations. His applications were rejected as no one could understand what he had scribbled in his notebooks. Luckily, Director of Madras Port Trust, Francis Spring, understood his capability and he gave Ramanujan a clerical job on a monthly salary of ?25. Later on, some teachers and educationists helped him to get a research fellowship. Thus, on 1st May 1913, the University of Madras granted him a fellowship of ?75 a month, though he had no qualifying degree.
Meanwhile, he had sent a letter to the great mathematician G.H. Hardy of Cambridge University, in which he set out 120 theorems and formulae. Among them was what is known as the Reimann Series, a topic in the definite integral of Calculus. These letters made G.H. Hardy and his colleague J.E. Littlewood realize that they bad discovered a rare mathematical genius. They made quick arrangements for his passage and stay at Cambridge University. Ramanujan sailed to Britain on March 17, 1914.
Even though he found it difficult to adapt to the new environment, he continued his research in Mathematics with determination. He forgot his hardships in the company of Hardy and Littlewood. Ramanujan was elected Fellow of the Royal Society on February 28, 1918. He was the youngest Indian to receive this distinguished fellowship. In October that year, he became the first Indian to be elected Fellow of Trinity College, Cambridge.
Ramanujan continued his works even though Tuberculosis, then an incurable disease, was devouring him. When his friends found him pale, exhausted and emaciated, they sent him back to India. He continued to play with numbers until his death. Apart from a mathematician, Ramanujan was an astrologer of repute and a good speaker.
From Zero to Infinity Glossary: | 6,487 | 28,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-22 | latest | en | 0.917835 |
https://tomcircle.wordpress.com/2013/04/02/relationship-mapping-inverse-rmi/ | 1,627,067,012,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150000.59/warc/CC-MAIN-20210723175111-20210723205111-00608.warc.gz | 590,143,264 | 25,521 | # Relationship-Mapping-Inverse (RMI)
Relationship-Mapping-Inverse (RMI)
(invented by Prof Xu Lizhi 徐利治 中国数学家 http://baike.baidu.com/view/6383.htm)
Find Z = a*b
By RMI Technique:
Let f Homomorphism: f(a*b) = f(a)+f(b)
Let f = log
log: R+ –> R
=> log (a*b) = log a + log b
1. Calculate log a (=X), log b (=Y)
2. X+Y = log (a*b)
3. Find Inverse log (a*b)
Prove:
$\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= 2$
1. Take f = log for Mapping:
$\log\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}$
$= \sqrt{2}\log\sqrt{2}^{\sqrt{2}}$
$= \sqrt{2}\sqrt{2}\log\sqrt{2}$
$= 2\log\sqrt{2}$
$= \log (\sqrt{2})^2$
$= \log 2$
2. Inverse of log (bijective):
$\log \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= \log 2$
$\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= 2$ | 312 | 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": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.548377 |
https://studysoup.com/tsg/11431/calculus-early-transcendentals-1-edition-chapter-3-1-problem-60ae | 1,632,597,387,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00634.warc.gz | 589,210,868 | 12,846 | ×
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 3.1 - Problem 60ae
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 3.1 - Problem 60ae
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Solved: One-sided derivatives The left-hand and right-hand
ISBN: 9780321570567 2
Solution for problem 60AE Chapter 3.1
Calculus: Early Transcendentals | 1st Edition
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Problem 60AE
One-sided derivatives? ?The left-hand and right-hand derivatives of a function at a point a are given by ? f(a+h)?f(a) f+(a) = lim + h h?0 and f (a) = lim f(a+h)?f(a) ? h?0 + h provided these limits exist. The derivative f??(?a?) ?exists if and only if ?. f+(a) = f (?) a.? ?Sketch the following functions. ? ? b.? ?Compute f (+)and f (a)at?the given point a. c.? ?Is f continuous at a? Is f differentiable at a? ;a=1
Step-by-Step Solution:
Step 1 of 3
SOLUTION Given and a=1 STEP 1 (a). Sketch the following functions STEP 2 (b). Compute f +a)and f ()at the given point a. The left and the right derivatives of the function are f(a+h)f(a) f(a+h)f(a) f +a) = lim + h And f(a) = lim + h h0 h0 Provided a=1. Therefore 2 2 2 (4(1+h) )(41 ) (4(1+h +2h)4+1 h 2h f +1) = lim + h = lim + h = lim h = lim h 2 = 2 h0 h0 h0 h0 Thus we got f (1) = 2 + 2(1+h)+1(2+1) 2+2h+121 2h Then, f (1 = lim h = lim h = lim h = 2 h0 h0 h0 Thus we got f (1) = 2 STEP 3 (c) Is f continuous at a Is f differentiable at a Now we need to verify whether the function is continuous at a=2 For this we have to show lim f(x) + lim f(x) x1 x1 lim f(x) = lim 4 x = lim 4 1 = 3 + + + x1 x1 x1 lim f(x) = lim 2x + 1 = 2 + 1 = 3 x1 x1 Thus the condition is satisfied. Thus the function f(x) is continuous at a=1. It is already known that the derivative f(a) exists if and only if . f+(a) = f () Since ,in this case f (a) + / f(a) the derivative does not exist. Therefore the given function f(x)is not differentiable at a=1.
Step 2 of 3
Step 3 of 3
Related chapters
Unlock Textbook Solution | 740 | 2,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.759323 |
https://www.jiskha.com/display.cgi?id=1253144697 | 1,503,176,337,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105922.73/warc/CC-MAIN-20170819201404-20170819221404-00265.warc.gz | 930,986,528 | 4,119 | # Algebra
posted by .
How come when you find the square of the binomial (2x-5y)^2, the answer is 4x^2-20xy+25y^2 (according to my book)? I would think that the middle term, 2(2x-5y) would be 4x-10y, but my book says it is -20xy. But I thought you 2x and 5y were separate terms so you can't combine them like that by multiplying them together to form -20xy. Please explain why this is. Thanks!
• Algebra -
FOIL: (x+y)^2= x^2+2xy+y^2
## Similar Questions
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More Similar Questions | 849 | 2,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.871396 |
http://math.stackexchange.com/questions/258842/how-is-this-equation-called/258875 | 1,469,702,481,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828010.65/warc/CC-MAIN-20160723071028-00145-ip-10-185-27-174.ec2.internal.warc.gz | 155,816,303 | 17,731 | # How is this equation called?
I'm trying to figure out some math problems.
In particular I have this "In an office you have 6 clerks. How many ways can you select a team of 3 clerks?" and the solution given is: $$\binom{6}{3} = \frac{6!}{3!*3!} = 20.$$
I don't really understand how that works, so I'm trying to find some tutorial, but how exactly is this equation even called? I can't search for it, because I have no idea what I'm searching :)
-
It would help the site if you re-wrote the title of your question to something more specific and descriptive so that it would be helpful later on when someone is searching for something similar to this. – JohnD Dec 14 '12 at 20:23
What you're referring to is called the "binomial coefficient", spoken "$n$ choose $k$" for natural numbers $n, k$ with $k\le n$.
The equation you posted simply makes explicit how to compute the binomial coefficient of
"$6$ choose $3$," which is denoted $\;\binom{6}{3}$. In this problem, $n = 6, k = 3$.
In general, to compute the number of ways to select $k$ items from $n$ items, without replacement, we compute:
$$\binom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{n(n-1)...(n-k+1)}{k!}.$$
For your problem, we have that $\displaystyle \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac {6!}{3!3!} = \frac{6\cdot 5\cdot 4}{3\cdot2 \cdot 1} = \frac{120}{6} = 20.$
-
It's the binomial coefficient, and you should google combination and permutation as a starter.
-
Thank you! I will. :) – NewProger Dec 14 '12 at 18:57 | 459 | 1,494 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2016-30 | latest | en | 0.929881 |
https://jp.mathworks.com/matlabcentral/cody/problems/42340-fibonacci-decomposition/solutions/2023930 | 1,579,766,287,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250609478.50/warc/CC-MAIN-20200123071220-20200123100220-00331.warc.gz | 505,648,830 | 17,252 | Cody
# Problem 42340. Fibonacci Decomposition
Solution 2023930
Submitted on 16 Nov 2019 by Doan Quang Manh
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
n = 1; f_correct = 1; assert(isequal(fib_decomposition(n),f_correct))
f = 1 y = 1 1 k = 2 f = 1 y = 1 1 1 k = 3 f = 2 y = 1 1 1 2 k = 4 a = 0 b = 1 f = 1
2 Pass
n = 4; f_correct = [1 3]; assert(isequal(fib_decomposition(n),f_correct))
f = 1 y = 1 1 k = 2 f = 1 y = 1 1 1 k = 3 f = 2 y = 1 1 1 2 k = 4 f = 3.0000 y = 1 1 1 2 3 k = 5 f = 5.0000 y = 1 1 1 2 3 5 k = 6 a = 1 b = 3 n = 1 y = 1 1 1 a = 0 b = 3 1 f = 1 3
3 Pass
n = 6; f_correct = [1 5]; assert(isequal(fib_decomposition(n),f_correct))
f = 1 y = 1 1 k = 2 f = 1 y = 1 1 1 k = 3 f = 2 y = 1 1 1 2 k = 4 f = 3.0000 y = 1 1 1 2 3 k = 5 f = 5.0000 y = 1 1 1 2 3 5 k = 6 f = 8.0000 y = 1 1 1 2 3 5 8 k = 7 a = 1 b = 5 n = 1 y = 1 1 1 a = 0 b = 5 1 f = 1 5
4 Pass
n = 2010; f_correct = [2 34 377 1597]; assert(isequal(fib_decomposition(n),f_correct))
f = 1 y = 1 1 k = 2 f = 1 y = 1 1 1 k = 3 f = 2 y = 1 1 1 2 k = 4 f = 3.0000 y = 1 1 1 2 3 k = 5 f = 5.0000 y = 1 1 1 2 3 5 k = 6 f = 8.0000 y = 1 1 1 2 3 5 8 k = 7 f = 13.0000 y = 1 1 1 2 3 5 8 13 k = 8 f = 21.0000 y = 1 1 1 2 3 5 8 13 21 k = 9 f = 34.0000 y = 1 1 1 2 3 5 8 13 21 34 k = 10 f = 55.0000 y = 1 1 1 2 3 5 8 13 21 34 55 k = 11 f = 89.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 k = 12 f = 144.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 k = 13 f = 233.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 k = 14 f = 377.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 k = 15 f = 610.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 k = 16 f = 987.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 k = 17 f = 1.5970e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 18 610 987 1597 k = 18 f = 2.5840e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 19 610 987 1597 2584 k = 19 a = 413 b = 1597 n = 413 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 a = 36 b = 1597 377 n = 36 y = 1 1 1 2 3 5 8 13 21 34 a = 2 b = 1597 377 34 n = 2 y = 1 1 1 2 a = 0 b = 1597 377 34 2 f = 2 34 377 1597
5 Pass
n = 35601; f_correct = [1 34 144 6765 28657]; assert(isequal(fib_decomposition(n),f_correct))
f = 1 y = 1 1 k = 2 f = 1 y = 1 1 1 k = 3 f = 2 y = 1 1 1 2 k = 4 f = 3.0000 y = 1 1 1 2 3 k = 5 f = 5.0000 y = 1 1 1 2 3 5 k = 6 f = 8.0000 y = 1 1 1 2 3 5 8 k = 7 f = 13.0000 y = 1 1 1 2 3 5 8 13 k = 8 f = 21.0000 y = 1 1 1 2 3 5 8 13 21 k = 9 f = 34.0000 y = 1 1 1 2 3 5 8 13 21 34 k = 10 f = 55.0000 y = 1 1 1 2 3 5 8 13 21 34 55 k = 11 f = 89.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 k = 12 f = 144.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 k = 13 f = 233.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 k = 14 f = 377.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 k = 15 f = 610.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 k = 16 f = 987.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 k = 17 f = 1.5970e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 18 610 987 1597 k = 18 f = 2.5840e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 19 610 987 1597 2584 k = 19 f = 4.1810e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 20 610 987 1597 2584 4181 k = 20 f = 6.7650e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 21 610 987 1597 2584 4181 6765 k = 21 f = 1.0946e+04 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 22 610 987 1597 2584 4181 6765 10946 k = 22 f = 1.7711e+04 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 23 610 987 1597 2584 4181 6765 10946 17711 k = 23 f = 2.8657e+04 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 24 610 987 1597 2584 4181 6765 10946 17711 28657 k = 24 f = 4.6368e+04 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 25 610 987 1597 2584 4181 6765 10946 17711 28657 46368 k = 25 a = 6944 b = 28657 n = 6944 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 21 610 987 1597 2584 4181 6765 a = 179 b = 28657 6765 n = 179 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 a = 35 b = 28657 6765 144 n = 35 y = 1 1 1 2 3 5 8 13 21 34 a = 1 b = 28657 6765 144 34 n = 1 y = 1 1 1 a = 0 b = 28657 6765 144 34 1 f = 1 34 144 6765 28657
6 Pass
n = 9227467; f_correct = [2 9227465]; assert(isequal(fib_decomposition(n),f_correct))
f = 1 y = 1 1 k = 2 f = 1 y = 1 1 1 k = 3 f = 2 y = 1 1 1 2 k = 4 f = 3.0000 y = 1 1 1 2 3 k = 5 f = 5.0000 y = 1 1 1 2 3 5 k = 6 f = 8.0000 y = 1 1 1 2 3 5 8 k = 7 f = 13.0000 y = 1 1 1 2 3 5 8 13 k = 8 f = 21.0000 y = 1 1 1 2 3 5 8 13 21 k = 9 f = 34.0000 y = 1 1 1 2 3 5 8 13 21 34 k = 10 f = 55.0000 y = 1 1 1 2 3 5 8 13 21 34 55 k = 11 f = 89.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 k = 12 f = 144.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 k = 13 f = 233.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 k = 14 f = 377.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 k = 15 f = 610.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 k = 16 f = 987.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 k = 17 f = 1.5970e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 18 610 987 1597 k = 18 f = 2.5840e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 19 610 987 1597 2584 k = 19 f = 4.1810e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 20 610 987 1597 2584 4181 k = 20 f = 6.7650e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 21 610 987 1597 2584 4181 6765 k = 21 f = 1.0946e+04 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 22 610 987 1597 2584 4181 6765 10946 k = 22 f = 1.7711e+04 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 23 610 987 1597 2584 4181 6765 10946 17711 k = 23 f = 2.8657e+04 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 24 610 987 1597 2584 4181 6765 10946 17711 28657 k = 24 f = 4.6368e+04 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 25 610 987 1597 2584 4181 6765 10946 17711 28657 46368 k = 25 f = 7.5025e+04 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 26 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 k = 26 f = 1.2139e+05 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 27 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 k = 27 f = 1.9642e+05 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 28 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 k = 28 f = 3.1781e+05 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 29 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 k = 29 f = 5.1423e+05 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 30 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 k = 30 f = 8.3204e+05 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 30 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 Column 31 832040 k = 31 f = 1.3463e+06 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 30 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 Columns 31 through 32 832040 1346269 k = 32 f = 2.1783e+06 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 30 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 Columns 31 through 33 832040 1346269 2178309 k = 33 f = 3.5246e+06 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 30 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 Columns 31 through 34 832040 1346269 2178309 3524578 k = 34 f = 5.7029e+06 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 30 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 Columns 31 through 35 832040 1346269 2178309 3524578 5702887 k = 35 f = 9.2275e+06 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 30 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 Columns 31 through 36 832040 1346269 2178309 3524578 5702887 9227465 k = 36 f = 1.4930e+07 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 30 610 987 1597 2584 4181 6765 10946 17711 28657 ...
7 Pass
n = 2015; f_correct = [2 5 34 377 1597]; assert(isequal(fib_decomposition(n),f_correct))
f = 1 y = 1 1 k = 2 f = 1 y = 1 1 1 k = 3 f = 2 y = 1 1 1 2 k = 4 f = 3.0000 y = 1 1 1 2 3 k = 5 f = 5.0000 y = 1 1 1 2 3 5 k = 6 f = 8.0000 y = 1 1 1 2 3 5 8 k = 7 f = 13.0000 y = 1 1 1 2 3 5 8 13 k = 8 f = 21.0000 y = 1 1 1 2 3 5 8 13 21 k = 9 f = 34.0000 y = 1 1 1 2 3 5 8 13 21 34 k = 10 f = 55.0000 y = 1 1 1 2 3 5 8 13 21 34 55 k = 11 f = 89.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 k = 12 f = 144.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 k = 13 f = 233.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 k = 14 f = 377.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 k = 15 f = 610.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 k = 16 f = 987.0000 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 k = 17 f = 1.5970e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 18 610 987 1597 k = 18 f = 2.5840e+03 y = Columns 1 through 15 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 Columns 16 through 19 610 987 1597 2584 k = 19 a = 418 b = 1597 n = 418 y = 1 1 1 2 3 5 8 13 21 34 55 89 144 233 377 a = 41 b = 1597 377 n = 41 y = 1 1 1 2 3 5 8 13 21 34 a = 7 b = 1597 377 34 n = 7 y = 1 1 1 2 3 5 a = 2 b = 1597 377 34 5 n = 2 y = 1 1 1 2 a = 0 b = 1597 377 34 5 2 f = 2 5 34 377 1597 | 5,944 | 10,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-05 | latest | en | 0.390781 |
https://www.bestdaixie.com/%E7%94%B5%E5%AD%90%E7%94%B5%E8%B7%AF%E4%BB%A3%E5%86%99-el-6463-advanced-hardware-design/ | 1,723,043,223,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694594.35/warc/CC-MAIN-20240807143134-20240807173134-00671.warc.gz | 519,317,369 | 20,016 | 电子电路代写 | EL 6463: Advanced Hardware Design
电子电路代写 | EL 6463: Advanced Hardware Design
Instructions
● Read and complete all questions carefully.
● This exam is to be completed individually. Collaboration/plagiarism will not be tolerated.
● Please sign the code of conduct (located at the bottom of this exam) and include it in your submission.
● Clarification questions can be posted on Slack in #finalexam, but do not put any specifics about your own answers in the channel.
● During the exam period, you will not be allowed to receive assistance from the teaching staff regarding the use of the tools or implementation/simulation.
Overview: In this examination you will design, implement, and simulate digital hardware with sequential and combinational logic to implement a pseudo-random number generator algorithm. This algorithm will be implemented via hardware known as a linear feedback shift register (LFSR). These have applications in generating simple randomness in gaming, cryptography, audio and signal processing.
An example of a 16-bit LFSR is shown below:
Image courtesy: CC0 https://en.wikipedia.org/wiki/File:LFSR-F16.svg
The depicted LFSR has XOR “taps” at positions [16, 14, 13, 11] (or zero-indexed: 15, 13, 12, 10).
In each clock cycle, the rightmost tap (which is known as the output bit) is XOR’d sequentially with the next right-most taps and the result is “shifted in” to the left-most bit (the most significant bit (MSB)).
In this final exam you will implement an LFSR with the following specifications (essential features).
Note: this is the same base specification as what was used in the mid-term exam.
1. It will be 32 bits long (from position 0 (MSB) to 31 (LSB)), with taps at positions as determined:
a. Each of the unique adjacent 2-digit numbers in your N number modulo 32.
b. The 32 bit (position 31, the LSB).
c. For example, if your N number is 01123497, you will have taps at positions
i. [01, 11, 12, 23, (34 % 32 =) 2, (49 % 32 =) 17, (97 % 32 =) 1] (“1” value is not unique) ii. = [1, 2, 11, 12, 17, 23, 31] (31 comes from point 1b) – remember these are zero-indexed
2. Top I/O:
a. clk: input
b. rstn: active-low reset signal
c. data_out: 8 bit output
d. get_random: input 1 bit control signal
f. 8-bit input “data_in”.
3. Unless otherwise specified, the LFSR will update (shift-in the XOR result) upon every clock cycle. The bits shift into position 0 (the MSB).
4. If “output” is requested by setting “get_random” to 1, the least significant 8 bits will be output to “data_out” in the next clock cycle, the next 8 in the next clock cycle, the next 8 in the next cycle, and the most significant 8 bits will be output in the final (fourth) cycle. No LFSR shifts will occur during this output sequence. At all other times the “data_out” will be all zeros.
a. NOTE:
i. If the get_random signal is held high for longer than 4 cycles then the LFSR would presumably immediately begin exporting the contents again, and so it wouldn’t perform any shifts, however there might be one clock cycle where the internals could shift. This part of the spec is upto to you.
ii. Similarly if get_random is high for less than 4 cycles, it should continue with the output until it is completed.
5. Reset signal “rstn” will reset the LFSR contents to 0x02468ACD
a. (Reset signal “rstn” should always take precedence over every other signal)
6. If “load_seed” is set to 1, the lower 8 bits will be loaded into the LFSR from “data_in” in the first clock cycle, the next 8 in the next, the next 8 in the next, and the upper 8 bits will be loaded in the final (fourth) cycle. No LFSR shifts will occur during loading. After loading, shifts will resume as normal. | 968 | 3,711 | {"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} | 3.0625 | 3 | CC-MAIN-2024-33 | latest | en | 0.900742 |
https://medium.com/@randerson112358/create-a-binary-search-tree-fff5d296b31d?source=---------9------------------ | 1,563,474,883,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525699.51/warc/CC-MAIN-20190718170249-20190718192249-00122.warc.gz | 479,114,933 | 23,085 | # Binary Search Trees
May 17 · 4 min read
How to create a binary search tree from an array
I’m going to discuss how to create a binary search tree from an array. This will be a basic integer array that contains 6 values that are unsorted.
Let’s begin by first establishing some rules for Binary Search Trees:
1. A parent node has, at most, 2 child nodes.
2. The left child node is always less than the parent node.
3. The right child node is always greater than or equal to the parent node.
The first value in the array is ‘7’, so it will be the root node in the tree. With the root node set, all of the remaining values will be children of this node. Using our rules from the beginning of this article we know the child nodes will be designated as the right or left child nodes depending on their value.
Next we insert the next value in the array. First we need to compare it with the parent node ‘7’. Since the next value is ‘2’ and ‘2’ is less than ‘7’ (it’s parent), then ‘2’ will go to the left of the node.
Insert the next value from the array into the tree, the value containing ‘9’. Compare the value with it’s parent node (the node containing the value ‘7’). Since ‘9’ is greater or equal to ‘7’, it will go to the right of its parent node.
Next insert the value ‘1’ in the tree. Since ‘1’ is less than ‘7’ (the root node), it will go to the left of that node. Next we need to compare the node that contains the value ‘2’ with our current value ‘1’. Since ‘1’ is less than ‘2’, it will be placed to the left of ‘2’. The node containing the value ‘2’ is its parent.
Insert the next value from the array, the value containing the number ‘5’. Compare ‘5’ with the value of the root node ‘7’. Since ‘5’ is less than ‘7’ it will go to the left of that node. Next we compare the ‘5’ with the node that contains the value ‘2’. Since ‘5’ is greater than or equal to ‘2’, it will go to the right of the node containing the value ‘2’. The node containing the value ‘2’ is its parent.
Finally we add the last value ‘14’ from the array into our Binary Search Tree. First we compare this value with the value in our root node. Since ‘14’ is greater than or equal to ‘7’, we will go to the right of out root node. Next we need to compare the node that contains the value ‘9’ with ‘14’. Since ‘14’ is greater than or equal to ‘9’, it will go to the right of that node. The node containing the value ‘9’ will be its parent.
You can watch a playlist on Binary Search Trees , or watch the playlist below:
If you are also interested in reading up a little bit more on Binary Search Trees and other data structures & algorithms, then I strongly recommend you check out the amazing book . This book goes over topics like binary search trees, sorting algorithms, heap sort, hash tables, red and black trees, divide and conquer algorithms, greedy algorithms and the list goes on !
Thanks for reading this article I hope its helpful to you all ! Keep up the learning, and if you would like more computer science and algorithm analysis videos please visit and subscribe to my channels (& )
# Check Out the following for content / videos on Computer Science, Algorithm Analysis, Programming and Logic:
randerson112358:
compsci112358:
Video Tutorials on Recurrence Relation:
Video Tutorial on Algorithm Analysis: | 780 | 3,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-30 | latest | en | 0.884933 |
http://technicalsymposium.com/GATE_Physics_syllabus.html | 1,502,963,372,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103167.97/warc/CC-MAIN-20170817092444-20170817112444-00172.warc.gz | 409,364,915 | 10,781 | ## /* Neon Lights Text By JavaScript Kit (http://javascriptkit.com) For this script, TOS, and 100s more DHTML scripts, Visit http://www.dynamicdrive.com */ var message="!TECHNICALSYMPOSIUM.COM!" var neonbasecolor="ORANGE" var neontextcolor="NAVY" var flashspeed=100 //in milliseconds ///No need to edit below this line///// var n=0 if (document.all||document.getElementById){ document.write('<font color="'+neonbasecolor+'">') for (m=0;m<message.length;m++) document.write('<span id="neonlight'+m+'">'+message.charAt(m)+'</span>') document.write('</font>') } else document.write(message) function crossref(number){ var crossobj=document.all? eval("document.all.neonlight"+number) : document.getElementById("neonlight"+number) return crossobj } function neon(){ //Change all letters to base color if (n==0){ for (m=0;m<message.length;m++) //eval("document.all.neonlight"+m).style.color=neonbasecolor crossref(m).style.color=neonbasecolor } //cycle through and change individual letters to neon color crossref(n).style.color=neontextcolor if (n<message.length-1) n++ else{ n=0 clearInterval(flashing) setTimeout("beginneon()",1500) return } } function beginneon(){ if (document.all||document.getElementById) flashing=setInterval("neon()",flashspeed) } beginneon()
What's New-Given Below
# Physics – PH
Mathematical Physics:
Linear vector space; matrices; vector calculus; linear differential equations; elements of complex analysis; Laplace transforms, Fourier analysis, elementary ideas about tensors.
Classical Mechanics:
Conservation laws; central forces, Kepler problem and planetary motion; collisions and scattering in laboratory and centre of mass frames; mechanics of system of particles; rigid body dynamics; moment of inertia tensor; noninertial frames and pseudo forces; variational principle; Lagrange’s and Hamilton’s formalisms; equation of motion, cyclic coordinates, Poisson bracket; periodic motion, small oscillations, normal modes; special theory of relativity – Lorentz transformations, relativistic kinematics, mass-energy equivalence.
Electromagnetic Theory:
Solution of electrostatic and magnetostatic problems including boundary value problems; dielectrics and conductors; Biot-Savart’s and Ampere’s laws; Faraday’s law; Maxwell’s equations; scalar and vector potentials; Coulomb and Lorentz gauges; Electromagnetic waves and their reflection, refraction, interference,diffraction and polarization. Poynting vector, Poynting theorem, energy and momentum of electromagnetic waves; radiation from a moving charge.
Quantum Mechanics:
Physical basis of quantum mechanics; uncertainty principle; Schrodinger equation; one, two and three dimensional potential problems; particle in a box, harmonic oscillator, hydrogen atom; linear vectors and operators in Hilbert space; angular momentum and spin; addition of angular momenta; time independent perturbation theory; elementary scattering theory.
Thermodynamics and Statistical Physics:
Laws of thermodynamics; macrostates and microstates; phase space; probability ensembles; partition function, free energy, calculation of thermodynamic quantities; classical and quantum statistics; degenerate Fermi gas; black body radiation and Planck’s distribution law; Bose-Einstein condensation; first and second order phase transitions, critical point.
Atomic and Molecular Physics:
Spectra of one- and many-electron atoms; LS and jj coupling; hyperfine structure; Zeeman and Stark effects; electric dipole transitions and selection rules; X-ray spectra; rotational and vibrational spectra of diatomic molecules; electronic transition in diatomic molecules, Franck-Condon principle; Raman effect; NMR and ESR; lasers.
Solid State Physics:
Elements of crystallography; diffraction methods for structure determination; bonding in solids; elastic properties of solids; defects in crystals; lattice vibrations and thermal properties of solids; free electron theory; band theory of solids; metals, semiconductors and insulators; transport properties; optical,dielectric and magnetic properties of solids; elements of superconductivity.
Nuclear and Particle Physics:
Nuclear radii and charge distributions, nuclear binding energy, Electric and magnetic moments; nuclear models, liquid drop model - semi-empirical mass formula, Fermi gas model of nucleus, nuclear shell model; nuclear force and two nucleon problem; Alpha decay, Beta-decay, electromagnetic transitions in nuclei; Rutherford scattering, nuclear reactions, conservation laws; fission and fusion; particle accelerators and detectors; elementary particles, photons, baryons, mesons and leptons; quark model.
Electronics:
Network analysis; semiconductor devices; Bipolar Junction Transistors, Field Effect Transistors, amplifier and oscillator circuits; operational amplifier, negative feedback circuits , active filters and oscillators; rectifier circuits, regulated power supplies; basic digital logic circuits, sequential circuits, flip-flops, counters, registers, A/D and D/A conversion. | 1,061 | 5,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-34 | longest | en | 0.446843 |
http://scienceuniverse101.blogspot.com/2012/03/measurements-of-gas-pressure.html | 1,466,805,342,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783391519.2/warc/CC-MAIN-20160624154951-00092-ip-10-164-35-72.ec2.internal.warc.gz | 258,914,502 | 40,791 | # Measurements of gas pressure. Aneroid barometer and altimeter
A liquid finds it own level:
When water or any other liquid is poured into the communicating tubes it stands at the same level in each tube. This illustrates the popular saying that, water finds its own level.
When the liquid is at rest in the vessel the pressure must be the same at all points along the same horizontal level, otherwise the liquid would move until the pressure were equalized. The fact that the liquid stands at the same vertical height in all tubes whatever their shape confirms that, for a given liquid, the pressure at a point within it varies only with the vertical depth of the point below the surface of the liquid.
The water barometer
# Measurement of gas pressure by the manometer
In the previous post ( Pressure in liquids and gases) it was stated that the atmospheric pressure was about 100 kilo newtons per m². Before explaining how this is measured let us make a short study of the manometer, an instrument for measuring the pressure of gas.
Manometer:
The manometer consists of a U-tube containing water. When both arms are open to the atmosphere the same atmospheric pressure is exerted on the water surfaces A and B, and these are at the same horizontal level.
In order to measure the pressure of the gas supply in the laboratory, the side A is connected to a gas-tap by a length of rubber tubing. When the tap is turned on the gas exerts pressure on the surface A, with the result that the level B rises until the pressure at C on the same horizontal level as A becomes equal to the gas pressure. Thus,
Pressure of gas = atmospheric pressure + pressure due to water column BC
It follows that the excess pressure, in N/m², of the gas above that of the atmosphere is given by the pressure of the water column BC, and is therefore equal to hρg as explained earlier.
The height, h, is called the head of water in the manometer and it is often convenient to express the excess pressure simply in terms of h only. In this case the units generally used are millimeters of water ( mm H2O ).
For measuring higher pressure than in the example above, mercury ( density 13.6 g /cm³) is used in the manometer, while for lower pressure a liquid such as xylene (density 0.88 g/cm³) is more suitable than water. We may therefore say,
excess pressure of gas supply = h ( in appropriate units).
## Torricelli’s experiment. Simple barometer
About the middle of the seventeenth century an Italian scientist named Torricelli, living at Pisa, suggested an experiment to discount the theory that nature abhorred vacuum. Torricelli believed that nature’s supposed horror of a vacuum was caused simply by atmospheric pressure.
In a famous experiment, first performed in 1643, he set up the first barometer,
an instrument for measuring the pressure of the air.
How can you do this experiment:
1- In the laboratory a simple barometer can be made by taking a stout-walled glass tube about a meter long and closed at one end, and filling it almost to the top with clean mercury.
2-This is done with the aid of a small glass funnel and short length of rubber tubing.
3- Small air bubbles will generally be noticed clinging to the walls of the tube, and these must be removed.
4- With the finger placed securely over its open end, the tube is inverted several times so that the large air bubble left at the top of the tube travels up and down, collecting the small bubbles on its way.
5- More mercury is then added so that the tube is completely full.
6- The finger is again placed over the open end of the tube, while is now inverted and placed vertically with its end well below the surface of some mercury in a dish.
7-The finger is then removed and the column of mercury in the tube falls until the vertical difference in level between the surfaces of the mercury in tube and dish is about 760 mm.
The vertical height of the mercury column remains constant even when the tube is tilted, unless the top of the tube is less than 760 mm above the level in the dish, in which case the mercury completely fills the tube.
Torricelli's experiment:
How Torricelli explained that
Torricelli explained that the column of mercury was supported in the tube by the atmospheric pressure acting on the surface of the mercury in the dish, and pointed out that small changes in the height of the column, which are noticed from day to day, are due to fluctuations in the atmospheric pressure.
The space above the mercury in the tube is called a Torricellian vacuum; it contains a little mercury vapor, and in this respect differs from a true vacuum.
### Pascal’s experiments with barometers
Torricelli died a few years after the barometer experiment had been performed, and did not live to see his explanation of it, in terms of atmospheric pressure, generally accepted among scientists.
After Torricelli’s death Pascal repeated the experiment in France and set up two barometers.
The first of these was placed at the foot of a mountain in Auvergne called the Puy-de-Dome, while the other was carried up the mountainside and the height of the column read at intervals on the way up.
Owing to the decreasing height of the atmosphere above this barometer, its mercury column showed a progressive fall due to the reduced atmospheric pressure.
The barometer at the foot of the mountain showed practically no change.
It was this final experiment which brought about the downfall of the theory that nature abhors a vacuum and established the principle that the atmosphere exerts a pressure.
#### The aneroid barometer and altimeter
Barometers of the aneroid ( without liquids) type are commonly used as weather glasses, the idea being that low pressure, or a sudden fall in pressure, generally indicates unsettled weather while a rising barometer or high pressure is associated with fine weather.
The essential part of an aneroid barometer is a flat cylindrical metal box of capsule, corrugated for strength, and hermetically sealed after having been partially causes the box to cave in slightly, while a decrease allows it to expand.
The movements of the box are magnified by a system of levers and transmitted to a fine chain wrapped round the spindle of a pointer.
The chain is kept taut by means of a hairspring attached to the spindle, while the pointer moves over a suitably calibrated scale. Aneroid barometer movements are also used in the construction of altimeters for aircraft. In these the scale is calibrated in meters of ascent.
Roughly speaking the pressure falls by 10 mmHg per 120 m of ascent in the lower atmosphere.
##### The water barometer
Since the density of mercury is 13.6 g /cm³, it follows that if water was used as the liquid in a simple barometer the water column would have to be 76 × 13.6 cm = 10.3 m long!.
Such a barometer was constructed in the seventeenth century by von Guericke and fixed on the outside wall of his house. The upper level of the column was indicated by a small wooden float inside the tube on the surface of the water. With the aid of this barometer von Guericke made the first recorded scientific weather forecast. Having noted a sudden fall in the height of the water column, he correctly predicted the imminence of a severe storm. | 1,548 | 7,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-26 | longest | en | 0.944185 |
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