url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://bengtmn.wordpress.com/2011/01/08/agm-related-limits/ | 1,542,453,224,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743353.54/warc/CC-MAIN-20181117102757-20181117124757-00077.warc.gz | 577,756,711 | 19,622 | ## AGM-related limits
Let there be two numbers a and b.
Take the AM and GM of those two call them respectively a_1 and b_1
Then take AM and GM of a_1 and b_1, and call them a_2 and b_2, … .
Continuing this way, if the limit of a_n and b_n as n goes to infinity is x and y
then find these two limit as n goes to infinity:
I) a_1 + a_2 + a_3 + … + a_nn·x
II) b_1 + b_2 + b_3 + … + b_nn·y
(Proposed by Sanchar Sharma)
Sachin points out x = y, the AGM of a and b, often denoted M(a,b). In fact, this is a well-known theorem (by Gauss I think).
_________________________________________________
Let s₁(n) = a₁ + a₂ + a₃ +… + a_n – n·x, and s₁(n) = b₁ + b₂ + b₃ +… + b_n – n·y.
Then 2s₁(n) + s₂(n) =
= 2a₁ + 2a₂ + 2a₃ +… + 2a_n – 2n·x + b₁ + b₂ + b₃ +… + b_n – n·y =
= 2a₁ + (a₁ + b₁) + (a₂ + b₂) +… + (a_(n-1) + b_(n-1)) + b₁ + b₂ + b₃ +… b_n – n·(2x + y) =
= 2a₁ + (a₁ + a₂ + … + a_(n-1)) + (b₁ + b₂ +… + b_(n-1)) + b₁ + b₂ + b₃ +… b_n – n·(2x + y) =
= 2a₁ + (s₁(n) – a_n + n·x) + (s₂(n) – b_n + n·y) + s₂(n) + n·y – n·(2x + y) =
= s₁(n) + 2s₂(n) + 2a₁ – a_n – b_n – n·(x – y).
This gives
2s₁(n) + s₁(n) = s₁(n) + 2s₂(n) + 2a₁ – a_n – b_n – n·(x – y),
s₁(n) – s₁(n) = + 2a₁ – a_n – b_n – n·(x – y)
Using x = y = M(a,b) this gives lim[n→∞] (s₁(n) – s₁(n)) = 2a – 2M(a,b).
So at least the difference s₁(n) – s₂(n) has a limit. Notice that x = y is necessary for this to exist! If the limit of one more (linear) combination could be found the problem would be solved. Unfortunately I haven’t been able to find more (yet). – Someone?
Of course we still dont know if the limits of s₁(n) and s₂(n) exist. But either both or none do!
__________________________________
Some numerical results.
Let a = x and b = 1 where x > 1. Then s₁(n) = Σ [k=1,n] (a_k – M(1,x)) and
s₂(n) = Σ [k=1,n] (b_k – M(1,x)) are funtions of x and n. Provided the limits exist,
Σ₁(x) ≡ lim [n→∞] Σ [k=1,n] (a_k – M(1,x)) and Σ₂(x) ≡ lim [n→∞] Σ [k=1,n] (b_k – M(1,x)) are functions of x.
Using Mathematica it seems like the sums for given x converge rapidly and using n = 10 gives almost the same value as larger n. For example (here I use the values for n = 10 and the four numbers are appr values of x, Σ₁(x), Σ₂(x), Σ₁(x)/Σ₂(x), 2a – 2M(a,b)).
{10, 7.0803, -4.41889, 11.4992, 11.4992}
{20, 16.4921, -9.17624, 25.6684, 25.6684}
{50, 47.353, -23.0023, 70.3553, 70.3553}
{70, 68.9654, -32.0087, 100.974, 100.974}
{100, 102.245, -45.3221, 147.567, 147.567}
{150, 159.184, -67.1502, 226.334, 226.334}
Notice that the last two values are equal confirming the earlier result.
_________________________________
Graphs for Σ₁(x) and Σ₂(x), the other the ratio Σ₁(x)/Σ₂(x):
Upper curve: y = Σ₁(x) ≡ lim [n→∞] Σ [k=1,n] (a_k – M(1,x))
Lower curve: y = Σ₂(x) ≡ lim [n→∞] Σ [k=1,n] (b_k – M(1,x))
M(1,x) is known; it’s equal to a certain elliptic integral where x is a parameter.
________________________
y = Σ₁(x)/Σ₂(x)
______________________________________
Convergence proof.
For definiteness we assume that a ≥ b > 0. Then from the theory of the AGM it is known that
(*) a = a₁ ≥ a₂ ≥ … ≥ a_n ≥ a_(n+1) ≥ … ≥ M ≥ … ≥ b_(n+1) ≥ b_n ≥ … ≥ b₂ ≥ b₁ = b > 0 , where M ≡ M(a,b), the AGM of a and b. Also, by induction,
(**) a_(n+1) – b_(n+1) ≤ (a_n + b_n)/2 – b_n = (a_n – b_n)/2 ≤ … ≤ (a – b)/2^n.
From (*) and (**) 0 ≤ a_n – M ≤ a_n – b_n ≤ (a – b)/2^(n-1).
Since Σ [n=1,∞] (a – b)/2^(n-1) is convergent it follows that Σ [n=1,∞] (a_n – M) is also convergent.
Annonser | 1,453 | 3,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-47 | longest | en | 0.751332 |
https://paramsingh.dev/blog/b81/ | 1,652,759,685,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515501.4/warc/CC-MAIN-20220517031843-20220517061843-00757.warc.gz | 525,908,635 | 2,850 | # Dealing with overplotting in a scatter plot
Saturday, Jun 5, 2021
Data Visualization R
Scatterplot is a good choice for plotting two continuous variable. It helps to visualize any interesting relationship between the two variables. Some of the information however could be masked due to overplotting. Overplotting refers to situation when multiple data points are plotted at the same location on the plot confounding some of the data points. Overplotting can occur due to:
1. Large number of data points
2. While the scale is continuous, the values available for the variable are actually discrete.
## Solution: Choosing more appropriate aesthetics for the point
Solution in these cases is to adjust the aesthetics of the point like color, size and alpha values.
### Less number of points
For few hundred points, large size, bright colors and no transparency is useful for the most part. In the following dataset, we have a little over a hundred data points, so we chose dark green solid circle of size 4.
air = datasets::airquality
plt = ggplot(air, aes(Wind, Ozone))
plt + geom_point(color = 'darkgreen', size = 4)
For even smaller number of points, I like to use larger circle with darker border and (somewhat) lighter color.
cars = datasets::cars
plt = ggplot(cars, aes(speed, dist))
plt + geom_point(color = 'darkred', size = 3, shape = 21, stroke = 2, fill = 'pink')
### Large number of points
In the data below, we have over 20,000 observations, so using solid circles masks most of the data making it harder to see any trend or data density information.
nyc = nycflights13::weather
plt2 = ggplot(nyc, aes(temp, pressure))
plt2 + geom_point(size = 2, shape = 19, alpha = 1, color = 'deepskyblue4')
So we could either set the alpha to a very low value:
plt2 + geom_point(size = 1,alpha = 0.1, color = 'deepskyblue3')
Or just represent data with a dot instead of circle:
plt2 + geom_point(shape = '.', color = 'red')
The above two plots represent the second problem highlighted earlier with vertical lines showing the discrete steps in temperature. One could use geom_jitter to add some noise to the data. Note that this is purely for visualizing data density and I think it is important to highlight this if sharing such a visualization with your collaborators or readers.
plt2 + geom_jitter(shape = '.', color = 'red') | 551 | 2,351 | {"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} | 2.65625 | 3 | CC-MAIN-2022-21 | longest | en | 0.840457 |
https://kjrose.blog/2015/01/15/forgetting-results/?share=google-plus-1 | 1,590,578,374,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394074.44/warc/CC-MAIN-20200527110649-20200527140649-00539.warc.gz | 431,194,049 | 18,999 | # Forgetting Results
This is a wonderful discussion on some pretty straightforward math.
Henry Smith was a mathematician of the 19th century who worked mainly in number theory. He especially did important work on the representation of numbers by various quadratic forms. We have discussed how even in something seemingly settled, like Joseph Lagrange’s theorem that every natural number is representable as a sum of four squares, new questions are always around—especially when one considers complexity.
Today Ken and I want to discuss a private slip of forgetfulness, and how often others may have done the same.
For a variety of reasons we recently were thinking about one of the most basic questions in linear algebra: the solvability of
\$latex displaystyle Ax = b, &fg=000000\$
where \$latex {A}&fg=000000\$ and \$latex {b}&fg=000000\$ are fixed and \$latex {x}&fg=000000\$ is to be determined. Over a field there is a polynomial time algorithm that determines whether there is a solution and finds one if there is. The…
View original post 2,062 more words | 232 | 1,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.952663 |
https://www.conwaylife.com/w/index.php?title=User:Saka/OCA_Spaceship_searches&diff=71881 | 1,597,446,594,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740343.48/warc/CC-MAIN-20200814215931-20200815005931-00096.warc.gz | 606,878,913 | 7,784 | # Difference between revisions of "User:Saka/OCA Spaceship searches"
This is a page to keep track of some spaceship searches I do in OCA rules to make sure I don't repeat them.
Notation for the "Partial" result:
[Partial length (full, as provided by search program)] - [# of columns that change after <period> gens]
## B3457/S4568 (Gems)
Velocity Symmetry Width Program Used Seedcolumns (if LSSS) Result
(1,0)c/6 Even 10 LSSS All Partial, 17-5 (00)
(1,0)c/6 Even 11 LSSS 00 Partial, 20-5
(1,0)c/6 Even 12 LSSS 00 Partial, 23-6
(1,0)c/4 Even 10 LSSS All Partial, 10-3 (00)
(1,0)c/4 Even 11 LSSS 00 Partial, 11-3
(1,0)c/4 Even 12 LSSS 00 Partial, 13-3
(1,0)c/4 Even 13 LSSS 00 Partial
(1,0)c/4 Even 14 LSSS 00 Partial, 15-3
## B35678/S5678 (Diamoeba)
Velocity Symmetry Width Program Used Seedcolumns (if LSSS) Result
(1,0)c/8 Even 10 LSSS All Partial, longest 00
(1,0)c/8 Even 11 LSSS 00 Partial
(1,0)c/8 Even 12 LSSS 00 Partial, 15-7
(1,0)c/8 Even 13 LSSS 00 Partial, 18-8
(1,0)c/7 Odd 10 LSSS 00 Partial, 15-6
(1,0)c/7 Odd 11 LSSS 00 Partial, 17-7
(1,0)c/7 Odd 12 LSSS 00 Ship![1]
(1,0)c/6 Even 12 LSSS 00-05 Partial, 10-5 (00)
(1,0)c/6 Even 13 LSSS 00 Partial, 11-3
(1,0)c/6 Odd 12 LSSS 00 Partial, 8-5
(1,0)c/6 Odd 13 LSSS 00 Partial, 11-4
(1,0)c/7 Gutter 12 LSSS 00- Partial
## B345/S5 (LongLife)
Velocity Symmetry Width Program Used Seedcolumns (if LSSS) Result
(1,0)c/5 Even 10 LSSS All Partial, longest 00
(1,0)c/5 Even 11 LSSS 00 Partial
(1,0)c/5 Even 12 LSSS 00 Partial
(1,0)c/5 Even 13 LSSS 00 Partial
(1,0)c/5 Even 13 LSSS 00 Partial
(1,0)c/5 Even 14 LSSS 00 Partial
## B3/S45678 (Coral)
Velocity Symmetry Width Program Used Seedcolumns (if LSSS) Result
(1,0)c/6 Even 10 LSSS 00-07 Partial, longest 00 (14-5), 05, 06 viable
(1,0)c/6 Even 11 LSSS 00 Partial, 14-5
(1,0)c/6 Even 12 LSSS 00 Partial, 14-5
## B35/S46 (From Eppstein's Most Wanted)
Velocity Symmetry Width Program Used Seedcolumns (if LSSS) Result
(1,0)c/5 Even 11 LSSS All Partial, longest 00 (21-4), 02-06 viable?
(1,0)c/5 Even 12 LSSS 00 Partial (23-5)
(1,0)c/5 Even 13 LSSS 00 Partial (25-4)
(1,0)c/5 Even 14 LSSS 00 Partial (26-4)
## B26/S125
Velocity Symmetry Width Program Used Seedcolumns (if LSSS) Result
(3,0)c/4 Odd 14 LSSS All Partial, best 04 and 07 it seems
(3,0)c/4 Odd 15 LSSS 04, 07 Partial
(3,0)c/4 Odd 16 LSSS 04, 07 Partial
(3,0)c/4 Odd 17 LSSS 04, 07 Partial
(3,0)c/4 Odd 18 LSSS 04, 07 Partial
(3,0)c/4 Odd 19 LSSS 04, 07 Partial
(3,0)c/4 Odd 20 LSSS 04, 07 Partial
(3,0)c/4 Odd 21 LSSS 04 Partial
(3,0)c/4 Odd 22 LSSS 04 Partial
(3,0)c/4 Odd 23 LSSS 04 Partial
(3,0)c/4 Odd 24 LSSS 04 Partial
(3,0)c/4 Odd 25 LSSS 04 Partial
(3,0)c/4 Odd 26 LSSS 04 Partial
(3,0)c/4 Odd 27 LSSS 04 Partial
(3,0)c/4 Odd 28 LSSS 04 Partial
## B27/S137
Velocity Symmetry Width Program Used Seedcolumns (if LSSS) Result
(2,0)c/3 Odd 14 LSSS 00, 01 Partial, best 00
(4,0)c/5 Odd 15 LSSS 02 Partial
(4,0)c/5 Odd 16 LSSS 02 Partial
(4,0)c/5 Odd 17 LSSS 02 Partial
(4,0)c/5 Odd 18 LSSS 02 Partial
(4,0)c/5 Odd 19 LSSS 02 Partial
(4,0)c/5 Odd 20 LSSS 02 Partial
(4,0)c/5 Odd 21 LSSS 02 Partial
(4,0)c/5 Odd 22 LSSS 02 Partial
(4,0)c/5 Odd 23 LSSS 02 Partial
(4,0)c/5 Odd 24 LSSS 02 Partial
(4,0)c/5 Odd 25 LSSS 02 Partial
## B25678/S4578
Velocity Symmetry Width Program Used Seedcolumns (if LSSS) Result
(3,0)c/5 Even 14 LSSS All Partial, best 05 (04 may be viable)
(3,0)c/5 Even 19 LSSS All Partial (Length 759!)
## B3/S0
Velocity Symmetry Width Program Used Seedcolumns (if LSSS) Result
(1,0)c/3 Even 10 LSSS All Nothing (Bad idea!)
## The Ships
1. Saka (18 April 2020). "Re: Spaceships in Life-like cellular automata". ConwayLife.com forums. Retrieved on 18 April 2020. | 1,602 | 3,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.767053 |
http://www.numbersaplenty.com/104853 | 1,580,188,999,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251773463.72/warc/CC-MAIN-20200128030221-20200128060221-00148.warc.gz | 246,779,038 | 3,414 | Search a number
104853 = 374993
BaseRepresentation
bin11001100110010101
312022211110
4121212111
511323403
62125233
7614460
oct314625
9168743
10104853
1171861
1250819
1338958
142a2d7
1521103
hex19995
104853 has 8 divisors (see below), whose sum is σ = 159808. Its totient is φ = 59904.
The previous prime is 104851. The next prime is 104869. The reversal of 104853 is 358401.
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 104853 - 21 = 104851 is a prime.
It is a Harshad number since it is a multiple of its sum of digits (21), and also a Moran number because the ratio is a prime number: 4993 = 104853 / (1 + 0 + 4 + 8 + 5 + 3).
It is a Duffinian number.
It is a Curzon number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (104851) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 2476 + ... + 2517.
It is an arithmetic number, because the mean of its divisors is an integer number (19976).
2104853 is an apocalyptic number.
It is an amenable number.
104853 is a deficient number, since it is larger than the sum of its proper divisors (54955).
104853 is an equidigital number, since it uses as much as digits as its factorization.
104853 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 5003.
The product of its (nonzero) digits is 480, while the sum is 21.
The square root of 104853 is about 323.8101295513. The cubic root of 104853 is about 47.1549136158.
The spelling of 104853 in words is "one hundred four thousand, eight hundred fifty-three".
Divisors: 1 3 7 21 4993 14979 34951 104853 | 535 | 1,755 | {"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} | 3.421875 | 3 | CC-MAIN-2020-05 | latest | en | 0.897459 |
https://www.edgedb.com/easy-edgedb/chapter15 | 1,709,090,721,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474690.22/warc/CC-MAIN-20240228012542-20240228042542-00118.warc.gz | 741,024,030 | 42,785 | Easy EdgeDB · Chapter 15
# The vampire hunt begins
Expression OnError Messages
It’s good that Jonathan is back, but he is still in shock. He doesn’t know if the experience with Dracula was real or not, and thinks he might be crazy. But then Jonathan meets Van Helsing who tells him that it was all true. Now that he knows everything was true, Jonathan becomes strong and confident again.
Our heroes begin the search for Dracula. The others learn that the Carfax mansion across from Dr. Seward’s asylum is the one that Dracula bought. So that’s why Renfield was so strongly affected!
The heroes search the house when the sun is up and find boxes of earth in which Dracula sleeps. They destroy them all in Carfax, but there are still many left in London. If they don’t destroy the other boxes, Dracula will be able to rest in them during the day and terrorize London every night when the sun goes down.
Our heroes learned something about vampires in this chapter: vampires need to sleep in coffins (boxes for dead people) with holy earth during the day. That’s why Dracula brought 50 of them over by ship on the Demeter. This is important for the mechanics of our game so we should create a type for this. And if we think about it:
• Each place in the world either has coffins or doesn’t have them,
• A place that has coffins is a place that vampires can enter and terrorize the people,
• If a place has coffins, we should know how many of them there are.
This sounds like a good case for an abstract type. Here it is:
Copy
```abstract type HasCoffins {
required coffins: int16 {
default := 0;
}
}
```
Most places will not have a special vampire coffin, so the default is 0. The `coffins` property is just an `int16`, and vampires can remain close to a place if the number is 1 or greater. In the mechanics of our game we would probably give vampires an activity radius of about 100 km from a place with a coffin. That’s because of the typical vampire schedule which is usually as follows:
• Wake up refreshed in the coffin after the sun goes down, get ready to leave by 8 pm to find people to terrorize.
• Feel a sense of freedom because the night has just begun, and start moving away from the safety of the coffins to find victims. A vampire might use a horse-driven carriage at 25 kph, which gives a pretty wide radius.
• Around 1 or 2 am, the vampire start to feel nervous. The sun will be up in about 5 hours. Is there enough time to get home?
So the part between 8 pm and 1 am is when the vampire is free to move away, and at 25 kph we get an activity radius of about 100 km around a coffin. At that distance, even the bravest vampire will start running back towards home by 2 am.
If we were building a more complex game, vampire terrorism on humans would be worse in the winter, when the activity radius might increase to about 150km due to longer nights… but we won’t get that detailed here.
With our abstract type done, we will want to have a lot of types `extending` this. First we can have `Place` extend it, and that gives it to all the other location types such as `City` and `OtherPlace`:
Copy
```abstract type Place extending HasCoffins {
required name: str {
delegated constraint exclusive;
};
modern_name: str;
important_places: array<str>;
}
```
Ships are also big enough to have coffins (the Demeter had 50 of them, after all) so we’ll extend for `Ship` as well:
Copy
```type Ship extending HasCoffins {
name: str;
multi sailors: Sailor;
multi crew: Crewman;
}
```
If we want, we can now make a quick function to test whether a vampire can enter a place:
Copy
```function can_enter(person_name: str, place: HasCoffins) -> optional str
using (
with vampire := assert_single((select Person filter .name = person_name)),
has_coffins := place.coffins > 0,
select vampire.name ++ ' can enter.'
if has_coffins else vampire.name ++ ' cannot enter.'
);
```
The function returns an `optional str` because it may return an empty set. You’ll also notice that `person_name` in this function actually just takes a str that it uses to select a `Person`. So technically it could say something like ‘Jonathan Harker cannot enter’.
If we can’t trust the user of the function to always enter a `Vampire` or `MinorVampire` object, there are some options:
• Overload the function to have two signatures, one for each type of Vampire:
Copy
```function can_enter(vampire: Vampire, place: HasCoffins) -> optional str
function can_enter(vampire: MinorVampire, place: HasCoffins) -> optional str
```
• Create an abstract type (like `type AnyVampire`) and extend it for `Vampire` and `MinorVampire`. Then `can_enter` can have this signature:
Copy
```function can_enter(vampire: AnyVampire, place: HasCoffins) -> optional str
```
Let’s learn a bit more about the `optional` keyword. Without it, you need to trust the users that the input argument will be there, because a function won’t be called if the input is empty. We can illustrate this point with this simple function:
Copy
```function try(place: City) -> str
using (
select 'Called!'
);
```
If we call it with this:
Copy
```select try((select City filter .name = 'London'));`
```
Then the output is `Called!` as we expected. The function requires a City as an argument, and then ignores it and returns ‘Called!’ instead.
So far so good, but the input is not optional so what happens if we type this instead?
Copy
```select try((select City filter .name = 'Beijing'));
```
Now the output will be {} because we’ve never inserted any data for the city ‘Beijing’ in our database (nobody in Bram Stoker’s Dracula ever goes to Beijing). So what if we want the function to be called in any case? We can put the keyword `optional` in front of the parameter like this:
Copy
```function try(place: optional City) -> str
using (
select 'Called!'
);
```
In this case we are still ignoring the argument `place` (the `City` type) but making it optional lets the function select ‘Called!’ regardless of whether it finds an argument or not. Having a `City` type and not having a `City` type are both acceptable in this case, and the function gets called in either case.
The documentation explains it like this:
```...the function is called normally when the corresponding argument is empty [...]
A notable example of a function that gets called on empty input is the coalescing operator.
```
Interesting! You’ll remember the coalescing operator `??` that we first saw in Chapter 11. And when we look at its signature, you can see the `optional` in there:
`optional anytype ?? set of anytype -> set of anytype`
So those are some ideas for how to set up your functions depending on how you think people might use them.
We’re coming up to an insert, so let’s migrate the schema. Here are all the schema changes from the discussion so for in this chapter:
Copy
```abstract type HasCoffins {
required coffins: int16 {
default := 0;
}
}
abstract type Place extending HasCoffins {
required name: str {
delegated constraint exclusive;
};
modern_name: str;
important_places: array<str>;
}
type Ship extending HasCoffins {
name: str;
multi sailors: Sailor;
multi crew: Crewman;
}
function can_enter(person_name: str, place: HasCoffins) -> optional str
using (
with vampire := assert_single((select Person filter .name = person_name)),
has_coffins := place.coffins > 0,
select vampire.name ++ ' can enter.' if has_coffins else vampire.name ++ ' cannot enter.'
);
```
Now let’s give London some coffins. According to the book, our heroes destroyed 29 coffins at Carfax that night, which leaves 21 in London.
Copy
```update City filter .name = 'London'
set {
coffins := 21
};
```
Now we can finally call up our function and see if it works:
Copy
```select can_enter('Count Dracula', (select City filter .name = 'London'));
```
We get `{'Count Dracula can enter.'}`.
Some other possible ideas for improvement later on for `can_enter()` are:
• Move the property `name` from `Place` and `Ship` over to `HasCoffins`. Then the user could just enter a string. The function would then use it to `select` the type and then display its name, giving a result like “Count Dracula can enter London.”
• Require a date in the function so that we can check if the vampire is dead or not first. For example, if we entered a date after Lucy died, it would just display something like the following:
```vampire.name ++ ' is already dead on ' ++ <str>.date
++ ' and cannot enter ' ++ city.name`
```
Let’s look at some more constraints. We’ve seen `exclusive` and `max_value` already, but there are some others that we can use as well.
There is one called `max_len_value` that makes sure that a string doesn’t go over a certain length. That could be good for our `PC` type. `NPC`s won’t need this constraint because their names are already decided by us the creators of the game, but `max_len_value()` is good for `PC`s to make sure that players don’t choose names that are too long to display. This constraint doesn’t exist on the original `Person` type, so we’ll also need to add the `overloaded` keyword here. The `PC` type with the new constraint now looks like this:
Copy
```type PC extending Person {
required class: Class;
required created_at: datetime {
default := datetime_of_statement()
}
required number: PCNumber {
default := sequence_next(introspect PCNumber);
}
constraint max_len_value(30);
}
}
```
And now `PC` objects like this one with names that are too long won’t be able to be inserted anymore:
Copy
```insert PC {
class := Class.Rogue,
name := "Oh man, let me tell you about this PC and his name. It all began one day when.."
};
```
The error is pretty nice and tells us everything we need to know:
```edgedb error: ConstraintViolationError: name must be no longer
than 30 characters.
Detail: `value of property 'name' of object type 'default`::`PC'`
must be no longer than 30 characters.
```
Another convenient constraint is called `one_of`, and is sort of like a quick enum. One place in our schema where we could use it is `title: str;` in our `Person` type. You’ll remember that we added that in case we wanted to generate names from various parts (first name, last name, title, degree…). This constraint could work to make sure that people don’t just make up their own titles:
Copy
```title: str {
constraint one_of('Mr.', 'Mrs.', 'Ms.', 'Lord');
}
```
For us it’s probably not worth it to add a `one_of` constraint though, as there are probably too many titles throughout the book (Count, the German Herr, Lady, Dr., Ph.D., etc.).
Another place you could imagine using a `one_of` is in the months, because the book only goes from May to October of the same year. If we had an object type generating a date then you could have this sort of constraint inside it:
Copy
```month: int64 {
constraint one_of(5, 6, 7, 8, 9, 10);
}
```
But that will depend on how the game works.
Now let’s learn about perhaps the most interesting constraint in EdgeDB: an expression that we create ourselves!
One particularly flexible constraint is called `expression on`, which lets us add any expression we want. After `expression on` you add the expression (in brackets) that must return `true` to create an object. In other words: “Create this object as long as (insert expression here)”.
Let’s say we need a type `Lord` for some reason later on, and all `Lord` types must have the word ‘Lord’ in their name. We can constrain the type to make sure that this is always the case. For this, we will use a function called `std::contains()` that looks like this:
Copy
```std::contains(haystack: str, needle: str) -> bool
```
This function returns `{true}` if the `haystack` (a string) contains the `needle` (usually a shorter string).
We can write the constraint with `expression on` and `contains()` like this:
Copy
```type Lord extending Person {
constraint expression on (
contains(__subject__.name, 'Lord')
);
}
```
`__subject__` there refers to the object itself.
Let’s do a migration here because a `Lord` type could be useful later.
Now when we try to insert a `Lord` without it, it won’t work:
Copy
```insert Lord {
name := 'Billy'
# Other stuff..
};
```
But if the `name` is ‘Lord Billy’ (or ‘Lord William’, or ‘Lord’ anything), it will work.
While we’re at it, let’s practice doing a `select` and `insert` at the same time so we see the output of our `insert` right away. We’ll change `Billy` to `Lord Billy` and say that Lord Billy (considering his great wealth) has visited every place in our database.
Copy
```select (
insert Lord {
name := 'Lord Billy',
places_visited := (select Place),
}
) {
name,
places_visited: {
name
}
};
```
Now that `.name` contains the substring `Lord`, it works like a charm:
```{
default::Lord {
name: 'Lord Billy',
places_visited: {
default::Country {name: 'Hungary'},
default::Country {name: 'Romania'},
default::Country {name: 'France'},
default::Country {name: 'Slovakia'},
default::Castle {name: 'Castle Dracula'},
default::City {name: 'Whitby'},
default::City {name: 'Munich'},
default::City {name: 'Buda-Pesth'},
default::City {name: 'Bistritz'},
default::City {name: 'Exeter'},
default::City {name: 'London'},
},
},
}
```
Since `expression on` is so flexible, you can use it in almost any way you can imagine. But that flexibility also means that there is no built-in way to let the user know how any `expression on` is supposed to work when a constraint is violated. After all, it’s anyone’s guess how a user might choose to use `expression on`. And that means that the automatically generated error message we have right now is not helping the user at all. Here’s the message we got when we tried to insert a `Lord` named `Billy`:
```edgedb error: ConstraintViolationError: invalid Lord
Detail: invalid value of object type 'default::Lord'
```
So there’s no way to tell that the problem is that `name` needs `'Lord'` inside it. Fortunately, all constraints allow you to set your own error message just by opening up a block with `{}` and specifying an `errmessage`, like this: `errmessage := "All lords need 'Lord' in their name."`
Let’s do that with our `Lord` type now:
Copy
```type Lord extending Person {
constraint expression on (contains(__subject__.name, 'Lord')) {
errmessage := "All lords need \'Lord\' in their name";
}
}
```
If you do a migration now and try to insert a `Lord` that is just called `Billy`, the error now tells us what to do:
````edgedb error: ConstraintViolationError: All lords need 'Lord' in their name
Detail: All lords need 'Lord' in their name`
```
Much better!
Practice Time
1. How would you create a type called Horse with a `required name: str` that can only be ‘Horse’?
2. How would you let the user know that it needs to be called ‘Horse’?
3. How would you make sure that `name` for type `NPC` is always between 5 and 30 characters in length?
Try it first with `expression on`.
4. How would you make a function called `display_coffins` that pulls up all the `HasCoffins` objects with more than 0 coffins?
5. How would you give the `Place` type a backlink to every `Person` type that visited it? | 3,684 | 15,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.93777 |
https://mathsimulationtechnology.wordpress.com/rules-of-integration/ | 1,501,267,252,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550977093.96/warc/CC-MAIN-20170728183650-20170728203650-00117.warc.gz | 669,705,128 | 33,009 | Linearity
Basic linearity properties of the integral $u(t)=\int_0^tf(s)\, ds$ follow directly from linearity of the underlying IVP $\dot u=f$:
• $\int_0^t(f(s)+g(s))\, ds=\int_0^tf(s)\, ds+\int_0^tg(s)\, ds, \int_0^t\alpha f(s)\, ds=\alpha\int_0^tf(s)\, ds$,
where $\alpha\in R$ is a constant. Further, for $a,
• $\int_a^bf(s)\, ds+\int_b^cf(s)\, ds=\int_a^cf(s)\, ds$,
which is extended to arbitrary limits $a,b$ and $c$, by defining for $b
• $\int_a^bf(s)\, ds=-\int_b^af(s)\, ds$.
Alternatively, these rules can be derived directly from the Riemann-sum representation of the integral.
Integration by Parts
By the Fundamental Theorem of Calculus, we have using the product rule for differentiation:
• $u(t)v(t)-u(0)v(0)=\int_0^t\frac{d}{ds}(u(s)v(s))ds=\int_0^t(u\dot v+u\dot v)ds$ ,
which can be written
• $\int_0^tu\dot v\, ds = [uv]_0^s - \int_0^tu\dot v\, ds$,
with $[uv]_0^s = u(t)v(t)-u(0)v(0)$. We see that we “can move the dot” from $u$ to $v$ if we change sign and take into account the difference of end-point values of $uv$.
Change of Integration Variable
If $w:[a,b]\rightarrow R$ is differentiable and $v:[w(a),w(b)]\rightarrow R$ is Lipschitz continuous, then
• $\int_{w(a)}^{w(b)}v(y)\, dy=\int_a^b v(w(x))w^\prime (x)dx$,
because with $y=w(x)$, we have $dy\equiv dw=w^\prime dx$.
On June 30 2007 The Swedish Parliament dismantled Integrationsverket, the Ministry for Integration, and replaced it by the Ministry for Time-Stepping. | 522 | 1,466 | {"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": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.723336 |
https://www.convert-measurement-units.com/convert+Light+minutes+to+Cubit+british.php | 1,721,516,620,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00169.warc.gz | 627,092,536 | 13,987 | Convert Light minutes to Cubit (british) (Distance)
## Light minutes into Cubit (british)
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Light+minutes+to+Cubit+british.php
# Convert Light minutes to Cubit (british) (Distance):
1. Choose the right category from the selection list, in this case 'Distance'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Light minutes'.
4. Finally choose the unit you want the value to be converted to, in this case 'Cubit (british)'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '642 Light minutes'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Distance'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '3 Light minutes to Cubit (british)' or '8 Light minutes into Cubit (british)' or '4 Light minutes -> Cubit (british)' or '5 Light minutes = Cubit (british)'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(14 * 15) Light minutes'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '12 Light minutes + 13 Cubit (british)' or '16mm x 17cm x 18dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 2.468 148 125 688 ×1020. For this form of presentation, the number will be segmented into an exponent, here 20, and the actual number, here 2.468 148 125 688. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 2.468 148 125 688 E+20. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 246 814 812 568 800 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 851 | 3,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-30 | latest | en | 0.864435 |
http://hackaday.com/tag/diy-capacitors/ | 1,500,928,443,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424910.80/warc/CC-MAIN-20170724202315-20170724222315-00273.warc.gz | 139,588,165 | 36,624 | # How to Measure the Dielectric Constant for DIY Capacitors
Every now and then you need to make your own capacitor. That includes choosing a dielectric for it, the insulating material that goes between the plates. One dielectric material that I use a lot is paraffin wax which can be found in art stores and is normally used for making candles. Another is resin, the easiest to find being automotive resin used for automotive body repairs.
The problem is that you sometimes need to do the calculations for the capacitor dimensions ahead of time, rather than just throwing something together. And that means you need to know the dielectric constant of the dielectric material. That’s something that the manufacturer of the paraffin wax that makes it for art stores won’t know, nor will the manufacturers of automotive body repair resin. The intended customers just don’t care.
It’s therefore left up to you to measure the dielectric constant yourself, and here I’ll talk about the method I use for doing that.
Once upon a time I was a real mad scientist. I was into non-conventional propulsion with the idea of somehow interacting with the quantum vacuum fluctuations, the zero point energy field. I was into it despite having only a vague understanding of what that was and without regard for how unlikely or impossible anyone said it was to interact with on a macro scale. But we all had to come from somewhere, and that was my introduction to the world of high voltages and homemade capacitors.
And along the way I made some pretty interesting, or different, capacitors which I’ll talk about here.
## Large Wax Cylindrical Capacitor
As the photos show, this capacitor is fairly large, appearing like a thick chunk of paraffin wax sandwiched between two wood disks. Inside, the lead wires go to two aluminum flashing disks that are the capacitor plates spaced 2.5cm (1 inch) apart. But in between them the dielectric consists of seven more aluminum flashing disks separated by plain cotton sheets immersed in more paraffin wax. See, I told you these capacitors were different.
I won’t go into the reasoning behind the construction — it was all shot-in-the-dark ideas, backed by hope, unicorn hairs, and practically no theory. The interesting thing here was the experiment itself. It worked!
I sat the capacitor on top of a tall 4″ diameter ABS pipe which in turn sat on a digital scale on the floor. High voltage in the tens of kilovolts was put across the capacitor through thickly insulated wires. The power supply contained a flyback transformer and Cockcroft-Walton voltage multiplier at the HV side. As I dialed up the voltage, the scale showed a reducing weight. I had weight-loss!
But after a few hours of reversing polarities and flipping the capacitor the other way around and taking plenty of notes, I found the cause. The weight-loss happened only when the feed wires were oriented with the top one feeding downward as shown in the diagram, but there was no weight change when the top wire was oriented horizontally. I’d seen high voltage wires moving before and here it was again, producing what looked like weight-loss on the scale.
But that’s only one of the interesting capacitors I’ve made. After the break we get into gravitators, polysulfide and even barium titanate.
# Plastic Plate Capacitors
We have been featuring some home made capacitors this week, and [Mike] wrote in to share his with us. While rolled capacitors are nice, they can be somewhat difficult to construct and grow to unwieldy sizes as capacitance and voltages increase. His solution is to stack the layers up using plastic plates.
In this forum post he explains that using disposable plastic plates and tinfoil you’re able to quickly make a capacitor, that for him was valued at around 12.2nF, using eleven layers . Applying pressure to the stack capacitance grew to about 14nF, though he is having a bit of trouble holding it with just glue.
Testing was conducted with high voltages charging the capacitor up, then its leads were shorted for a nice spark and a good pop. Definitely fun for the next family cook out, though we don’t know how some left over potato salad goo would effect the end results. | 886 | 4,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-30 | latest | en | 0.94649 |
https://www.mothering.com/threads/aluminum-math.1542929/ | 1,624,307,989,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488289268.76/warc/CC-MAIN-20210621181810-20210621211810-00628.warc.gz | 815,167,565 | 18,327 | 1 - 4 of 4 Posts
MarieBaxter
·
Registered
Joined
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2 Posts
Discussion Starter · ·
Okay, so I am doing some math on Aluminum Exposure. Help me understand.
Using CHOP's numbers, Formula has 0.225 mg/L. An infant drinks about one L per day & absorbs about 1% of that Aluminum. So the overall daily dose is about 0.00225 mg/day. So Prevnar has 0.125 mg. Daptacel & Pentacel have 0.330 mg. So it would take an infant something like 55.5 days to get the equivalent absorbed exposure of one Prevnar shot from drinking Formula.
So why is this just fine/not crazy loco? I am honestly asking, because it seems a little crazy to me. The baby would be excreting 50-70% of the dose by the next day? But that would still leave 25 days worth of Aluminum in the body.
Deborah
·
Registered
Joined
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18,833 Posts
Okay, so I am doing some math on Aluminum Exposure. Help me understand.
Using CHOP's numbers, Formula has 0.225 mg/L. An infant drinks about one L per day & absorbs about 1% of that Aluminum. So the overall daily dose is about 0.00225 mg/day. So Prevnar has 0.125 mg. Daptacel & Pentacel have 0.330 mg. So it would take an infant something like 55.5 days to get the equivalent absorbed exposure of one Prevnar shot from drinking Formula.
So why is this just fine/not crazy loco? I am honestly asking, because it seems a little crazy to me. The baby would be excreting 50-70% of the dose by the next day? But that would still leave 25 days worth of Aluminum in the body.
That is also assuming that all babies excrete at the same speed which is silly.
See my thread on problems with aluminum in addition to toxicity and other threads looking at recent science showing a broad range of problems with aluminum in living organisms.
applejuice
Tweety_Bird
·
Registered
Joined
·
1,570 Posts
There is a difference between ingesting or injecting a substance.
100% of injected aluminium has to be filtered by the kidneys. So, the amount of aluminium in vaccines is not just 'a drop in the bucket' compared to consumed aluminium.
.4 mcg - 5 mcg of aluminium from food sources pales in comparison to the 295 - 1225 mcg a child could receive in one day from vaccines (The exact amount depends on the brands given).
bannerd
·
Registered
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234 Posts
Okay, so I am doing some math on Aluminum Exposure. Help me understand.
Using CHOP's numbers, Formula has 0.225 mg/L. An infant drinks about one L per day & absorbs about 1% of that Aluminum. So the overall daily dose is about 0.00225 mg/day. So Prevnar has 0.125 mg. Daptacel & Pentacel have 0.330 mg. So it would take an infant something like 55.5 days to get the equivalent absorbed exposure of one Prevnar shot from drinking Formula.
So why is this just fine/not crazy loco? I am honestly asking, because it seems a little crazy to me. The baby would be excreting 50-70% of the dose by the next day? But that would still leave 25 days worth of Aluminum in the body.
Think about cholesterol, it builds up and hangs out because it forms in areas that don't have high blood pressure. One day that person decides to go for a run and the pressure builds, cholesterol breaks free and damage can happen to the body. If it breaks off on the upstream and goes to the brain or heart it can be fatal. Would have to be large in size for the heart but not so much for the brain. So what happens when you get injected and it ends up in the blood upstream which is direct to the brain? Aluminium right now is being tested by many researches, the body can not break it down fast enough even when it attaches acids to it.
There is a huge saying that vaccines don't enter the blood stream, yet blood comes out from most all injection sites. We've moved inoculations from the buttocks. Now the choice site is the arm, thigh, and calf which contain arteries and many (c)branch veins. So many that these areas are highly sensitive.. great for tickling your young one.. not so much for injection.
1 - 4 of 4 Posts | 987 | 3,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-25 | latest | en | 0.936321 |
https://blog.adnansiddiqi.me/tag/matplotlib/ | 1,719,071,551,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00666.warc.gz | 109,881,403 | 27,504 | • In this post I am going to discuss a Matplotlib feature which let you add multiple plots within a figure called subplots. Subplots are helpful when you want to show different data presentation in a single view, for instance Dashboards. There are multiple ways you can create subplots but I am here going to discuss the one which let you add graphs in grids by using subplot2grid method. subplot2grid takes two mandatory parameters, the first one is size and the next is location. A typical subplot2grid call will look like below: ax = plt.subplot2grid((2, 2), (0, 0)) The first parameter is the size that is a 2 x 2 grid, the 2nd is…
• In last post I discussed scatter, today I am going to discuss Pie charts. What are Pie Charts? An Emma chart (or a circle chart) is a circular statistical graphic which is divided into slices to illustrate numerical proportion. In a pie chart, the arc length of each slice (and consequently its central angle and area), is proportional to the quantity it represents. While it is named for its resemblance to a pie which has been sliced, there are variations on the way it can be presented. The earliest known pie chart is generally credited to William Playfair’s Statistical Breviary of 1801.[1][2] Pie charts are good to show proportional…
• In last post I talked about plotting histograms, in this post we are going to learn how to use scatter plots with data and why it could be useful. What is Scatter Plot? From Wikipedia: A scatter plot (also called a scatterplot, scatter graph, scatter chart, scattergram, or scatter diagram)[3] is a type of plot or mathematical diagram using Cartesian coordinates to display values for typically two variables for a set of data. If the points are color-coded, one additional variable can be displayed. The data are displayed as a collection of points, each having the value of one variable determining the position on the horizontal axis and the value…
• In the last post I talked about bar graphs and their implementation in Matplotlib. In this post I am going to discuss Histograms, a special kind of bar graphs. What is Histogram? From Wikipedia A histogram is an accurate graphical representation of the distribution of numerical data. It is an estimate of the probability distribution of a continuous variable (quantitative variable) and was first introduced by Karl Pearson. It is a kind of bar graph. To construct a histogram, the first step is to “bin” the range of values—that is, divide the entire range of values into a series of intervals—and then count how many values fall into each…
• In last post I covered line graph. In this post I am going to show how to draw bar graph by using Matplotlib. Bar Graph What is bar graph? According to Wikipedia A bar chart or bar graph is a chart or graph that presents categorical data with rectangular bars with heights or lengths proportional to the values that they represent. The bars can be plotted vertically or horizontally. So in short, bar graphs are good if you to want to present the data of different groups that are being compared with each other. For this post I have picked the data of movies genres released from 1995-2017. The source of data is here. import matplotlib.pyplot as plt import numpy…
• I recently covered data gathering via scraping. Now I am going to cover how the data can be visualized. The best way to do is to plot graphs. Graphs makes it easier to see the relation between a data variable with other. There are various kinds of graphs available: Line, Bar, Chart, Histogram etc. Since we are dealing in Python, it provides a very good library for plotting cool graphs. It’s called Matplotlib. From the Official Site: Matplotlib is a Python 2D plotting library which produces publication quality figures in a variety of hardcopy formats and interactive environments across platforms. Installation The best way to install it by using… | 842 | 3,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-26 | latest | en | 0.897198 |
https://homeworksol.com/this-is-a-math-project-can-someone-please-help/ | 1,656,769,916,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104141372.60/warc/CC-MAIN-20220702131941-20220702161941-00336.warc.gz | 343,933,704 | 10,766 | 4.8
4.9
4.7
GOAL: To redesign the bus route zones in a city based on circles and their equations.ROLE: You are the City Zoning Commissioner.AUDIENCE:
The bus routes need to be redesigned to be more cost effective. You
are trying to avoid cutting jobs by making the zones more efficient.
Many Port Authority workers are depending on you to save their jobs.SITUATION:
You have been chosen to determine the best way to redesign the bus
routes. Zone 1 will cover people living within 3 miles of the city
center. Zone 2 will cover people living between 3 and 7 miles from the
center of the city. Zone 3 will cover people living between 7 and 10
miles from the center of the city. Consider the center of the city the
origin on the coordinate plane and each unit on the coordinate plane
equal to one mile.PERFORMANCE: 1.
Determine the equation of the circles for Zone 1, Zone 2, and Zone 3.
Determine the area of each zone using the area formula for circles (A =
3.14xRadius^2). Show all work and explain how you arrived at your
answers. Round to the nearest whole number.3. Your house is located at (-1,3). What Zone do you live in? Show your work and explain how you arrived at your answer.
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We offer free unlimited revisions until your assignment is of your desired quality.. | 555 | 2,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.931407 |
https://freepdfbook.com/tag/finite-element-method-books-for-beginners/ | 1,695,406,598,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506421.14/warc/CC-MAIN-20230922170343-20230922200343-00248.warc.gz | 306,056,435 | 31,490 | Home Tags Finite element method books for beginners
# Tag: finite element method books for beginners
## The Finite Element methodology with An introduction with partial differential equations
Author(s) : A.J DaviesWriter : Oxford
Version : Second
Pages : 308
PDf dimension : 1.82 MB
##### Book Description:
The finite aspect methodology is a method for fixing issues in utilized science and engineering. The essence of this eBook is the appliance of the finite aspect methodology to the answer of boundary and initial-value issues posed when it comes to partial differential equations. The methodology is developed for the answer of Poisson’s equation, in a weighted-residual context, after which proceeds to time-dependent and nonlinear issues. The relationship with the variational strategy can be defined. The Finite Element Method with An introduction partial differential equations by A.J Davies e book is written at an introductory degree, creating all the mandatory ideas the place required. Consequently, it’s well-placed for use as a e book for a course in finite parts for last 12 months undergraduates, the standard place for finding out finite parts. There are lots of labored examples all through and every chapter has a set of workout routines with detailed options.
##### Book Contents:
1. Historic introduction
2. Weighted residual and variational methodology
3. The finite aspect methodology for elliptic issues
4. Greater-order parts: the isoparametric idea
5. Additional subjects within the finite aspect methodology
6. Convergence of the finite aspect methodology
7. The boundary aspect methodology
8. Computational points
Appendix A Partial differential equation fashions within the bodily sciences
Appendix B Some integral theorems of the vector calculus
Appendix C A method for integrating merchandise of space coordinates over a triangle Contents
Appendix D Numerical integration formulae
Appendix E Stehfest’s method and weights for numerical Laplace remodel inversion
References
Index
### The Finite Element Method: An Introduction with Partial Differential Equations PDF
Author(s): A. J. Davies
Publisher: Oxford University Press, USA, Year: 2011
ISBN: 0199609136
The Finite Element methodology with An introduction with partial differential equations | 461 | 2,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.816374 |
http://andrewgelman.com/2016/08/03/30833/ | 1,524,602,231,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947328.78/warc/CC-MAIN-20180424202213-20180424222213-00467.warc.gz | 19,650,696 | 22,072 | ## In policing (and elsewhere), regional variation in behavior can be huge, and perhaps give a clue about how to move forward.
Rajiv Sethi points to a discussion of Peter Moskos on the recent controversy over racial bias in police shootings.
Here’s Sethi:
Moskos is not arguing here that the police can do no wrong; he is arguing instead that in the aggregate, whites and blacks are about equally likely to be victims of bad shootings. . . .
Moskos offers another, quite different reason why bias in individual incidents might not be detected in aggregate data: large regional variations in the use of lethal force.
To see the argument, consider a simple example of two cities that I’ll call Eastville and Westchester. In each of the cities there are 500 police-citizen encounters annually, but the racial composition differs: 40% of Eastville encounters and 20% of Westchester encounters involve blacks. There are also large regional differences in the use of lethal force: in Eastville 1% of encounters result in a police killing while the corresponding percentage in Westchester is 5%. That’s a total of 30 killings, 5 in one city and 25 in the other.
Now suppose that there is racial bias in police use of lethal force in both cities. In Eastville, 60% of those killed are black (instead of the 40% we would see in the absence of bias). And in Westchester the corresponding proportion is 24% (instead of the no-bias benchmark of 20%). Then we would see 3 blacks killed in one city and 6 in the other. That’s a total of 9 black victims out of 30. The black share of those killed is 30%, which is precisely the black share of total encounters. Looking at the aggregate data, we see no bias. And yet, by construction, the rate of killing per encounter reflects bias in both cities.
This is just a simple example to make a logical point. Does it have empirical relevance? Are regional variations in killings large enough to have such an effect? Here is Moskos again:
Last year in California, police shot and killed 188 people. That’s a rate of 4.8 per million. New York, Michigan, and Pennsylvania collectively have 3.4 million more people than California (and 3.85 million more African Americans). In these three states, police shot and killed… 53 people. That’s a rate of 1.2 per million. That’s a big difference.
Were police in California able to lower their rate of lethal force to the level of New York, Michigan, and Pennsylvania… 139 fewer people would be killed by police. And this is just in California… If we could bring the national rate of people shot and killed by police (3 per million) down to the level found in, say, New York City… we’d reduce the total number of people killed by police 77 percent, from 990 to 231!
This is a staggeringly large effect.
Additional evidence for large regional variations comes from a recent report by the Center for Policing Equity. The analysis there is based on data provided voluntarily by a dozen (unnamed) departments. Take a close look at Table 6 in that document, which reports use of force rates per thousand arrests. The medians for lethal force are 0.29 and 0.18 for blacks and whites respectively, but the largest recorded rates are much higher: 1.35 for blacks and 3.91 for whites. There is at least one law enforcement agency that is killing whites at a rate more than 20 times greater than that of the median agency.
On the reasons for these disparities, one can only speculate:
I really don’t know what some departments and states are doing right and others wrong. But it’s hard for me to believe that the residents of California are so much more violent and threatening to cops than the good people of New York or Pennsylvania. I suspect lower rates of lethal force has a lot to do with recruitment, training, verbal skills, deescalation techniques, not policing alone, and more restrictive gun laws.
This is all important in its own right but I also wanted to highlight it as an example of a more general principle about different levels of variation when considering policy interventions.
One of my favorite examples here is smoking: it’s really hard to have an individual-level intervention to help people quit smoking. But aggregate interventions, such as banning indoor smoking, seem to work. This seems a bit paradoxical: after all, aggregate changes are nothing but aggregations of individual changes, so how could it be easier to change the smoking behavior of many thousands of people, than to change behaviors one at a time? But that’s how it is. Individual decisions are not so individual, as is most obvious, perhaps, in the variation across populations and across eras in family size: nowadays, it’s trendy in the U.S. to have 3 kids; a couple decades back, 2 was the standard; and a few decades earlier, 4-child families were common. We make our individual choices based on what other people are doing. And, again, it’s really hard to quit smoking, which can make it seem like smoking is as inevitable as death or taxes, but smoking rates vary a lot by country, and by state within this country.
To return to the policing example, we’ve had lots of discussion about whether or not particular cops or particular police departments are racially biased—lots of comparisons within cities—but Moskos argues we have not been thinking hard enough about comparisons between cities. An interesting point, and it would be good to see it on the agenda.
1. Ibn says:
Are Eastville and Westchester located in Simpsons County?
• Rahul says:
+1. Kinda strange to see the whole discussion without even mentioning Simpson.
2. Tom Passin says:
I looked into robbery vs murder rates a while ago, to see whether there seemed to be a relationship. Using FBI reports, over the period 1985-2010 (the most uniform coverage I could find), I found that for large and medium size cities I looked at, there was pretty nearly a linear relationship with surprisingly small dispersion (that is, as viewed on a log-log plot). The actual rates varied more than an order of magnitude. Remember that this period covers a large drop in crime rates from a peak near 1985 or 1990 to the present.
Papers I found on the number of murders committed during robberies had those rates far too low to explain the relationship. Although large cities like New York or Chicago did tend to have higher rates (of both murder and robbery), there was not clean or obvious division into large vs smaller cities. And they all were clearly on the same regression line (by eye on a log-log plot).
OTOH, plotting the same kind of data for rural states – states like Montana, South Dakota, or Maine, where there is no large city to distort the results, the relationship is entirely different. By eye, it appears flat – robbery rates don’t increase with the murder rates – with far lower robbery rates, though the murder rates overlap the murder rates of the cities.
There is clearly a big difference between rural states and cities in their murder vs robbery behavior. It would not be a surprise if there were big differences in policing behavior too.
3. Z says:
“Individual decisions are not so individual, as is most obvious, perhaps, in the variation across populations and across eras in family size: nowadays, it’s trendy in the U.S. to have 3 kids; a couple decades back, 2 was the standard; and a few decades earlier, 4-child families were common. We make our individual choices based on what other people are doing.”
Is there evidence that people emulating each other is a better explanation than common external forces (e.g. the economy) for the existence of a number of children mode?
• Andrew says:
Z:
I assume it’s a combination of economic factors, demographic factors, and people emulating each other. But I’ve not seen any research on the topic, so this is just my general impression.
Maybe some expert is reading deep enough into this thread to notice this discussion and can add some information?
• Ed Hagen says:
Not an expert, but the “demographic transition”, from relatively uniform high total fertility rates a couple of centuries ago to the low total fertility rates seen in most countries today has been the subject of countless studies for more than century (with no consensus in sight):
There is widespread agreement that this transition occurred roughly around the time that child mortality rates dropped dramatically, but my impression is that the causal role of reduced child mortality is still debated.
In fact, there is now widespread concern that fertility rates are dropping below replacement (2) in many countries, the so-called “second demographic transition.”
Copying others is one idea out there. This paper references some of that stuff:
http://www.pnas.org/content/107/Supplement_2/8985.full
4. “it’s trendy in the U.S. to have 3 kids;” Data?
• Andrew says:
Raghuveer:
Good point. It appears that having 3 kids is no more or less popular in the U.S. than it was a decade ago. The difference is in who is having the larger families.
Here’s what’s up. I googled *percentage of americans with 3 kids over time*, and the first link was Family size in America: Are large families back?:
With celebrities breaking the two-kid barrier, big families suddenly seem as trendy as jumbo-sized sunglasses and handbags.
And if you read the news reports, you might think it’s not just Hollywood but regular American families that are going super-sized. “Get ready for the new baby boom,” proclaims a recent headline from Life magazine. “For more parents, three kids are a charm,” says USA Today. We decided to cut through the buzz and find out whether big families really are on the upswing . . .
Here’s what they found:
Not really, says Steve Martin, a sociologist at the University of Maryland. When Martin crunched the numbers from a 2004 government survey — the most recent available — he found that 28 percent of women age 35 to 44, who are winding up their childbearing years, have three kids or more. Ten years ago, it was 29 percent. The numbers for younger women haven’t budged much, either. . . .
So why does it suddenly seem like you can’t walk down the street without tripping over a double-wide stroller and a few toddlers?
Despite the nationwide numbers, big broods could be a trend in certain areas, says S. Philip Morgan, sociologist and demographer at Duke University. “You do get clusters of behavior that are very real,” he says. “But it’s not appropriate to generalize them across the country, because there are other pockets that are behaving very differently.”
Also:
While the percentage of moms having Brady Bunch-sized broods has held steady, the women who make up their ranks have changed somewhat. . . . Professional moms have twice as many kids at home, on average, than their high-powered counterparts did back in 1977, according to a 2002 report from the Families and Work Institute. And in a 2000 study, sociologist Martin found that college-educated women who put off motherhood until their 30s are suddenly having families almost as big as everyone else’s. “That’s historically unprecedented,” he says.
And:
Wealthier families in general seem to be warming up to the idea of moving past a tasteful two. “Our survey from 2002 found that 12 percent of higher-income women had three or more children,” says Anjani Chandra, a researcher at the National Center for Health Statistics. “The figure from 1995 is only about 3 percent.”
This is all consistent with Z’s comment above regarding economic factors, in that the increase in #kids for rich families is coinciding with an increase in income and wealth among richer people. Still, an increase from 3% to 12% is a big deal no matter how you slice it.
In summary, I made the classic pundit error of generalizing from my social class to “the U.S.” as a whole.
• Thanks, especially since I realize that this is a tangent from your actual post.
The quote “You do get clusters of behavior that are very real” does raise red flags of chasing noise, though! (Randomness routinely leads to clusters that look “very real,” since people have a terrible intuition for randomness.) Agreed on the 3-12% shift being large, though, assuming the data are correct.
On your last line of “generalizing from my social class” — I will now update my mental picture of you as having “jumbo-sized sunglasses and handbags.”
• elin says:
The growth in multiple births is very real.
• Andrew says:
Elin:
Good point about the multiple births from fertility treatments. That explains some of the larger families right there.
5. Rahul says:
The smoking analogy, is that really about individual vs aggregate? Seems more like a nudge vs a hard-law difference.
e.g. If you passed draconian laws targeting specific smokers trying to quit where they wore a sensor-bracelet that gave them a \$1000 fine or a night in jail each time they smoked it could be an individual-level intervention that works?
6. jrc says:
re: “within” and “between” comparisons:
I find one of the most annoying aspects of observational social science research to be the reliance on a single “preferred” point estimate, one that often relies exclusively on either “within” or “between” variation, or occasionally some type of “difference in difference” that utilizes both. But it seems to me a mistake that we don’t try to learn from all the sources of variation whenever they are available.
I just think this within/between comparisons distinction is a really powerful conceptual tool that is fundamentally intuitive, easy to implement, and interpretable to lay-people. So maybe that’s why academics don’t love it – it’s too easy! But I would much prefer that when people have some sort of repeated cross-sections or panel data, and they use “time” and/or “individual/group” fixed-effects, they would stop writing “this controls for…” and start writing “this forces comparisons…”
And now I guess you could chime in that multi-level Bayesian modeling does a great job at trading-off identifying from within/between group variation, and all you need is 7 schools. But I want to see the pure-within and pure-between estimates too.
• Keith O'Rourke says:
> why academics don’t love it – it’s too easy!
I think it more that academics what to be tough guys and figure out the one best (or few best) way to definitively analyse the data.
As an example, in an earlier draft of this paper http://cid.oxfordjournals.org/content/28/4/800.full.pdf, I was suggesting numerous non-definitive analyses and the tough guy statistical group from the south insisted they could find the best model and only include that in the paper. I was asked to drop out as an author and a draft was submitted. Fortunately the journal reviewers were so negative that I was asked to rejoin and a paper that more reasonably reflected the insurmountable challenges of analyzing an observation intervention study resulted.
Remember discussing the experience with some more experienced statisticians at a JSM and indicating getting this into the paper:
“Because of uncertainty about optimal model selection given the small sample size, we looked at the estimated IVIG effect for all possible models. The minimum estimated odds ratio for survival associated with an IVIG effect was 4.3.”
One was very surprised I _got away with that_ and asked that I mail them a copy.
• jrc says:
First – note to self: don’t get streptococcal toxic shock syndrome.
Second – re: getting away with stuff…. Here’s hoping I get away with the 50 or so p-values I provide for a particular test in my new paper. It is a new test, and no one has experience with exactly how to specify it, so I specify it a bunch of ways and show all the results in a graph (along with the exact same specifications of a placebo test where I know it should fail to reject). Seems much less like “getting away with something” than just showing the one p-value most conducive to whichever interpretation I prefer.
Third – I get your point about “tough guy” statistics. But can’t a tough guy also recognize that different “models” are actually identified using different comparisons and that the differences in the point estimates are telling us important information about who is affected and how much they respond? I mean, once you accept that different “models” are actually estimating different aspects of the world (and not all trying to get at the same aspect of the world, the same “parameter”) then there is no reason to look for the “right” analysis anymore, it is about using all the information to make interesting, nuanced inferences from your different analyses. I guess my point is just that the motivation to be a “tough guy” who gets the “right” or “best” answer is prefaced on the false ideas that there a) is one answer and b) that all models estimating some coefficient are actually estimating the same parameter. Maybe this is more important with a concept like “Labor Supply” than with “Ability of Drug to Keep You Alive”.
7. Tom Passin says:
Right, because how can you really know how much to weigh the two extreme cases to develop an overall estimate? Which is more or less equivalent to asking how well you can support one prior or another, absent any other information. At least if you can see the range of the kinds of estimates, you may find some plausible way to weigh one more heavily.
8. Steve Sailer says:
I did some investigation into the two law enforcement killings in my Los Angeles neighborhood in this decade:
One was a classic suicide-by-cop: a white guy sits down on Ventura Blvd’s sidewalk and starts firing into the air until LAPD kills him. In retrospect, he was clearly not trying to kill bystanders, just get himself killed. I feel bad for the cop who had to shoot him. Perhaps in the future we could develop methods for averting suicide-by-cop?
The other police killing was a chain of screw-ups that ended with a federal agent killing an 18-year-old violist (white). Law enforcement tried to spin it as nefarious elements hanging out in the upscale parking lot, but I ran into the bereaved mother looking for clues at the scene and told her that as a long time local, the the cops’ story sounded phony and she should sue. She did and won \$3 million from a judge.
Both dead guys were white so nobody much cared about these shootings outside of friends and family.
But both seem like the kind of system problems that research and training could make less likely. But all the energy is focused on proving that white racism is the culprit in police shootings rather than on improving police systems universally across races. So not much gets accomplished because the Obama Administration and the media want to obsess over white racism rather than law enforcement not having the ideals methods for dealing with difficult situations.
9. Steve Sailer says:
I wonder if the rise of video cameras inclines cops to use less non-lethal force.
For example, the LAPD traditionally used a chokehold to subdue violent resistors, occasionally killing somebody through asphyxiation. That was banned, so they switched to beating recalcitrants with night sticks, such as Rodney King. That looked really bad on video, especially because the shakey first few seconds of the video establishing what King did were edited out of broadcast. In the retrial of the cops, the jury eventually convicted, but three jurors told the newspapers that only the last of about 60 blows could not be justified by King’s resistance.
Maybe post Rodney King it seemed faster and less likely to wind up on nightly news just to shoot somebody and get it over with quick than to non-lethally restrain them will billy clubs or choke holds. But now cameraphones are omnipresent, so even quick shootings are recorded.
10. Tom says:
The example illustrates how individual bias can hide in aggregate stats, but then that example is applied to actual data. The regional differences can also be explained by regional crime differences which is not discussed. You need a measure of excess deaths. This can be a threshold design, i.e. once an individual acts such that lethal force is justified, does a white individual linger in that zone longer or have to do something more outrageous to be killed or etc.? Excess deaths are then relative to the lower threshold. Deaths per police counter is an entirely inadequate measure for such a complex decision.
Really you would need to code the circumstances of each death according to whether it was justified. You would then need to see whether there are regional differences in justified police lethal force. Then we can argue about about changing those rules. Of course if it is unjustified, there’s a cover up, the reports are fabricated, etc then that is already punishable. Perhaps the regional variation can suggest deeper inquiry but they are not dispositive. As I recall Fryer could only attempt such a coding in Houston, which says better measurement is sorely needed.
11. Krzys says:
The problem with the story is that the aggregate bias is substantially in the other direction. It would take pretty wild patterns of bias and differential lethal force use to save anti black bias from the aggregate data. Blacks and whites are arrested roughly equally for violent crimes ( in terms of absolute numbers), yet, judging by Guardian numbers, blacks are only a quarter of victims of police shootings, while whites constitute 50% of the sample. Again, there is no evidence that composition effects can come even close to explaining this pattern.
The only hope Rajiv and his ilk have is to argue that the arrest data are biased. In other words substantial numbers of blacks are being railroaded by the cops to skew the numbers. Obviously, this would require pretty massive conspiracy. Even worse, you also have the homicide data, which are a tad hard to fake. They show that roughly 50% of homicide victims are black. Given that violent crime is known to correlate well with homicide, it’s very hard to believe that the arrest data are in any way substantially skewed. Unless, you imagine, that posses of KKK, unbeknownst to anybody, are riding inner cities in search for deadly entertainment.
• Andrew says:
Krzys:
I don’t know what’s Rajeev’s ilk. I just think the numbers that he is relaying from Moskos are interesting: There does seem to be some huge geographic variation.
• Krzys says:
Yes, the numbers are interesting, but the high variability is not likely to eliminate (the negative) aggregate bias. The US has a problem with the levels of fatal police violence. However, there is little evidence so far that we have a problem with racist police.
Rajeev’s ilk are people who jump to conclusions based on anecdotes and prejudice, and search for racism wherever they can.
12. elin says:
Moskos doesn’t really mean that there are regional variations in the sense that region is the real variable in question. He means there are variations from organization to organization, and therefore the way to approach the issue of reducing police shootings is to work at the organizational level. As he says, if everywhere would go as low as NYPD in terms of police shootings per capita the total would be much lower. So the thing is to look at how, say, NYPD does training of officers, what policies it enforces, and how the environment (e.g. lower gun availability and a much lower violent crime rate than most other big cities) combine to create this. The point being that, for example, LA could change not that somehow the weather in LA produces more police discharges.
• I think something that shouldn’t be underestimated is in fact the weather. How many people commit crimes in the middle of a Pennsylvania sleeting/freezing-rain/blizzard? In LA every single day you can be out committing crimes.
Also, another thing that shouldn’t be underestimated is the role of borders and drug importation. What fraction of Mexican cartel drugs comes through So Cal? It’s gotta be a LOT.
The endemic violent gang structure in south-central LA is legendary, quite literally. It’s imbued into a whole industry of “gangsta rap” storytelling. It’s become a self-perpetuating nightmare of violence.
In my trips to NY and Philly, I’ve also noticed a relatively strong and growing black middle class. People working at restaurants or the Port Authority or other transit areas etc where travelers are include notably large populations of multiple races. In Los Angeles, the equivalent is a large latino population, doing yard care, construction, restaurant work, etc. I don’t know that I’ve ever once seen a black person waiting a table, or doing a UPS delivery (in Pasadena, Altadena, Glendale, Burbank, etc). There are whole neighborhoods near me full of immigrant Chinese, Vietnamese, Korean, Armenian, and Latino ethnic groups. They don’t have the “no fly zone” around them. But South Central is pretty much “Stay out” unless you’re driving through on the 50 foot raised 110 freeway at 80 MPH.
I don’t dispute that training methods and soforth could help a lot, but I also think there are just some inevitably difficult demographic, geographic, climate, and other factors that actually have an important role which can’t be transported between locations the way training can.
• For those that aren’t from this area, LA County is one enormous urban area, with the borders between cities, or districts noticeable only because of the street sign that says you’re entering location X. Literally you could drive on surface streets in one direction averaging 25 MPH for like 5 hours and never leave an urban area (though you’d probably get out of LA county into San Bernardino or Orange, depending on which direction you decided to go)
Because of this there’s a tendency to think of LA as the whole county of Los Angeles. But, when it comes to policing, the LAPD handles the city of LA only. The city of LA is highly variable, but for the most part, people who have any kind of wealth just stay out. In that respect it’s a little like Philly, surrounded by suburbs where the wealth lives.
• Also, I should clarify, it’s not that I’ve never seen any successful say upper-middle class black people in this area. In fact I know professors, TV writers, a fireman, and several other black families who are quite successful. What I haven’t seen here is a path where poorer black people can get into a middle-class environment. Most of the middle and upper middle class black families I know came from the midwest, or east coast, or other parts of CA. That stuck-in-place situation seems to breed a lot of violent encounters with the police.
• Krzys says:
This speaks to a larger issue. The level of fatal police violence in the US is simply shockingly high, and not justified at all by the higher crime rates. There’s clearly a problem with accountability. | 5,643 | 26,868 | {"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} | 2.515625 | 3 | CC-MAIN-2018-17 | latest | en | 0.93852 |
https://www.hackmath.net/en/math-problem/3679?tag_id=101_64 | 1,611,144,507,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703520883.15/warc/CC-MAIN-20210120120242-20210120150242-00269.warc.gz | 795,874,558 | 14,047 | Glass door
What is the weight of glass door panel 5 mm thick height 2.1 meters and a width of 65 cm and 1 cubic dm of glass weighs 2.5 kg?
Correct result:
m = 17.0625 kg
Solution:
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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http://www.fixya.com/support/t24206565-measured_freezer_in_inches_20_30_20 | 1,508,550,484,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824537.24/warc/CC-MAIN-20171021005202-20171021025202-00607.warc.gz | 472,914,965 | 33,616 | Question about Office Equipment & Supplies
I measured a freezer in inches which Are 20 30 20 Please calculate in cibic feet
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• Office Equip... Master
20 inches times 30 inches times 20 inches is 12,000 cubic inches. There are 12 inches in a foot, so one foot is 12^3 or 1728 cubic feet. 12,000 cubic inches divided by 1728 cubic inches per cubic foot is about 6.9 cubic feet.
Posted on Apr 22, 2014
Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).
goodluck!
Posted on Jan 02, 2017
SOURCE: conversion
Heres one called science tools and it has sig figs, vectors and unit conversions and one other. http://education.ti.com/educationportal/downloadcenter/SoftwareDetail.do?website=US&tabId=1&appId=205
Posted on Sep 13, 2008
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Related Questions:
Calculating cubic ft
In order to calculate cubic feet you must multiply the LxWxH (where L=length, W=width and H=height or depth) making sure all three numbers are in feet. For example: a box 3' long by 2' wide by 1.5' high would contain 9 cubic feet of space. (3x2x1.5). If one or more of the measurements is in inches then you must convert it to feet or the others to inches. If all three measurements are in inches multiply them and then divide your answer by 1728 to give cubic feet because their are 1728 cubic inches in a cubic foot. (12x12x12). For example the same problem above in inches would be: 36x24x18=15,552cubic inches. Divide 15,552 by 1728 = 9 cubic feet
Apr 25, 2017 | Office Equipment & Supplies
What is 28x42x9.5 in cubic feet
I could give you a direct answer but that would only help you once. I prefer to give you the tools to use in the future.
If the above measurements are in feet, just multiply 28 x 42 x 9.5. The result is cubic feet (multiplying 2 measurements would be square feet). If the measurements are in inches (resulting in cu inch result) Google is great at both calculations and conversions. Take the result of the multiplication and go to the Google home page Google Type convert #### cu in to cu ft.
May 13, 2016 | Office Equipment & Supplies
How do I determine the cubic feet of my GR2SHKXKQ02 Type C22TFA02 Top Freezer purchased in 2002? It is running but is not cooling. I would like to replace it but do not know its size. Susan
Hi Susan:
Consider measuring it.
- Height (from floor to top of lid)
- Depth (from back to front)
- Width ( from left to right)
.* take all measurements in inches
* Divide all measurement by 12 (that gets feet and decimals)
* Multiply the 3 results (2 decimal places is plenty)
What you end up with is volume of entire freezer including thickness of bottom, sides, and top.
I t will be slightly larger than the cubic foot capacity of the freezer, but will help you find one that will fit in the same space.
*** If you want the capacity of the freezer, do the same calculations only measure the inside of the freezer.
Cheers.
Feb 28, 2016 | Whirlpool Refrigerators
How many cubic feet
a cibic foot is a measurement of a 1 ft X1 ft X 1 ft ( length X width X height)
measure your item and that will be the answer in cubic feet
you are in the washing machine category and they are not measured in cubic capacity but in pounds wash so it will be a 4 lb was or 5.5 lbs or 6 or what ever is on the model plate attached to the machine.
May 22, 2015 | Washing Machines
How do I calculate cubic feet of refrigerator
Normally refrigerators have stickers inside that state the cubic feet. If that's missing you must measure the width X depth X height to get cubic feet. WxDxH= cubic feet.
Jan 10, 2015 | Refrigerators
WHAT IS THE CUBIC FEET OF JVM1339WB-002 GE MICROWAVE SPACE SAVER?
Determining the cubic feet of any given appliance, Microwave, Oven, Freezer or Refrigerator is quite simple.
Open the door of your microwave and measure the length, width and height in inches of the interior cavity.
Multiply the length by the width by the height using a calculator to get the total in cubic inches.
Divide the cubic inches by 1,728, which is the volume of a cubic foot. The resulting number will be your microwave's measurement in cubic feet. You can also use an online conversion calculator to simplify matters
Hope that helps
Sep 12, 2014 | General Electric Microwave Ovens
How do I do feet and inches calculations on the Casio FX-115ES Plus? For example, how can I do calculations like 10'-3" + 5'-4" + 11'-11" without having to convert the inches to feet?
Calculators work to base '10' not base '12', so you'll have to convert. the S.I. unit of measurement is used for calculators which is metres and they are base '10'. The metre was calculated by a french bloke who got the calculation wrong, but they still use it. Feet and inches are better.
Jul 22, 2014 | Casio fx-115ES Plus Scientific Calculator
How to calculate cubic feet?
Multiply 21 inches by 14 inches by 10 inches to get 2940 cubic inches. There are 12 inches in a foot, so there are 12^3=1728 cubic inches in a cubic foot. Divide 2940 cubic inches by 1728 cubic inches per cubic foot to get about 1.7 cubic feet.
Alternatively, first convert all measurements to feet. 21 inches is 1 3/4 feet, 14 inches is 1 1/6 feet, 10 inches is 5/6 feet. Multiply them together to get 245/144 cubic feet, which again is about 1.7 cubic feet.
It's always nice when doing a problem two different ways gives the same answer.
Jul 04, 2014 | Office Equipment & Supplies
I need the capacity (total, fridge and freezer each) and the dimensions of this refrigerator.
Fzr and Ref are measured in cubic feet. The simple answer is to measure and multiply hieght by width by depth try to use the closest ft measurement or you will have to calculate from inches to get the square footage
Aug 08, 2010 | Refrigerators
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Level 3 Expert | 1,626 | 6,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-43 | latest | en | 0.904735 |
http://programming.empathy.works/examples/sinecosine.html | 1,713,027,813,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816820.63/warc/CC-MAIN-20240413144933-20240413174933-00530.warc.gz | 25,385,825 | 3,299 | # Sine Cosine.
Sine Cosine.
Linear movement with sin() and cos(). Numbers between 0 and PI*2 (TWO_PI which angles roughly 6.28) are put into these functions and numbers between -1 and 1 are returned. These values are then scaled to produce larger movements.
```
float x1, x2, y1, y2;
float angle1, angle2;
float scalar = 70;
void setup() {
size(640, 360);
noStroke();
rectMode(CENTER);
}
void draw() {
background(0);
x1 = width/2 + (scalar * cos(ang1));
x2 = width/2 + (scalar * cos(ang2));
y1 = height/2 + (scalar * sin(ang1));
y2 = height/2 + (scalar * sin(ang2));
fill(255);
rect(width*0.5, height*0.5, 140, 140);
fill(0, 102, 153);
ellipse(x1, height*0.5 - 120, scalar, scalar);
ellipse(x2, height*0.5 + 120, scalar, scalar);
fill(255, 204, 0);
ellipse(width*0.5 - 120, y1, scalar, scalar);
ellipse(width*0.5 + 120, y2, scalar, scalar);
angle1 += 2;
angle2 += 3;
}
```
## Functions Used
fill()
Sets the color used to fill shapes
size()
Defines the dimension of the display window width and height in units of pixels
ellipse()
Draws an ellipse (oval) to the screen
sin()
Calculates the sine of an angle
rectMode()
Modifies the location from which rectangles are drawn by changing the way in which parameters given to rect() are intepreted
rect()
Draws a rectangle to the screen
setup()
The setup() function is run once, when the program starts
cos()
Calculates the cosine of an angle
draw()
Called directly after setup(), the draw() function continuously executes the lines of code contained inside its block until the program is stopped or noLoop() is called
Converts a degree measurement to its corresponding value in radians | 466 | 1,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.729166 |
http://www.docstoc.com/docs/109554216/Probability-of-Simple-Games-Coin-Toss-A-Single-Coin-Toss-Have | 1,418,956,071,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768169.20/warc/CC-MAIN-20141217075248-00137-ip-10-231-17-201.ec2.internal.warc.gz | 473,769,813 | 13,932 | # Probability of Simple Games Coin Toss A. Single Coin Toss Have
Document Sample
``` Probability of Simple Games
Coin Toss:
A. Single Coin Toss: Have students discuss why football teams use a
single coin toss at the beginning of a game to determine which team kicks or
receives. Is this a fair way to determine this?
Have students work in pairs and toss a single coin 50 times each. They
should record their results using tally marks on the attached sheet and report
them to the class when done.
After students discuss their results and compare all groups results, they
need to find the mean of the entire class’ results.
B. Two Coin Toss:Ask how many possible outcomes there are to tossing
two coins?
Have students work in pairs and toss two coins 50 times each. Record
results using tally marks on the attached sheet and report them to the class when
done. Students could also find the mean of the entire class’ results.
After students discuss their results and compare all groups results,
complete the two tree diagrams on coin tossing and discuss the difference in
theoretical and experimental probability.
Dice:
Have students recall single die or pair of dice outcomes that are
considered lucky or unlucky. For example: rolling a “lucky seven” using a pair of
dice. Have students complete an outcome table (attached). List the total number
of ways each sum can be obtained.
1.) Is seven is really a lucky or likely number or is another number more
likely?
2.) Which numbers are less likely?
3.) How likely is it to roll doubles, “snake-eyes”?
4.) What are the chances of rolling doubles two times in a row?
5.) What are the chances of rolling doubles three times in a row, and losing a
turn in some games?
6.) Are even sums more likely to result than odd sums?
Single Coin Toss Recording Sheet
Experimental
Student 1
Name:
Student 2
Name:
Totals
Class Totals
Write out totals as simplified fractions, decimals and percents.
Two Coin Toss
Experimental
Student 1
Name:
Student 2
Name:
totals
Class Totals
Write out totals as simplified fractions, decimals and percents.
Tree Diagrams:
Single Coin Toss
Theoretical
Two Coin Toss
Theoretical Outcomes for Tossing a Pair of Dice
value from
each die
1 2 3 4 5 6
1
2
3
4
5
6
```
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views: 11 posted: 12/24/2011 language: English pages: 5 | 576 | 2,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | longest | en | 0.912774 |
https://www.dataunitconverter.com/gigabit-per-minute-to-tebibyte-per-hour | 1,716,389,756,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00163.warc.gz | 632,300,297 | 17,253 | # Gbit/Min to TiB/Hr → CONVERT Gigabits per Minute to Tebibytes per Hour
expand_more
info 1 Gbit/Min is equal to 0.00682121026329696178436279296875 TiB/Hr
S = Second, M = Minute, H = Hour, D = Day
Sec
Min
Hr
Day
Sec
Min
Hr
Day
Gbit/Min
## Gigabits per Minute (Gbit/Min) Versus Tebibytes per Hour (TiB/Hr) - Comparison
Gigabits per Minute and Tebibytes per Hour are units of digital information used to measure storage capacity and data transfer rate.
Gigabits per Minute is a "decimal" unit where as Tebibytes per Hour is a "binary" unit. One Gigabit is equal to 1000^3 bits. One Tebibyte is equal to 1024^4 bytes. There are 8,796.093022208 Gigabit in one Tebibyte. Find more details on below table.
Gigabits per Minute (Gbit/Min) Tebibytes per Hour (TiB/Hr)
Gigabits per Minute (Gbit/Min) is a unit of measurement for data transfer bandwidth. It measures the number of Gigabits that can be transferred in one Minute. Tebibytes per Hour (TiB/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Tebibytes that can be transferred in one Hour.
## Gigabits per Minute (Gbit/Min) to Tebibytes per Hour (TiB/Hr) Conversion - Formula & Steps
The Gbit/Min to TiB/Hr Calculator Tool provides a convenient solution for effortlessly converting data rates from Gigabits per Minute (Gbit/Min) to Tebibytes per Hour (TiB/Hr). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Gigabit) and target (Tebibyte) data units.
Source Data Unit Target Data Unit
Equal to 1000^3 bits
(Decimal Unit)
Equal to 1024^4 bytes
(Binary Unit)
The conversion from Data per Minute to Hour can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
The formula for converting the Gigabits per Minute (Gbit/Min) to Tebibytes per Hour (TiB/Hr) can be expressed as follows:
diamond CONVERSION FORMULA TiB/Hr = Gbit/Min x 10003 ÷ (8x10244) x 60
Now, let's apply the aforementioned formula and explore the manual conversion process from Gigabits per Minute (Gbit/Min) to Tebibytes per Hour (TiB/Hr). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Tebibytes per Hour = Gigabits per Minute x 10003 ÷ (8x10244) x 60
STEP 1
Tebibytes per Hour = Gigabits per Minute x (1000x1000x1000) ÷ (8x1024x1024x1024x1024) x 60
STEP 2
Tebibytes per Hour = Gigabits per Minute x 1000000000 ÷ 8796093022208 x 60
STEP 3
Tebibytes per Hour = Gigabits per Minute x 0.0001136868377216160297393798828125 x 60
STEP 4
Tebibytes per Hour = Gigabits per Minute x 0.00682121026329696178436279296875
Example : By applying the previously mentioned formula and steps, the conversion from 1 Gigabits per Minute (Gbit/Min) to Tebibytes per Hour (TiB/Hr) can be processed as outlined below.
1. = 1 x 10003 ÷ (8x10244) x 60
2. = 1 x (1000x1000x1000) ÷ (8x1024x1024x1024x1024) x 60
3. = 1 x 1000000000 ÷ 8796093022208 x 60
4. = 1 x 0.0001136868377216160297393798828125 x 60
5. = 1 x 0.00682121026329696178436279296875
6. = 0.00682121026329696178436279296875
7. i.e. 1 Gbit/Min is equal to 0.00682121026329696178436279296875 TiB/Hr.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Gigabits per Minute to Tebibytes per Hour using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Gigabit ?
A Gigabit (Gb or Gbit) is a decimal unit of digital information that is equal to 1,000,000,000 bits and it is commonly used to express data transfer speeds, such as the speed of an internet connection and to measure the size of a file. In the context of data storage and memory, the binary-based unit of gibibit (Gibit) is used instead.
arrow_downward
#### What is Tebibyte ?
A Tebibyte (TiB) is a binary unit of digital information that is equal to 1,099,511,627,776 bytes (or 8,796,093,022,208 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'tebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'terabyte' (TB). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Gigabits per Minute (Gbit/Min) to Tebibytes per Hour (TiB/Hr)
Apply the formula as shown below to convert from 1 Gigabits per Minute (Gbit/Min) to Tebibytes per Hour (TiB/Hr).
A B C
1 Gigabits per Minute (Gbit/Min) Tebibytes per Hour (TiB/Hr)
2 1 =A2 * 0.0001136868377216160297393798828125 * 60
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Gigabits per Minute (Gbit/Min) to Tebibytes per Hour (TiB/Hr) Conversion
You can use below code to convert any value in Gigabits per Minute (Gbit/Min) to Gigabits per Minute (Gbit/Min) in Python.
gigabitsperMinute = int(input("Enter Gigabits per Minute: "))
tebibytesperHour = gigabitsperMinute * (1000*1000*1000) / (8*1024*1024*1024*1024) * 60
print("{} Gigabits per Minute = {} Tebibytes per Hour".format(gigabitsperMinute,tebibytesperHour))
The first line of code will prompt the user to enter the Gigabits per Minute (Gbit/Min) as an input. The value of Tebibytes per Hour (TiB/Hr) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Tebibytes(TiB) are there in a Gigabit(Gbit)?expand_more
There are 0.0001136868377216160297393798828125 Tebibytes in a Gigabit.
#### What is the formula to convert Gigabit(Gbit) to Tebibyte(TiB)?expand_more
Use the formula TiB = Gbit x 10003 / (8x10244) to convert Gigabit to Tebibyte.
#### How many Gigabits(Gbit) are there in a Tebibyte(TiB)?expand_more
There are 8796.093022208 Gigabits in a Tebibyte.
#### What is the formula to convert Tebibyte(TiB) to Gigabit(Gbit)?expand_more
Use the formula Gbit = TiB x (8x10244) / 10003 to convert Tebibyte to Gigabit.
#### Which is bigger, Tebibyte(TiB) or Gigabit(Gbit)?expand_more
Tebibyte is bigger than Gigabit. One Tebibyte contains 8796.093022208 Gigabits.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,973 | 6,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-22 | latest | en | 0.744032 |
https://metanumbers.com/2696 | 1,709,562,770,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476452.25/warc/CC-MAIN-20240304133241-20240304163241-00792.warc.gz | 375,500,183 | 7,395 | # 2696 (number)
2696 is an even four-digits composite number following 2695 and preceding 2697. In scientific notation, it is written as 2.696 × 103. The sum of its digits is 23. It has a total of 4 prime factors and 8 positive divisors. There are 1,344 positive integers (up to 2696) that are relatively prime to 2696.
## Basic properties
• Is Prime? no
• Number parity even
• Number length 4
• Sum of Digits 23
• Digital Root 5
## Name
Name two thousand six hundred ninety-six
## Notation
Scientific notation 2.696 × 103 2.696 × 103
## Prime Factorization of 2696
Prime Factorization 23 × 337
Composite number
Distinct Factors Total Factors Radical ω 2 Total number of distinct prime factors Ω 4 Total number of prime factors rad 674 Product of the distinct prime numbers λ 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 2696 is 23 × 337. Since it has a total of 4 prime factors, 2696 is a composite number.
## Divisors of 2696
1, 2, 4, 8, 337, 674, 1348, 2696
8 divisors
Even divisors 6 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ 8 Total number of the positive divisors of n σ 5070 Sum of all the positive divisors of n s 2374 Sum of the proper positive divisors of n A 633.75 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 51.923 Returns the nth root of the product of n divisors H 4.25404 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 2696 can be divided by 8 positive divisors (out of which 6 are even, and 2 are odd). The sum of these divisors (counting 2696) is 5070, the average is 633.75.
## Other Arithmetic Functions (n = 2696)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 1344 Total number of positive integers not greater than n that are coprime to n λ 336 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 395 Total number of primes less than or equal to n r2 8 The number of ways n can be represented as the sum of 2 squares
There are 1,344 positive integers (less than 2696) that are coprime with 2696. And there are approximately 395 prime numbers less than or equal to 2696.
## Divisibility of 2696
m n mod m
2 0
3 2
4 0
5 1
6 2
7 1
8 0
9 5
The number 2696 is divisible by 2, 4 and 8.
• Refactorable
• Deficient
• Polite
## Base conversion 2696
Base System Value
2 Binary 101010001000
3 Ternary 10200212
4 Quaternary 222020
5 Quinary 41241
6 Senary 20252
8 Octal 5210
10 Decimal 2696
12 Duodecimal 1688
20 Vigesimal 6eg
36 Base36 22w
## Basic calculations (n = 2696)
### Multiplication
n×y
n×2 5392 8088 10784 13480
### Division
n÷y
n÷2 1348 898.666 674 539.2
### Exponentiation
ny
n2 7268416 19595649536 52829871149056 142429332617854976
### Nth Root
y√n
2√n 51.923 13.9179 7.20576 4.85449
## 2696 as geometric shapes
### Circle
Diameter 5392 16939.5 2.28344e+07
### Sphere
Volume 8.20821e+10 9.13376e+07 16939.5
### Square
Length = n
Perimeter 10784 7.26842e+06 3812.72
### Cube
Length = n
Surface area 4.36105e+07 1.95956e+10 4669.61
### Equilateral Triangle
Length = n
Perimeter 8088 3.14732e+06 2334.8
### Triangular Pyramid
Length = n
Surface area 1.25893e+07 2.30937e+09 2201.27 | 1,181 | 3,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-10 | latest | en | 0.830048 |
http://reference.wolfram.com/legacy/v5/Built-inFunctions/ListsAndMatrices/ListConstruction/FurtherExamples/SparseArray.html | 1,506,371,497,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693363.77/warc/CC-MAIN-20170925201601-20170925221601-00405.warc.gz | 286,890,022 | 7,622 | This is documentation for Mathematica 5, which was
based on an earlier version of the Wolfram Language.
Further Examples: SparseArray Here are some rules suitable for constructing a sparse array. In[1]:= Here is a sparse array corresponding to rules. In[2]:= Out[2]= This constructs the same sparse array using one rule. In[3]:= Out[3]= In[4]:= Out[4]= This shows the ordinary representation of ss. In[5]:= Out[5]//MatrixForm= This yields a sparse array in which unspecified elements are taken to have value instead of . In[6]:= Out[6]//MatrixForm= The usual commands for matrices work with sparse arrays. In[7]:= Out[7]= In[8]:= Out[8]//MatrixForm= This constructs a sparse array representing a banded matrix. In[9]:= Out[9]= In[10]:= Out[10]//MatrixForm= The full form of a sparse array contains the necessary data stored in an efficient form. In[11]:= Out[11]//FullForm= Evaluating the next two commands displays the structure of bb. In[12]:= In[13]:= This shows the structure of another sparse array. In[14]:= This defines the function BlockDiagonalMatrix which creates a block diagonal matrix from a list of matrices. For another efficient implementation see the Further Examples for PadRight. In[15]:= Here are four matrices. In[16]:= Out[16]//MatrixForm= In[17]:= Out[17]//MatrixForm= In[18]:= Out[18]//MatrixForm= In[19]:= Out[19]//MatrixForm= Here is the block diagonal matrix constructed from the four matrices. In[20]:= Out[20]//MatrixForm= In[21]:= | 376 | 1,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-39 | longest | en | 0.671714 |
https://ericlippert.com/2014/10/16/producing-combinations-part-two/?replytocom=29580 | 1,632,182,488,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057119.85/warc/CC-MAIN-20210920221430-20210921011430-00609.warc.gz | 282,678,488 | 34,729 | # Producing combinations, part two
Last time we saw that producing all the subsequences of a given size k from a given sequence of size n is essentially the same problem as computing all the sequences of n Booleans where exactly k of them are true. How do we do that?
An approach that I often see is “enumerate all sequences of n Booleans, count the number of on bits, and discard those which have too many or too few”. Though that works, it’s not ideal. Suppose we are seeking to enumerate the combinations of 16 items chosen from a set of 32. There are over four billion possible combinations of 32 bits, and of those over three billion of them have more or fewer than 16 true bits, so that’s a lot of counting and discarding to do. We can do better! To do so, we’ll use a combination of all my favourite techniques:
• Immutable data structures
• Abstract classes with derived nested classes
• Recursion
Long-time readers of this blog will have seen me use these techniques before, but for new readers, a quick introduction is in order.
The idea of an immutable collection is that the collection does not change when you add or remove something from it. That seems contradictory, but really it makes a lot of sense. The number 3 does not change into 7 when you add 4 to it; the number 3 is eternal and unchanging. Rather, adding 4 to it produces a new, entirely different number called 7. Similarly, adding an item to a collection does not change the original collection; it produces a new collection.
That might sound inefficient, but actually it is very efficient: because the original collection is immutable, we can use all or some of the original collection data structure in the new data structure.
The data structure I need to solve this problem is an immutable stack of Booleans, so let’s whip one of those up. (Of course we could use data structures in the BCL immutable collections, but for pedagogic and amusement purposes, let’s just make our own.) The operations I’m going to need are just pushing and enumerating, but for completeness let’s add public members for popping and emptiness checking as well. I’ll use one of my favourite techniques of an abstract base class whose derived classes are nested within it:
```using System;
using System.Collections;
using System.Collections.Generic;
abstract class ImmutableStack<T>: IEnumerable<T>
{
public static readonly ImmutableStack<T> Empty = new EmptyStack();
private ImmutableStack() {}
public abstract ImmutableStack<T> Pop();
public abstract T Top { get; }
public abstract bool IsEmpty { get; }
public IEnumerator<T> GetEnumerator()
{
var current = this;
while(!current.IsEmpty)
{
yield return current.Top;
current = current.Pop();
}
}
IEnumerator IEnumerable.GetEnumerator()
{
return this.GetEnumerator();
}
public ImmutableStack<T> Push(T value)
{
return new NonEmptyStack(value, this);
}
private class EmptyStack: ImmutableStack<T>
{
public override ImmutableStack<T> Pop()
{
throw new InvalidOperationException();
}
public override T Top
{
get { throw new InvalidOperationException(); }
}
public override bool IsEmpty { get { return true; } }
}
private class NonEmptyStack : ImmutableStack<T>
{
{
this.tail = tail;
}
public override ImmutableStack<T> Pop() { return this.tail; }
public override T Top { get { return this.head; } }
public override bool IsEmpty { get { return false; } }
}
}
```
We can create one of these like this:
```ImmutableStack<bool> myStack =
ImmutableStack<bool>.Empty
.Push(true)
.Push(false)
.Push(false);
```
And hey, now we’ve got the stack `false, false, true`. Again, remember, pushing on `Empty` did not change Empty at all. It’s the same as it ever was.
Our technique is going to be to recursively build longer bit sequences out of shorter bit sequences. Since pushing onto an immutable stack is cheap, and we don’t ever have to worry about the stack being changed by someone else, we’ll see that the algorithm is pretty straightforward.
Next time: Now that we have our helper class, we’ll actually enumerate the combinations.
## 69 thoughts on “Producing combinations, part two”
1. Two derived private classes within an abstract public class that returns instances of the private classes cast as the public class? Mind = blown. (The lack of needing any reflection is just icing on top of the cake.) This code is poetry.
• Thanks! It is a lovely technique that I wish more people used.
• This pattern is also really useful for emulating Java’s enums – you create a fixed set of instances of private derived classes, and expose those via the public base class.
You can prevent *other* code from subclassing by giving the public class a private constructor (and no other constructors). That way the *only* derived classes can be the nested ones.
• (And of course Eric’s code has a private constructor, presumably for this reason. It’s easy to miss that out, but it’s a really important part of the pattern.)
• While a good principle in general, making the constructor private is less important in this case, because in this case there’s no public method on the abstract class that takes an ImmutableStack, so even if you created your own class derived from ImmutableStack you couldn’t “inject” it.
• It’s a variation of the factory pattern.
• Unless you can override `Finalize` with a sealed method–something C# won’t allow (and which would be ugly in any case), I don’t think there’s any way to prevent an evil class from inheriting from an unsealed concrete class, defining two constructors which chain to each other while passing a parameter whose evaluation will throw an exception, and overriding `Finalize` to resurrect the object whose construction failed.
It is possible to protect an abstract class from outside inheritance with a technique I think I first saw from you, i.e. including an internal abstract method. Whereas a callable base constructor is only needed if one wishes to “properly” construct an object instance, one can’t even *define* a derived type–must less create any instances–without overriding all abstract methods. I don’t know any way to use this trick with concrete types, but I think it might be usable here.
• That’s correct, and it is an unfortunate flaw in the design of .NET. I’ve been meaning to blog about this for some time; thanks for the reminder.
• I look forward to it. I often find myself thinking that what’s really “needed” is a means by which a class can separate their “public-facing”, “parent-facing”, and “descendant-facing” faces. The question of who should be allowed to use one of Foo’s constructor to make an instance of Foo should be separate from the question of who should be allowed to use it to construct derived-class instances. Likewise the fact that a class overrides a public method does not mean that the override should be usable by derived-class instances.
In most cases, if a class overrides base method Foo(), it would be proper for the class to expose Foo() as part of its public face, and allow derived classes to use the inherited method without having to override it themselves. In the case of something like your Pop() method, however, there’s really no reason it should be exposed as a member of EmptyStack. If ImmutableStack were an interface, then EmptyStack could implement ImmutableStack.Pop without having to include Pop in its public face. Conceptually, it would seem like such usage should be no less reasonable when overriding an abstract method.
• Oh good heavens. I hadn’t thought of that. I’ll race Eric to blog about it – though no doubt in a less thoughtful way 😉 It’s a shame that the internal abstract method approach won’t work inside the same assembly… really, you want a private abstract method, but that’s not allowed (even though it does just about make sense).
• Why wouldn’t a private abstract method make sense, given that it would be accessible to, and could be overridden by, nested classes?
• (I can’t reply to supercat’s message, presumably due to a nesting limit.)
It *does* make sense – but only in this very limited situation. That’s my point – it would be useful, but the language disallows it.
• When you said “just about [makes sense]” I interpreted that as “doesn’t quite”. There are a number of places where C# seems to go out of its way to disallow things which would be legal in CIL because the implementers of C# didn’t see a particular usage case; an example you’ve written about IIRC about is the rule against using `System.Enum` as a generic constraint. It would be interesting to know which inheritance-related rules are imposed by the Runtime, and C# enforces them because failure to do so would yield types that can’t be loaded, or would otherwise be broken, and which rules are imposed purely by the C# compiler and, if waived, would result in code that worked exactly as normal semantics would suggest that it should.but for the existence of the rule.
• Trying to use language features to prevent “evil” classes is misdirected effort resulting in a false sense of security. Access restrictions are a tool for detecting encapsulation errors, nothing more.
• Nothing is cast … they’re derived classes. I can’t see anything mindblowing … the derived classes are simply made private and stuck inside the base class because there’s no need to instantiate them elsewhere and they’re simply implementation details.
2. I really like your abstract base class with private constructor. I see the parallels with functional programming data types. Except one thing stands out to me as different: in functional programming the type cases are normally publicly accessible, and can be used for pattern matching (I guess the C# equivalent would be something like `if (mystack is EmptyStack)`). I understand making the derived classes private (and instead exposing an `IsEmpty` method) if you think of the classes as an implementation detail, but I can’t immediately see the advantage of calling them implementation details and not public surface area. Would you mind sharing your thoughts about how you decided to do it this way?
• Good question. On the one hand I agree that in languages with pattern matching it can be nice to do matching as you describe. On the other hand, doing so then limits the ability of the author of the class to change the implementation details while preserving the abstract data type. Suppose for performance reasons I decided to use bit twiddling on an int to represent my stack of bits instead of a linked list as I am doing here. Code which depends on the implementation details is then broken; code which depends only on the public surface area is not.
• A linked list exposes its node types. A stack implemented as a linked list should not. This is true in functional programming too. There is absolutely no reason do pattern matching on the nodes of a stack (unless you’re the implementer).
3. The nested classes remind me of what we called “variant records” in a programming language class I once took. In C-speak, a variant record is a union of structs. That is, you have one container-like union that can take the form of different struct types. Actually, we used Scheme with some custom data structure libraries, but the effect was the same.
They’re pretty directly parallel, which is pretty satisfying to me. I’m almost sad that I didn’t think to use a more functional-style approach in my previous solution!
• It also reflects a very interesting OOP concept: the type of an object is useful for more than just polymorphism, because the type itself is meaningful independently of its members. I’m not sure I can quite put my wonder into words; it’s just a very intriguing idea to me.
• Indeed, it is possible to use variant records — often also called “discriminated unions” — as an implementation detail of functional languages to obtain this effect.
4. Doesnt the emptystack violate Liskov substitution Principle?
But yea, i dont know, since its private and you are the only user of it…but still I would like expert’s opinion 🙂
Thank you
• Suppose I have a base class Integer, and a derived class ColoredInteger. It has all the magnitude of an integer, but also tracks its color, whatever that is. The LSP suggests that I ought to be able to use a blue ColoredInteger in any context in which an Integer is required without changing the correctness of the program. So does that mean that if the program creates an Integer of value 123, I should be able to change the program to instead create a blue ColoredInteger of value 456 without changing the meaning of the program? I cannot imagine that this is what Liskov had in mind when she proposed her principle. You can substitute an empty stack for any context that takes *any stack*, but you can’t substitute an empty stack *for any particular stack*.
Here’s another way to look at it: the LSP can be stated as “if a theorem can be proved about objects of base type B then the theorem should be provable about objects of derived type D”. If you believe that the LSP has been violated then please prove a theorem about objects of type ImmutableStack that cannot be proved about objects of type EmptyStack.
I can prove the following theorems about ImmutableStack: (1) you can always push on any stack, (2) if IsEmpty is false then you can always pop, (3) if IsEmpty is true then popping an empty stack throws, (4) Suppose pushing item X onto immutable stack Y produces immutable stack Z. Z is not empty, the top of Z is X and popping Z produces Y. I could go on stating theorems all day. All those theorems can be proven about both the derived classes. So why would either be a violation of the LSP?
• In code that you have you are of course right.
But let’s say I have method that takes ImmutableStack. It promises to work correctly with instances of that type. But thanks to parameter contravariance you can pass into it instances of EmpyStack, which would throw once the method tries to do something. So what’s the solution ? Do you test inside method if you have correct instance of this or that and then do logic based on that ? Doesn’t that break the whole concept of polymorphism? The instance of EmptyStack doesn’t deliver what it promises…
I know we can try/catch and all…but according to wikipedia:
– No new exceptions should be thrown by methods of the subtype, except where those exceptions are themselves subtypes of exceptions thrown by the methods of the supertype.
Let’s say you have no control over what comes into your method call, and if there were 5 subtypes, each with own exception, you would have to know internals of them and think of it…
I understand, that in your piece of code, there is no violation, but to me, this instantly popped into my mind…what if…? I am trying to understand it all, so thank you for this article and for reply.
• I may be wrong, but don’t believe it’s a form of contravariance if a function accepts a parameter of type `ImmutableStack` and is given an argument of type `EmptyStack`. I would just call that “polymorphism” (or “subtype polymorphism”). The contravariance would come into play when you consider the type of the function itself, for example a value of type `Action` can be assigned to a variable of type `Action`.
But regarding substitution, if a function accepts a parameter of type `ImmutableStack` then nothing is violated by giving it an `EmptyStack`. The implicit contract of any `ImmutableStack` is that it can only be “popped” if `IsEmpty` returns false. The `EmptyStack` fulfills this perfectly because it returns true for `IsEmpty` and so can do whatever it wants when it’s popped – because the caller was never meant to call pop on it in the first place. It’s helpful to not have it do “whatever it wants” but instead throw a helpful exception to say “you aren’t supposed to pop *any* ImmutableStack that claims to be empty – no matter how it’s implemented”.
To put it another way, it’s a property of any `ImmutableStack` that it can be popped if it returns false for `IsEmpty`. Both `EmptyStack` and `NonEmptyStack` fullfil this property (and the other 2 properties Eric mentioned) and so they can be substituted safely wherever an `ImmutableStack` is required. I believe this is what Eric was saying in the previous comment, and so LSP is upheld this case.
• Your scenario contradicts itself. The method cannot *both* “promise to work correctly with instances of that type” and throw when the method tries to pop an empty stack. The contract of the ImmutableStack is “thou shalt not pop an empty stack”, and the method violates that. The contract of the ImmutableStack is “I will throw if you pop an empty stack”, and EmptyStack fulfills that contract. There’s no violation.
• “which would throw once the method tries to do something”
Only if you did something that violated the semantics of ImmutableStack, which says that you can’t pop an empty stack.
“No new exceptions should be thrown by methods of the subtype, except where those exceptions are themselves subtypes of exceptions thrown by the methods of the supertype.”
And these exceptions ARE thrown by the methods of UmmutableStack, under the exact same circumstances … an empty stack.
I think your intuitions are not well formed. It *should* be intuitively obvious that an EmptyStack is simply an ImmutableStack that is empty, and that it has exactly the same semantics.
• “But thanks to parameter contravariance you can pass into it instances of EmpyStack”
No, that has nothing to do with contravariance. Contravariance applies to type parameters, such as the T in ImmutableStack. If a method can return an ImmutableStack, then an override that returns an ImmutableStack would be contravariant.
• Arggh! How can a C# guy have a blog where < (less-than) and >( greater-than) are eaten in comments? And there’s no preview and no other help for commenters? Hopefully this will work but I won’t try again if it doesn’t:
Contravariance applies to type parameters, such as the T in ImmutableStack<T>. If a method can return an ImmutableStack<object>, then an override that returns an ImmutableStack<string> would be contravariant.
• [Sorry to come so late to the party. I’m bingereading 3 years of fabulousness.]
I think your comment was on the right track but not exact. IIRC, contravariance applies to type parameters, as you say, but it involves the substitution of _less_ derived types. So suppose ImmutableStack had a method
static void NoOp(ImmutableStack < T > source){ }0
Then this is legal b/c Action < > is contravariant:
Action < EmptyStack > actor = NoOp;
It’s safe because actor will only ever invoke with an EmptyStack, and every EmptyStack is an ImmutableStack < T >.
“If a method can return an ImmutableStack…”
Return types are the province of covariance, not contravariance. Covariant methods can have more-derived return types, so a covariant override of your method could specify that it will return only EmptyStack. (Er, well, it could if your method were an interface, but that’s even farther into the weeds.)
My mnemonic: “co” rhymes with O for “out” and with “low” for more-derived [I think of base classes as higher, as in “high-level abstraction”, though I know some people think the opposite]; “contra” is its opposite.
5. It took me some time to figure out how to create and use one of these things:
ImmutableStack stack = ImmutableStack.Empty;
As opposed to:
ImmutableStack stack = new ImmutableStack();
Which of course doesn’t compile. If I had seen this thing in the wild, I would have balked at that bug-a-boo.
• That’s a good point. Though it is a nice property that an empty immutable stack is a singleton; they’re all the same, so why bother ever making more than one?
• Are you saying that your first stack should be created with ImmutableStack.Empty.Push()? How does one obtain an initial stack otherwise?
• “How does one obtain an initial stack otherwise?”
That’s a good question. How does one create a stack without putting objects on that stack? Of course, if you already have a stack, you can use some or all of that. I suppose you could write a convenience method that accepts an IEnumerable and pushes them all onto a stack, but that’s not really something the stack needs to worry about.
• If one wishes to have a variable that behaves as a mutable stack of Thing [the normal scenario] one would start by saying: ImmutableStack<Thing:> myStack = ImmutableStack<Thing:>.Empty; and would push each thing onto it via myStack=myStack.Push(newThing); without having to worry about whether the stack was empty or not.
Note that the whole idea being ImmutableStack is that references to it can be stored in mutable variables, and those variables themselves will then behave as mutable stacks. Copying a variable will effectively copy the state of the stack encapsulated thereby to a another variable which would behave as an independent mutable stack, whereas copying a variable of a MutableStack type would merely create another “live view” of the same stack. It might perhaps help to think of the type as really being “ImmutableStackState”.
• No, your first stack is obtained with ImmutableStack.Empty. An empty stack is a stack.
• “Are you saying that your first stack should be created with ImmutableStack.Empty.Push()? How does one obtain an initial stack otherwise?”
Of course not … Push() takes an argument. You already got it right:
ImmutableStack stack = ImmutableStack.Empty;
As for
ImmutableStack stack = new ImmutableStack();
the availability of new is part of a class’s contract. I actually think that publicly exposing new is a bad idea, as it’s an implementation detail that can’t be later changed. All construction should be through factories, and that’s how it works in some well-designed languages.
6. Shouldn’t a pop operation return the head (a single element) instead of the tail, according to “standard” stack semantics? Is there a way to build an immutable stack which complies with this (perhaps an out parameter)?
• Yuck. There are logically two operations: look at the top of a stack, and produce a popped stack. I see no reason why those ought to be combined into one operation.
• I agree they shouldn’t be combined into a single method, but the other point stands. The Pop operation normally returns T, not a collection of T. It’s not that it’s incorrect to do so, but just it goes against the general usage of a stack. I’m sure you’re solution will be complete and concise, but this is definitely an unusual implementation of Pop().
• The usual stack has a Pop function that returns the top element and modifies the stack it’s called on. Since this is an immutable stack, the stack it’s called on cannot be modified. It would be possible to create a T Pop(out ImmutableStack s) function that simultaneously returns the top element and the resulting stack, but I’m not at all convinced that it would be a better adaptation to an immutable stack than having separate methods.
• I just think the Pop method should have a different name. Top in this case is equivalent to Peek, but Pop is not equivalent to Stack.Pop. I think Pop should be called something else. GetTail() but that’s not a great name either.
• @briantobin
A naming convention I’ve seen, that I quite like, is to use past-tense for operations that return a new instance rather than mutating an existing one. For example, if I see the code `myList.Append(x)` I would assume that `myList` is mutated, but if I see `myList.AppendedTo(x)` I would assume that `myList` is not mutated and that the result is a new list with x at the end of it. If that convention were used here, then the method would be called “Popped”. The word “pop” could be seen as an imperative command, as in “[You must] pop [now]!”, but the word “popped” could be seen as an *adjective*, as in “[the] popped [stack]” (I think the term may be “deverbal adjective”). In other words, they *describe* the result (in terms of the subject) rather than dictating an action.
Another example that comes to mind is Python’s “sort” command, which sorts a list by mutating it, vs its “sorted” function, which returns a “sorted list” based on the original list.
In the case of an immutable stack there is no ambiguity, so calling it “pop” seems fine, but it may illuminate some confusion if it were named in the imperative form.
• Think of it this way, in a mutable stack the Pop() method will mutate the stack to be the equivalent of the tail. So it makes sense that the Pop() method of the immutable stack should return the tail.
• Why does the Pop operation normally return the element on top? It’s because the stack itself mutates to be “popped”; instead of returning void, it’s much more C-style (stuff as much into one line as possible) to return something. The purpose of Pop is not to return an element; Peek/Top already does that! The purpose of Pop is to remove an item from the top of the stack. The only way the user can get to the resulting stack is to return it. In other words, the popped stack is truly the result of the pop operation, and should be the return value.
• ” I’m sure you’re solution will be complete and concise, but this is definitely an unusual implementation of Pop().”
It’s not at all an unusual implementation of an immutable stack … it’s the usual way.
Functional programming and its idioms have been around since at least 1958 … it’s about time that the programming population at large became familiar with them and didn’t think of them as “unusual”.
• Push and Pop methods which took a storage location of type ImmutableStack as a ref parameter could be written to provide lock-free thread-safety when used simultaneously by any number of “pushing” threads and one “popping” thread, but for Pop to be thread-safe in the presence of other threads pushing, it must in one operation update the stack reference and return either the old reference or the item contained therein. Having to use static members and pass `ref` parameters is a bit syntactically ugly (I wish there were a syntax which would allow something like myStack.:Pop() to be shorthand for ImmutableStack.Pop(ref myStack); the use of “.:” rather than “.” should serve as an adequate clue that the method is acting upon the *reference* itself, rather than upon the object identified thereby.)
• Immutable containers are thread-safe … that’s one of the big motivators for them. If you want to act on a stack in multiple threads, put the stack behind a thread-safe accessor.
• ” Is there a way to build an immutable stack which complies with this (perhaps an out parameter)?”
What a horrid idea, just to try to get an immutable stack to return what a mutable stack typically does. How could you possibly prefer
ImmutableStack<Thing> tempStack;
var thing = stack.Pop(out tempStack);
stack = tempStack;
over
var thing = stack.Top();
stack = stack.Pop();
?
• I didn’t say I preferred it. I just wanted to see what other people thought about Pop; and got a couple of nice answers.
• You asked “Is there a way to build an immutable stack which complies with this (perhaps an out parameter)?” … My question is, why would you *want* that, given how horrid the interface is? If is of course possible … how could it not be?
• And as for “nice answers” … Eric said “Yuck” but I bothered to show you just how awful the usage would be.
7. Would it simplify or complicate matters if popping an EmptyStack (without checking .IsEmpty) returned the same instance of EmptyStack rather than throwing an exception? Likewise, calling EmptyStack.Top could return Default(T)?
My hunch is that while it would remove the implicit contract that you cannot pop an EmptyStack, it would only lead to sloppy recursion code that could open the door to a stack overflow in the case of Pop, and a potential Null Reference exception in the case of Top. The exceptions as they are would properly inform the developer that they’re not using the class in the way it was designed to be used, but I’m curious if you had other reasons to throw those exceptions (such as a design principle).
• If you had to walk up to some programmer and ask them what it means to pop a stack, most people might say something like “removes the top item off the stack”, rather than “removes the top item off a stack, unless it’s empty – then no item would be removed”. If you had to further inquire, “What about when the stack is empty?” they might say something like “well then there are no items which can be removed, so it doesn’t make sense to remove the top item”. The same is true of getting the `Top` – if the stack is empty then there is no top to get, so it’s not valid to get the top of an empty stack. Getting the top of an empty stack is like getting the 5th element of a 4-element array – it’s simply not valid.
Apart from that, it just seems more helpful to the caller to “fail-fast” if they mistakenly attempt to do something stupid like removing an item from something that was already empty, since it’s best you’re alerted to the likelihood of there being a mistake in the caller’s algorithm as soon as possible rather than being left wondering where exactly that “null” value originated from somewhere else in the program.
• Good question. My opinion is that good reusable code has the property that it complains early and loudly when it is misused. Popping an empty stack is simply wrong; it indicates a fundamental bug in the program. A program which pops an empty stack is arbitrarily badly broken, and the right thing to do when you find yourself in a hole is *stop digging*. A program that is so broken that it pops an empty stack is so broken that it could be doing arbitrary harm to the user, so stop the program immediately before it does something really bad.
8. i don’t get something
when I was a CS student we learned that referencing objects is the basic of OOD, and what is the point of referencing something that can’t be changed (beyond performance)?
I do realize that changing an object state can cause unexpected results, and even if not it might point SRP violation. but we always use cloning instead of referencing (again, ignoring the performance) and then enjoy both worlds.
so my question is why to use a class if it is immutable. why not just use structs (even without immutability as every change will affect only a single instance copy).
• Well, you do arithmetic, probably. What’s the point of using the number 12 if the number 12 never changes?
And cloning makes the problem worse, not better. Cloning is expensive, cloning can be deep or shallow and getting it wrong makes bugs, and cloning wastes memory. In an immutable world, cloning is just copying a reference.
The reason not to use structs is because structs cannot be recursively defined in C#.
• Thanks!
• Generic structs can be recursively defined, actually, but the only way to use them without boxing would be to have a method that receives a generic struct give make a method call which passes a bigger struct that contains the first. Not terribly useful. | 6,677 | 31,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.921282 |
http://dec41.user.srcf.net/h/III_L/modular_forms_and_l_functions/0 | 1,575,590,678,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540482954.0/warc/CC-MAIN-20191206000309-20191206024309-00321.warc.gz | 40,075,050 | 72,595 | 0Introduction
III Modular Forms and L-functions
0 Introduction
One of the big problems in number theory is the so-called Langlands programme,
which is relates “arithmetic objects” such as representations of the Galois group
and elliptic curves over
Q
, with “analytic objects” such as modular forms and
more generally automorphic forms and representations.
Example. y
2
+
y
=
x
3
x
is an elliptic curve, and we can associate to it the
function
f(z) = q
Y
n1
(1 q
n
)
2
(1 q
11n
)
2
=
X
n=1
a
n
q
n
, q = e
2πiz
,
where we assume
Im z >
0, so that
|q| <
1. The relation between these two
objects is that the number of points of
E
over
F
p
is equal to 1 +
p a
p
, for
p 6
= 11. This strange function
f
is a modular form, and is actually cooked up
from the slightly easier function
η(z) = q
1/24
Y
n=1
(1 q
n
)
by
f(z) = (η(z)η(11z))
2
.
This function
η
is called the Dedekind eta function, and is one of the simplest
examples of a modular forms (in the sense that we can write it down easily).
This satisfies the following two identities:
η(z + 1) = e
/12
η(z), η
1
z
=
r
z
i
η(z).
The first is clear, and the second takes some work to show. These transformation
laws are exactly what makes this thing a modular form.
Another way to link E and f is via the L-series
L(E, s) =
X
n=1
a
n
n
s
,
which is a generalization of the Riemann ζ-function
ζ(s) =
X
n=1
1
n
s
.
We are in fact not going to study elliptic curves, as there is another course
on that, but we are going to study the modular forms and these
L
-series. We are
going to do this in a fairly classical way, without using algebraic number theory. | 515 | 1,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-51 | latest | en | 0.924181 |
https://web2.0calc.com/questions/help-meee_15 | 1,618,440,113,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078900.34/warc/CC-MAIN-20210414215842-20210415005842-00408.warc.gz | 680,977,639 | 6,013 | +0
# help meee
-1
83
3
+229
In the diagram, two pairs of identical isosceles triangles are cut off of square $ABCD$, leaving rectangle $PQRS$. The total area cut off is $200 \text{ m}^2$. What is the length of $PR$, in meters?
Mar 3, 2021
#1
+352
+2
we can use pythagorean theorum to solve this problem, so PR^2 = PS^2+SR^2.
those two lengths are the hypotenuses of triagnles APS and SDR, so
PS^2+SR^2 = 2AP^2+2SD^2.
we also know that AP^2+SD^2=200, so AP^2+2SD^2=2(200)=400.
this means that PR^2=400, so PR=$\boxed{20}$
Mar 3, 2021
edited by SparklingWater2 Mar 3, 2021
#3
+1164
+1
I was guesstimating...
jugoslav Mar 3, 2021
#2
+1164
+1
PB => x AP => 2x AB => 3x
x2 + (2x)2 = 200 x = 2√10
PR = sqrt[x2 + (3x)2] = √400 = 20
Mar 3, 2021 | 343 | 780 | {"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 | 4 | CC-MAIN-2021-17 | longest | en | 0.751056 |
https://betterlesson.com/lesson/resource/3142741/78178/entrance-ticket-docx | 1,527,177,680,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866511.32/warc/CC-MAIN-20180524151157-20180524171157-00365.warc.gz | 521,857,763 | 18,103 | ## Entrance Ticket.docx - Section 2: New Info / Exploration 1
Entrance Ticket.docx
# Comparing Linear and Exponential Functions
Unit 7: Functions
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# Finally done with GMAT 760--Q50,V42
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Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 01:17
2
KUDOS
Hey guys i gave GMAT yesterday and got a 760........
This was my second attempt got a 680 in the first one .
Thx to all the Gmatclub members for your support and invaluable suggestions without which i wouldn't have scored this much
Regards
Ambar
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 01:54
Very good score. Looking for a debrief.
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 02:00
maliyeci wrote:
Very good score. Looking for a debrief.
Congratulations and second the debrief request, after the celebrations end of course
Thank you for sharing the good news!
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 02:03
from 680 to 760, thats certainly a huge jump and requires lot of patience and hard work. Congratulations to you, and all the best for your applications.
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 02:21
Congratulations !!! Wish you all the best with your apps
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 08:43
1
KUDOS
First of all i m not a good writer and i m also tired of grammar rules ,so pls bear with me ...
I gave GMAT for the first time on 3rd august and got 680 Q50, v30. I messed up the verbal section badly. Wht i realised after my first attempt was that a good night sleep before the test is very very important. Before my first attempt i couldn't sleep on the night before the exam. Juz got 3-4 hours of sleep and beleive me I was not able to concentrate on any verbal question. I was juz guessing and rushing through the questions. So a humble request, pls try to maintain a schedule atleast 15-20 days before the exam. However i still didnt learn from my past mistakes and didnt maintained a schedule, so i had to take a pill on the night before the gmat. Thank God, it worked and i got a decent sleep. Again if u want to take a pill, try it out alteast once before the exam to be sure that u would not feel sleepy during the exam.
Anyways feeling fresh and charged up,I got to the test center 30 mins early, after completing all the formalities the lady at the center alloted me a system.
After some instructional screens the game began, Argument was pretty ok i wrote a lot in argument, but issue was kind of
weird but i beleive i managed to write enough to get atleast a 5.0 in AWA
Quant section: Being an engineer, i am pretty good in quant. Started of well in the quant section . It was going well all
through the section uptill the end where i got 2-3 very very difficult D.S problems. I didnt have too much time left so
have to rush through them. Finished the section just on tym. About the quant section i beleive that inequalities and absolute value are two concept where GMAT can cause big problems. Although its my personal opinion but ill recommend that if u r not that comfortable with these concepts than juz try to go through them properly.
During the break, I was bit disappointed feeling that i screwed up the quant section but somehow managed to organise myself for the verbal section. Ate something, had a sip of redbull and i was ready to go.
Verbal started, Last time what i did wrong was i was really rushing through the questions in the Verbal sections .And i
finished my Verbal section last time with 4-5 minutes to spare. What i have analysed through practice tests was that i
was not doing that good in initial 10-12 questions. So i decided ill give these initial 10-12 questions a little more time. .
Verbal section was pretty easy. Other than 2-3 questions, sentence correction was pretty straight forward. About C.R, got some paradoxy questions and a Bold face question in the middle. The Bold face question increased my confidence that i m doing pretty well in the section. 3 RC's were pretty easy and one that i got at the end was really a monster. Although the last RC was easy to comprehend but the questions were quite difficult. Anyways finished my verbal section just on time.
Was very nervous after finishing my test. Just rushed through all the screens to see my score. And boom, it was 760 i couldnt beleive that i got a 760. In Quant i was in the range of 50-51 in all my practice tests. But i was not expecting that much in verbal. I didnt even pressed the next button after seeing the score juz rushed through to the invigilator at the center. She had to press the next button and finish my test. While getting the unofficial report card i was literally shivering and what more, the lady out there asked me Have u cancelled ur score???? I was lyk are u serious i juz got a 760 why would i cancel my score. Anyways she apologized and handed over my report. I had to look at the score card for 10-12 time to confirm that i really got 760. Only after calling my parents did i stopped shivering. And it took me about half an hour to settle down.
Anyways I was very happy and partied all night long yesterday.
About the material : For AWA, How to get 6.0 AWA....my guide by chinese democracy is a must read for everyone. For Quant, I dint worked that much on Quant section. Took 2 GMATClub Quant challenges, by far the hardest questions u could get for ur Quant preparation. Other than that i just concentrated on improving my verbal skills. Improving the verbal score was really a daunting challenge for me. For sc, 1000sc and OG along with Manhattan Gmat Sc guide are more than enough. 1000 sc is a very good resource specially for someone who is not that good in SC. About RC and CR, I found Kaplan Solved LSAT papers a very good resource and they helped me a lot specially in improving my Reading Comprehension. However the best resouce are these online forums, i improved my verbal skills juz bcoz of GmatClub and other forums.
I would like to thank everyone who has responded to my post although i didnt posted that much. I have been a silent visitor out here at GMAT club but i have learnt a lot by veiwing people posts . Good luck to everyone appearing for GMAT - beleive me its not that difficult. With some hard work and faith in yourself u can rock on G-DAY.
Regards
Ambar
Last edited by amb2512 on 08 Sep 2009, 09:48, edited 1 time in total.
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 09:29
Nice job! What are your target schools?
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 10:06
great job on the test! and i agree 100% with your strategy about taking more time at the start of verbal - it worked for me too. good luck on all your apps
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 10:22
amb2512,
Congrats on great score and thanks for the debrief.
goodluck with the app process.
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 11:30
Superb score ... all the best for your apps
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 11:32
That's a great score...congrats.
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 12:40
Congratulations on getting a wonderful score.
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 12:52
Congrats for a superb score! What were your practice scores like?
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 13:07
amb2512 wrote:
First of all i m not a good writer and i m also tired of grammar rules ,so pls bear with me ...
I gave GMAT for the first time on 3rd august and got 680 Q50, v30. I messed up the verbal section badly. Wht i realised after my first attempt was that a good night sleep before the test is very very important. Before my first attempt i couldn't sleep on the night before the exam. Juz got 3-4 hours of sleep and beleive me I was not able to concentrate on any verbal question. I was juz guessing and rushing through the questions. So a humble request, pls try to maintain a schedule atleast 15-20 days before the exam. However i still didnt learn from my past mistakes and didnt maintained a schedule, so i had to take a pill on the night before the gmat. Thank God, it worked and i got a decent sleep. Again if u want to take a pill, try it out alteast once before the exam to be sure that u would not feel sleepy during the exam.
Anyways feeling fresh and charged up,I got to the test center 30 mins early, after completing all the formalities the lady at the center alloted me a system.
After some instructional screens the game began, Argument was pretty ok i wrote a lot in argument, but issue was kind of
weird but i beleive i managed to write enough to get atleast a 5.0 in AWA
Quant section: Being an engineer, i am pretty good in quant. Started of well in the quant section . It was going well all
through the section uptill the end where i got 2-3 very very difficult D.S problems. I didnt have too much time left so
have to rush through them. Finished the section just on tym. About the quant section i beleive that inequalities and absolute value are two concept where GMAT can cause big problems. Although its my personal opinion but ill recommend that if u r not that comfortable with these concepts than juz try to go through them properly.
During the break, I was bit disappointed feeling that i screwed up the quant section but somehow managed to organise myself for the verbal section. Ate something, had a sip of redbull and i was ready to go.
Verbal started, Last time what i did wrong was i was really rushing through the questions in the Verbal sections .And i
finished my Verbal section last time with 4-5 minutes to spare. What i have analysed through practice tests was that i
was not doing that good in initial 10-12 questions. So i decided ill give these initial 10-12 questions a little more time. .
Verbal section was pretty easy. Other than 2-3 questions, sentence correction was pretty straight forward. About C.R, got some paradoxy questions and a Bold face question in the middle. The Bold face question increased my confidence that i m doing pretty well in the section. 3 RC's were pretty easy and one that i got at the end was really a monster. Although the last RC was easy to comprehend but the questions were quite difficult. Anyways finished my verbal section just on time.
Was very nervous after finishing my test. Just rushed through all the screens to see my score. And boom, it was 760 i couldnt beleive that i got a 760. In Quant i was in the range of 50-51 in all my practice tests. But i was not expecting that much in verbal. I didnt even pressed the next button after seeing the score juz rushed through to the invigilator at the center. She had to press the next button and finish my test. While getting the unofficial report card i was literally shivering and what more, the lady out there asked me Have u cancelled ur score???? I was lyk are u serious i juz got a 760 why would i cancel my score. Anyways she apologized and handed over my report. I had to look at the score card for 10-12 time to confirm that i really got 760. Only after calling my parents did i stopped shivering. And it took me about half an hour to settle down.
Anyways I was very happy and partied all night long yesterday.
About the material : For AWA, How to get 6.0 AWA....my guide by chinese democracy is a must read for everyone. For Quant, I dint worked that much on Quant section. Took 2 GMATClub Quant challenges, by far the hardest questions u could get for ur Quant preparation. Other than that i just concentrated on improving my verbal skills. Improving the verbal score was really a daunting challenge for me. For sc, 1000sc and OG along with Manhattan Gmat Sc guide are more than enough. 1000 sc is a very good resource specially for someone who is not that good in SC. About RC and CR, I found Kaplan Solved LSAT papers a very good resource and they helped me a lot specially in improving my Reading Comprehension. However the best resouce are these online forums, i improved my verbal skills juz bcoz of GmatClub and other forums.
I would like to thank everyone who has responded to my post although i didnt posted that much. I have been a silent visitor out here at GMAT club but i have learnt a lot by veiwing people posts . Good luck to everyone appearing for GMAT - beleive me its not that difficult. With some hard work and faith in yourself u can rock on G-DAY.
Regards
Ambar
Hi dude,
thanks so much for sharing your experience, and I just wanted to ask you a couple of questions since I think we have things in common.
I also got a 680 (Q50 V31- AWA 4.5) last August in my first attempt, and I´m planning to take it again in December, so I have 3 months to go.
Like you, I am an engineer, and I feel very comfortable with quant section. Nevertheless, I´m a not native speaker, and verbal section is a big monster for me. How did you improve so much in so little time???
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 13:33
Incredible score and progression,
congratulation and thank you for the inspiring words!
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 14:08
Quote:
Hi dude,
thanks so much for sharing your experience, and I just wanted to ask you a couple of questions since I think we have things in common.
I also got a 680 (Q50 V31- AWA 4.5) last August in my first attempt, and I´m planning to take it again in December, so I have 3 months to go.
Like you, I am an engineer, and I feel very comfortable with quant section. Nevertheless, I´m a not native speaker, and verbal section is a big monster for me. How did you improve so much in so little time???
hey nuburu
In order to improve you have to figure out what went wrong during your last attempt and than make sure that you improve on that particlular thing this time around.
As far as RC's and CR's are concerned i worked on Rc's for 1 month and 36 official lsat test papers are a great resource for them. Each passage has been explained very nicely so i beleive if u go through them than u can get a good hold of RC's. For Cr's also these papers are worth doing. LSAT RC's and CR's are a bit difficult than GMAT one's so they can prepare you for difficult GMAT questions .
About SC other than OG i did 1000 sc and Brutal SC's. SC will improve with practice only. Prepare your own notes ,write down each concept you learn from different forums or books and the mistakes u generally make and go through them on regular basis.
HTH
Ambar
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 14:10
getmba wrote:
Congrats for a superb score! What were your practice scores like?
Thx getmba these were my practice scores
Manhattan GMAT
Cat1-650 Q47,V32
Cat2-680 Q48,V34
cat3-740 Q50,V41
cat4-740 Q51,V40
cat5-770 Q50,V44
Cat6-730 Q51,V38
Kaplan Cat1-650 q51 v35
Kaplan Cat2- 660 q50, v 38
Kaplan Cat3-650 q50,v36
Princeton Cat1-720 q51 v 38
Princeton Cat2-730 q51,v39
Gmat prep1-720 q51,v38---repeat 760
Gmat prep2-740 q51,v40---repeat 770
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 17:37
amb512, Can you please tell me what pill you took previous night to sleep.....but biggest fear is that I won't be able to sleep well the previous night........and I don't want ruin the test just because of lack of sleep.
I understand that the pill might not have the same effect on everyone, and I wont hold you accountable for any mishaps...
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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08 Sep 2009, 20:39
Awsome score and goodluck with ur applications.
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Re: Finally done with GMAT 760--Q50,V42 [#permalink]
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09 Sep 2009, 02:42
sdrandom1 wrote:
amb512, Can you please tell me what pill you took previous night to sleep.....but biggest fear is that I won't be able to sleep well the previous night........and I don't want ruin the test just because of lack of sleep.
I understand that the pill might not have the same effect on everyone, and I wont hold you accountable for any mishaps...
I took a 25 mg tranquilizer. I m from india so i dont think that u'll be getting the same pill at ur end. Ask any doctor about some tranquilizer.
Dude but remember to try it out atleast once before the exam..
All the best for ur Exam...
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Re: Finally done with GMAT 760--Q50,V42 [#permalink] 09 Sep 2009, 02:42
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# Finally done with GMAT 760--Q50,V42
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®. | 5,971 | 21,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-47 | latest | en | 0.859053 |
https://www.shaalaa.com/question-bank-solutions/put-equation-x-y-b-1-slope-intercept-form-find-its-slope-y-intercept-general-equation-of-a-line_58855 | 1,618,905,718,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039379601.74/warc/CC-MAIN-20210420060507-20210420090507-00150.warc.gz | 1,070,431,646 | 9,584 | Put the Equation X a + Y B = 1 to the Slope Intercept Form and Find Its Slope and Y-intercept. - Mathematics
Put the equation $\frac{x}{a} + \frac{y}{b} = 1$ to the slope intercept form and find its slope and y-intercept.
Solution
The given equation is $\frac{x}{a} + \frac{y}{b} = 1$
$bx + ay = ab$
$\Rightarrow ay = - bx + ab$
$\Rightarrow y = - \frac{b}{a}x + b$
This is the slope intercept form of the given line.
∴ Slope = $- \frac{b}{a}$ and y-intercept = b
Is there an error in this question or solution?
APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.9 | Q 3 | Page 72 | 200 | 632 | {"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} | 4.1875 | 4 | CC-MAIN-2021-17 | latest | en | 0.775257 |
https://www.physicsforums.com/threads/year12-physics-need-to-conduct-own-physics-experiment-what-experiment-should-i-do.298564/ | 1,553,558,368,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204736.6/warc/CC-MAIN-20190325234449-20190326020449-00167.warc.gz | 843,361,855 | 14,479 | # Year12 Physics - Need to conduct own physics experiment -What experiment should i do?
#### krzysiek
1. The problem statement, all variables and given/known data
Hey,
Basically my EPI (Extended Practical Investigation) is coming up and I need to conduct an experiment in class and research at home. The topic is Motion, and I can pretty much do anything I want - as long as i can work something out, or prove a formula. I was initially thinking of doing proving the Drag Coefficient of a balloon, but to do that, I would essentially need the terminal velocity of the balloon to work out the Drag Coefficient - which I can't do effectively.
What i mean, for example, is ie - Taking 5 springs that are identical. Use one spring to calculate the K constant for those springs. Then, join the springs in parallel AND in series to work out if the K constant changes when they work together, and try and establish a relationship. That's just an example for you guys to see what i'm trying to say, basically just wondering if anyone here has any ideas on what type of topic to do for the Prac / Experiment.
I will be extremely greatful for any help or ideas in regards to what to test or experiment on!
#### kreil
Gold Member
Re: Year12 Physics - Need to conduct own physics experiment -What experiment should i
I think a topic involving the simple harmonic oscillator would be good. For example, you could show how the equations of motion for a mass connected to a spring and for a pendulum are similar.
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• Solo and co-op problem solving | 394 | 1,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-13 | latest | en | 0.923685 |
https://metanumbers.com/255046800000001 | 1,627,706,898,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154053.17/warc/CC-MAIN-20210731043043-20210731073043-00082.warc.gz | 396,318,097 | 11,141 | 255046800000001
255,046,800,000,001 (two hundred fifty-five trillion forty-six billion eight hundred million one) is an odd fifteen-digits composite number following 255046800000000 and preceding 255046800000002. In scientific notation, it is written as 2.55046800000001 × 1014. The sum of its digits is 31. It has a total of 2 prime factors and 4 positive divisors. There are 254,755,982,667,312 positive integers (up to 255046800000001) that are relatively prime to 255046800000001.
Basic properties
• Is Prime? No
• Number parity Odd
• Number length 15
• Sum of Digits 31
• Digital Root 4
Name
Short name 255 trillion 46 billion 800 million 1 two hundred fifty-five trillion forty-six billion eight hundred million one
Notation
Scientific notation 2.55046800000001 × 1014 255.046800000001 × 1012
Prime Factorization of 255046800000001
Prime Factorization 877 × 290817331813
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 255046800000001 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 255,046,800,000,001 is 877 × 290817331813. Since it has a total of 2 prime factors, 255,046,800,000,001 is a composite number.
Divisors of 255046800000001
4 divisors
Even divisors 0 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 2.55338e+14 Sum of all the positive divisors of n s(n) 2.90817e+11 Sum of the proper positive divisors of n A(n) 6.38344e+13 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.59702e+07 Returns the nth root of the product of n divisors H(n) 3.99544 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 255,046,800,000,001 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 255,046,800,000,001) is 255,337,617,332,692, the average is 63,834,404,333,173.
Other Arithmetic Functions (n = 255046800000001)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 254755982667312 Total number of positive integers not greater than n that are coprime to n λ(n) 21229665222276 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7933874523642 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 254,755,982,667,312 positive integers (less than 255,046,800,000,001) that are coprime with 255,046,800,000,001. And there are approximately 7,933,874,523,642 prime numbers less than or equal to 255,046,800,000,001.
Divisibility of 255046800000001
m n mod m 2 3 4 5 6 7 8 9 1 1 1 1 1 2 1 4
255,046,800,000,001 is not divisible by any number less than or equal to 9.
Classification of 255046800000001
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
Other numbers
• LucasCarmichael
Base conversion (255046800000001)
Base System Value
2 Binary 111001111111011010110101111010100100010000000001
3 Ternary 1020110001020101000100202222211
4 Quaternary 321333122311322210100001
5 Quinary 231412141321300000001
6 Senary 2302234514333502121
8 Octal 7177326572442001
10 Decimal 255046800000001
12 Duodecimal 24731941341941
20 Vigesimal 14i2f6500001
36 Base36 2iemvrludd
Basic calculations (n = 255046800000001)
Multiplication
n×i
n×2 510093600000002 765140400000003 1020187200000004 1275234000000005
Division
ni
n⁄2 1.27523e+14 8.50156e+13 6.37617e+13 5.10094e+13
Exponentiation
ni
n2 65048870190240510093600000001 16590506185636298378610570720765140400000001 4231355513026760455816000144803221221141441020187200000001 1079193683259837799977925252492054018144361682808701902401275234000000001
Nth Root
i√n
2√n 1.59702e+07 63417.1 3996.27 760.894
255046800000001 as geometric shapes
Circle
Diameter 5.10094e+14 1.60251e+15 2.04357e+29
Sphere
Volume 6.94941e+43 8.17428e+29 1.60251e+15
Square
Length = n
Perimeter 1.02019e+15 6.50489e+28 3.60691e+14
Cube
Length = n
Surface area 3.90293e+29 1.65905e+43 4.41754e+14
Equilateral Triangle
Length = n
Perimeter 7.6514e+14 2.8167e+28 2.20877e+14
Triangular Pyramid
Length = n
Surface area 1.12668e+29 1.95521e+42 2.08245e+14 | 1,579 | 4,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.78224 |
https://www.physicsforums.com/threads/question-about-the-velocity-of-the-center-of-mass-reference-frame.1055168/ | 1,722,817,475,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00508.warc.gz | 740,840,010 | 25,492 | # Question about the velocity of the center of mass reference frame
• I
• gionole
gionole
I'm looking into center of mass and I saw the derivation of:
## V = \frac{\sum\limits_{i = 1}^{n} m_iv_i}{\sum\limits_{i = 1}^{n} m_i} ##
I understand how it's derived, so no need to explain this further. It's a velocity of the frame in which total momentum of our objects is zero. Forget about the center of mass as well and let's focus on this ##V## only.
Imagine, we got 2 objects moving both to the right with masses 2 and 3 kg with the speeds: 10 and 25. If there is a train moving to the right by ##19m/s##, in this frame, total momentum would be 0, but what's so special about this that is different if train had moved by ##20m/s## ? I know that total momentum wouldn't be 0, but so what ?
Let's analyze this example.
Case 1: train moving by ##19## and see the distances travelled by objects and our tain.
After 1sec, 10 -> 19 -> 25 ( 9 - train - 6)
After 2sec, 20 -> 38 -> 50 ( 18- train - 12)
After 3sec, 30 -> 57 -> 75 (27- train - 18)
Case 2: train moving by ##20## and see the distances travelled by objects and our tain.
After 1sec, 10 -> 20 -> 25 ( 10 - train - 5)
After 2sec, 20 -> 40 -> 50 ( 20- train - 10)
After 3sec, 30 -> 60 -> 75 (30- train - 15)
What's so special in this example ##19m/s## moving, that for ##20m/s## train, it's not special anymore ?
gionole said:
I'm looking into center of mass and I saw the derivation of:
## V = \frac{\sum\limits_{i = 1}^{n} m_iv_i}{\sum\limits_{i = 1}^{n} m_i} ##
I understand how it's derived, so no need to explain this further. It's a velocity of the frame in which total momentum of our objects is zero. Forget about the center of mass as well and let's focus on this ##V## only.
Imagine, we got 2 objects moving both to the right with masses 2 and 3 kg with the speeds: 10 and 25. If there is a train moving to the right by ##19m/s##, in this frame, total momentum would be 0, but what's so special about this that is different if train had moved by ##20m/s## ? I know that total momentum wouldn't be 0, but so what ?
Let's analyze this example.
Case 1: train moving by ##19## and see the distances travelled by objects and our tain.
After 1sec, 10 -> 19 -> 25 ( 9 - train - 6)
After 2sec, 20 -> 38 -> 50 ( 18- train - 12)
After 3sec, 30 -> 57 -> 75 (27- train - 18)
Case 2: train moving by ##20## and see the distances travelled by objects and our tain.
After 1sec, 10 -> 20 -> 25 ( 10 - train - 5)
After 2sec, 20 -> 40 -> 50 ( 20- train - 10)
After 3sec, 30 -> 60 -> 75 (30- train - 15)
What's so special in this example ##19m/s## moving, that for ##20m/s## train, it's not special anymore ?
1) Do you care about the momentum of the train here, or do you just mean a reference frame used for analysis? If you don't care about the momentum of the train, why introduce a physical train object at all?
2) What do you mean by "special"? All inertial frames are equivalent in terms the laws of motion. But depending on the scenario, the math can be simpler in some frames than in others. For example, when analyzing a two body collision in the CoM frame, the momenta pre and post collision are equal but opposite.
Last edited:
vanhees71
A.T. said:
1) Do you care about the momentum of the train here, or do you just mean a reference frame used for analysis? If you don't care about the momentum of the train, why introduce a physical train object at all?
2) What do you mean by "special"? All inertial frames are equivalent in terms the laws of motion. But depending on the scenario, the math can be simpler in some frames than in others. For example, when analyzing a two body collision in the CoM frame, in momenta pre and post collision are equal but opposite.
1. No, i don't care about momentum of train, I brought it as an example. The thing is the formula derivation is for the frame in which total momentum is zero. I get that, but why do we care about determining the velocity of such frame ?
2. I showed the example where velocity of the frame is 19 and then 20. We determined 19 is the velocity of such frame where total momentum is 0. Why did we even care about determining velocity of such frame ? Thats the question I guess, what makes 19 differ from 20 in my example(other than total momentum is zero ? )
@A.T. said it:
A.T. said:
All inertial frames are equivalent in terms the laws of motion. But depending on the scenario, the math can be simpler in some frames than in others.
DrClaude said:
@A.T. said it:
gionole said:
Yes, where you basically repeated what you said in #1.
Imagine now that your two masses are connected by a spring (massless, linear). Which frame would you rather use to calculate the motion of the two masses?
DrClaude said:
Yes, where you basically repeated what you said in #1.
Imagine now that your two masses are connected by a spring (massless, linear). Which frame would you rather use to calculate the motion of the two masses?
I don't know. I'm asking why we chose/decided to use the frame specifically in which momentum is zero..
gionole said:
I don't know. I'm asking why we chose/decided to use the frame specifically in which momentum is zero..
Because this is the reference frame in which you can separate the centre-of-mass motion from the relative motion. In the example I gave above, the relative motion of the two masses reduces to the problem of a single mass connected to a spring. This is easy to solve and one can find the motion of the two masses in any other frame from this solution.
Juanda
gionole said:
I'm asking why we chose/decided to use the frame specifically in which momentum is zero..
A.T. said:
...depending on the scenario, the math can be simpler in some frames than in others.
DrClaude said:
Because this is the reference frame in which you can separate the centre-of-mass motion from the relative motion. In the example I gave above, the relative motion of the two masses reduces to the problem of a single mass connected to a spring. This is easy to solve and one can find the motion of the two masses in any other frame from this solution.
In my example in the #1, ##V = 19## is such frame. Can you explain by example for my example how ##19## makes things easier and ##20## wouldn't do ?
gionole said:
In my example in the #1, ##V = 19## is such frame. Can you explain by example for my example how ##19## makes things easier and ##20## wouldn't do ?
What are "things"? It depends on the scenario and the goal, if the CoM frame makes it easier. I gave you an example in post #2:
A.T. said:
For example, when analyzing a two body collision in the CoM frame, in momenta pre and post collision are equal but opposite.
A.T. said:
For example, when analyzing a two body collision in the CoM frame, in momenta pre and post collision are equal but opposite.
Wouldn't the pre momentum be 0 in the CoM frame ? in which case is this why you mean "simpler" ?
Update: ah, you mean, individual momenta of these 2 bodies.
gionole said:
Update: ah, you mean, individual momenta of these 2 bodies.
Yes.
gionole said:
What's so special in this example 19m/s moving, that for 20m/s train, it's not special anymore ?
It isn’t “special”, but it is easier for some calculations. Especially collisions. In perfectly elastic collisions each object keeps the same speed before and after and just changes direction. In perfectly inelastic collisions their final speed is 0. Easy.
gionole
Dale said:
It isn’t “special”, but it is easier for some calculations. Especially collisions. In perfectly elastic collisions each object keeps the same speed before and after and just changes direction. In perfectly inelastic collisions their final speed is 0. Easy.
Perfectly elastic collisions:
Without zero momentum frame, to find after speeds, one has to do:
1. write conservation of momentum
2. write conservation of kinetic energy
3. solve system of equations
with zero momentum frame, to find after speeds, one has to do:
1. find V of zero momentum frame
2. find initial velocities of each object in this frame
3. reverse the signs from (2) to find after speeds
4. move back to original frame(lab frame) - i.e transform velocities
Question 1: How do you asses that zero momentum frame makes things easy in terms of calculations ?
Question 2: If the collision is not perfect(since not perfect, doesn't matter whether it's elastic or inelastic, there definitely was a loss of kinetic energy and we can't solve/find after velocities even in zero momentum frame. Then what ?
gionole said:
How do you asses that zero momentum frame makes things easy in terms of calculations ?
By doing the calculations. And if you don't find it easier, nobody is forcing you to use a particular frame.
gionole said:
... we can't solve/find after velocities even in zero momentum frame. Then what ?
If you can't solve it then you can't solve it. You need additional information for partially elastic collisions.
Last edited:
gionole
A.T. said:
By doing the calculations. And if you don't find it easier, nobody is forcing you to use a particular frame.If you can't solve it then you can't solve it. You need addition information for partially elastic collisions.
I agree, I'm just asking why you find it easier to do calculations in zero momentum frame ? maybe then I will see why it's easier once you tell me why you find it easier.
gionole said:
How do you asses that zero momentum frame makes things easy in terms of calculations ?
I can do a sign reversal in my head. I usually need to open Mathematica to solve a system of equations.
gionole said:
If the collision is not perfect(since not perfect, doesn't matter whether it's elastic or inelastic, there definitely was a loss of kinetic energy and we can't solve/find after velocities even in zero momentum frame. Then what ?
Then the benefit is less.
gionole
gionole said:
maybe then I will see why it's easier
Just do the calculations in different frames and decide for yourself.
Dale
I found one use case. For perfectly elastic collision, If 3 bodies collide, without zero momentum frame, we got 2 equations(kinetic energy conservation + momentum conservation), but for 3 bodies, these 2 equations are not enough and we would need to try to find 3rd(who knows what this 3rd could be). In ZMF though, we don't need this headache. Thoughts ?
Dale
A.T. said:
And if you don't find it easier, nobody is forcing you to use a particular frame.
@gionole this is key. You can use whatever coordinates you like. Sometimes changing coordinates helps a lot. Sometimes it doesn’t. It is your choice what to use for a given problem.
gionole said:
Question 1: How do you asses that zero momentum frame makes things easy in terms of calculations ?
This is a bit like asking why we do long multiplication even though it requires several multiplication steps where directly doing the multiplication only requires one. Sure, but the steps are easy, you don't ever need anything higher than your nine times table, and there's a standard way of handling carried digits.
Similarly, the ZMF approach gives a series of steps that can be carried out relatively easily. Your "solve system of equations" step involves at least one matrix inversion. The ZMF approach requires nothing more complex than division and vector addition.
If you prefer the matrix inversion approach, use that.
SammyS
## What is the velocity of the center of mass reference frame?
The velocity of the center of mass reference frame is the velocity at which the center of mass of a system of particles or objects moves. It is calculated as the total momentum of the system divided by the total mass of the system.
## How do you calculate the velocity of the center of mass for a system of particles?
To calculate the velocity of the center of mass for a system of particles, you sum the product of the mass and velocity of each individual particle, and then divide by the total mass of the system. Mathematically, it is given by:
$$\mathbf{V}_{cm} = \frac{\sum_{i} m_i \mathbf{v}_i}{\sum_{i} m_i}$$, where $$m_i$$ and $$\mathbf{v}_i$$ are the mass and velocity of the i-th particle, respectively.
## Why is the center of mass reference frame important in physics?
The center of mass reference frame is important because it simplifies the analysis of motion for a system of particles or a rigid body. In this frame, the motion can often be described more easily, and certain physical laws, such as conservation of momentum, can be applied more conveniently.
## Does the velocity of the center of mass change in an isolated system?
In an isolated system, where no external forces are acting, the velocity of the center of mass remains constant. This is a consequence of the conservation of momentum, which states that the total momentum of an isolated system is conserved.
## How does the velocity of the center of mass relate to collisions?
In collisions, the velocity of the center of mass of the system can provide insight into the behavior of the colliding bodies. For example, in an inelastic collision, while individual velocities of the colliding bodies change, the velocity of the center of mass remains constant if no external forces are involved. This helps in understanding the overall motion and energy distribution after the collision.
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1K | 3,380 | 13,796 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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} | 4.125 | 4 | CC-MAIN-2024-33 | latest | en | 0.93446 |
https://io.zouht.com/46.html | 1,653,330,754,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00783.warc.gz | 365,020,254 | 7,022 | # 【题目】Keep Connect
UNIQUE VISION Programming Contest 2022（AtCoder Beginner Contest 248）
F - Keep Connect
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : $500$ points
### Problem Statement
You are given an integer $N$ greater than or equal to $2$ and a prime $P$.
Consider the graph $G$ with $2N$ vertices and $(3N-2)$ edges shown in the figure below.
More specifically, the edges connect the vertices as follows, where the vertices are labeled as Vertex $1$, Vertex $2$, $\ldots$, Vertex $2N$, and the edges are labeled as Edge $1$, Edge $2$, $\ldots$, Edge $(3N-2)$.
• For each $1\leq i\leq N-1$, Edge $i$ connects Vertex $i$ and Vertex $i+1$.
• For each $1\leq i\leq N-1$, Edge $(N-1+i)$ connects Vertex $N+i$ and Vertex $N+i+1$.
• For each $1\leq i\leq N$, Edge $(2N-2+i)$ connects Vertex $i$ and Vertex $N+i$.
For each $i=1,2,\ldots ,N-1$, solve the following problem.
Find the number of ways, modulo $P$, to remove exactly $i$ of the $3N-2$ edges of $G$ so that the resulting graph is still connected.
### Constraints
• $2 \leq N \leq 3000$
• $9\times 10^8 \leq P \leq 10^9$
• $N$ is an integer.
• $P$ is a prime.
### Input
Input is given from Standard Input in the following format:
$N$ $P$
### Output
Print $N-1$ integers, the $i$-th of which is the answer for $i=k$, separated by spaces.
3 998244353
### Sample Output 1
7 15
In the case $N=3$, there are $7$ ways, shown below, to remove exactly one edge so that the resulting graph is still connected.
There are $15$ ways, shown below, to remove exactly two edges so that the resulting graph is still connected.
Thus, these numbers modulo $P=998244353$ should be printed: $7$ and $15$, in this order.
16 999999937
### Sample Output 2
46 1016 14288 143044 1079816 6349672 29622112 110569766 330377828 784245480 453609503 38603306 44981526 314279703 408855776
Be sure to print the numbers modulo $P$.
### 我的笔记
• $E_{a_1},E_{a_2},\dots,E_{a_{N-1}}$，即顶部的边，编号 $i=1,2,\dots,N-1$，如下图红边；
• $E_{b_1},E_{b_2},\dots,E_{b_{N-1}}$，即底部的边，编号 $i=1,2,\dots,N-1$，如下图蓝边；
• $E_{c_0},E_{c_1},\dots,E_{c_{N-1}}$，即中间的边，编号 $i=0,1,\dots,N-1$，如下图绿边。
• $V_{a_0},V_{a_1},\dots,V_{a_{N-1}}$，即上面的一排点；
• $V_{b_0},V_{b_1},\dots,V_{b_{N-1}}$，即下面的一排点。
• 状态 $0$：图联通，如下图中上侧图。
• 状态 $1$：图不连通，但有两个联通子图，并且两子图分别包含 $V_{a_i}$、$V_{b_i}$，如下图中下侧图。
• 如果新加的三条边 $E_{a_i},E_{b_i},E_{c_i}$ 都保留，那么这个图会变成状态 $0$，如下图上面的情况（红色为新加的边）
• 递推式：$dp[i][j][0]$ += $dp[i-1][j][1]$
• 如果新加的三条边 $E_{a_i},E_{b_i},E_{c_i}$ 去掉 $E_{c_i}$，那么这个图仍然为状态 $1$，如下图下面的情况
• 递推式：$dp[i][j+1][1]$ += $dp[i-1][j][1]$
• 如果新加的三条边 $E_{a_i},E_{b_i},E_{c_i}$ 都保留，那么这个图仍然为状态 $0$，如下图 $1$ 号情况（红色为新加的边）
• 递推式：$dp[i][j][0]$ += $dp[i-1][j][0]$
• 如果新加的三条边 $E_{a_i},E_{b_i},E_{c_i}$ 去掉 $E_{a_i}$，那么这个图仍然为状态 $0$，如下图 $2$ 号情况
• 如果新加的三条边 $E_{a_i},E_{b_i},E_{c_i}$ 去掉 $E_{c_i}$，那么这个图仍然为状态 $0$，如下图 $3$ 号情况
• 如果新加的三条边 $E_{a_i},E_{b_i},E_{c_i}$ 去掉 $E_{b_i}$，那么这个图仍然为状态 $0$，如下图 $4$ 号情况
• 递推式均为：$dp[i][j+1][0]$ += $dp[i-1][j][0]$
• 如果新加的三条边 $E_{a_i},E_{b_i},E_{c_i}$ 去掉 $E_{a_i},E_{c_i}$，那么这个图会变成状态 $1$，如下图 $5$ 号情况
• 如果新加的三条边 $E_{a_i},E_{b_i},E_{c_i}$ 去掉 $E_{b_i},E_{c_i}$，那么这个图会变成状态 $1$，如下图 $6$ 号情况
• 递推式均为：$dp[i][j+2][1]$ += $dp[i-1][j][0]$
### 代码
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3100;
long long dp[MAXN][MAXN][2];
int N, P;
int main()
{
dp[0][0][0] = 1;
dp[0][1][1] = 1;
cin >> N >> P;
for (int i = 1; i < N; i++)
{
for (int j = 0; j < N; j++)
{
dp[i][j][0] += dp[i - 1][j][1];
dp[i][j][0] %= P;
dp[i][j + 1][1] += dp[i - 1][j][1];
dp[i][j + 1][1] %= P;
dp[i][j][0] += dp[i - 1][j][0];
dp[i][j][0] %= P;
dp[i][j + 1][0] += 3 * dp[i - 1][j][0];
dp[i][j + 1][0] %= P;
dp[i][j + 2][1] += 2 * dp[i - 1][j][0];
dp[i][j + 2][1] %= P;
}
}
for (int i = 1; i < N; i++)
cout << dp[N - 1][i][0] << ' ';
return 0;
} | 1,778 | 3,730 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 3.125 | 3 | CC-MAIN-2022-21 | longest | en | 0.637852 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-12-exponential-functions-and-logarithmic-functions-mid-chapter-review-mixed-review-page-812/11 | 1,575,619,905,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540486979.4/warc/CC-MAIN-20191206073120-20191206101120-00418.warc.gz | 735,213,395 | 13,167 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$19$
$\log_{8}(8^{19})=\quad$... Apply the property $\log_{a}a^{k}=k$ $=19$ | 49 | 154 | {"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} | 3.609375 | 4 | CC-MAIN-2019-51 | latest | en | 0.538272 |
https://plainmath.org/integral-calculus/24586-calculus | 1,716,577,926,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00149.warc.gz | 391,060,688 | 28,469 | 2021-09-07
Calculus: Line Integrals
Jeffrey Jordon
Explanation
Line integrals can be used to evaluate integrals of two- or three-dimensional curves
The curve C is expressed by the following parametric equations given by
$x={t}^{2}$ and $y=2t$
Thus, to evaluate the following integral given by
$I={\int }_{c}xyds$
we use the following line integral formula given by
${\int }_{c}f\left(x,y\right)ds={\int }_{a}^{b}f\left(x\left(t\right),y\left(t\right)\right)|{r}^{\prime }|dt$
$={\int }_{a}^{b}f\left(x\left(t\right),y\left(t\right)\right)|\frac{dr}{dt}|dt$
=$={\int }_{a}^{b}f\left(x\left(t\right),y\left(t\right)\right)\sqrt{\left(\frac{dx}{xt}{\right)}^{2}+\left(\frac{dy}{xt}{\right)}^{2}}dt$
Since $x\left(t\right)={t}^{2}$ and $y\left(t\right)=2t$
We want to evaluate ${\int }_{c}xyds$ as follows
Firstly, we must find the integrand $f\left(x\left(t\right),y\left(t\right)\right)$ as follows
$f\left(x\left(t\right),y\left(t\right)\right)=xy$
$={t}^{2}\left(2t\right)$
=$2{t}^{3}$
Also, the derivatives of $x\left(t\right)$ and $y\left(t\right)$ must be
$\frac{d}{dt}x\left(t\right)=\frac{d}{dt}\left({t}^{2}\right)=2t$ and $\frac{d}{dt}y\left(t\right)=\frac{d}{dt}\left(2t\right)=2$
Since $0\le t\le 5$, so that the upper and lower limits of the integral must be
$a=0$ and $b=5$
Thus, using the line integral formula, we get
${\int }_{c}xyds={\int }_{a}^{b}f\left(x\left(t\right),y\left(t\right)\right)\sqrt{\left(\frac{dx}{xt}{\right)}^{2}+\left(\frac{dy}{xt}{\right)}^{2}}dt$
$={\int }_{0}^{5}\left(2{t}^{3}\right)\sqrt{\left(2t{\right)}^{2}+\left(2{\right)}^{2}}dt$
$=2{\int }_{0}^{5}{t}^{3}\sqrt{4{t}^{2}+4}dt=2{\int }_{0}^{5}{t}^{3}\sqrt{4\left({t}^{2}+1\right)}dt$
$=2{\int }_{0}^{5}2{t}^{3}\sqrt{\left({t}^{2}+1\right)}dt$
$=4{\int }_{0}^{5}{t}^{3}\sqrt{\left({t}^{2}+1\right)}dt$
To evaluate the following integral ${\int }_{0}^{5}{t}^{3}\sqrt{\left({t}^{2}+1\right)}dt$, we use the method of substitution as follows.
Let $u={t}^{2}+1$, so that
Differentiable both sides of the following equation $u={t}^{2}+1$, implies that
$du=2tdt$ and $tdt=\frac{du}{2}$
Also, we have ${t}^{2}=u-1$
Also, we find that the upper and lower limits of the integral must be
$t=0⇒u={0}^{2}+1=1$ and $t=5⇒u={5}^{2}+1=26$
Thus, the line integral becomes
${\int }_{c}xyds=4{\int }_{t=0}^{5}{t}^{2}\sqrt{{t}^{2}+1}tdt$
$4{\int }_{t=0}^{26}\left(u-1\right)\sqrt{u}\left(\frac{du}{2}\right)$
$=\frac{4}{2}{\int }_{t=0}^{26}\left(u-1\right)\sqrt{u}du$
$=2{\int }_{t=0}^{26}\left({u}^{\frac{3}{2}-\frac{1}{2}}\right)du$
Thus, the following line integral implies that
${\int }_{c}xyds=2\left[\frac{{u}^{\frac{5}{2}}}{5/2}-\frac{{u}^{\frac{3}{2}}}{3/2}{\right]}_{u=1}^{26}$
$=2\left[\frac{2{u}^{\frac{5}{2}}}{5}-\frac{2{u}^{\frac{3}{2}}}{3}{\right]}_{u=1}^{26}$
$=2\left[\left(\frac{2\left(26{\right)}^{\frac{5}{2}}}{5}-\frac{2\left(26{\right)}^{\frac{3}{2}}}{3}\right)-\left(\frac{2\left(1{\right)}^{\frac{5}{2}}}{5}-\frac{2\left(1{\right)}^{\frac{3}{2}}}{3}\right)\right]$
$=\frac{8}{15}\left(949\sqrt{26}+1\right)$
Do you have a similar question? | 1,343 | 3,078 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 51, "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} | 4.5625 | 5 | CC-MAIN-2024-22 | latest | en | 0.522205 |
https://fr.slideserve.com/Jims/lesson-20-rounding-numbers-to-the-nearest-ten-rounding-to-nearest-dollar | 1,669,516,482,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710155.67/warc/CC-MAIN-20221127005113-20221127035113-00109.warc.gz | 306,818,029 | 23,354 | Chemistry 1 – McGill Chapter 3 Scientific Measurement
# Chemistry 1 – McGill Chapter 3 Scientific Measurement
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## Chemistry 1 – McGill Chapter 3 Scientific Measurement
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##### Presentation Transcript
1. Chemistry 1 – McGillChapter 3Scientific Measurement WARNING: Learn it now...it will be used all year in Chemistry!!!
2. 3.1 Qualitative measurements • measurements that give results in a descriptive, non-numerical form. Examples: He is tall Electrons are tiny
3. Quantitative measurements • measurement that gives results in a definite form, usually as numbers and units. Examples: He is 2.2 m tall Electrons are 1/1840 times the mass of a proton
4. What is Scientific Notation? • Scientific notation is a way of expressing really big numbers or really small numbers. • It is most often used in “scientific” calculations where the analysis must be very precise. • For very large and very small numbers, scientific notation is more concise.
5. Scientific notation consists of two parts: • A number between 1 and 10 • A power of 10 N x 10x
6. To change standard form to scientific notation… • Place the decimal point so that there is one non-zero digit to the left of the decimal point. • Count the number of decimal places the decimal point has “moved” from the original number. This will be the exponent on the 10. • If the original number was less than 1, then the exponent is negative. If the original number was greater than 1, then the exponent is positive.
7. Scientific Notation • a number is written as the product of two numbers: a coefficient and 10 raised to a power. Examples: 567000 = 5.67 X 105 0.00231 = 2.31 X 10-3
8. Examples: Convert to or from Scientific Notation: 2.41 x 102 6.015 x 103 1.62 x 10-2 5.12 x 10-1 662 .0034 241 = 6015 = .0162 = .512 = 6.62 x 102 = 3.4 x 10-3 =
9. Learning Check • Express these numbers in Scientific Notation: • 405789 • 0.003872 • 3000000000 • 2 • 0.478260 4.05789 X 105 3.872 X 10-3 3 X 109 2 X 100 4.78260 X 10-1
10. FYI: “EE” button on calc= typing “X10^” Scientific Notation Cont. (This is important to master!!!) 6.25 x 103 - 2.01 x 102 = (2.15 x 103)(6.1 x 105)(5.0 x 10-6) = 3.25 x 108 = 3.6 x 107 6.05 x 103 6.6 x 103 9.03
11. Accuracy Vs. Precision What do you think the differences are? Ideas anyone???
12. 3.2 Accuracy • the measure of how close a measurement comes to the actual or true value of whatever is measured. • how close a measured value is to the accepted value. Precision • the measure of how close a series of measurements are to one another.
13. Can you hit the bull's-eye? Three targets with three arrows each to shoot. How do they compare? Both accurate and precise Precise but not accurate Neither accurate nor precise Can you define accuracy and precision?
14. Let’s use a golf analogy
15. Accurate? No Precise? Yes 10
16. Precise? Yes Accurate? Yes 12
17. Accurate? Maybe? Precise? No 13
18. Precise? We cant say! Accurate? Yes 18
19. In terms of measurement • Three students measure the room to be 10.2 m, 10.3 m and 10.4 m across. • Were they precise? • Were they accurate?
20. Percent Error Formula: % Error = accepted value- experimental value x 100 accepted value *always a positive number- indicated by the absolute value sign* You will use this formula when checking the accuracy of your experiment.
22. Significant Figures – includes all of the digits that are known plus a last digit that is estimated. ! FYI: These rules are not fun, but they will save you many points in the future if you learn them NOW!
23. Rules for determining Significant Figures1. All non-zero digits are significant. 1, 2, 3, 4, 5, 6, 7, 8, 9
24. 2. Zeros between non-zero digits are significant. (AKA captive zeros) 102 7002
25. 3. Leading zeros (zeros at the beginning of a measurement) are NEVER significant. 00542 0.0152
26. 4. Trailing zeros (zeros after last integer) are significant only if the number contains a decimal point. 210.0 0.860 210
27. 5. All digits are significant in scientific notation. 2.1 x 10-5 6.02 x 1023 Time to practice!!
28. Exact numbers have unlimited Significant Figures Do not use these when you are figuring out sig figs… Examples: 1 dozen = exactly 12 29 people in this room
29. Examples:How many significant digits do each of the following numbers contain: a) 1.2 d) 4600b) 2.0 e) 23.450c) 3.002 f) 6.02 x 1023 2 2 2 5 3 4
30. Learning Check A. Which answers contain 3 significant figures? • 0.4760 2) 0.00476 3) 4760 B. All the zeros are significant in 1) 0.00307 2) 25.300 3) 2.050 x 103 C. 534,675 rounded to 3 significant figures is 1) 535 2) 535,000 3) 5.35 x 105
31. Solution A. Which answers contain 3 significant figures? 1) 0.47602) 0.00476 3) 4760 B. All the zeros are significant in 1) 0.003072) 25.300 3) 2.050 x 103 C. 534,675 rounded to 3 significant figures is 1) 5352) 535,000 3) 5.35 x 105
32. Learning Check In which set(s) do both numbers contain the samenumber of significant figures? 1) 22.0 and 22.00 2) 400.0 and 40 3) 0.000015 and 150,000
33. Solution In which set(s) do both numbers contain the samenumber of significant figures? 3) 0.000015 and 150,000
34. Learning Check State the number of significant figures in each of the following: A. 0.030 m 1 2 3 B. 4.050 L 2 3 4 C. 0.0008 g 1 2 4 D. 3.00 m 1 2 3 E. 2,080,000 bees 3 5 7
35. Learning Check State the number of significant figures in each of the following: A. 0.030 m 1 2 3 B. 4.050 L 2 3 4 C. 0.0008 g 1 2 4 D. 3.00 m 1 2 3 E. 2,080,000 bees 3 5 7
36. Rounding Rules: 5 round up < 5 round down (don’t change) Examples: Round 42.63 to 1 significant digit = Round 61.57 to 3 sig. digs.= Round 0.01621 to 2 = Round 65,002 to 2 sig. digs. = 40 61.6 0.016 65,000
37. Addition and Subtraction The measurement with the fewest significant figures to the right of the decimal point determines the number of significant figures in the answer.
38. Examples: Solve using correct significant figures 45.756 m + 62.1 m = 75.263 m + 1123.93 m = 107.9 1199.19
39. Multiplying and Dividing Measurements The measurement with the fewest significant figures determines the number of significant figures in the answer.
40. Examples:Solve using correct significant figures: 3.43 m X 6.4253 m = 45.756 m X 1.2 m = 45.01 m / 2.2 m = 22.0 m2 55 m2 20. ***Why did the “m” unit go away on the last example?*** Noticethe decimal!
41. Uncertainty In lab, you record all numbers you know for sure plus the first uncertain digit. The last digit is estimated and is said to be uncertain but still considered significant. • Graduated cylinders have markings to the nearest mL (milliliter) and you will determine volume to the nearest 0.1 mL… because that is ONE DIGIT OF UNCERTAINTY.
42. International System of Units • revised version of the metric system • abbreviated SI All units, their meanings and values can be found on pgs. 63,64,65. Meter (m) – Liter (L) – Gram (g) – SI unit for length SI unit for volume SI unit for mass
43. Some Tools for Measurement Which tool(s) would you use to measure: A. temperature B. volume C. time D. weight
44. Solution A. temperature thermometer B.volume measuring cup, graduated cylinder C.time watch D. weightscale
45. Learning Check Match L) length M) mass V) volume ____ A. A bag of tomatoes is 4.6 kg. ____ B. A person is 2.0 m tall. ____ C. A medication contains 0.50 g Aspirin. ____ D. A bottle contains 1.5 L of water. M L M V
46. Learning Check What are some U.S. units that are used to measure each of the following? A. length B. volume C. weight D. temperature
47. Solution Some possible answers are A.length inch, foot, yard, mile B. volume cup, teaspoon, gallon, pint, quart C. weight ounce, pound (lb), ton D. temperature °F
48. Metric Prefixes • Kilo- means 1000 of that unit • 1 kilometer (km) = 1000 meters (m) • Centi- means 1/100 of that unit • 1 meter (m) = 100 centimeters (cm) • 1 dollar = 100 cents • Milli- means 1/1000 of that unit • 1 Liter (L) = 1000 milliliters (mL)
49. Metric Prefixes
50. Metric Prefixes | 2,486 | 8,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-49 | latest | en | 0.823348 |
http://mathhelpforum.com/algebra/171735-hard-fraction-2-print.html | 1,527,063,214,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865456.57/warc/CC-MAIN-20180523063435-20180523083435-00403.warc.gz | 190,415,271 | 4,135 | # hard fraction!
Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• Feb 18th 2011, 02:00 PM
andyboy179
Quote:
Originally Posted by harish21
No it does not. You are subtracting x and 1 from both sides of the equation.
$\displaystyle (x+1)=(6x^2-5x+1)$
subtract 1 from both the sides:
$\displaystyle (x+1)-1=(6x^2-5x+1)-1$
that leaves you with:
$\displaystyle x = 6x^2-5x$
now you subtract x from both the sides. this gives you:
$\displaystyle x-x=(6x^2-5x)-x$
which gives you:
$\displaystyle 0=6x^2-5x-x$
therefore $\displaystyle 0=6x^2-6x$
************************************************** *************
Note: $\displaystyle \boxed{-5x-x=-6x} \;\;and \;\;\boxed{-5x+x=-4x}$
In the first box, you are subtracting x from -5x. In the second box, you are adding x to -5x. This is where your confusion lies.
If you still dont understand it, please refer to your books/notes/instructor to avoid trouble in the future.
Please refer to post Number 2 by Unknown008
your question was $\displaystyle \dfrac{3}{2x - 1} - \dfrac{4}{3x-1}=1$
then Unknown008's method (which you've been told multiple number of times) takes you :
$\displaystyle \dfrac{x+1}{6x^2 - 5x + 1}=1$
multiplying both sides of this equation by $\displaystyle 6x^2 - 5x + 1$.gives you:
$\displaystyle x+1=6x^2 - 5x + 1$.
Have fun!
okay i think i get it now, so would 0=6x^2 -6x be the answer?
• Feb 18th 2011, 02:03 PM
harish21
Does your question tell you to find the value of x?
• Feb 18th 2011, 02:11 PM
andyboy179
usually with questions like these i usually just find X via quadratic equation or by dividing a number by the amount of X's.
how would i work out x with this question?
• Feb 18th 2011, 02:17 PM
harish21
You have:
$\displaystyle 0=6x^2-6x$
Find x from here..
• Feb 18th 2011, 02:22 PM
andyboy179
would i square root 6x^2=6
0=6-6
plus 1 to each side to get:
6=6
6/6=1
x=1?
• Feb 18th 2011, 04:34 PM
Quote:
Originally Posted by andyboy179
would i square root 6x^2=6
0=6-6
plus 1 to each side to get:
6=6
6/6=1
x=1?
No, to find all answers, write as a multiplication of factors...
$\displaystyle 6x^2-6x=0$
$\displaystyle 6xx-6x=0$
$\displaystyle (x)6x+(-1)6x=0$
6x is common to both, so factor...
$\displaystyle 6x(x-1)=0$
6 is not 0, so $\displaystyle x(x-1)=0$
Those factors now tell you the answers because any multiple of zero is zero.
• Feb 18th 2011, 09:18 PM
Unknown008
Yes, x = 1 but you forgot 0
If you so much want to use the formula, you can make this:
$\displaystyle 0 = ax^2 + bx + c$
$\displaystyle x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Here, you have $\displaystyle 0 = 6x^2 - 6x$
So, a = 6, b = -6, c = 0
• Feb 19th 2011, 01:21 AM
andyboy179
So by doing the quadratic equation I would end with the answers of 1 and 0 for x??
• Feb 19th 2011, 01:27 AM
Unknown008
Did you get something else than 1 and 0? (Wondering)
• Feb 19th 2011, 01:51 AM
andyboy179
No I did the quadratic and get 1 and 0.
To get it I did:
X=6+ (square root)36-4x6x0/12=1
X=6- (square root)36-4x6x0/12=0
Show 40 post(s) from this thread on one page
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# MGMAT 3rd version and OG12
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MGMAT 3rd version and OG12 [#permalink]
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20 Jan 2010, 22:55
Hi
I hav OG12 and have a deal for MGMAT 3rd version full 8 book set.
Though there is an official mapping of OG 11 to OG 12 for MGMAT usage purpses, just wanted to know how effective is this?
Is this fine or should I better go for MGMAT 4th version?
Thanks
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21 Jan 2010, 09:59
post2pras wrote:
Hi
I hav OG12 and have a deal for MGMAT 3rd version full 8 book set.
Though there is an official mapping of OG 11 to OG 12 for MGMAT usage purpses, just wanted to know how effective is this?
Is this fine or should I better go for MGMAT 4th version?
Thanks
Realistically, there is no difference and your materials will do as good of a job. As far as I understand (and feel free to correct me), the 4th edition was to update the 3rd edition to the OG 12.
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Re: MGMAT 3rd version and OG12 [#permalink]
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21 Jan 2010, 10:29
Hi post2pras,
We actually have a page dedicated to the differences between our fourth and third editions, which you can see here: http://www.manhattangmat.com/fourth-ed-info.cfm
Two key differences are:
1. the reference to the new OG (12th edition) in our fourth edition
2. the split between basic and advanced content in our fourth edition
If you do buy the third edition and you're using the 12th ed OG, you'll need to download the conversion documents on that page.
We always recommend that people buy the latest versions of our books if possible, because we do make content updates with each revision, but you could probably do all right with our third editions.
The only other thing to keep in mind is that if you're buying them used, you won't have access to our online CAT exams and question banks (which usually come with our books) as that 1-year access would have been registered with the original owner of the books.
Let me know if you have any other questions!
Best Wishes,
Caitlin Clay
Student Services Associate
Manhattan GMAT
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Re: MGMAT 3rd version and OG12 [#permalink]
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21 Jan 2010, 13:00
Thanks for the info..
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Re: MGMAT 3rd version and OG12 [#permalink] 21 Jan 2010, 13:00
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# MGMAT 3rd version and OG12
Moderator: HiLine
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®. | 1,089 | 3,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-43 | latest | en | 0.912101 |
https://convertoctopus.com/13-6-feet-per-second-to-miles-per-hour | 1,628,072,193,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154798.45/warc/CC-MAIN-20210804080449-20210804110449-00553.warc.gz | 190,393,190 | 8,068 | ## Conversion formula
The conversion factor from feet per second to miles per hour is 0.68181818181818, which means that 1 foot per second is equal to 0.68181818181818 miles per hour:
1 ft/s = 0.68181818181818 mph
To convert 13.6 feet per second into miles per hour we have to multiply 13.6 by the conversion factor in order to get the velocity amount from feet per second to miles per hour. We can also form a simple proportion to calculate the result:
1 ft/s → 0.68181818181818 mph
13.6 ft/s → V(mph)
Solve the above proportion to obtain the velocity V in miles per hour:
V(mph) = 13.6 ft/s × 0.68181818181818 mph
V(mph) = 9.2727272727273 mph
The final result is:
13.6 ft/s → 9.2727272727273 mph
We conclude that 13.6 feet per second is equivalent to 9.2727272727273 miles per hour:
13.6 feet per second = 9.2727272727273 miles per hour
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 mile per hour is equal to 0.1078431372549 × 13.6 feet per second.
Another way is saying that 13.6 feet per second is equal to 1 ÷ 0.1078431372549 miles per hour.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that thirteen point six feet per second is approximately nine point two seven three miles per hour:
13.6 ft/s ≅ 9.273 mph
An alternative is also that one mile per hour is approximately zero point one zero eight times thirteen point six feet per second.
## Conversion table
### feet per second to miles per hour chart
For quick reference purposes, below is the conversion table you can use to convert from feet per second to miles per hour
feet per second (ft/s) miles per hour (mph)
14.6 feet per second 9.955 miles per hour
15.6 feet per second 10.636 miles per hour
16.6 feet per second 11.318 miles per hour
17.6 feet per second 12 miles per hour
18.6 feet per second 12.682 miles per hour
19.6 feet per second 13.364 miles per hour
20.6 feet per second 14.045 miles per hour
21.6 feet per second 14.727 miles per hour
22.6 feet per second 15.409 miles per hour
23.6 feet per second 16.091 miles per hour | 601 | 2,170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.763226 |
https://studydaddy.com/question/density-hcl-1-19-g-cm-3-mass-percent-of-solute-38-what-is-the-molarity-molality | 1,555,629,350,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526904.20/warc/CC-MAIN-20190418221425-20190419003425-00216.warc.gz | 560,570,977 | 10,375 | Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
# Density HCl 1.19 g/cm^3 Mass Percent of Solute 38 what is the molarity, molality, and mole fraction?
is defined as the mass of the , which in your case is "HCl", divided by the total mass of the solution and multiplied by 100.
The easiest way to approach such solution problems is by picking a convenient sample of the solution to base the calculations on. In this case, let's pick a "1-L" sample of your "38% concentration by mass solution.
The first thing you need to do is determine is how much the "1-L" sample weighs. Since was given to you "g/cm"^3, it'll be best to transform it in "g/dm"^3, since "1 L = 1 dm"^3
1.19 "g"/"cm"^3 * ("1000 cm"^3)/("1 dm"^3) = "1190 g/dm"^3
Now focus on finding out how much "HCl" you have in this much solution.
"38%" = m_("[solute](http://socratic.org/chemistry/solutions-and-their-behavior/solute)")/m_("solution") * 100 => m_("solute") = (m_("solution") * 38)/100
m_("solute") = (38 * 1190)/100 = "452.2 g HCl"
For , you need moles of solute per liter of solution. Use "HCl"'s molar mass to determine how many moles you have
"452.2 g" * ("1 mole")/("36.5 g") = "12.40 moles HCl"
Therefore,
C = n/V = "12.40 moles"/"1 L" = "12.4 M"
will be moles of solute per kilogram of solution, so
"b" = n_("solute")/m_("solution") = ("12.40 moles")/(1190 * 10^(-3)"kg") = "10.4 molal"
For mole fraction you first need to determine the total number of moles you have in the sample. Find the number of moles of water by
m_("water") = "1190 g" - "452.2 g" = "737.8 g"
"737.8 g" * ("1 mole")/("18.0 g") = "41.0 moles water"
The total number of moles will be
n_("total") = n_("water") + n_("solute") = 41.0 + 12.40 = "53.4 moles"
Therefore, fraction for "HCl" is
"mole fraction" = n_("solute")/n_("total") = 0.232 | 616 | 1,883 | {"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} | 3.84375 | 4 | CC-MAIN-2019-18 | longest | en | 0.904052 |
https://goprep.co/q40-7-for-what-value-of-x-shall-we-have-l-m-a-x-50-b-x-70-c-i-1njxiu | 1,611,323,108,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529331.99/warc/CC-MAIN-20210122113332-20210122143332-00778.warc.gz | 370,185,806 | 51,080 | Q. 405.0( 1 Vote )
# For what value of x shall we have l || m?A. x = 50B. x = 70C. x = 60D. x = 45
X + 20 = 2x – 30(Corresponding Angles)
2x –x = 30 + 20
X = 50o
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Near Trivial Quiver Varieties
So, today I started learning the definition of a quiver variety, and wanted to make sure I'm understanding things right, so first, my setup:
I've been looking at the simplest case that didn't look completely trivial: two vertices with one directed edge. Now, my understanding is that then we have two vector spaces $V$ and $W$ of dimensions $n$ and $m$, and the quiver variety is $\hom(V,W)\oplus \hom(W,V)/GL(V)\times GL(W)$, with the action of each group on the domain or codomain, as applicable.
Now, a naive dimension count (this is one place that I may be very, very wrong) is that this is a $2nm-n^2-m^2=-(m-n)^2$ dimensional space. So presumably if $m=n$, the GIT quotient is a single point.
Now, what if $n\neq m$? Presumably, we don't really get anything for the GIT quotient, either a point or the empty set (I don't know GIT very well, so perhaps no stable points?)
Finally, does this object have interesting geometry as a stack? It seems obvious that this should be a smooth Artin stack (if my intuition for them is even vaguely accurate) which just happens to be negative dimensional, but what kinds of properties does it have?
-
Do you mean two directed edges in opposite directions? One directed edges would just give you hom(V,W) and not the other one. – Greg Kuperberg Dec 10 2009 at 0:00
The definition I'd been given was that you take a digraph, get $\hom(V_i,V_f)$ where $V_i$ is the initial space and $V_f$ the final one, then you take the cotangent bundle (which in this case is the dual, and thus gives a $\hom(V_f,V_i)$ and then mod out. Is that not the standard definition? Perhaps I was supposed to use symplectic reduction rather than GIT quotient with this? – Charles Siegel Dec 10 2009 at 0:05
My misconcption, I guess. I suppose that "quiver variety" usually means a Nakajima quiver variety, which is more complicated than just a moduli space of quiver representations as in Gabriel's theorem. – Greg Kuperberg Dec 10 2009 at 0:16
...ok...I knew about half those words. VERY new to the subject, and am trying to figure out the simplest case. I take it that either this is a nonstandard definition, or else I'm understanding things wrong? – Charles Siegel Dec 10 2009 at 0:36
No, I think that you are right and I am wrong. Some of it is new to me too. The Nakajima quiver variety is defined as you said, according to several references. There is also a representation variety of a quiver, but that is a simpler object. It so happens in this one case that the Nakajima quiver variety is the representation variety of another relatively tractable quiver. – Greg Kuperberg Dec 10 2009 at 0:54
show 1 more comment
First, one important point: people who study quiver varieties seem not to usually take stack quotients, but rather GIT quotients (though there ARE very good reasons to do this if you like geometric representation theory) which leads them to come up different dimension formulae from you. The reason is that you are think of a fine moduli space, whose stack dimension is the geometric dimension of the underlying variety minus the dimension of the automorphism group of the object, and modules over an algebra ALWAYS have automorphisms (if nothing else, they have multiplication by constants).
Now, the answer proper: You seem to have mixed up two of the more popular notions of a quiver variety (leading your confusing dialogue with Greg in the comments), and Kevin seems to have mixed in a third, which may or may not actually be the one you had in mind (all of which are, of course, closely related). If you have two vertices and one arrow, then there are two things you can do.
You can take the moduli space of quiver representations of the path algebra of that quiver, which is given by $\mathrm{Hom}(V,W)/GL(V)\times GL(W)$. This has a very small dimension $(nm-n^2-m^2)$ and should probably be thought of as finitely many points (indeed, this quiver only has finitely many representations of any given dimension. The indecomposibles look like $k\to 0$, $0\to k$ and $k\to k$), all of which have a bunch of automorphisms.
Now it sounds like what you intended to do was take the "hyperkählerization" of this quiver variety. What you should do for this is take the cotangent bundle of $\mathrm{Hom}(V,W)$, but before you mod out, you have to impose a moment map condition. The reason this is a good idea is that you want a resulting variety which is a holomorphic symplectic result (just like the cotangent bundle), which you can also think of as hyperkähler by picking a hermitian metric on $\mathrm{Hom}(V,W)$. This moment map condition is basically that both possible compositions of maps along your arrows are 0 (note: I think this is not a flat map, so I think if you want to really think properly about this story, you should probably take the derived fiber). Then take the quotient of that.
What Kevin is referring to is probably the most popular definition of quiver varieties for geometric representation theory. This is yet another definition, where he interpreted one of your vertices as a shadow vertex. This extra layer of confusion results from some (IMHO poor) notational choices of Nakajima.
-
I can just tell you what the space looks like up to the action. The orbit classification of $\text{hom}(V,W) \oplus \text{hom}(W,V)$ looks a lot like the orbit classification of $\text{hom}(V,V)$, which as you know from linear algebra is given by Jordan canonical form. In fact, if you compose the two homs, the classification is clearly at least as complicated as JCF. There is just a modest amount more structure because the nilpotent part is more complicated. Although the formal dimension of the quotient is indeed negative or 0, its geometric dimension is strictly positive when $V$ and $W$ are non-trivial.
Let $$f:V \to W \qquad g:W \to V$$ be the two linear maps. Then $f$ has a kernel, $g \circ f$ could have a larger kernel, etc. Define the stable kernel $V_0$ and the stable image $V_1$ of $g \circ f$ to be the direct limits of the kernel and the image of $(g \circ f)^n$. As in one proof of Jordan canonical form, $V = V_0 \oplus V_1$. Similarly $W = W_0 \oplus W_1$. The pair $(f,g)$ canonically splits into two pairs, $(f_0,g_0)$ and $(f_1,g_1)$. The pair $(f_0,g_0)$ is nilpotent, while the pair $(f_1,g_1)$ is invertible and establishes an isomorhism $V_1 \cong W_1$.
Because of the isomorphisms between $V_1$ and $W_1$, the invariant information in the pair $(f_1,g_1)$ is the Jordan canonical form of $g_1 \circ f_1$, which is the same as the JCF of the other composition. In other words, either $f_1$ or $g_1$ can be any isomorphism, and then the other one can be chosen to establish a prescribed Jordan canonical form. Any eigenvalue can appear other than 0.
The nilpotent pair $(f_0,g_0)$ is a little more interesting. It looks like an Ouroboros. An indecomposable nilpotent pair is any finite chain $$0 \to k \to k \to \cdots \to k \to 0,$$ rolled up from $\mathbb{Z}$-graded to $\mathbb{Z}/2$-graded. (The connecting maps in the middle are all isomorphisms, not differentials.) The chains can have odd length, so that $V_1$ and $W_1$ don't have to have the same dimension. A chain of any length can also descend in two different ways to $V_1$ and $W_1$.
I gather from Ben's comment that this is a right answer to a wrong question. It is a good description of the representation variety of a cyclic quiver; there is nothing special about cycle length 2 in the analysis. But the $A_2$ quiver variety is something else.
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Ahh, so in particular, my naive dimension count fails. And that's because the adjoint action doesn't have a dense orbit. Got it. Does the stackiness of the quotient give us anything interesting? – Charles Siegel Dec 10 2009 at 1:12 I hardly understand stacks in the sense of algebraic geometry. But just in the sense of group actions: If $g \circ f$ is diagonalizable and either both annihilate the nilpotent part or one is an isomorphism on the nilpotent part, then the stabilizer is semisimple. Otherwise the stabilizer is a parabolic subgroup. The computation is similar to deriving the centralizer of an endomorphism of a vector space, or the non-uniqueness of a JCF basis. – Greg Kuperberg Dec 10 2009 at 1:25
As the comments say, one of the problems here is the multiplicity of things that get called "quiver varieties". The example you are looking at would correspond in Nakajima's context to a quiver variety attached to $SL_2$, and maybe that's worth making explicit. Nakajima starts with a quiver Q, and does two things: first he adds a "shadow vertex" for each original vertex, and an edge from that vertex to it's corresponding vertex in the original quiver, then he take the contangent bundle of everything, which in graph terms means adding for each edge you now have another edge in the opposite direction. Thus for $SL_2$ the original quiver is just a single vertex with no edges. The Nakajima procedure adds another vertex and two edges one in each direction.
The quotients that Nakajima would then consider would be with respect to the action of the general linear groups on the original vertices: in our case just one of them, say $V$. For the GIT quotient, you would use some choice of stability, which in this case could be just that the map $x \colon W \to V$ is surjective. Then you need the moment map for the action of the group $\text{GL}(V)$: this is just the map to the composite $xy \in \text{End}(V)$ (where $y\colon V \to W$ is the other linear map). The symplectic quotient is then the quotient by the action of $\text{GL}(V)$ on the stable part of the zero locus of the moment map: so pairs $(x,y)$ where $x$ is surjective, and the image of $y$ lies in the kernel of $x$. The surjectivity of $x$ means that the $\text{GL}(V)$ action is free, and the quotient can then be identified with the cotangent bundle to a Grassmannian of $W$: it is the image of the map $(x,y) \mapsto (\text{ker}(x),yx)$. The affine quotient is the image of the map $(x,y) \mapsto yx$, which gives you the closure of a nilpotent orbit in $\text{End}(V)$.
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This answer is merely in relation to the naive dimension count calculated in the question. When taking a GIT quotient, one cannot merely subtract the dimensions to get the dimension of the quotient. For example if you quotient gln by GLn with the adjoint action, then the GIT quotient is n-dimensional. (the quotient here is actually affine n-space and the map sends a matrix to its characteristic polynomial).
- | 2,698 | 10,680 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2013-20 | latest | en | 0.96215 |
https://software.intel.com/de-de/forums/topic/507074 | 1,438,495,399,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988962.66/warc/CC-MAIN-20150728002308-00278-ip-10-236-191-2.ec2.internal.warc.gz | 883,934,723 | 15,360 | # problem in computation of 2D FFT using two seperate real and imaginary arrays
## problem in computation of 2D FFT using two seperate real and imaginary arrays
Hi,
i want to compute FFT of a complex 2D array. So first i have tried it in 2 ways.
1) Directly using complex 2D array(real and imaginary interleaved) .
2) 2 seperate arrays(where real and imaginary are deinterleaved into 2 seperate arrays).
i found the output is different in both cases. can some one tell me if i do some thing wrong in the code.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <float.h>
#include <complex>
#define MKL_Complex8 std::complex<float>
#include "mkl_dfti.h"
int main()
{
printf("hi\n");
int N1 = 4;
int N2 = 4;
MKL_LONG status = 0;
MKL_Complex8 *x = 0;
float* real;
float* imag;
DFTI_DESCRIPTOR_HANDLE hand = 0;
DFTI_DESCRIPTOR_HANDLE hand1 =0;
MKL_LONG N[2]; N[0] = N1; N[1] = N2;
status = DftiCreateDescriptor(&hand, DFTI_SINGLE, DFTI_COMPLEX, 2, N);
if (0 != status) printf("failed\n");
status = DftiSetValue( hand, DFTI_PLACEMENT, DFTI_INPLACE );
if (0 != status) printf("failed\n");
status = DftiCommitDescriptor(hand);
if (0 != status) printf("failed\n");
x = (MKL_Complex8*)malloc(N1*N2*sizeof(MKL_Complex8));
real = (float*)malloc(N1*N2*sizeof(float));
imag = (float*)malloc(N1*N2*sizeof(float));
status = DftiCreateDescriptor(&hand1, DFTI_SINGLE, DFTI_COMPLEX, 2,N);
if (0 != status) printf("failed\n");
status = DftiSetValue(hand1, DFTI_COMPLEX_STORAGE, DFTI_REAL_REAL);
if (0 != status) printf("failed\n");
status = DftiCommitDescriptor(hand1);
if (0 != status) printf("failed\n");
for(int i = 0; i < N1*N2; i++)
{
x[i].real() =(float) 1;
real[i] = 1;
imag[i] = 1;
x[i].imag() = (float)1;
}
printf("before printng\n");
for(int i =0 ; i < N1*N2; i++)
{
printf(" %f %f----------%f %f\n",x[i].real(),x[i].imag(),real[i],imag[i]);
}
status = DftiComputeForward(hand, x);
if (0 != status) printf("failed\n");
status = DftiComputeForward(hand,real, imag);
printf("after printng\n");
for(int i =0 ; i < N1*N2; i++)
{
printf(" %f %f--------%f %f\n",x[i].real(),x[i].imag(),real[i],imag[i]);
}
}
Thanks
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Nähere Informationen zur Compiler-Optimierung finden Sie in unserem Optimierungshinweis.
Hi, looks like you made a typo in line 52:
` status = DftiComputeForward(hand,real, imag); `
should be replaced with
` status = DftiComputeForward(hand1,real, imag); `
Then the output is the same.
Thank you Evarist Fomenko.
## Kommentar hinterlassen
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The ancient Greek philosopher Plato described 27 as the most revered number.
27 Kabbalistic letters and 27 Hebrew letters (when extended from the normal 22 letters).
27000 people die when a wall falls on them. (Bible, “Kings: 20:30”). The 27000 who die might be just 27, as the word for ‘thousands’ and ‘armed soldiers’ is similar, some say.
27 links to the phi ratio (aka the golden ratio = 1.618 approx) indirectly via 54 (27*2). In trigonometry, Sine 54 degrees = phi ratio/2
27*(2/3) = 18, whereas 27*2 = 54. If you place the 18 alongside the 54, you get 1854.
1854 = 927 * 2 = 618 * 3
618 is in the phi ratio (approx 1.618).
God lives in Room 2700 on the 27th floor, in the film “Oh, God!” (1977), written and produced by Jewish writers. God is played by Jewish actor/comedian GEORGE BURNS. The film’s director, CARL REINER was Jewish, so too the screenplay writer LARRY GELBART (co-creator and producer of “MASH”). The pair were part of the network of numerous, successful Jewish writers that dominated American comedy, e.g. SID CAESAR, MEL BROOKS, WOODY ALLEN, NEIL SIMON. The film’s Jewish producer JERRY WEINTRAUB had worked at MCA (Music Corporation of America) as assistant to MCA manager LEW WASSERMAN (jewish). MCA’s co-founder JULES STEIN was Jewish.
27 is a Freemasonic number, because it contains the holy trinity: 3 cubed (3 raised to the power of 3). The Kaaba stone is a cube. 3 cubed can be thought of as 33, the highest degree officially in Freemasonry. 27 (and 9) can be thought of as the ‘living arch’, formed by 3 masons linking their left hands, and their right hands, and arranging their right feet as a triangle.
27 guests at the fictional hotel “Fawlty Towers” in the TV episode “The Kipper and The Corpse”. The kipper could be the fish that symbolises Jesus, while the corpse is his body before his resurrection.
“Fawlty Towers” co-creator/writer JOHN CLEESE plays an architect desperate to join the Freemasons in the “Monty Python” sketch, “The Architects Sketch”.
Cleese read law at Cambridge University; was Rector of the University of St. Andrews. Other Rectors there were Freemason James Barrie (Peter Pan); Freemason and novelist RUDYARD KIPLING;
ANDREW CARNEGIE (wealthy steel industrialist);
ARTHUR BALFOUR (freemason; British Prime Minister; wrote the “Balfour Declaration” to Walter Rothschild);
5th Earl of ROSEBERY (Prime Minister); Field Marshal Sir DOUGLAS HAIG (senior military officer during world war one);
JAN SMUTS, South African Prime Minister; supported Zionism and the Balfour Declaration; helped create the UN and its predecessor the LON.
27 seconds (26.6 to be precise) is the duration of the famous Zapruder film footage of the murder of J.F. Kennedy. Abraham Zapruder was a 33rd degree Freemason and Jewish. There are 486 frames in the footage. 486 = 27*18.
27 and 18 join to become 2.718, which approximates the mathematical constant ‘e’ (2.71828).
27 has popped up in numerous episodes of “The Simpsons”, e.g. “The Lard of The Dance”, in which Homer spends \$27 on some bacon. That episode was written by JANE O’BRIEN, sister of famous US TV host CONAN O’BRIEN, who had himself been a writer/producer on “The Simpsons” (1991-1993). Conan studied at Harvard University where he became President of the “Harvard Lampoon” humor magazine. At this time, his future boss (President) at NBC- JEFF ZUCKER- was President of Harvard’s student newspaper “The Harvard Crimson”. Zucker was in 2015 the President of CNN Worldwide. Can I say that Zucker is Jewish? I’m not anti-semitic, honest!
The number 27 is represented visually and speaks its number to Lisa Simpson in the episode “Girls Just Want to Have Sums” (ep.19, season 17).
\$27000 is mentioned in the “Simpsons” episode “The Bob Next Door”, in which the ‘perfect crime’ takes place at FIVE CORNERS, where 5 U.S. states meet in the form of a pentagon, an occult symbol and the shape of the US military HQ of the same name. The perfect crime could be an occult reference to the ‘Great Work’, which I believe involves a rebellious mind attempting to detach from karma/’god’/time.
27 seasons of DVD commentaries is to be recorded by “Simpsons” character Krusty the Clown in the episode “A Star Is Torn”. The episode’s ‘showrunner’ was AL JEAN, who as a Harvard University undergraduate befriended MIKE REISS (atheist but Jewish family), another “Simpsons” writer. The duo at Harvard collaborated at the “Harvard Lampoon”, to which Reiss served as co-President with future “Simpsons” writer ROBERT VITTI. Vitti’s wife ANN was the sister of “Harvard Lampoon” President and “Simpsons” writer GEORGE MEYER.
As you’ve probably gathered, “Harvard Lampoon” and “The Simpsons” have a close link., e.g. JEFF MARTIN.
Season 27 of “The Simpsons” was first broadcast on the 27th September 2015.
“27 dresses” (2008) is a comedy film about a woman who has been a bridesmaid at 27 weddings, hence 27 wedding dresses. The film’s writer, ALINE MCKENNA, graduated from Harvard University.
The number 27 is emblazoned on a helicopter in the movie “A Good Day To Die Hard” (2013), one of the “Die Hard” series of films about foiling terrorist plots.
27 seconds is the amount of time that a contestant has to speak for in order to win points in the BBC Radio 4 panel show “Don’t make me laugh” (2nd series began in April 2016), created and hosted by Jewish comedian/writer DAVID BADDIEL. The game’s goal is to speak without making the audience laugh. Baddiel is a Jewish atheist, which sounds like a contradiction to me, but officially isn’t. Religious atheism is – IMO – all part of the Ruling Elite’s plan to merge science with religion.
“…my perfect 27 years” of development and growth, says GREY NAMER in the sci-fi novel “Something More” (2001) by PAUL CORNELL (author of some “Doctor Who” stories). His novel is about immortality, aliens, Jesus and God. Cornell wrote the “Shadow Police” series of novels incorporating the occult/magic/supernatural.
The number 27 is on the front of a T-shirt worn by a character in the 2008 sci-fi movie “Jumper”, about people who can teleport, based on the novel of the same name. The film has Jewish connections, as usual, most noticeably one of the producers: ARNON MILCHAN, a prolific producer of independent films and a billionaire who grew up in Israel and worked for the Israeli intelligence organisations MOSSAD and LEKEM where he was an arms dealer and worked on the Israeli nuclear weapons program.
“Jumper” director DOUGLAS LIMAN (raised Jewish) was the son of Jewish lawyer ARTHUR LIMAN who graduated at Yale Law School and Harvard University. Arthur Liman served as Chief Counsel for the US Senate hearings on the Iran-Contra affair. He helped his son Douglas co-found the now defunct NACB (National Association of College Broadcasters), funded by the CBS Foundation. Former CBS news anchorman WALTER CRONKITE gave the NACB’s inaugural keynote address.
Douglas Liman directed “The Bourne Identity” (2002).
“Jumper” co-producer and co-screenplay writer SIMON KINBERG (Jewish) worked in his early career with STEVEN SPIELBERG (Jewish) and TV/film producer JERRY BRUCKHEIMER (Jewish parents).
“Jumper” co-screenplay writer DAVID GOYER had a Jewish mother and he attended Hebrew School.
27 hours, the length of time that the robot BENDER spends listening to “The Eagles” music, in “Futurama: Law And Oracle”, written by JOSH WEINSTEIN, born JOSHUA MOSES WEINSTEIN, an executive producer and writer for “The Simpsons”. Weinstein’s Jewish father HARRIS WEINSTEIN was a lawyer at the US Supreme Court and the inaugural chairman of the JEWISH FUNERAL PRACTISES COMMITTEE of Greater Washington. Harris was a member of the MIT Corporation (MIT’s governing body). Josh Weinstein’s mother ROSA was a Director of the HIMMELFARB MOBILE UNIVERSITY, run by the JSSA (Jewish Social Service Agency).
Josh Weinstein went to a Protestant/Episcopal private school where he became lifelong friends with writing partner BILL OAKLEY. Weinstein studied at Stanford University where he was the editor-in-chief of the humor magazine “Stanford Chaparral”. Weinstein and Oakley’s partnership at “The Simpsons” stemmed from them submitting a speculative script for the comedy “Seinfeld”, created by Jewish writers/actors JERRY SEINFELD and LARRY DAVID.
Weinstein and Oakley worked on several shows together, including “22 Birthdays” (pilot in 2007, but not developed). A remake was supposed to have involved DOUG LIMAN. The number 22 is an occult number and is in “The Simpsons” episode “22 Short Films About Springfield”, co-written by Weinstein, Oakley and others, including:
1)DAVID X. COHEN, co-developed “Futurama”; President of Harvard’s “Harvard Lampoon”, of which Weinstein was an honorary member.
2)GREG DANIELS, writer for the “Harvard Lampoon” and friends at Harvard with “Harvard Lampoon” President CONAN O’BRIEN (see above). Daniels and O’Brien were recruited by the co-creator/producer of “Saturday Night Live” LORNE MICHAELS (Jewish parents). Lorne Michaels was born on a Kibbutz in pre-independence Israel as LORNE LIPOWITZ.
Greg Daniels’ father AARON DANIELS was President of ABC Radio Networks in New York, and a consultant to Capital Cities-ABC. Aaron was a close friend of FRED WEINHAUS, President of Capital Cities-ABC.
Two of the 3 “Simpsons” co-creators were sons of Jewish parents: 1)SAM SIMON whose father was of Russian Jewish heritage. His family lived opposite Jewish comedy legend GROUCHO MARX.
2)JAMES L. BROOKS, whose father was born EDWARD BERNSTEIN.
Weinstein and Oakley were ‘showrunners’ (show overseers and main producers) of several series of “The Simpsons”, including the episode when Homer and his family go to New York and 9/11 is seemingly prophesized (“The City Of New York versus Homer Simpson”). Bart visits the offices of the satirical magazine “MAD”, which has Jewish links. It was founded in 1952 by its’ Editor HARVEY KURTZ (Russian Jewish parents) and the Publisher WILLIAM GAINES, son of Jewish father MAX GAINES (born MAXWELL GINSBURG or GINZBERG).
Max Gaines was a pioneering figure in the creation of the modern comic book, as were Jews in general, as documented in “How The Jews Created The Comic Book Industry”, by ARIE KAPLAN. Max Gaines helped put “Superman” into publication. Superman has religious/occult overtones and was created by 2 Jewish writers. Superman is the son of JOR-EL. Gaines’ EDUCATIONAL COMICS was taken over by his son William.
Bart sees “MAD” magazine’s iconic mascot, ALFRED E. NEUMAN (formerly MEL HANEY and MELVIN COWZNOFSKI) the boy with – dare I say it- ‘Jewish’ stereotypical features of reddish hair, white skin and freckles. This image was spotted by Kurtz, he said, on a postcard belonging to “MAD” contributor BERNARD SHIR-CLIFF, an Editor at BALLANTINE BOOKS. Ballantine co-founder IAN BALLANTINE also co-founded BANTAM BOOKS. Ian Ballantine’s mother was the niece of the Anarchist EMMA GOLDMAN (Russian Jew).
Kurtzman was succeeded as editor by ALBERT FELDSTEIN for 30 years (1956- 1985).
27000 people rallied for US Democrat BERNIE SANDERS in New York ahead of a primary (17th April 2016). See link: Bernie Sanders
27 people died from a disease in the “X-Files” episode “Our Town” (ep 24, season 2, 1995), written by FRANK SPOTNITZ, later an executive producer of the series. He also founded a TV production company that produced shows such as “The Man In The High Castle”, based on the novel by sci-fi legend P.K. Dick. Spotnitz was an executive producer of that series, along with renowned Director RIDLEY SCOTT.
DANA SCULLY’S colleague FOX MULDER resides at apartment number 42, possibly a reference to the sci-fi classic “The Hitchhiker’s Guide To The Galaxy”, in which the number 42 has great cosmic significance, possibly (no-one is sure). 42 also pops up in the fantasy writings of LEWIS CARROLL. 42 could refer to the the 4th and 2nd letters of the alphabet- D & B, the initials of “X-Files” character DUANE BARRY, an alleged alien abductee. Barry’s character was based on the real medical case of PHINEAS CAGE, who suffered a personality change after an injury to his brain. Consequently, his left eyelid drooped to cover his eye, thus suitable for use for occult symbolism because a one-eyed person/creature is an old occult/religious/mythological symbol.
27 is the number of the apartment of a key character in the “X-Files” episode “2Shy” (episode 6, season 3), while 47 is the number of murder victims killed by a parasitic murderer. Parasitism/vampirism are occult themes because the ‘Rebel’ is living off its’ own inner resources- feeding off itself, so to speak (The ouroboros snake eats its own tail). A key character in the episode is called VIRGIL INCANTO, probably alluding to the Roman poet Virgil who is also the fictional guide of Italian poet DANTE as he travels through Hell (“The Divine Comedy”). Virgil Incanto translates Italian literature and knows medieval poetry. The name ‘Incanto’ is Italian for “magic, enchantment, spell, charm”.
27 books in the New Testament bible, with the 27th and last being the Book of Revelation.
27 city blocks need to be deprived of electricity in order for the chosen one, NEO (in the movie “The Matrix Reloaded”), to find his way to the source of the Matrix (computer generated virtual realm).
27 years is the amount of time that 2 sisters have spent apart from each other, in the British TV drama “Pat and Margaret” (1994), written by the late comedian/actor/director VICTORIA WOOD, who spent many years married to the TV magician GEOFFREY DURHAM (aka “The Great Soprendo”), a member of the British “Magic Circle”, and a member of its prestigious ‘Inner Magic Circle”. He won their MASKELYNE Award in 2002 and was a Quaker.
“Pat and Margaret” is about a successful English actress (Pat) who after 27 years is reunited with her sister MARGARET MOTTERSHEAD (initials M.M.) on the TV show “Magic Moments” (also MM). MM is the initials of MARY MAGDALENE (supposedly Jesus’ lover) and actress MARILYN MONROE. Pat is ashamed of her past and her ‘less successful’ sister Margaret- a roadside cook. Separation and reunion are occult themes.
27 Pound note: silly counterfeit money forged by Eddie in the BBC comedy sitcom “Bottom” (episode “Dough”), first broadcast on the 27th (Jan, 1995). The show starred, and was written by, RIK MAYALL and ADRIAN EDMONDSON, who met at Manchester University, where they also met their future “The Young Ones” writing colleagues BEN ELTON and LISE MAYER.
27 times, the number of times that a million pound lottery win is counted, according to one of the ‘Old Gits’ in an episode of the BBC comedy sketch show “Harry Enfield’s Television Show” (not sure which episode). Harry Enfield’s father EDWARD was a BBC TV/Radio presenter and a magazine writer, e.g. a reporter on the BBC’s “Watchdog” consumer show. Edward oversaw the privatisation of school meals and cleaning whilst working at West Sussex County Council.
27000 people (approx) signed a petition to cancel or delay the 2016 British Referendum on European Union. Click on the following link: 2016 referendum
Was the murderer of Jo Cox a ‘manchurian candidate’ (mind- controlled) designed to tip the vote in favour of the Remain camp? The timing of the vote was noteworthy, bearing in mind that we want as few distractions as possible. The Referendum vote came in the middle of the European football championship, so the inevitable hooliganism was used to highlight ‘racist, xenophobic, right-wing nationalists’. Football distracts sports fans and patriots. The June exams distract young voters at college/university. June is holiday time for many.
27 storeys – the height of GRENFELL TOWER in London, in which many died in a high profile fire, although the mass media couldn’t make up its mind whether it was 27 or 24 storeys. The figure in Wikipedia is 24, but 6 months after the fire I saw the number 27 on the BBC’s News channel. Was this confusion deliberate? Here are some of the reports: Grenfell, ITV This morning
Grenfell, Evening Standard
Grenfell IB Times
27 million miles to freedom: in an advert that I saw in 2001 for the “Economist” magazine. The advert related to a man jailed in South Africa (I think), and showed a fake road signpost with the words “Freedom: 27 million miles”. Note- I haven’t been able to verify this advert.
27000 light years, the approximate distance of the Earth to the centre of our galaxy.
27 million degrees Fahrenheit: the approximate temperature of the core of our Sun.
Batch 27, a group of simulant droids in an episode of the sci-fi comedy series “Red Dwarf”, which I believe is full of occult/religious themes. In this episode (“Twentica”, episode one, series 11), time travel features. The simulants want the ‘casket of Cronos’ (a reference to the Greek time-god Kronos). The “Red Dwarf” chief writer DOUG NAYLOR also wrote comedy sketch shows on BBC Radio, and was a writer on the popular British satirical TV series “Spitting Image”.
27 years old is the age of Bart Simpson’s Fat Camp leader in “The Simpsons” episode “The Heartbroke Kid”, written by IAN MAXTONE-GRAHAM who was also a writer for “Saturday Night Live” and for the short-lived comedy magazine “Army Man”, a recruiting ground for “The Simpsons”. “Army Man” was created by GEORGE MEYER, previously a President of the “Harvard Lampoon”. Maxtone-Graham’s father, John, was a naval historian and a Broadway stage manager.
27 million to one: the odds of a chance meeting between some twins, according to “X-Files” FBI agent DANA SCULLY, in the episode “Fight Club” (Series 7, episode 20), written by the show’s creator Chris Carter. The episode revolves around twins and look-alikes: an occult theme, e.g. the ‘evil twin’. Karma is a ‘double’ in the sense that it reflects/mirrors our true state. The Biblical Adam is a double in the sense that he is made in the image of God.
27 robberies are reported to have been committed in the “X-Files” episode “The Pine Bluff Variant” (series 5, episode 18).
The 27th July was not only the date of the 2012 London Olympics opening ceremony and the annual Satanic “Grand Climax”, but is also a key date in the fictional sci-fi film “K-Pax” (2001), starring KEVIN SPACEY as a man claiming to be an alien from an idyllic planet called K-PAX, where there are no wars, disease, etc. Although the word PAX is Latin for PEACE, the name K-PAX could be a coded version of Spacey’s name. K=Kevin; PAX = PACS = anagram of SPAC = abbreviation of SPACEY. More word play is in the name of the author of the original novel “K-Pax”, written by a geneticist called GENE BREWER (a brewer of genes?). Brewer wrote the novel “Becoming Human”, about a human-like artificial brain created by scientists. Our universe is controlled by an artificial intelligence, I reckon.
Is the K-PAX alien real or a human mental patient with multiple personality disorder?, wonders a psychiatrist.
Approximately 2.7 million tickets were offered and sold for the 2012 London Paralympics (according to the official website of the Paralympic movement), thus maintaining the London Olympics connection to 27. See the link: https://www.paralympic.org/the-ipc/paralympic-games
2 years before the Olympics, was the following website headline: “The Olympic Delivery Authority (ODA) is to make £27 million in procurement savings…” (“Supply Management” 21/July/2010). Also, “Olympics Delivery Authority returns £27 million in savings” (“The Daily Telegraph”, 19th July, 2010). See link: £27 million in savings
£27 million in savings “The Telegraph”
The opening ceremony cost £27 million, featured a giant bell that weighed 27 tonnes, and was watched on TV by “Almost 27 million Britons” (Reuters News, 28th July, 2012).
The ceremony featured Harry Potter author J.K Rowling reading a passage from “Peter Pan”, a story that contains the number 27, the number of the house that Wendy’s parents are visiting when Peter Pan lures Wendy and the other children away to the island of Neverland. Peter Pan author James M. Barrie was a Freemason, while 27 is a Freemasonic number. Barrie has been suspected of being a pedophile. In the olympic ceremony, there was the ‘child catcher’ from “Chitty Chitty Bang Bang”.
Child sex leads onto the day of the opening ceremony- the 27th day of July, which to Satanists is the final day of the annual ‘Great Climax’, when blood and sex rituals take place. These are not actual death sacrifices, say the ‘good’ Satanists, just sex and blood practises designed to release energies, in conjunction with the ‘law of synchronicity’.
If the 27th July is a significant day in the Earth’s energy field, then what happened to David Icke on the night of the 26th/27th July in 1993 when he and others visited Stonehenge stone circle? In his book “Heal The World” (Page 101), David writes that he channeled some very powerful energies. There were reports the next day of power cuts, bright lights in the sky, psychics seeing vortexes, etc, some of it in the USA.
If you add half a year (6 months) to 27th July, you reach 27th January, which is now “International Holocaust Remembrance Day” (prisoners were liberated from Auschwitz on the 27th Jan 1945). Perhaps the 2 dates are harmonically linked (resonate with each other).
It’s probably coincidence, but in 2016 there were approx. 27 million households in Britain, as has been made clear in the British media. See link: households
Up to 27 million people are living in slavery, according to US Secretary of State Hillary Clinton. See link 27 million people
There are 27 cut-out-and-keep route cards in each edition of “Country Walking” – Britain’s best-selling walking magazine. See link: Country walking
270,000 was the viewing figures of Piers Morgan’s CNN chat show when it was axed – down from the initial 2 million. See link: chat show
2700 victims of fuel poverty predicted in 2011 for Britain. See link: fuel poverty
2700 killed over 25 years in religious festival, and 270 stoned. See link: religious festival
2700 kills by top US soldier. See link: soldier
In the recent British government’s spending plans, the Chancellor announced a £27 billion windfall. See link: windfall
270 schoolgirls were kidnapped in 2014 by the Islamic extremist group Boko Haram. See link: Boko Haram
“The Law of Takeaway No.27” was in the TV advert for Hungryhouse.co.uk, sponsors of the “Big Bang Theory”, on 26/11/2015 Channel E4.
27 links to the phi ratio (golden ratio) indirectly via 54 (27*2). In trigonometry, Sine 54 degrees = phi ratio/2
27 minutes is the amount of time that someone spends in a ‘cryo-chamber’ (for freezing) in an episode of the TV comedy horror “Scream Queens” (episode 11, season 1). The freezing could be an occult reference to the freezing of time, and to immortality, and to Peter Pan who doesn’t want to grow up.
“Scream Queens” revolves around a serial killer who dresses up as a red devil.
One of the series’ co-creators was Brad Falchuk, a co-creator of hit TV show “Glee”. His mother Nancy was President of the influential Jewish/Zionist/feminist organisation HADASSAH. Falchuk has been dating actress Gwyneth Paltrow whose father was Jewish. Paltrow raised her children in the Jewish tradition.
The other “Scream Queens” co-creators were Catholics.
270 votes are needed to win the US Presidential election, at least from the year 1964 onward. Prior to that the number was less, e.g. 269 or 266, etc.
£2.7 billion is “waiting to be saved”, according to a “Go Compare” TV advert that I saw on the 15th/Jan/2016.
27 British ships fought against 33 Spanish & French ships in the 1805 Battle of Trafalgar. 33 is another occult number. There were other ships, but not ‘of the line’. Including the others, there were 33 British ships and 41 Spanish/French.
27 = 9+9+9. 999 is the American police code for “Officer Down”, as in the film “Triple Nine” (2016), about criminals blackmailed by the Russian Mafia.
27000 ravers at a legendary concert by The Stone Roses, dramatized in the film “Spike Island”. See link: Stone Roses | 6,039 | 24,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-50 | longest | en | 0.944928 |
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### Long Division in Binary
```
Date: 05/16/2000 at 12:14:54
From: Gina Ames
Subject: Base 2 division
Dear Dr. Math,
My problem is 1011 base 2 divided by 11 base 2. I only know how to
divide in base 10.
Gina Ames
```
```
Date: 05/18/2000 at 11:43:07
From: Doctor TWE
Subject: Re: Base 2 division
Hi Gina - thanks for writing to Dr. Math.
You can use the same algorithm as long division in decimal, but the
values will go in either one time or 0 times. Let's do a similar
example: 1000101 / 1100 (this is 69/12 in decimal.)
First, we write it in "long division form":
__________
1100 ) 1000101
Since 1100 has four bits (digits), we look at the first four bits of
1000101. Can 1100 go into 1000...? No, so we move one digit to the
right. Can 1100 go into 10001...? Yes, exactly once. So we put a 1 in
the quotient above the 1100, subtract, and carry down the next bit. We
now have:
______1___
1100 ) 1000101
1100
----
101
Note that we have to do some borrowing to do the subtraction. If
you're not familiar with binary subtraction, look at the following
pages from our archives:
Binary Subtraction
http://mathforum.org/dr.math/problems/houston.7.25.96.html
Complement of a Number
http://mathforum.org/dr.math/problems/steve.7.10.96.html
Next, we bring down the next digit and move one place to the right.
Can 1100 go into 1010? No, so we record a 0 in the quotient and carry
down the next bit, like this:
______10__
1100 ) 1000101
1100
------
10101
Moving over another place, can 1100 go into 10101? Yes, so we record a
1 in the quotient and subtract:
______101_
1100 ) 1000101
1100
------
10101
1100
----
1001
At this point, if we want an integer (whole number) answer in quotient
and remainder form, we'll write it as 101 r 1001. That's 5 remainder 9
in decimal. We can also write that as 101 1001/1100 (5 9/12 decimal)
by putting the remainder over the divisor. That fraction can be
reduced to 101 11/100 (5 3/4 decimal) by dividing the numerator and
denominator by 3.
If we want a "radix" answer - the binary equivalent of 5.75 - we can
continue the long division process by adding a radix point (the binary
equivalent of a decimal point) and some trailing zeros on the
dividend. We'll put a radix point in the quotient directly above the
divisor's as well:
______101.____
1100 ) 1000101.000
1100
------
10101
1100
----
1001
Now we can continue the long division process. We carry down the next
digit (the first of our trailing zeros) and check: Can 1100 go into
10010? Note that for the subtractions, we'll ignore the radix point.
Again, the answer is yes, so we record a 1 (this one is to the right
of the radix point in the quotient) and subtract:
______101.1___
1100 ) 1000101.000
1100
------
10101
1100
------
1001 0
110 0
------
11 00
We'll carry down the next digit one more time and move over another
place. Can 1100 go into 1100? Again, the answer is yes, so we record
another 1 and subtract. Since the result of this subtraction is zero,
we're done and we have an exact answer:
______101.11__
1100 ) 1000101.000
1100
------
10101
1100
------
1001 0
110 0
------
11 00
11 00
-----
0
Just as with decimal, some values won't divide evenly, and we'll get a
repeating fractional part. We can stop at any time, but realize that
we've only found an approximation, not an exact value.
One final note: If the divisor (the 1100 in this example) has a radix
point in it, move the radix points in BOTH the dividend and divisor to
the right an equal number of places sufficient to remove it from the
divisor. For example, to divide 11001.1 by 11.001, first move both
radix points right 3 places (you'll have to add zeros to the
dividend.) Then divide 11001100 by 11001.
For some more examples and explanations, look at these answers in our
archives:
Binary Operations
http://mathforum.org/dr.math/problems/matt4.7.97.html
Multiplying and Dividing Computer Style
http://mathforum.org/dr.math/problems/hodsdon9.8.98.html
I hope this helps. If you have any more questions, write back.
- Doctor TWE, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 1,304 | 4,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | longest | en | 0.809397 |
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Topic review
### Re: Scrabble island detection
I had already talked about a mutable version of this algorithm for flood filling island detection here. This time I went immutable with Clojure The algorithm is a bit different, this we will not have a variable in the outer scope but pass a new instance of a board (which is efficiently managed by Clojure) to the function for each iteration/recursion.
The full source is at this repo
there a few helper functions
first-char-on-board scans the board to find the first tile that is not empty split this will split a string such as "abc" into a vector as ["a" "b" "c"]
the basic idea is to keep a list of next slots to visit on the board as we visit slots. the next-frontier function will return a list of next slots. This is basically BFS search on a graph.
Code: Select all
``````(defn flood-fill
"flood fill the board starting from the first non empty slot.
the flooded slots will contain '!' empty slots '.'. If there
are islands they will contain letters.
"
[board]
(let [[y x] (first-char-on-board board)
board (mapv split board)
]
(loop [board board
frontier [[x y]]]
(if (-> frontier count zero?)
board
(let [front (first frontier)
x (first front)
y (second front)]
(recur
(assoc-in board [y x] "!")
(concat (next frontier) (next-frontier board [x y]))))))))
``````
the next-frontier code is also worth mentioning
Code: Select all
``````(defn next-frontier
"return the coordinates of candidate neighbors for flood filling"
[board [x y]]
(for [i [(dec y) y (inc y)]
j [(dec x) x (inc x)]
:when (and (>= i 0)
(>= j 0)
(< i 15)
(< j 15)
(or (= x j) (= y i))
(not (and (= x j) (= y i)))
(not= "!" (get-in board [i j]))
(not= "." (get-in board [i j])))]
[j i]))
``````
flood-fill will mark visited slots with ! and next-frontier will return a list of the candidate slot from the 8 neighbors of the current slot. if the neighbor is visited or empty it's not included in the list.
if the starting board was like this
Code: Select all
``````...a...
...bc..
..de...
``````
the flood would start with a and the next-frontier would return only "b" from the neighbors and mark "a" as "!". "b" would return "c" and "e" and "d". "a" would be marked as "!" so it would not be included. After the fill, the board would be like
Code: Select all
``````...!...
...!!..
..!!...
``````
this algorithm is useful when detecting if the scrabble board is connected.
### Scrabble island detection
there is a game app called “kelimelik” which is a Turkish version of scrabble that I’ve been playing for a while now which is quite fun. So I thought it might be a nice way to waste a couple of hours trying to write a program that would give the highest scoring word given a board state. Most of the program is quite trivial in that it just has to generate permutations of the given letters and place them in all possible combinations on the board and then check if the board is valid. E.g that all the words on the board are in a dictionary and all the tiles are connected. The second part of the validity check is interesting. The first thing that popped into my mind was using a flood-fill algorithm starting from the tile nearest to the top-left corner. After the fill if there are any tiles on the board that are unmarked then there is more than 1 island on the board and the board is invalid. So how does a flood-fill algorithm work? Quite simple actually. You mark the current tile you are on at (x,y) and recurse to the neighbors if they are not empty, or have not been marked. Here is a simple bit code that does this:
Code: Select all
``````private boolean connectedCheck(int i, int j, String[][] board) {
board[i][j] = "!";
boolean resh = true;
boolean resv = true;
boolean resh2 = true;
boolean resv2 = true;
if (i < 14 && !board[i + 1][j].equals("*") && !board[i + 1][j].equals("!")) {
connectedCheck(i + 1, j, board);
}
if (i > 0 && !board[i - 1][j].equals("*") && !board[i - 1][j].equals("!")) {
connectedCheck(i - 1, j, board);
}
if (j < 14 && !board[i][j + 1].equals("*") && !board[i][j + 1].equals("!")) {
connectedCheck(i, j + 1, board);
}
if (j > 0 && !board[i][j - 1].equals("*") && !board[i][j - 1].equals("!")) {
connectedCheck(i, j - 1, board);
}
}
`````` | 1,178 | 4,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-26 | latest | en | 0.916827 |
https://studyqas.com/the-fences-will-be-aligned-with-the-exterior-angles-angle1/ | 1,679,788,354,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945376.29/warc/CC-MAIN-20230325222822-20230326012822-00037.warc.gz | 617,666,812 | 41,175 | # The fences will be aligned with the exterior angles angle1 and angle2 What are some other relationships you can see between angle1
The fences will be aligned with the exterior angles angle1 and angle2 What are some other relationships you can see between angle1 , , and angle b?
## This Post Has 4 Comments
1. Expert says:
24 students walk to school
$Agroup of 80 students was asked to share how they get to school most of the time. the following circ$
2. Expert says:
c
step-by-step explanation:
$Me i really don't feel like doing this ; )$
3. Expert says:
step-by-step explanation:
4. Expert says:
the area of the rectangle is 35units^2.
step-by-step explanation:
graph it and find the distance.
it is a 5*7 rectangle.
therefore the area is 35units^2. | 188 | 772 | {"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} | 2.921875 | 3 | CC-MAIN-2023-14 | latest | en | 0.935673 |
http://codeforces.com/topic/68885/en1 | 1,575,949,178,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525781.64/warc/CC-MAIN-20191210013645-20191210041645-00550.warc.gz | 31,604,254 | 13,522 | Codenation Hiring Test Discussion
Revision en1, by codenationtest, 2019-07-16 02:47:57
Codenation test was conducted on Hackerrank on 15th July. There were 3 questions in the hiring contest. Q1) Given a NxM (NxM <= 1e5) matrix with each cell having a value mod(Aij) <= 1e9. Job is to rank each element by the following rules: 1) Ranks start from 1. 2) Equal elements in the same row or same column should have the same rank. 3) Greater elements in the same row or same column should have a greater rank. 4) Maximum Rank should be minimized. Print the final rank matrix.
Q2) N employees are given, you have to divide them into teams of two. If n is odd the last person will form a one man team. Problem is all employees are not compatible with each other. and if both are compatible with each other then only a team can be formed. Aij ='C' if employee i is compatible with employee j, Aij = 'X' otherwise. Aii ='C' if employee i can work as one man team, Aii = 'X' otherwise. find the maximum no of valid teams that can be formed. constraints: T <=20, N<=18.
Q3) Given a Directed graph with N nodes and M edges. each node has a population Ai. you have to open outlets in some or all nodes following the conditions in decreasing priority: 1) Every Node must have access to some outlet either directly or through some path. 2) Maximize the average sales per outlet. sales are linearly proportional to the no of people that visit that location. Average sales per outlet = (sum of all outlet sales)/(total number of outlets) 3) Minimize the number of outlets. Print how many outlets will you open and in which node. Constraints : T <= 100, N <= 1e3, M <= min( 1e5, N(N-1)), Ai <= 1e3. | 445 | 1,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-51 | latest | en | 0.917178 |
https://eyebulb.com/how-do-you-find-the-perpendicular-bisector-of-a-line-with-two-points/ | 1,709,336,462,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475711.57/warc/CC-MAIN-20240301225031-20240302015031-00345.warc.gz | 227,802,315 | 36,530 | How do you find the perpendicular bisector of a line with two points?
How do you find the perpendicular bisector of a line with two points?
A perpendicular bisector is a line that cuts a line segment connecting two points exactly in half at a 90 degree angle. To find the perpendicular bisector of two points, all you need to do is find their midpoint and negative reciprocal, and plug these answers into the equation for a line in slope-intercept form.
What is the perpendicular bisector of a line segment?
A perpendicular bisector is defined as a line or a line segment that divides a given line segment into two parts of equal measurement. ‘Bisect’ is the term used to describe dividing equally. Perpendicular bisectors intersect the line segment that they bisect and make four angles of 90° each on both sides.
Where does a perpendicular bisector cross a line segment?
With a perpendicular bisector, the bisector always crosses the line segment at right angles (90°). In the figure above, the segment PQ is being cut into two equal lengths (PF and FQ) by the bisector line AB, and does so at 90°.
What is the equation for perpendicular bisector?
Perpendicular bisector will pass through the points A and B i.e. point M. In this case, the perpendicular bisector is eventually a line passing through point M(5,3) and having slope m2=1. Thus the equation of the perpendicular bisector is x−y−2=0.
What is the perpendicular bisector equation?
⇒m1×m2=−1, where m2 is the slope of the perpendicular bisector. Perpendicular bisector will pass through the points A and B i.e. point M. In this case, the perpendicular bisector is eventually a line passing through point M(5,3) and having slope m2=1. Thus the equation of the perpendicular bisector is x−y−2=0.
Does perpendicular bisector pass through midpoint?
The perpendicular bisector of a side of a triangle is a line perpendicular to the side and passing through its midpoint. The three perpendicular bisectors of the sides of a triangle meet in a single point, called the circumcenter .
Does a perpendicular bisector always go through a vertex?
A perpendicular bisector (always, sometimes, never) has a vertex as an endpoint. The angle bisectors of a triangle (always, sometimes, never) intersect at a single point. A perpendicular bisector can also be an altitude.
Does line segment have many bisectors?
Points, lines, segments, and rays are all types of segment bisectors. If either a ray or a line serves as a segment bisector, it will be infinite. A segment may have many bisectors at the same time.
Which segment is perpendicular at midpoint?
A line, ray, or line segment (referred to as segment) that is perpendicular to a given segment at its midpoint is called a perpendicular bisector . To bisect means to cut or divide the given segment into two congruent segments.
What is a bisected line segment?
Line segment is bisected means cut in half. At the intersection point (the midpoint of the segment), the two lines will be perpendicular. This means their slopes will be negative reciprocals of each other.
Can a right triangle form a perpendicular bisector?
So, right triangle can form perpendicular bisector with base and height. Parallel lines can also form perpendicular bisector if a line is drawn from mid point of one line to mid point of another line.
06/12/2020 | 736 | 3,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.905315 |
https://www.mentorforbankexams.com/2017/04/reasoning-syllogism-new-pattern-for-sbi-po.html | 1,606,779,168,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141515751.74/warc/CC-MAIN-20201130222609-20201201012609-00125.warc.gz | 745,610,390 | 56,227 | Reasoning Syllogism (New Pattern) for SBI PO
Reasoning Syllogism (New Pattern) for SBI PO
Directions (1 – 5): In the following questions, certain conclusions are given and five statements are given as 1, 2, 3, 4 and 5. Find out from which statement the following conclusions can be drawn.
1. Conclusion: No snake is reptile; some reptiles are animal.
Statements:
a) Some reptiles are snake; no snake is animal; some animal are birds.
b) Some snakes are reptiles; all reptiles are animal; no snake is a bird.
c) No reptile is an animal; no animal is a snake; all snakes are birds.
d) No snake is a bird; all reptiles are birds; some animal are reptile.
e) All reptiles are animal; all animal are snake; all snakes are birds.
2. Conclusion: Some silver are platinum. Some diamonds are silver.
Statements:
a) All silver are gold. Some gold are diamond. No diamond is platinum.
b) All silvers are diamond. No gold is silver. Some diamond is platinum.
c) Some silvers are gold. All gold are diamond. All diamond is platinum.
d) All silvers are gold. No gold is diamond. All diamond is platinum
e) All platinum are silver. Some gold is silver. No gold is diamond.
3. Conclusion: buildings are definitely not houses; all houses are gates.
Statements:
a) Some houses are gates; all gates are buildings; no road is a house.
b) All gates are roads; all roads are houses; no road is a building.
c) All houses are roads; all roads are gates; no road is a buildings.
d) Some houses are buildings; all buildings are gates; some gates are definitely not road.
e) Some houses are definitely not buildings; all houses are roads; some buildings are gates.
4. Conclusion: some bags are definitely not tables; some bags are pens.
Statements:
a) No table is a pencil; some pencils are bags; all pens are bags.
b) All tables are bags; no bag is a pen; all pens are pencils.
c) All pens are tables; no table is a bag; all bags are pencils.
d) All bags are tables; all tables are pens; all pens are pencils.
e) No table is a pencil; all bags are tables; all pens are pencils.
5. Conclusion: all dogs are cats; no cat is a lion.
Statements:
a) All dogs are lions; no lion is a cat; all cats are leopards.
b) No leopard is a lion; all cats are leopards; all dogs are lions.
c) All dogs are leopards; all leopards are cats; no lion is a cat.
d) No cat is a leopard; all dogs are leopards; some lions are cats.
e) All lions are leopards; all leopards are dogs; no dog is a cat.
Directions (6 – 10): Each of the following questions consists of six statements followed by options consisting of three statements put together in a specific order. Choose the options that indicates a combination where the third statement can be logically deduced from the first two statements and that option will be your answer.
6. i. Some X is Y. ii. All Y is Z. iii. No Z is W iv.Some W are Y v. All Y are T. vi. Some T are X.
a) [ii, iii, iv]
b) [vi, i, v]
c) [iv, ii, iii]
d) [iv, iii, ii]
e) None is correct
7. i. Some A is B. ii. Some C is D. iii. No D is E. iv. Some Y is A. v. Some E is A. vi. All B is Y.
a) [i, v, iii]
b) [v, iv, vi]
c) [i, vi, iv]
d)[iv, vi, i]
e) None is correct
Directions (8 – 10): In the questions below are given two conclusions followed by five set of statements. You have to choose the correct set of statements that logically satisfies given conclusions either definitely or possibly. Assume the given statements to be true even if they seem to be at variance from commonly known facts.
8. Conclusions: Some air is system. Some road is control.
Statements:
(a)All air is control. Some control is system. All system is idea. Some idea is traffic. All traffic is road.
(b)All air is road. Some air is traffic. All traffic is control. All control is system. Some system is idea.
(c)Some air is road. All road is idea. Some idea is traffic. All traffic is system. All system is control.
(d) All air is traffic. Some traffic is control. All control is system. Some system is idea. All idea is road.
(e) None of these.
9. Conclusions: Some man is not doll. Some toy is man.
Statements:
(a)All toy is boy. Some boy is child. No child is man. All man is girl. All girl is doll.
(b)Some toy is man. All man is doll. No doll is girl. All boy is girl. Some girl is child.
(c)Some toy is child. All child is man. Some man is girl. No girl is doll. All boy is doll.
(d)Some toy is girl. All girl is doll. No doll is child. All boy is child. Some child is man.
(e) None of these.
10. Conclusions: Some red is white. Some yellow is red.
Statements:
(a)All white is black. All black is red. Some red is violet. All violet is yellow. Some yellow is green.
(b)All white is violet. Some violet is red. All red is black. Some black is yellow. All yellow is green.
(c)All white is green. Some green is red. All red is black. Some black is violet. All violet is yellow.
(d) All white is yellow. Some yellow is red. All red is black. Some black is violet. All violet is green.
(e) None of these.
Solutions:
1. D)
2. C)
3. C)
4. A)
5. C)
6. E) There is no option in which third statement is conclusion of first two statements. So option (e) None of these is correct.
7. C) We can conclude the third statement (iv) from the first two statements (i, vi).
8. B)
9. C)
10. A) | 1,409 | 5,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.959941 |
https://scicomp.stackexchange.com/questions/43374/optimal-quadrature-rule-for-heavy-tail-measure/43377 | 1,723,652,546,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641118845.90/warc/CC-MAIN-20240814155004-20240814185004-00568.warc.gz | 398,781,569 | 41,786 | Optimal quadrature rule for heavy tail measure
I'm looking for a well-thought quadrature rule for this measure $$d\mu(t)=\frac{dt}{t^s}$$ for $$s\in(0,1)$$, the underlying motivation is to compute this integral $$\lambda^{s-1}=\frac{1}{\Gamma(1-s)}\int_{0}^{\infty} e^{-t\lambda}\frac{dt}{t^s}$$ with $$\lambda > 0$$, see this reference, Page 5, equation (5), https://arxiv.org/abs/1808.05159. So I'm looking for a way of dealing with the singularity at $$t=0$$ as well as the limit to $$\infty$$. Observe that $$\int_{T}^\infty \frac{dt}{t^s} = \infty$$ I was using the QuadGK.jl package and it is working just fine
λ = 2; s = 0.8
λs = λ^(s-1)
f(t) = exp(-λ*t)/t^s/gamma(1-s)
The predicted error and the real error is the same order of magnitude. However, when using the package for another formula $$\lambda^s = \frac{1}{\Gamma(-s)}\int_0^\infty (e^{-t\lambda}-1)\frac{dt}{t^{1+s}}$$ same reference as before. I again do the quadrature
λs = λ^s
g(t) = (exp(-λ*t)-1)/t^(1+s)/gamma(-s)
it works just fine, but this time the real error is of order 1e-3 while the predicted error is around 1e-8. And if I put a positive matrix instead of a scalar
A = # some positive definite matrix
h(t) = (exp(-A.*t)-I)/t^(1+s)/gamma(-s)
I get an ERROR: DomainError with 0.5: integrand produced NaN in the interval (0.0, 1.0), and, to my understanding, quadgk does not evaluate the function at $0$ as per the documentation https://juliamath.github.io/QuadGK.jl/stable/. And if I write
res, err = quadgk(h,1e-50,Inf)
the integral blows up. If I augment the lower bound to 1e-20 then I get a norm2 error of 0.24 w.r.t the fractional matrix, if I keep augmenting the lower bound (1e-10, 1e-5), the norm2 error keeps augmenting. I computed the error using the exact eigenvalues and eigenvectors. Has somebody got an intuition to where the instability behind the scenes might be? I would also like to know whether the Gauss-Kronrod quadrature is optimal for this integral, and if I'm doing fine, as I'm relatively new to this :)
EDIT (following a comment): unfortunately does not make a lot of difference, I leave the fully reproducible code here
using QuadGK
using SpecialFunctions
s = 0.8
nx = 3; Δx = 1/(nx+1) # change nx to whatever
A = -[2. -1 0; -1 2 -1; 0 -1 2]/Δx^2
Χ = zeros(Float64,nx,nx)
for p=1:nx
Χ[:,p] = sin.(p*pi*(1:nx)*Δx)
end
Λ = diagm(2*(cos.((1:nx)*pi*Δx).-1)/Δx^2)
As = -Χ*(diagm(diag(-Λ).^s))/Χ
h(t) = (exp(A.*t)-I)/t^(1+s)/gamma(-s)
res2 = res21 + res22; err2 = err21 + err22
@show res2
@show norm(res2+As)/norm(As)
@show norm(res1+As)/norm(As)
@show norm(res1-res2)
You can split the integral up such that you don't get the singularity at $$t=0$$. Something like this
using QuadGK
using SpecialFunctions
λ = 2.0; s = 0.8
λs = λ^(s-1)
f(t) = exp(-λ*t)/t^s/gamma(1-s)
λs = λ^s
g(t) = (exp(-λ*t)-1)/t^(1+s)/gamma(-s)
A = [2 -1 0; -1 2 -1; 0 -1 2] # some positive definite matrix
h(t) = (exp(-A.*t)-I)/t^(1+s)/gamma(-s)
singularity_integral, singularity_error = quadgk(h, 1e-10, 1e-5)
regular_integral, regular_error = quadgk(h, 1e-5, Inf)
@show result = singularity_integral + regular_integral
@show error = singularity_error + regular_error
which will print
(res, err) = quadgk(f, 0, Inf) = (0.8705505365052336, 1.2164780075652479e-8)
(res, err) = quadgk(g, 0, Inf) = (1.7402976255082268, 2.515036776166618e-8)
result = singularity_integral + regular_integral = [1.6837899364901885 -0.7050512093960599 -0.03988519645256732; -0.7050512093960599 1.6439047399317526 -0.7050512098966073; -0.03988519645256732 -0.7050512098966073 1.683789936502625]
error = singularity_error + regular_error = 3.822238569033589e-8
Alternatively, you can also look into something like Cauchy principal values and this example
--Edit: update to new information--
I'm afraid I can't reproduce the same error. When I run your code (I've added using LinearAlgebra for diagm) I get
res2 = [15.354655632950173 -6.419790221101635 -0.3665284746983265; -6.419790221101635 14.988127158828647 -6.4197902215468154; -0.3665284746983265 -6.4197902215468154 15.354655632887033]
norm(res2 + As) / norm(As) = 0.018647374162762107
norm(res1 + As) / norm(As) = 0.01864737561229984
norm(res1 - res2) = 4.9872522079524005e-8
4.9872522079524005e-8 | 1,519 | 4,235 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-33 | latest | en | 0.817548 |
https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.21/share/doc/Macaulay2/NCAlgebra/html/___N__C__Ring.html | 1,685,440,676,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2023-23/segments/1685224645595.10/warc/CC-MAIN-20230530095645-20230530125645-00678.warc.gz | 295,169,699 | 4,239 | NCRing -- Type of a noncommutative ring
Description
All noncommutative rings have this as an ancestor type. It is the parent of the types NCPolynomialRing and NCQuotientRing.
In addition to defining a ring as a quotient of a NCPolynomialRing, some common ways to create NCRings include skewPolynomialRing, endomorphismRing, and oreExtension.
Let's consider a three dimensional Sklyanin algebra. We first define the tensor algebra:
i1 : A = QQ{x,y,z} o1 = A o1 : NCPolynomialRing
Then input the defining relations, and put them in an ideal:
i2 : f = y*z + z*y - x^2 2 o2 = zy+yz-x o2 : A i3 : g = x*z + z*x - y^2 2 o3 = zx-y +xz o3 : A i4 : h = z^2 - x*y - y*x 2 o4 = z -yx-xy o4 : A i5 : I=ncIdeal{f,g,h} 2 2 2 o5 = Two-sided ideal {zy+yz-x , zx-y +xz, z -yx-xy} o5 : NCIdeal
Next, define the quotient ring (as well as try a few functions on this new ring). Note that when the quotient ring is defined, a call is made to Bergman to compute the Groebner basis of I (out to a certain degree, should the Groebner basis be infinite).
i6 : B=A/I --Calling Bergman for NCGB calculation. Complete! o6 = B o6 : NCQuotientRing i7 : generators B o7 = {x, y, z} o7 : List i8 : numgens B o8 = 3 i9 : isCommutative B o9 = false i10 : coefficientRing B o10 = QQ o10 : Ring
As we can see, x is an element of B.
i11 : x o11 = x o11 : B
If we define a new ring containing x, x is now part of that new ring
i12 : C = skewPolynomialRing(QQ,(-1)_QQ,{x,y,z,w}) --Calling Bergman for NCGB calculation. Complete! o12 = C o12 : NCQuotientRing i13 : x o13 = x o13 : C
We can 'go back' to B using the command use(NCRing).
i14 : use B o14 = B o14 : NCQuotientRing i15 : x o15 = x o15 : B i16 : use C o16 = C o16 : NCQuotientRing
We can also create an Ore extension. First define a NCRingMap with ncMap.
i17 : sigma = ncMap(C,C,{y,z,w,x}) o17 = NCRingMap C <--- C o17 : NCRingMap
Then call the command oreExtension.
i18 : D = oreExtension(C,sigma,a) --Calling Bergman for NCGB calculation. Complete! o18 = D o18 : NCQuotientRing i19 : generators D o19 = {x, y, z, w, a} o19 : List i20 : numgens D o20 = 5
Methods that use an object of class NCRing :
• basis(ZZ,NCRing) -- Returns a basis of an NCRing in a particular degree.
• "centralElements(NCRing,ZZ)" -- see centralElements -- Finds central elements in a given degree
• coefficientRing(NCRing) -- Returns the base ring of an NCRing
• "envelopingAlgebra(NCRing,Symbol)" -- see envelopingAlgebra -- Create the enveloping algebra
• "freeProduct(NCRing,NCRing)" -- see freeProduct -- Define the free product of two algebras
• generators(NCRing) -- The list of algebra generators of an NCRing
• hilbertSeries(NCRing) -- Computes the Hilbert series of an NCRing
• isCommutative(NCRing) -- Returns whether an NCRing is commutative
• "isExterior(NCRing)" -- see isCommutative(NCRing) -- Returns whether an NCRing is commutative
• "isHomogeneous(NCRing)" -- see isHomogeneous(NCIdeal) -- Determines whether the input defines a homogeneous object
• "ncMap(NCRing,NCRing,List)" -- see ncMap -- Make a map to or from an NCRing
• "ncMap(NCRing,Ring,List)" -- see ncMap -- Make a map to or from an NCRing
• "ncMap(Ring,NCRing,List)" -- see ncMap -- Make a map to or from an NCRing
• "ncMatrix(NCRing,List,List)" -- see ncMatrix -- Create an NCMatrix
• numgens(NCRing) -- The number of algebra generators of an NCRing
• "oppositeRing(NCRing)" -- see oppositeRing -- Creates the opposite ring of a noncommutative ring
• "oreExtension(NCRing,NCRingMap,NCRingElement)" -- see oreExtension -- Creates an Ore extension of a noncommutative ring
• "oreExtension(NCRing,NCRingMap,NCRingMap,NCRingElement)" -- see oreExtension -- Creates an Ore extension of a noncommutative ring
• "oreExtension(NCRing,NCRingMap,NCRingMap,Symbol)" -- see oreExtension -- Creates an Ore extension of a noncommutative ring
• "oreExtension(NCRing,NCRingMap,Symbol)" -- see oreExtension -- Creates an Ore extension of a noncommutative ring
• "oreIdeal(NCRing,NCRingMap,NCRingElement)" -- see oreIdeal -- Creates the defining ideal of an Ore extension of a noncommutative ring
• "oreIdeal(NCRing,NCRingMap,NCRingMap,NCRingElement)" -- see oreIdeal -- Creates the defining ideal of an Ore extension of a noncommutative ring
• "oreIdeal(NCRing,NCRingMap,NCRingMap,Symbol)" -- see oreIdeal -- Creates the defining ideal of an Ore extension of a noncommutative ring
• "oreIdeal(NCRing,NCRingMap,Symbol)" -- see oreIdeal -- Creates the defining ideal of an Ore extension of a noncommutative ring
• "NCRing ** NCRing" -- see qTensorProduct -- Define the (q-)commuting tensor product
• "qTensorProduct(NCRing,NCRing,QQ)" -- see qTensorProduct -- Define the (q-)commuting tensor product
• "qTensorProduct(NCRing,NCRing,RingElement)" -- see qTensorProduct -- Define the (q-)commuting tensor product
• "qTensorProduct(NCRing,NCRing,ZZ)" -- see qTensorProduct -- Define the (q-)commuting tensor product
• "setWeights(NCRing,List)" -- see setWeights -- Set a nonstandard grading for a NCRing.
• "toM2Ring(NCRing)" -- see toM2Ring -- Compute the abelianization of an NCRing and returns a Ring.
• use(NCRing) -- Brings the variables of a particular NCRing in scope
For the programmer
The object NCRing is a type, with ancestor classes Ring < Type < MutableHashTable < HashTable < Thing. | 1,612 | 5,285 | {"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} | 3.40625 | 3 | CC-MAIN-2023-23 | latest | en | 0.673271 |
https://figshare.com/articles/_a_Spectrum_of_Bogoliubov_excitations_red_dots_for_a_homogeneous_system_with_sharp_boundaries_calcul/1012015/1 | 1,547,788,349,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659890.6/warc/CC-MAIN-20190118045835-20190118071835-00042.warc.gz | 504,970,851 | 8,629 | (a) Spectrum of Bogoliubov excitations (red dots) for a homogeneous system with sharp boundaries, calculated for <em>J<sub>z</sub></em> = δ and <em>J</em> = 2 Δ<sup>(2)</sup>
2013-06-24T00:00:00Z (GMT)
<p><strong>Figure 4.</strong> (a) Spectrum of Bogoliubov excitations (red dots) for a homogeneous system with sharp boundaries, calculated for <em>J<sub>z</sub></em> = δ and <em>J</em> = 2 Δ<sup>(2)</sup>. It exhibits a bulk gap Δ<sub>bulk</sub> = 0.18 <em>E<sub>r</sub></em> and a pair of zero-energy Majorana states with a residual splitting Δ<sub><em>s</em></sub> ~ 10<sup>−12</sup> <em>E<sub>r</sub></em>. (b) Evolution of the bulk gap amplitude Δ<sub>bulk</sub> as a function of the ratio <em>J<sub>z</sub></em>/δ (red line), for <em>J</em> = Δ<sup>(2)</sup>, with Δ<sup>(2)</sup> given by equation (<a href="http://iopscience.iop.org/0953-4075/46/13/134005/article#jpb448206eqn20" target="_blank">20</a>). The black line represents the prediction of second-order perturbation theory Δ<sub>bulk</sub> = 2 Δ<sup>(2)</sup>. (c) Density distribution along <em>x</em> of a zero-energy Majorana state, in planes <em>A</em> (red line) and <em>B</em> (blue line, offset for clarity). In the strong coupling regime <em>J</em> ~ δ, the population in <em>B</em> is not negligible. (d) Total density distribution along <em>x</em> calculated at zero temperature. Majorana states are not visible in this almost uniform density profile.</p> <p><strong>Abstract</strong></p> <p>We propose an experimental implementation of a topological superfluid with ultracold fermionic atoms. An optical superlattice is used to juxtapose a 1D gas of fermionic atoms and a 2D conventional superfluid of condensed Feshbach molecules. The latter acts as a Cooper pair reservoir and effectively induces a superfluid gap in the 1D system. Combined with a spin-dependent optical lattice along the 1D tube and laser-induced atom tunnelling, we obtain a topological superfluid phase. In the regime of weak couplings to the molecular field and for a uniform gas, the atomic system is equivalent to Kitaev's model of a p-wave superfluid. Using a numerical calculation, we show that the topological superfluidity is robust beyond the perturbative limit and in the presence of a harmonic trap. Finally, we describe how to investigate some physical properties of the Majorana fermions located at the topological superfluid boundaries. In particular, we discuss how to prepare and detect a given Majorana edge state.</p> | 697 | 2,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | longest | en | 0.722356 |
https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_14 | 1,708,758,197,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474523.8/warc/CC-MAIN-20240224044749-20240224074749-00629.warc.gz | 103,502,337 | 11,017 | # 1984 AHSME Problems/Problem 14
## Problem
The product of all real roots of the equation $x^{\log_{10}{x}}=10$ is
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }-1 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ }10^{-1} \qquad \mathrm{(E) \ } \text{None of these}$
## Solution
Take the logarithm base $x$ of both sides to get $\log_{10}x=\log_x10$. Use the change of base formula to get $\log_{10}x=\frac{\log_{10}10}{\log_{10}x}=\frac{1}{\log_{10}x}$. Therefore, $(\log_{10}x)^2=1$ and $\log_{10}x=\pm1$. From $\log_{10}x=1$, we have $x=10^1=10$ and from $\log_{10}x=-1$ we have $x=10^{-1}=\frac{1}{10}$. The product is thus $10\times\frac{1}{10}=1, \boxed{\text{A}}$. | 274 | 668 | {"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": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.492881 |
https://www.jiskha.com/questions/1249456/a-parabola-has-a-vertex-v-6-4-and-a-focus-f-6-1-what-is-the-equation | 1,638,145,744,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00428.warc.gz | 931,006,610 | 4,833 | # algebra 2
a parabola has a vertex v=(6,4) and a focus f=(6,1) what is the equation
1. 👍
2. 👎
3. 👁
1. from first principles, the directrix is y = 7
let P(x,y) be any point
then PF = PD, where D is the vertical distance from P to the directrix
√( (x-6)^2 + (y-1)^2 ) = √(x-x)^2 + (y-7)^2 )
square and expand
x^2 - 12x + 4 + y^2 - 2y + 1 = y^2 - 14y + 49
x^2 - 12x - 44 = -14y
verification:
http://www.wolframalpha.com/input/?i=x%5E2+-+12x+-+44+%3D+-14y
1. 👍
2. 👎
2. Or, knowing that the equation of a parabola with focus at y=(0,0) and focus at y = -p is
x^2 = 4py
we have a shifted parabola with p=-3, so the equation is
(x-6)^2 = 4(-3)(y-4)
x^2-12x+36 = -12y+48
x^2-12x+84 = -12y
Verify at
http://www.wolframalpha.com/input/?i=parabola+%28x-6%29^2+%3D+4%28-3%29%28y-4%29
Oops, Reiny - if you add the word parabola to your function at wolframalpha, you will see where the focus and vertex lie. Must have had a typo somewhere in the math.
1. 👍
2. 👎
## Similar Questions
1. ### algebra
Suppose a parabola has a vertex (-4,7) and also passes through the point (-3,8) Write the equation of the parabola in vertex form. f(x)=a(x-h)^2+k I believe h=-4 k=7 Not sure what to do from here.
2. ### algebra
what is an equation of a parobola with the given vertex and focus? vertex(5,4) and focus(8,4) Can someone show me the steps to get the equation please?
3. ### Algebra
An engineer designs a satellite dish with a parabolic cross section. The dish is 10 ft wide at the opening, and the focus is placed 8 ft from the vertex. a) Position a coordinate system with the origin at the vertex and
4. ### Algebra II
what is an equation of a parabola with the given vertex and focus? Vertex: (-2,5); focus: (-2,6)
1. ### Math
Write the equation of a parabola with a vertex at the origin and a focus at (4, 0).
an engineer designs a satellite dish with a parabolic cross section. The dish is 15 ft wide at the opening and the focus is placed 4 ft from the vertex. find an equation of the parabola. I know how to work this problem, but how do
3. ### algebra
Suppose a parabola has vertex (-4,7) and also passes through the point (-3,8), write the equation of the parabola in vertex form.
4. ### algebra
A few more question Id like for someone to check please. 1) what are the vertex, focus, and directrix of the parabola with the given equation? x^2-8x-28y-124=0 vertex (4,-5) focus (0,7) directrix y=-12 2) write an equation of a
1. ### Trigonometry
1. Write the equation of a parabola with a vertex at the origin and a focus at (4, 0).
2. ### math
Using the given information write the equation of parabolas: The parabola has focus (0, 6) and vertex at the origin;
3. ### Algebra
What is an equation of a parabola with the given vertex and focus? Vertex: (5, 4) Focus: (8, 4) a. x = 1/2 (y - 4)^2 + 5 b. y = 1/2 (x + 4)^2 - 5 c. x = 1/2 (y + 4)^2 - 5 d. y = 1/2 (x - 4)^2 + 5
4. ### Math/Algebra
Find the vertex, Focus,and Directrix of the parabola. Graph the equation. y^2=12x | 1,036 | 3,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.875746 |
http://mathsticks.com/resource/1203/addition-and-subtraction-eggs-nest | 1,409,556,594,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535917463.1/warc/CC-MAIN-20140901014517-00373-ip-10-180-136-8.ec2.internal.warc.gz | 352,211,646 | 13,379 | # Addition and Subtraction - Eggs in the Nest
Here's a quirky little game idea with an Easter/Egg theme.
The focus is on addition and subtraction of small numbers, and the activities are perfect for early Key Stage 1 children.
The addition game uses a pair of dice and an egg and nest game board. Two players sit opposite each other and place 10 counters on the numbered eggs on their side of the board. They now take it in turns to roll a pair of dice. The sum of the dots shown on the dice indicate a numbered egg on their board. If they have a counter on that egg, they place it in the nest.
The winner is the first to get all of their egg-counters in the nest.
Throughout the game the children are counting dots on the dice, determining the sum and also identifying the numeral that represents this value. If you want to introduce a little more strategic thinking into the game, then let the children placed their ten counters on any combination of the numbers at the beginning. At this point (if children are familiar with using dice) they may choose to place more counters on the middle numbers; 5, 6, 7, 8, 9 and avoid the ones that are more difficult to roll such as 2, 3, 11 and 12.
The subtraction, or difference game uses a similar egg and nest game board and a set of 0 - 10 number cards. As before, the players place 10 counters on the eggs on their side of the board. This time, however instead of rolling dice they draw two number cards at random, turn them over and calculate the difference. It is this value which indicates an egg. If the player has a counter on that egg, he or she can place it in the nest. The winner is the first to get all of their egg-counters in the nest.
Both of these games also offer excellent assessment opportunities. Find them on the attached pdf file below.
Mathsticks+ members can access three additional boards, "Doubles"; "Add 1 more" and "Odd & Even". See the link below for details.
If you enjoy using these maths resources please help by telling your friends. You can use the +1 button at the top of this description to tell Google too. You can also follow mathsticks.com on Google+ , or use the social network buttons below. Thanks, your help is very much appreciated.
## Comment viewing options
### Love the graphics on this
Love the graphics on this game. It is just right for my class of 5 year olds. Looking forward to using it just before Easter but we will also use it to consolidate addition and subtraction understandings next term.
1 user has voted.
### Lovely graphics - looks
Lovely graphics - looks simple and easy to modify for quicker thinkers - can't wait to use it!
0 users have voted.
### So simple but just the sort
So simple but just the sort of thing my Year 1s love. Do you think I can get away with playing it after the Easter holidays??
0 users have voted.
### Great resource. The children
Great resource. The children loved it. We used 10 sided dice instead of cards for the subtraction.
1 user has voted.
### This looks really good and I
This looks really good and I am going to use this with my year 1's. It would be great if there was a game that would be compatible with a 20 sided die.
0 users have voted.
### This looks really good and I
This looks really good and I am going to use this with my year 1's. It would be great if there was a game that would be compatible with a 20 sided die.
0 users have voted.
### Cannot wait to use this game
Cannot wait to use this game in class next week. Love the simple concepts and lovely graphics -Will appeal to al my children in Foundation Stage.
1 user has voted.
### Thanks. Teaching the
Thanks. Teaching the difference is always tricky so it is nice to find something that is fun and appealing. I am also glad I found this just before Easter! I think I may prepare some subtraction calculation cards to turn over so that I can differentiate this for my more able year 2.
0 users have voted.
### Really good for the kids to
Really good for the kids to play initially under supervision and then together whilst I deal with others. Teaches addition, subtraction, taking turns, being a good sport and is a good way to introduce different addition and subtraction language, ie. sum of, total, difference etc
0 users have voted.
### Great game, just a small
Great game, just a small thing, but it would be good if the addition board had an egg with 12 on it.
0 users have voted.
### Yes, that's a good point - we
Yes, that's a good point - we may adapt the board to include '12' in the future. Thank you for your comment.
0 users have voted.
### This is a great idea to
This is a great idea to practise number bonds!
0 users have voted.
### I will be sending this game
I will be sending this game home as homework for my year 1 class over the Easter holiday so that they keep practising addition.
1 user has voted.
### Great resource, just what I
Great resource, just what I need for the last week of term :)
0 users have voted.
### Great! Very clear & easy to
Great! Very clear & easy to follow!
0 users have voted.
### This will be great for my low
This will be great for my low level students. Thanks so much.
0 users have voted.
### Really good idea...will be
Really good idea...will be useful for homework too!! Thanks.
0 users have voted.
### This game looks amazing!
This game looks amazing! I'll use it tomorrow with a little girl that I am working with (5 years old) who is having great difficulty with addition and subtraction. I am sure she'll be motivated by it! Thank you!!
2 users have voted.
### Will be fab for my Easter
Will be fab for my Easter themed maths morning tomorrow!
1 user has voted.
### Thanks for this, it looks
Thanks for this, it looks great and will be an excellent resource to use with my Reception class as we are currently doing addition and subtraction. It is also particularly useful for encouraging turn taking and playing a game fairly as many of the children in my class are finding this difficult at the moment. :-)
1 user has voted.
### This game looks good for my
This game looks good for my less advanced group in maths and is just right for this time of year. Thank you.
1 user has voted.
### This game looks great. Will
This game looks great. Will really help my lower ability with simple addition.
2 users have voted. | 1,421 | 6,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2014-35 | latest | en | 0.960416 |
https://ronnybergmann.net/mvirt/manifolds/Sn/parallelTransport.html | 1,553,018,607,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202003.56/warc/CC-MAIN-20190319163636-20190319185636-00378.warc.gz | 621,783,540 | 4,985 | $\DeclareMathOperator{\arccosh}{arccosh} \DeclareMathOperator*{\argmin}{arg\,min} \DeclareMathOperator{\Exp}{Exp} \newcommand{\geo}[2]{\gamma_{\overset{\frown}{#1,#2}}} \newcommand{\geoS}{\gamma} \newcommand{\geoD}[2]{\gamma_} \newcommand{\geoL}[2]{\gamma(#2; #1)} \newcommand{\gradM}{\nabla_{\M}} \newcommand{\gradMComp}[1]{\nabla_{\M,#1}} \newcommand{\Grid}{\mathcal G} \DeclareMathOperator{\Log}{Log} \newcommand{\M}{\mathcal M} \newcommand{\N}{\mathcal N} \newcommand{\mat}[1]{\mathbf{#1}} \DeclareMathOperator{\prox}{prox} \newcommand{\PT}[3]{\mathrm{PT}_{#1\to#2}#3} \newcommand{\R}{\mathbb R} \newcommand{\SPD}[1]{\mathcal{P}(#1)} \DeclareMathOperator{\Tr}{Tr} \newcommand{\tT}{\mathrm{T}} \newcommand{\vect}[1]{\mathbf{#1}}$
# The parallel transport the sphere
This function parallel transports a vector $\xi\in T_x\mathbb S^n$ along a geodesic $\geo{x}{y}$ (where uniqueness is determined) by the logarithmic map implementation) by
$P_{x\to y}(\xi) = \xi - \frac{\langle \log_xy,\xi\rangle_x}{d_{\mathbb S^n}(x,y)} \bigl( \log_xy+\log_yx \bigr)$
This formula is taken from [1,2] and can be interpreted as follows: all components of $\xi$ that share no part with the direction $\log_xy$ are left unchanged. For the remaining part (second term) we have to perform a correction.
### Matlab Documentation
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% eta = parallelTransport(x,y,xi) transport xi from TxM parallel to TyM
%
% INPUT
% x,y : two (sets of) points on the manifold
% xi : a (set of) vectors from TxM
%
% OUTPUT
% eta : the parallel transported vectors in TyM
% ---
% Manifold-valued Image Restoration Toolbox 1.2
% R. Bergmann | 2018-03-01 | 572 | 1,648 | {"found_math": true, "script_math_tex": 5, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-13 | longest | en | 0.466603 |
https://caspoc.com/help/handson/greenenergy/wind/ | 1,642,567,144,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301263.50/warc/CC-MAIN-20220119033421-20220119063421-00678.warc.gz | 191,882,766 | 25,533 | ## Wind Power
Wind power is one type of renewable energy. The wind can blow the blades of a wind turbine and then drive the main shaft which connects with the rotor of a generator. Depending on the wind speed, the angular speed of the main shaft changes and generates a corresponding electrical current. In this simple example, we practice two examples of wind power and using some sensors to measure the power loss in each phase.
### Wind turbine
In the first example, we integrate a wind turbine into the main grid of 3-phase AC voltage source as below. We need a wind turbine, a 3-phase AC voltage source, a generator which transforms the mechanical power to the electrical power, and an AC/AC converter which transforms the voltage or frequency level. For getting the power loss, we also use a mechanical power sensor and an AC circuit power sensor.
To put a wind turbine, left-click Components/Library/GreenEnergy/Wind/WindTurbine (step 1) and release the mouse. The wind turbine follows the cursor and left-click to put it on the workscreen. Right-click on the wind turbine (step 2) to change its parameters. In the pop-window, we can see the parameters as below (step 3):
• A = 4500 m˛ (turbine swept area)
• Friction = 250 Nm/Rad/s (friction of the rotor)
• Inertia = 6MEG Kgm˛ (inertia of the rotor)
• Initial Rotor Speed = 0 Rad/s (initial speed of the turbine)
• Rho = 1.2 (air density)
Click OK to save the setting (step 4).
Then give a generator to transform the mechanical power to the electrical power. Left-click Components/Library/ElectricalMachines/System/Generator (step 1) and release the mouse. The generator follows the cursor and left-click to put it in the right side of the wind turbine (step 2). Notice to keep a bit space between the wind turbine and the generator in order to insert a power sensor later. Right-click the generator (step 2) to modify its parameters if needed. In the pop-window, we can see the parameters as below (step 3):
• Efficiency[from 0 to 1] = 0.95 (efficiency of the generator)
• Minimum Angular Speed = 1 Rad/s (if the angular speed is below this minimum angular speed value, the power consumption changes from constant power into constant braking torque)
• Power factor[from 0 to 1] = 1 (the power factor or cos(phi) between the voltage and current vector because the generator has only sinusoidal outputs)
Click OK to save the setting (step 4).
To let the AC/AC converter adjust its parameters according to the mechanical power, we add a power optimizer which measures the angular speed of the main shaft and then calculate the power value for the AC/AC converter.
Left-click Components/Library/GreenEnergy/Wind/OptimumPower (step 1) and release the mouse. The power optimizer follows the cursor and left-click to put it below the generator (step 2). Right-click the power optimizer to change its parameter if needed. The parameter we have here is the constant value K (step 3). Click ‘Help Lib’ to see how K works for calculating the power value (step 4). In the pop-window, we can see that the torque value from the main shaft T= K * &omega * &omega, so that the power = torque * angular speed = T * &omega. Click OK to save the setting if needed (step 5).
Connect the power optimizer and the AC/AC converter as the following setting (step 6). Its input is to sense the angular speed of the main shaft; its output is to calculate the power value and then send this value to the AC/AC converter.
To connect the AC/AC converter with the main grid, we put a 3-phase voltage source in the right side of the converter. Left-click Components/Library/Source/Circuit/ThreePhase/VoltageSourceRMS (step 1) and release the mouse. The voltage source follows the cursor and left-click to put it in the left side of the AC/AC converter (step 2). Right-click on the voltage source to change its parameters. In the pop-window, we can see its parameters as below (step 3):
• Frequency = 50 Hz
• Vrms = 230 (RMS voltage)
• Phase [from 0 to 360] = 0 (phase in degrees)
• Select ‘Flip Vertical’ (step 4) in the rotation setting
Click OK to save the setting (step 5).
Left-click Components/Library/Sensor/Rotational/PowerTorque (step 1) and release the mouse. Left-click on the workscreen to connect the power torque sensor with the generator (step 2).
Left-click Components/Library/Sensor/Electrical/PowerAC3Phase (step 3) and release the mouse. Left-click on the workscreen to connect the AC power sensor with the AC/AC converter (step 4).
Connect the wind turbine and the power torque sensor (step 1). Connect the voltage source and the AC power sensor (step 2). Currently we have not set an electric ground so that right-click on the neutral input of the voltage source to add a ground label (step 3). In the pop-window, enter label = Ground (step 4) and click OK to save the setting (step 5).
Make sure there is a ground label in the neutral input of the voltage source (step 1).
Add an UI2RST sensor to read the RST signals of the voltage’s and the current’s vectors. Left-click Components/Library/sensor/Electrical/ui2rst (step 2) and release the mouse. Left-click on the workscreen to put the UI2RST sensor below the AC power sensor (step 3).
Then define the wind speed for the wind turbine. Right-click the wind speed input (step 4) and in the pop-window, enter label = 8 (which stands for the wind speed = 8m/s) (step 5). Click OK to save the setting (step 6).
Then give a scope to read the power values from these power sensors. Click the scope icon in the experience bar (step 1) and release the mouse. Left-click to put the scope above the AC power sensor. Left-click the right-bottom corner of the scope and hold down the mouse to enlarge this scope (step 2).
Connect the fourth output (P[kW]) of the power torque sensor to the first blue input trace of the first scope (step 1). Also the power output of the AC power sensor to the second red input trace. Right-click the first scope (step 2) and in the pop-window, select Trace/Trace1 (step 3). In the pop-window of the Trace1, set the Scale in the right side (step 4) and click OK to save the setting (step 5).
Click the scope icon in the experience bar (step 6) and release the mouse. Left-click to put the second scope in right side of the UI2RST sensor. Make sure that the first blue input trace is connected with the first output Ur of the UI2RST sensor. Left-click the right-bottom corner of the scope and hold down the mouse to enlarge this scope (step 7). Then the 6 input trace of the second scope shall be well connected to Ur, Us, Ut, Ir, Is and It. If not, adjust the position of second scope to make sure that all the connections are well done.
Click the short-cut of simulation parameter (step 1). In the pop-window, select Euler for the Numerical Integration Method (step 2), Tscreen = 100, dt = 100u (step 3) and then click OK to save the setting (step 4).
Click the short-cut of start simulation (step 1). Right-click on the first scope (step 2) to see the details of the simulation result. Click the shortcut of listbox (step 3) to enable the numeric display of simulation result. Click Right or Left key to find the precise values according to the time (step 4) and then we can find that the relation between the mechanical power and the AC electrical power depends on the efficiency of the generator (step 6): 677.867 kW * 0.95 = 643.97365 kW ~ 643.656kW.
### 3-phase rectifier for a wind turbine
In the second example, we remove the main grid but add a RL circuit to be an electrical load for the wind power. As well as the previous example, we use a generator namely AC motor to convert the mechanical power to electrical power. Then convert the alternating current to direct current by using a rectifier, that is, AC/DC converter. We also add power sensors to measure the power loss in each phase as the following configuration.
Continue the second example by the previous example. Left-click and draw the mouse to select all the components except the wind turbine and the power torque sensor (step 1). Click the Delete key to remove the components we selected (step 2).
Right-click the wind turbine (step 1). In the pop-window, modify the parameters of the wind turbine as below (step 2):
• A = 6.6 m˛ (turbine swept area)
• Friction = 1m Nm/Rad/s (friction of the rotor)
• Inertia = 1 Kgm˛ (inertia of the rotor)
• Initial Rotor Speed = 0 Rad/s (initial speed of the turbine)
• Rho = 1.2 (air density)
Click OK to save the setting (step 3).
Left-click Components/Library/ElectricalMachines/PMSM/PMSM (step 1) and release the mouse. The PMSM follows the cursor and left-click on the workscreen to put it in the right side of the power torque sensor (step 2). Right-click the PMSM to modify its parameters. In the pop-window, we can see the parameters as below:
• Friction = 1m Nm/Rad/s (friction of the bearing)
• JR = 100m Kgm˛ (rotor inertia)
• Ke = 1 [Vpeak/(Rad/s) = Nm/A] (torque / back EMF constant)
• LS = 15mH (winding inductance per phase)
• RS = 10m ohm (winding resistance per phase)
• Rm = 1000 (the resistance in parallel with the magnetizing inductance for modeling the eddy current and hysteresis losses in the electrical machine)
• Polepair = 12 (number of pole pairs) (step 3)
• Select ‘Flip Vertical’ (step 4) in the rotation setting
Click OK to save the setting (step 5).
Connect the PMSM and the power torque sensor (step 1). Left-click Components/Library/Sensor/Electrical/PowerAC3Phase (step 2) and release the mouse. The AC power sensor follows the cursor and left-click to put it on the workscreen and connect with the PMSM (step 3).
Now use a rectifier to convert the alternating current. Left-click Components/Library/PowerConverters/Rectifier3Phase/b6 (step 1) and release the mouse. The rectifier follows the cursor and left-click to put it in the right side of the AC power sensor (step 2). Right-click the rectifier to see its parameters as below (step 3):
• BV = 100k Volt (break down voltage per diode)
• Roff = 1e6 ohm (blocking resistance of the diode)
• Ron = 1m ohm (on resistance of the diode)
• Von = 0.6 volt (conduction on state voltage)
Click ‘Help Lib’ to see the explanation of these parameters (step 4). If any parameter is modified, click OK to save the setting (step 5).
Before giving a RL circuit, we place a DC power sensor in the right side of the rectifier. Left-click Components/Library/Sensor/Electrical/PowerDC (step 1) and release the mouse. Left-click on the workscreen to put the DC power sensor in the right side of the rectifier (step 2). Make sure the upper DC input of the DC power sensor is well connected with the DC output of the rectifier. Connect the left ground of the DC power sensor and the ground of the rectifier (step 3).
Left-click Components/Circuit/RLC/L (step 1) and release the mouse. The inductor follows the cursor and right-click to change its direction if needed (step 2). Until the anode of the inductor (indicated by the little dot) is upward, left-click to put it in the right side of the DC power sensor. Right-click the inductor to modify its value (step 3). In the pop-window, enter value = 100m (step 4) and click OK to save the setting (step 5).
Left-click Components/Circuit/RLC/R (step 1) and release the mouse. The resistor follows the cursor and right-click to change its direction if needed (step 2). Until the anode of the resistor (indicated by the little dot) is upward, left-click to put it below the inductor. Right-click on the resistor to modify its value (step 3). In the pop-window, enter value = 200 (step 4) and click OK to save the setting (step 5).
Connect the RL circuit with the DC power sensor with the following configuration (step 1). Currently we have not set an electrical ground. Right-click the cathode of the RL circuit to add a ground label (step 2). In the pop-window, enter label = Ground (step 3) and click OK to save the setting (step 4).
Left-click the power value output of the power torque sensor (step 1) to add a label. In the pop-window, enter label = P_m (step 2) and click OK to save the setting (step 3). Left-click the power value output of the AC power sensor (step 4) to add a label. In the pop-window, enter label = P_ac (step 5) and click OK to save the setting (step 6). Left-click the power value output of the DC power sensor (step 7) to add a label. In the pop-window, enter label = P_dc (step 8) and click OK to save the setting (step 9).
Now take some scopes to record the simulation result. Click the scope icon in the experience bar (step 1) to put a scope below the wind turbine. Left-click the right-bottom corner of the scope and hold down the mouse to enlarge this scope (step 2). Likewise, click the scope icon in the experience bar (step 3) to put a scope below the AC power sensor. Left-click the right-bottom corner of the scope and hold down the mouse to enlarge this scope (step 4).
Connect the T phase of the AC motor with the first blue input trace of the second scope (step 1) so that the voltage value of T phase will be shown in this scope.
In the first scope, left-click the first blue input trace and draw it leftward a bit to extend its length (step 2). Repeat the same steps with the second red and the third azure input traces. Right-click the first input trace (step 3) to add a label. In the pop-window, enter label = P_m (step 4) and click OK to save the setting (step 5). Right-click the second input trace (step 6) to add a label. In the pop-window, enter label = P_ac (step 7) and click OK to save the setting (step 8). Right-click the third input trace (step 9) to add a label. In the pop-window, enter label = P_dc (step 10) and click OK to save the setting (step 11).
Click the scope icon in the experience bar (step 1) to put a scope in the right side of the inductor (step 2). Connect the first input trace of the third scope with the anode of the inductor (step 3). Click the scope icon in the experience bar (step 4) to put a scope in the right side of the resistor (step 5). Right-click its first input trace (step 6). In the pop-window, enter Trace=Current through: R1 (step 7) and click OK to save the setting (step 8).
click the short-cut of simulation parameter (step 1). In the pop-window, select Euler for the Numerical Integration Method (step 2), Tscreen = 10, dt = 100u (step 3) and then click OK to save the setting (step 4).
Click the short-cut of start simulation (step 1). Right-click on the first scope (step 2) to see the details of the simulation result. Click the shortcut of listbox (step 3) to enable the numeric display of simulation result. Click Right or Left key to find the precise values according to the time (step 4). Then we can see clearly the power loss after the AC motor and the rectifier (step 5). | 3,683 | 14,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-05 | longest | en | 0.835321 |
https://www.lse.ac.uk/resources/calendar2017-2018/courseGuides/MA/2017_MA319.htm | 1,680,142,303,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00458.warc.gz | 928,879,984 | 4,565 | ## MA319Half UnitPartial Differential Equations
This information is for the 2017/18 session.
Teacher responsible
Dr Amol Sasane
Availability
This course is available on the BSc in Business Mathematics and Statistics, BSc in Mathematics and Economics, BSc in Mathematics with Economics and BSc in Statistics with Finance. This course is available with permission as an outside option to students on other programmes where regulations permit and to General Course students.
Pre-requisites
Students must have completed Further Mathematical Methods (MA212) and Real Analysis (MA203).
Course content
The aim of the course is the study of partial differential equations. The focus will be on first order quasilinear equations, and second order linear equations. The method of characteristics for solving first order quasilinear equations will be discussed. The three main types of linear second order partial differential equations will be considered: parabolic (diffusion equation), elliptic (Laplace equation), and hyperbolic (wave equation). Techniques for solving these for various initial and boundary value problems on bounded and unbounded domains, using eigenfunction expansions (separation of variables, and elementary Fourier series), and integral transform methods (Fourier and Laplace transforms) will be treated. Elementary distributional calculus and the notion of weak solutions will also be considered. Applications and examples, such as the solution technique for Black-Scholes option pricing, will be discussed throughout the course.
Teaching
22 hours of lectures and 10 hours of classes in the LT.
Formative coursework
Students will be expected to produce 10 problem sets in the LT.
Written answers to set problems will be expected on a weekly basis.
1. S.J. Farlow. Partial Differential Equations for Scientists and Engineers. Dover, 1993.
2. J.D. Logan. Applied Partial Differential Equations. Second Edition. Springer, 2004.
3. W. Strauss. Partial Differential Equations. An Introduction. Second Edition. John Wiley, 2008.
Lecture notes will be provided.
Assessment
Exam (100%, duration: 2 hours) in the main exam period.
Key facts
Department: Mathematics
Total students 2016/17: 13
Average class size 2016/17: 13
Capped 2016/17: No
Lecture capture used 2016/17: Yes (LT)
Value: Half Unit
Guidelines for interpreting course guide information
PDAM skills
• Self-management
• Problem solving
• Communication
• Application of numeracy skills
• Specialist skills | 527 | 2,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-14 | latest | en | 0.884877 |
https://cran.stat.sfu.ca/web/packages/ptools/vignettes/spat-feateng.html | 1,726,181,559,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00078.warc.gz | 168,852,504 | 99,512 | # Spatial Feature Engineering with ptools
#### 2023-02-07
This vignette goes over the different functions the ptools package has to generate spatial features common to criminological stype spatial covariate analysis. These include creating spatial units of analysis (grid cells, hexagons). And calculating different spatial feature engineering, such as the distance to nearest and kernel density estimates for covariate values.
## Creating Spatial Units of Analysis
I will start off with an example loading in the data that is included in the package, and I load the sp library to use its simple plotting functions.
library(ptools)
library(sp) #for plotting functions
# Plot shootings in NYC
plot(nyc_bor)
We can create a set of vector grid cells over the study area. The projection is in feet, so here I make half-square mile grid cells over the NYC borough outline.
hm_grid <- prep_grid(nyc_bor,5280/2)
# Plot shootings in NYC
plot(nyc_bor)
I have provided two arguments to the functions to limit the vector areas returned. One is sometimes we only want areas that have at least one observation for certain analyses (so at least one shooting over the 15+ year period in the current sample).
lim_grid <- prep_grid(nyc_bor,5280/2,point_over=nyc_shoot)
# Plot shootings in NYC
plot(nyc_bor)
So you can see it is not as filled in for Staten Island. Note that there is an additional argument, point_n, if you wanted to limit grid cells even more, so at least 2, 3, etc. points are in a grid cell.
A second way we can limit the grid cells are to not have dongle grid cells – cells that only cover a small portion of the city. Here I only include grid cells that have over 30% of their area inside of the border of the city. Note though that this will cause some areas inside the jurisdiction to not be covered. But you can pretty clearly see the reduced number of areas compared to the original grid cells.
lim_grid2 <- prep_grid(nyc_bor,5280/2,clip_level = 0.3)
# Plot shootings in NYC
plot(nyc_bor)
# See how many fewer than original
print(nrow(hm_grid)) #original
#> [1] 1493
print(nrow(lim_grid2)) #no dongles less than 30%
#> [1] 1290
print(nrow(lim_grid)) #only overlap 1 shooting
#> [1] 874
I also have a similar function, prep_hexgrid, that prepares spatial units of analysis in hexagons instead of square grid cells. One difference is that instead of specifying the size of the cell per length of a side, it specifies it per unit area. So it make it a square mile, use 5280^2 as the area argument. (Here I make it a half square mile.)
hex_grid <- prep_hexgrid(nyc_bor,(5280/2)^2,clip_level = 0.3)
#> Warning: attribute variables are assumed to be spatially constant throughout all
#> geometries
# Plot shootings in NYC
plot(hex_grid,col='lightblue',border='white')
The prep_hexgrid function also has the same point_over and point_n arguments as does the prep_grid function. Here I limit to areas with more than 4 shootings.
hex_lim <- prep_hexgrid(nyc_bor,(5280/2)^2,point_over=nyc_shoot,point_n=4)
# Plot shootings in NYC
plot(hex_lim,col='lightblue',border='white')
Note that the way I wrote these functions, the clip_level for hexagons can be somewhat slow. Doing point_over though should be reasonably fast. prep_grid converts from raster to vector, so if you have very tiny cells, you may consider just keeping the original raster format for analysis. The conversion to vector will always creep up your memory usage and computation time.
Finally, I have a function that if you have a set of origin points, you can create voronoi polygons within an outline area. Sometimes if working with line or address based features it is simpler to convert to polygons and do subsequent geospatial operations. But note this can take a bit of time as well for any really dense point set (same as really tiny grid cells or hexagons).
# Make a smaller sample of the liquor stores in NYC
liq_samp <- nyc_liq[sample(rownames(nyc_liq@data),20),]
# Buffer, it is hard to view all the little islands
nyc_buff <- buff_sp(nyc_bor,5000)
# Now create Voronoi/Thiessen areas
liq_vor <- vor_sp(nyc_buff,liq_samp)
#> Warning in sp::proj4string(outline): CRS object has comment, which is lost in output; in tests, see
#> https://cran.r-project.org/web/packages/sp/vignettes/CRS_warnings.html
# Plot to show off
plot(liq_vor,col='tan',border='white')
## Spatial Feature Engineering
The spatial feature engineering functions are not only for creating vector spatial units, but assigning typical values we may be interested in. So lets reuse our hex_grid over the city, and calculate the number of shootings in each hexgrid:
hex_grid$shoot_cnt <- count_xy(hex_grid,nyc_shoot) spplot(hex_grid,zcol='shoot_cnt') Remember since hex_grid eliminated some dongles, we may be interested to see if we happened to lose some shootings on the edges of NYC. print(sum(hex_grid$shoot_cnt))
#> [1] 27243
print(nrow(nyc_shoot))
#> [1] 27312
So we ended up losing 37 shootings using this grid. We can also use weights in these functions. So here, say I wanted to do a pre/post analysis. A simple way to do that is to weight the observations in the original point pattern, and then use that field as a weight.
covid_date <- as.Date('03/22/2020', format="%m/%d/%Y")
nyc_shoot$pre <- 1*(nyc_shoot$OCCUR_DATE < covid_date)
nyc_shoot$post <- 1*(nyc_shoot$OCCUR_DATE >= covid_date)
hex_grid$shoot_pre <- count_xy(hex_grid,nyc_shoot,weight='pre') hex_grid$shoot_post <- count_xy(hex_grid,nyc_shoot,weight='post')
plot(hex_grid$shoot_pre,hex_grid$shoot_post)
So you can see that the pre and post shooting counts are highly correlated and a few maybe outliers.
Sometimes we want other feature engineering RTM or Egohood style, with density of nearby features. So we can calculate the density (assuming a Gaussian kernel), for liquor stores at the centroid of our polygon features. Since the area of our hexgrid is half a square mile, (5280/2)^2, this means that the vertex to vertex diameter of our hexagon cells is hex_dim((5280/2)^2), not quite 3300 feet (see also the hex_area and hex_wd functions to convert between different hexagon sizes). So there is not much point of doing a KDE bandwidth for under this (will basically be the same as just counting up the total inside of the hexagon. So here I do a bandwidth of 1 mile.
hex_grid$kliq_1mile <- kern_xy(hex_grid,nyc_liq,5280) spplot(hex_grid,zcol='kliq_1mile') I also have similar functions to kernel density for bisquare weights, bisq_xy, and for inverse distance weighting, idw_xy. And for these you can pass in weights the same as for the counts I showed above. I don’t have any demographic data here, but for egohoods you may have census data at centroids, such as the total number households in poverty, and use that as a weight (or a mean estimate). One function that is slightly different than others is the distance between two point sets. So here is the distance to the nearest NYC outdoor cafe (note that this data source has no outdoor cafes in Staten Island): hex_grid$dist_cafe <- dist_xy(hex_grid,nyc_cafe)
spplot(hex_grid,zcol='dist_cafe')
Unlike the other functions, you can pass in two point sets here, under the hood it just grabs the X/Y coordinates for each set. So here is the closest shooting to a cafe.
dist_shoot <- dist_xy(nyc_cafe,nyc_shoot,fxy=c('X_COORD_CD','Y_COORD_CD'))
summary(dist_shoot)
#> Min. 1st Qu. Median Mean 3rd Qu. Max.
#> 33.23 267.98 494.16 596.37 835.83 5765.32
Not shown, but if the base polygons/features do not have X/Y coordinates. You can do something like coords <- coordinates(base); base$x <- coords[,1] and ditto the second column for y (assuming base is a sp class object). This function does not have a weight function, as that does not make sense. The final feature engineering point to note is the dcount_xy function. Unlike the other functions that operate on the centroids of the polygons (or whatever X/Y coordinates you pass in), this counts the number of points within a specified distance to the polygon border. This buffers the feature dataset and then calculates the overlap, so with very large features can take a bit (whereas kernel density calc timing is mostly a function of how many base polygons you have): hex_grid$dcnt_cafe1mile <- dcount_xy(hex_grid,nyc_cafe,5280)
plot(hex_grid$dist_cafe,hex_grid$dcnt_cafe1mile,xlim=c(0,10000))
abline(v=5280) # not perfect, because of difference between polygon distance vs centroid distance
Again you can use a weight function in the dcount_xy function. So if you wanted to say count the number of children attending school in 1 kilometer of your base polygons, this would be the function you use. | 2,161 | 8,682 | {"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} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.904403 |
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Topic: Hex Win Proof?
Replies: 41 Last Post: Mar 24, 2004 6:39 PM
Messages: [ Previous | Next ]
Brian Chandler Posts: 1,899 Registered: 12/6/04
Re: Hex Win Proof?
Posted: Mar 19, 2004 2:09 AM
w.taylor@math.canterbury.ac.nz (Bill Taylor) wrote in message news:<716e06f5.0403181938.72a82f90@posting.google.com>...
> It is an old theorem that in Hex, once the board has been completely
> filled in with two colours, there *must* be a winning path for one
> or other of them.
>
> Now, I can prove this easily enough mathematically, but I'm wondering if
> there is a simple proof, or proof outline, that would be understandable
> and reasonably convincing to the intelligent layman.
>
> Can anyone help out please?
Imagine the board tipped up, so player A goes from top to bottom,
player B from left to right. Suppose the board is actually two sheets
of clear plastic, and two players have "stones" of white sugar (A) and
red granite (B), and moreover these stones "plug" the hexagon cells
(in the obvious way).
Now pour water in at the top. Either it dissolves through the sugar
and reaches the bottom, in which case A has a path, or it doesn't, in
which case there must be a continuous boundary of impervious granite,
and B has a path.
(Does this help? I think it does (slightly) use some facts about the
behaviour of liquids to help the imagination along.)
Brian Chandler
http://imaginatorium.org
Date Subject Author
3/18/04 Bill Taylor
3/18/04 Tim Brauch
3/19/04 Brian Chandler
3/19/04 Jonathan Welton
3/19/04 Tim Brauch
3/19/04 Richard Henry
3/20/04 Chan-Ho Suh
3/21/04 Arthur J. O'Dwyer
3/19/04 Bob Harris
3/19/04 Tim Smith
3/19/04 Dvd Avins
3/20/04 Nate Smith
3/20/04 Chan-Ho Suh
3/20/04 G. A. Edgar
3/19/04 Richard Henry
3/19/04 Steven Meyers
3/20/04 Nate Smith
3/20/04 Larry Hammick
3/20/04 Tim Smith
3/21/04 Steven Meyers
3/22/04 Torben Mogensen
3/22/04 Chan-Ho Suh
3/22/04 Torben Mogensen
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3/23/04 Torben Mogensen
3/23/04 Robin Chapman
3/23/04 Chan-Ho Suh
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3/24/04 Tim Smith
3/24/04 Robin Chapman
3/24/04 Tim Smith
3/24/04 Jon Haugsand
3/22/04 Andrzej Kolowski
3/23/04 Alexander Malkis
3/23/04 Chan-Ho Suh
3/23/04 Dr. Eric Wingler
3/24/04 Danny Purvis
3/24/04 Danny Purvis | 777 | 2,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.902442 |
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Save your time and search cool lifehacks here!
## What is the difference of architectural and engineering scale?
Although similar in appearance to an architect’s scale, the engineering scale is designed to be more precise and has a decimal scaling scheme whereas an architect’s scale uses fractional scaling. It is designed to only be read from left to right. An architect’s scale can be read from either the left or right side.
What is an engineer’s ruler used for?
An engineering ruler is a straight edge designed to measure objects on a construction plan to scale. The engineering ruler has six different scales printed on its prongs; each scale represents a different conversion factor.
What is an engineering ruler called?
scale ruler
A scale ruler is a tool for measuring lengths and transferring measurements at a fixed ratio of length; two common examples are an architect’s scale and engineer’s scale. In scientific and engineering terminology, a device to measure linear distance and create proportional linear measurements is called a scale.
### What is an architect ruler used for?
An Architect’s or scale ruler is designed for use in determining the actual dimensions of a distance on a scaled drawing. Most architectural, construction and engineering drawings and blueprints are scaled to allow for large areas, structures or items to conveniently fit on a reasonable size of paper.
Why is scale important in design?
Scale defines the size of an item in relationship to something else. An important guideline in creating any deliverable, whether it’s a building, a space, or an image – scale is often the secret ingredient that destroys or refines a project.
How do you read architectural rulers?
How to Read an Architecture Ruler
1. Lay out the blueprint or drawing that you want to read with the architectural ruler.
2. Locate the scale on the drawing or blueprint.
3. Locate the corresponding scale on the architectural ruler, i.e. 1/4″ = 1′ (1/4 inch = 1 foot).
#### What is a 3 sided ruler called?
A scale ruler is the three-sided ruler used by architects and readers of blueprints to convert between scaled drawings and the actual dimensions without having to resort to any mathematical calculations. An architect uses the scale ruler to convert dimensions into a smaller drawing of a building plan.
What is an engineer’s tape measure?
The design of an engineer’s tape measure allows you to measure a great distance, and it can be easily carried in your pocket or toolkit. It also allows you to measure around curves and corners, which can be an advantage for many contractors and builders.
What is the best type of ruler?
The Best Ruler – 2021
1. The Best Ruler.
2. Westcott Stainless Steel Cork Base Ruler, 12-Inch.
3. Arteza Triangular Architect Scale Ruler, 12-Inch.
4. BAZIC Jeweltones Color Ruler, 12-Inch, 4-Pack.
5. Fiskars 01-005358 Wooden Ruler, 12-Inch.
6. Soraco Triangular Architectural Ruler, 12-Inch.
7. JAM PAPER Stainless Steel Ruler, 12-Inch.
## What is scale architecture?
The term scale is also used to describe the relationship between a depiction of a building, object, area of land etc compared to its actual size. Drawing accurately to scale, and being able to shift between scales, is a fundamental skill of architectural drawing and spatial design.
What is scale and why is it important?
Scale is important simply because the magnitude of the problems faced in areas such as poverty reduction, the environment, gender issues and healthcare require solutions at scale. By their nature they are often cross-border or not focused solely on one location.
When to use an architectural or scale ruler?
An Architect’s or scale ruler is designed for use in determining the actual dimensions of a distance on a scaled drawing. Most architectural, construction and engineering drawings and blueprints are scaled to allow for large areas, structures or items to conveniently fit on a reasonable size of paper.
### Which is the best architect ruler to buy?
The Arteza Triangular Architect Scale coming in 12” of size, is a color-coded and of great quality. It comes up first if you search “Architect ruler” on Amazon’s search engine. At least that’s what we found! It has a nice look to it, not on the heavy side yet still manages to be sturdy.
When to use an architect’s or engineer’s scale?
An architect’s scale is typically used for smaller or residential projects, when a plan needs to show things in a greater amount of detail, while an engineer’s scale is very useful for larger public parks, infrastructure projects, or general land planning purposes.
What does 1 / 8 on the ruler mean?
For example, 1/8 on the ruler is in fact a scale that converts 1/8 inch on the drawing to 1 foot. This would represent a drawing with a scale of 1/8″ = 1 foot. Be careful when selecting the scale on the ruler, there are two scales on each edge. | 1,041 | 4,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-50 | longest | en | 0.926416 |
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# Reducing the Vertex Cover Number via Edge Contractions
## File
LIPIcs.MFCS.2022.69.pdf
• Filesize: 0.87 MB
• 14 pages
## Acknowledgements
The last author would like to thank Roohani Sharma for insightful discussions.
## Cite As
Paloma T. Lima, Vinicius F. dos Santos, Ignasi Sau, Uéverton S. Souza, and Prafullkumar Tale. Reducing the Vertex Cover Number via Edge Contractions. In 47th International Symposium on Mathematical Foundations of Computer Science (MFCS 2022). Leibniz International Proceedings in Informatics (LIPIcs), Volume 241, pp. 69:1-69:14, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2022)
https://doi.org/10.4230/LIPIcs.MFCS.2022.69
## Abstract
The Contraction(vc) problem takes as input a graph G on n vertices and two integers k and d, and asks whether one can contract at most k edges to reduce the size of a minimum vertex cover of G by at least d. Recently, Lima et al. [MFCS 2020, JCSS 2021] proved, among other results, that unlike most of the so-called blocker problems, Contraction(vc) admits an XP algorithm running in time f(d) ⋅ n^O(d). They left open the question of whether this problem is FPT under this parameterization. In this article, we continue this line of research and prove the following results: - Contraction(vc) is W[1]-hard parameterized by k + d. Moreover, unless the ETH fails, the problem does not admit an algorithm running in time f(k + d) ⋅ n^o(k + d) for any function f. In particular, this answers the open question stated in Lima et al. [MFCS 2020] in the negative. - It is NP-hard to decide whether an instance (G, k, d) of {Contraction(vc)} is a Yes-instance even when k = d, hence enhancing our understanding of the classical complexity of the problem. - Contraction(vc) can be solved in time 2^O(d) ⋅ n^{k - d + O(1)}. This XP algorithm improves the one of Lima et al. [MFCS 2020], which uses Courcelle’s theorem as a subroutine and hence, the f(d)-factor in the running time is non-explicit and probably very large. On the other hand, this shows that when k = d, the problem is FPT parameterized by d (or by k).
## Subject Classification
##### ACM Subject Classification
• Theory of computation → Parameterized complexity and exact algorithms
##### Keywords
• Blocker problems
• edge contraction
• vertex cover
• parameterized complexity
## Metrics
• Access Statistics
• Total Accesses (updated on a weekly basis)
0
## References
1. Hans L. Bodlaender, Pinar Heggernes, and Daniel Lokshtanov. Graph Modification Problems (Dagstuhl Seminar 14071). Dagstuhl Reports, 4(2):38-59, 2014. URL: https://doi.org/10.4230/DagRep.4.2.38.
2. Bruno Courcelle. The Monadic Second-Order Logic of Graphs. I. Recognizable Sets of Finite Graphs. Information and Computation, 85(1):12-75, 1990. URL: https://doi.org/10.1016/0890-5401(90)90043-H.
3. Christophe Crespelle, Pål Grønås Drange, Fedor V. Fomin, and Petr A. Golovach. A survey of parameterized algorithms and the complexity of edge modification, 2020. URL: http://arxiv.org/abs/2001.06867.
4. Marek Cygan, Fedor V. Fomin, Lukasz Kowalik, Daniel Lokshtanov, Dániel Marx, Marcin Pilipczuk, Michal Pilipczuk, and Saket Saurabh. Parameterized Algorithms. Springer, 2015. URL: https://doi.org/10.1007/978-3-319-21275-3.
5. Reinhard Diestel. Graph Theory, 4th Edition, volume 173 of Graduate texts in mathematics. Springer, 2012. URL: https://dblp.org/rec/books/daglib/0030488.bib.
6. Öznur Yaşar Diner, Daniël Paulusma, Christophe Picouleau, and Bernard Ries. Contraction and deletion blockers for perfect graphs and H-free graphs. Theoretical Computer Science, 746:49-72, 2018. URL: https://doi.org/10.1016/j.tcs.2018.06.023.
7. Esther Galby, Paloma T. Lima, Felix Mann, and Bernard Ries. Using edge contractions to reduce the semitotal domination number, 2021. URL: http://arxiv.org/abs/2107.03755.
8. Esther Galby, Paloma T. Lima, and Bernard Ries. Reducing the domination number of graphs via edge contractions and vertex deletions. Discrete Mathematics, 344(1):112169, 2021. URL: https://doi.org/10.1016/j.disc.2020.112169.
9. Esther Galby, Felix Mann, and Bernard Ries. Blocking total dominating sets via edge contractions. Theoretical Computer Science, 877:18-35, 2021. URL: https://doi.org/10.1016/j.tcs.2021.03.028.
10. Esther Galby, Felix Mann, and Bernard Ries. Reducing the domination number of (P₃+kP₂)-free graphs via one edge contraction. Discrete Applied Mathematics, 305:205-210, 2021. URL: https://doi.org/10.1016/j.dam.2021.09.009.
11. Subhash Khot and Venkatesh Raman. Parameterized complexity of finding subgraphs with hereditary properties. Theoretical Computer Science, 289(2):997-1008, 2002. URL: https://doi.org/10.1016/S0304-3975(01)00414-5.
12. Paloma T. Lima, Vinícius Fernandes dos Santos, Ignasi Sau, and Uéverton S. Souza. Reducing graph transversals via edge contractions. Journal of Computer and System Sciences, 120:62-74, 2021. URL: https://doi.org/10.1016/j.jcss.2021.03.003.
13. Daniël Paulusma, Christophe Picouleau, and Bernard Ries. Critical vertices and edges in H-free graphs. Discrete Applied Mathematics, 257:361-367, 2019. URL: https://doi.org/10.1016/j.dam.2018.08.016.
14. Saket Saurabh, Uéverton dos Santos Souza, and Prafullkumar Tale. On the parameterized complexity of grid contraction. In Proc. of the 17th Scandinavian Symposium and Workshops on Algorithm Theory (SWAT), volume 162 of LIPIcs, pages 34:1-34:17, 2020. URL: https://doi.org/10.4230/LIPIcs.SWAT.2020.34.
15. Toshimasa Watanabe, Tadashi Ae, and Akira Nakamura. On the NP-hardness of edge-deletion and -contraction problems. Discrete Applied Mathematics, 6(1):63-78, 1983. URL: https://doi.org/10.1016/0166-218X(83)90101-4.
16. Mihalis Yannakakis. Node- and Edge-Deletion NP-Complete Problems. In Proc. of the 10th Annual ACM Symposium on Theory of Computing (STOC), pages 253-264, 1978. URL: https://doi.org/10.1145/800133.804355.
X
Feedback for Dagstuhl Publishing | 1,800 | 5,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.836612 |
http://www.shuati123.com/blog/2015/02/07/reservoir-sampling/ | 1,529,350,377,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860776.63/warc/CC-MAIN-20180618183714-20180618203714-00199.warc.gz | 500,557,818 | 3,743 | # [Question] Reservoir Sampling
Reservoir sampling is a family of randomized algorithms for randomly choosing k samples from a list of n items, where n is either a very large number. Typically n is large enough that the list doesn’t fit into main memory. For example, a list of search queries in Google and Facebook.
### Question
given a big array (or stream) of numbers (to simplify), and we need to write an efficient function to randomly select k numbers where 1 <= k <= n. Let the input array be stream[].
### Solution
O(n) time!
1. Create an array sample[0..k-1] and copy first k items of stream[] to it.
2. Now one by one consider all items from (k+1)th item to nth item.
1. Generate a random number ‘j’ from 0 to i where i is index of current item in stream[].
2. If j is in range 0 to k-1, replace sample[j] with stream[i]
not written | 217 | 854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-26 | latest | en | 0.878552 |
https://www.jiskha.com/display.cgi?id=1310927630 | 1,502,965,353,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103167.97/warc/CC-MAIN-20170817092444-20170817112444-00285.warc.gz | 905,332,407 | 3,577 | # math 209
posted by .
The perimeter of a rectangle is 44ft the length is 4ft longer than the width. Find the demensions. Write a system of linear equations and solve the resulting system. Let x be the length and y be the width.
• math 209 -
perimeter=2L+2W
L-4=W
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More Similar Questions | 675 | 2,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-34 | latest | en | 0.827936 |
https://www.jiskha.com/archives/2017/07/17 | 1,601,542,427,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402124756.81/warc/CC-MAIN-20201001062039-20201001092039-00476.warc.gz | 823,645,095 | 7,368 | # Questions Asked onJuly 17, 2017
1. ## decreased by twice a number
50 decreased by twice a number
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The table below represents a linear function f(x) and the equation represents a function g(x): x f(x) −1 −5 0 −1 1 3 g(x) g(x) = 2x − 7 Part A: Write a sentence to compare the slope of the two functions and show the steps you used to determine the
3. ## math
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If the instantaneous rate of change of f(x) at (2,-4) is 5, write the equation of the line tangent to the graph of f(x) at x=2.
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Luke is tiling a square patio in his backyard. He has 121 square-shaped tiles that will cover the entire patio. The maximum number of tiles that Luke can fit along one side of the patio is
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Insert parentheses to make the equality a true statement:12+3*8−8÷4=7
9. ## Teaching
21) T/F. Reflection is an important element for children in an early childhood portfolio. 22) T/F. Rotating and changing areas and materials in the room creates insecurity in the children. 23) T/F. Rubrics are not an appropriate use of self-assessment in
10. ## Calculus
How do you find the original function given the equation of a tangent line and two points?
11. ## maths
An arithmetic progression has 10 terms. Sum of the 10 terms is 220. Sum of the odd terms is 100. Find the first term and common difference. pl give me the answer.
12. ## math
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13. ## Teaching
5) T/F. Assessment tests should be used to determine grade or level placement for children. 6) T/F. At least daily children should have a block of time for spontaneous, free play. 7) Authentic assessment I. is often referred to as performance assessment
14. ## Algebra
Solve for x in the following matrix equation. [5x-8 1] = [2 1] [3 4m-1] [3 7m+6] How do you do this? I hardly know anything about this... The packet I got that was supposed to help me understand matrices and stuff just made me more confused.
15. ## SCHM012
diagram all possible pathways for deexciting from n=4 to n=1
16. ## Algebra
The table below represents the distance of an airplane from its destination as a function of time: Time (hours) x Distance (miles) y 0 2,700 1 2,160 2 1,620 3 1,080 4 540 Part A: What is the y-intercept of the function, and what does this tell you about
17. ## algebra
a rocket is launched at the top of a building. the height of the rocket (in meters) written in terms of time (seconds) can be modeled by h(t)=-4.9t^2+9.8t+73.5. what is the maximum height?
18. ## math
A square 4 inches on a side is cut up into smaller squares 1 inche on a side. What is the maximum number of such squares that can be formed?
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20. ## math
0.7 of which is 3.43
21. ## math
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22. ## Teaching
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23. ## Teaching
11) T/F. Friedrich Froebel originated the first kindergarten in Germany. 12) T/F. Group-administered standardized tests are recommended once children are in 1st grade. 13) T/F. How a child feels about himself/herself is reflected in his/her behavior. 14)
24. ## Algebra 2
What is the solution for the determinant? |1 7 | |0 -3| How do you find the solution? I'm probably going to feel really stupid because it's probably simple but I'm just having some problems. Okay, a lot.
25. ## PELAGIE
what is the time necessary for 99.9%of the atoms of a radioactive compound to be disintegrated? t½=5times.
26. ## PELAGIE
how many electrons are found in the electron cloud of an atom whose K,L and M shells are full?
27. ## Chemistry
How are the three types of intermomecular forces similar? How are they different?
28. ## Math
A certain grometry class has 36 students. if two-thirds are boys and three-fourths of the boys are under six feet tall, how many boys in the class are under 6 feet tall?
29. ## Physics
How long does a car with an acceleration of 2.0m/s2 take to go from 10m/s?
30. ## Math
Write as a fraction: 4.5%
31. ## math
Write 2/5% as a decimal.
32. ## Math
What is 62 1/2% of 80?
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34. ## Math
80 is 40% of what number?
35. ## Math
c is 83 1/3% of what number?
36. ## math
What is 112% of 80?
37. ## math
what percent of 60 is 72?
38. ## math
If a trapezium was drawn such that it has got two right angled triangles,why the area of that trapezium will differ if someone is considering the areas of two right angled triangles lastly he/she add to obtain the total areas from the one who will use the
39. ## Calculus
How do you find the original function given a point (a,b) and the equation of the line tangent to the graph of f(x) at (a,b). For example: The point is (4,-11) and 7x-3y=61 is the equation of the line tangent to the graph of f(x)
40. ## One-to-one function
Explanation of why functions must be one-to-one functions to have an inverse?
41. ## math
If a trapezium was drawn such that it has got two right angled triangles,why the area of that trapezium will differ if someone is considering the areas of two right angled triangles lastly he/she add to obtain the total areas from the one who will use the
42. ## algebra
Miranda worked two part-time jobs last week. The first job paid \$8 an hour, and the second job paid \$11 an hour. She worked a total of 23 hours last week and earned \$208. How many hours did Miranda work at the first job?
43. ## Math
Determine which is greater 99^100 or 100^99 ? Explain
44. ## KSU
Suppose you are to prepare 1 liter of 70% ethyl alcohol.What is the ratio needed to prepare the solution?
45. ## Chemistry
When diluting the commercial vinegar, what would have been the effect on acetic acid molairty of failing to rinse the pooped with commercial vinegar, assuming there was some distilled water left in the piper from a previous student? ( think about how the
46. ## math
A plane is flying due east at 600 km/h at a constant altitude. From an observation point P on the ground, the plane is sighted on a bearing of 320◦. One minute later, the bearing of the plane is 75◦ and its angle of elevation is 25◦. Show that the
47. ## English
Are the below given sentences written correctly? 1. He's slow as a tortoise. 2. It's ugly as monkey. 3. She's as weak as a kitty. 4. She's fast as a horse. Kindly send me the correctly written sentences .
48. ## Maths
In the country Fivethree, they have only \$3 and \$5 notes. How many amounts, from \$1 to \$100 inclusive, cannot be made up exactly with these notes? A. 2 B. 3 C. 4 D. 7 E. 21 | 2,061 | 7,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-40 | latest | en | 0.898547 |
https://www.homeimproverx.com/what-is-rechargeable-battery-voltage/ | 1,674,923,083,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499646.23/warc/CC-MAIN-20230128153513-20230128183513-00799.warc.gz | 813,943,707 | 22,767 | # What Is Rechargeable Battery Voltage?
Standard size single-use batteries have a nominal voltages of 1.5 and 1.2 volts, whereas rechargeable batteries have a nominal voltages of 1.5 and 1.2.
## What is the difference between 1.2 V and 1.5 V rechargeable batteries?
The main difference between the two is that the alkaline battery starts at 1.5 volts and gradually drops to less than 1.0 volts. NiMH batteries have a discharge cycle that lasts for most of the time. The lower the capacity of the battery, the lower the voltage.
## Are all rechargeable batteries 1.2 V?
A few chemistries that are popular have a nominal voltages of 1.2V or less. The cell voltages of the batteries are 1.2V.
## What is the voltage of a rechargeable AA battery?
The batteries come with a built-in battery management system that makes sure the required 1.5 volts is reached.
## Why there are no 1.5 V rechargeable batteries?
The NOMINAL voltage is used to rate the cells. They will either have a higher or lower voltages if they are fully charged. There are cells in a battery. The answer is based on chemistry.
## How do I choose a good rechargeable battery?
If you want to know the amount of energy a battery will hold, you should look for themAh. The longer a device is powered up, the easier it will be to charge it again. The battery is being used for a device.
## Do 1.5 V rechargeable batteries exist?
Is there a standard size for rechargeable batteries like AA,AAA, C or D? They’re finally available. There are a few Chinese companies that make them. The AA andAAA batteries have 1.5 volts of power.
## Can I use 600mAh instead of 300mAh?
It’s possible to replace your 300mAh battery with a 600mAh battery if it’s the same voltage. It’s important to make sure the replacement battery is the right size for your device.
## How do you know if a battery is rechargeable?
The charging capacity of a rechargeable battery is shown on the screen. You can see the number on the packaging and on the battery.
## At what voltage is a 1.5 volt battery bad?
The AA andAAA batteries have 1.5 and 1.25 volts, respectively. The 1.5 V battery is dead and you might be wondering what the voltage is. If the battery tests less than 1.3 volts in the multimeter, it’s time to replace it.
## Are all batteries 1.5 V?
The amount of electric current that can be produced by AA, C, D batteries is not the same as the amount of electric current that can be produced byAAA, AA, C, D batteries.
## Is it OK to leave rechargeable batteries in the charger?
After charging is complete, be sure to remove the battery from the charging point. Overcharging can shorten the life of a battery.
## What is the difference between regular batteries and rechargeable batteries?
A simple battery has many cells attached to it. A cell is made up of three parts, two of which are located in the same place. The chemical reaction can be reversed with the help of a rechargeable battery.
## Are all rechargeable batteries the same?
The main batteries people use are the AA, C, D, Sc and 9V which are the same sizes as the Rechargeable batteries.
## Are all AA batteries 1.2 volts?
There are three chemistries of the AA size battery. There are different nominal voltages for each of them. NiCad batteries have a nominal voltages of 1.2 and 1.2, while NiMH batteries have a nominal voltages of 1.2 and 1.2.
## Are all NiMH batteries 1.2 V?
The NiMH batteries are rated to run at 1.2 V, which is the same as every other battery. They have a capacity of 1,100 mAh, which is similar to most of theAAAs we tested. We averaged the results after we charged and discharged each set ofAAAs three times.
## What is the difference in rechargeable batteries?
A simple battery has many cells attached to it. A cell is made up of three parts, two of which are located in the same place. The chemical reaction can be reversed with the help of a rechargeable battery. | 944 | 3,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-06 | latest | en | 0.937873 |
https://www.chicagotribune.com/business/ct-xpm-2012-02-07-ct-biz-0207-problem-connolly-20120207-story.html | 1,628,160,045,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155529.97/warc/CC-MAIN-20210805095314-20210805125314-00136.warc.gz | 701,027,298 | 27,660 | # Problem Solver: Swim in hotel pool proves costly
For Michelle Connolly, it seemed an ingenious plan. Her three children were off for Martin Luther King Day, and she was looking for a fun way to spend the day.
Using Marriott rewards points, the Park Ridge mom booked a room at The Courtyard by Marriott in Glenview for Jan. 15. She had no intention of actually staying overnight at the hotel — she just wanted access to the pool.
The idea was a hit. The kids and two of their friends swam for about an hour on Jan. 15, went home to sleep, then returned the next morning for a hotel breakfast and more splashing around.
Everything seemed fine, until she went to check out that afternoon.
The hotel clerk asked her if she should put the \$122.25 room charge on her husband's credit card.
Connolly said no, the room was already paid for with points. When the clerk disagreed, Connolly went to a computer in the hotel's lobby and printed out proof the 10,000 points had been deducted from her account.
As she left, Connolly thought the issue was resolved, but when she later checked her husband's credit card account online, she saw the \$122.25 had, in fact, been charged.
During the next several weeks, Connolly called the hotel repeatedly and received different explanations about the charge, she said. After much back and forth with the hotel staff, Connolly was finally told she made two reservations: one with her points, the other which was to be charged to the credit card, she said.
Connolly was flabbergasted.
"I did not make two reservations," she said. "I made just one reservation, online."
Unable to persuade the Courtyard by Marriott to refund her money, Connolly gave an ultimatum: Give back the \$122.25 or she would email What's Your Problem?
With the charge still on her husband's credit card Jan. 29, she followed through on her threat.
"They just have no intention of refunding it," Connolly told the Problem Solver. "I don't think \$122.25 is a small chunk of change."
The Problem Solver contacted Laurie Goldstein, a spokeswoman for Marriott, and forwarded her copies of Connolly's credit card bills, which clearly showed the \$122.25 was charged and not returned. The Problem Solver also forwarded proof that Connolly used rewards points to book the room.
Initially, the hotel's manager told Marriott's headquarters that there was some confusion about the name on the reservation, with the system showing two reservations under slightly different spellings. The manager maintained that the extra charge had, in fact, been refunded, Goldstein said.
After verifying that the money was not returned to the credit card, the manager decided to take out a money order in the amount of \$122.25 and send it to Connolly via FedEx on Friday afternoon.
"There were some issues with the computer system that wouldn't allow them to credit it directly through the credit card," Goldstein said. "A money order was the quickest way to get the money back to her."
Connolly said that as of midday Monday the money had not arrived.
The Problem Solver will provide an update as warranted. | 662 | 3,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-31 | latest | en | 0.988521 |
https://www.proprofs.com/quiz-school/story.php?title=maps-mapmaking | 1,582,590,320,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145989.45/warc/CC-MAIN-20200224224431-20200225014431-00305.warc.gz | 853,798,958 | 25,494 | # Maps + Mapmaking
24 Questions
Settings
GIS/Cartography intro quiz.
Related Topics
• 1.
How do you make qualitative differences visible (symbols)?
• A.
Distance, orientation, shape
• B.
Color, Shape, Area
• C.
Shape, Orientation, Color
• 2.
How many UTM zones does Texas have?
• A.
3
• B.
4
• C.
5
• D.
6
• 3.
Texas Statewide Mapping System (TSMS): What's the projection?
• A.
Mercator Cylindrical
• B.
Lambert Conformal Conic
• C.
Oblique Mercator
• 4.
TSMS: Which datum?
• A.
• B.
• 5.
Which three conformal projections does state plane coordinate system have? Pick 3
• A.
Lambert Conformal Conic
• B.
Transverse Mercator
• C.
Equatorial Planar
• D.
Oblique Mercator
• 6.
What are the three important projections?
• A.
Lambert Conformal Conic
• B.
Equatorial Planar
• C.
Universal Transverse Mercator
• D.
Oblique Mercator
• E.
Albers Equal Area Conic
• 7.
What are the four aspects of map projections?
• A.
Polar, Equatorial, Transverse, Oblique
• B.
Mercator, Polar, Transverse, Lambert
• C.
Oblique, Polar, Equatorial, Lambert
• D.
Transverse, Equatorial, Oblique, Mercator
• 8.
What is Tangency? and 2 types.
• A.
Tangent case and secant case
• B.
The closeness to the surface of an oboject
• C.
The location or locations that a projection surface touches or cuts through the globe. ie: standard parallel.
• 9.
What are the types of map distortions?
• A.
Shape, scale, area, distance
• B.
Scale, length, area, shape
• C.
Shape, area, distance, direction
• D.
Direction, scale, shape, and area
• 10.
What is the best way to define earth?
• A.
An oblate ellipsoid
• B.
A spheroid
• C.
A geoid
• D.
An oblate geoid
• 11.
What are the four most important characteristics of cartographic maps?
• A.
Scale
• B.
Vertical or oblique views
• C.
Symbolized representations
• D.
Map projection
• E.
Coordinate system
• 12.
Which is not an important characteristic of mental maps?
• A.
Not to scale
• B.
Subjective and inconsistent
• C.
They are false
• D.
Might not include cardinal directions
• E.
No geometrical reference framework
• 13.
What is not a function of reference maps?
• A.
Communication model
• B.
Natural and cultural features
• C.
Nautical and aeronautical charts
• D.
The USGS (United States Geological Survey
• E.
Location for distances/directions
• 14.
What do thematic maps include? 1 answer
• A.
Thematic content and spatial location
• B.
Usgs
• C.
Elevation
• D.
Spatial attributes
• 15.
What are the most important characteristics of maps?
• A.
Area, distance, scale, size
• B.
Area, distance, direction, size
• C.
Symbology and scale
• D.
Vertical/oblique views, scale, map projection, and symbols
• 16.
What is the equation for converting scale?
• 17.
What is the equation for converting to decimal degrees?
• 18.
What does URF and KRF mean? All lower case, type exactly like: ________ and _________
• 19.
How to you figure out if a map is good? What equation do you use?
• 20.
What does zi and zt stand for?
• A.
Zi= interpolated elevation (predicted zt= true elevation
• B.
Ziti
• C.
Zitiiiii!
• 21.
What is a datum?
• A.
A mathematical model of the earth
• B.
The mathematical model related to real world features
• C.
Real world features projected with minimum distortion from a round earth to flat map
• D.
A system of coordinates
• 22.
What is an ellipsoid
• A.
A mathematical model of earth
• B.
A system of coordinates
• C.
Earth as a sphere
• 23.
Who uses the WGS84?
• A.
Department of geography
• B.
Department of education
• C.
Department of defense and geospatial intelligence agency
• 24.
Where is the zero located for the NAD27 | 1,038 | 3,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-10 | latest | en | 0.728136 |
https://sport-topics.com/how-many-yards-is-a-nba-basketball-court | 1,643,195,321,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00300.warc.gz | 585,537,274 | 14,036 | # How many yards is a nba basketball court?
11
Asked By: Marian Reichert
Date created: Sun, May 2, 2021 10:08 AM
Date updated: Wed, Jan 26, 2022 10:59 AM
Content
Video answer: How many yards long is a basketball court?
## Top best answers to the question «How many yards is a nba basketball court»
A basketball court at the collegiate and professional levels measures 94 feet in length and 50 feet in width, or approximately 31.3 yards long and 16.6 yards wide. High school courts are slightly smaller in length, measuring 84 feet, or 28 yards.
FAQ
Those who are looking for an answer to the question «How many yards is a nba basketball court?» often ask the following questions:
### âť“ How many yards in basketball court?
#### Basketball court dimensions in yards
A basketball court is 30.62 yards long, and 16.4 yards wide.
### âť“ How many yards is basketball court?
#### 30.62 yards
A basketball court is 30.62 yards long, and 16.4 yards wide.
### âť“ How many yards is a basketball court!?
In calculating the dimensions of an NCAA and NBA basketball court in yards, convert 94 feet and 50 feet by multiplying each to â…“. Hence, college and professional basketball courts measure 31.3 yards by 16.67 yards.
Court Size Overall: The Professional NBA and (WNBA) Court is 94 feet long and 50 feet wide. Note: WNBA Basketball Court Dimensions are the same in every respect as the NBA. The Foul Line: For all courts including the NBA, the “foul line” distance is 15 feet from the foul line to the front of the backboard. The “foul line” is 18 feet 10 ...
An NBA basketball court is 94 feet long and 50 feet wide. In meters, it’s 28.65 m long and 15.24 m wide. The WNBA and NCAA use an identical court size—94 feet long by 50 feet wide. In meters ...
Regulation basketball court dimensions are 94 feet long by 50 feet wide. Basketball court size varies depending on the league and level of play. For NBA court dimensions, as well as for WNBA and college, the court measures 94 feet long by 50 feet wide. Note the paint area – the free throw lane – is 16 feet across.
A basketball court at the collegiate and professional levels measures 94 feet in length and 50 feet in width, or approximately 31.3 yards long and 16.6 yards wide. High school courts are slightly smaller in length, measuring 84 feet, or 28 yards. The placement of the three-point line on the court is different for each level of play.
An NBA court is 94 feet long by 50 feet wide, or 31.333 yards long by 16.667 yards wide. What is the width of a basketball court? A basketball court is 50 feet wide which is also: 600 inches 16.6...
Dimensions. Basketball courts come in many different sizes. In the National Basketball Association (NBA), the court is 94 by 50 feet (28.7 by 15.2 m). Under International Basketball Federation (FIBA) rules, the court is slightly smaller, measuring 28 by 15 meters (91.9 by 49.2 ft). In amateur basketball, court sizes vary widely. Many older high school gyms were 84 feet (26 m) or even 74 feet (23 m) in length.
A basketball court is 30.62 yards long, and 16.4 yards wide. The total playing area for a court is 502.31yds² and 810yds² if run-offs and sidelines are required.
In calculating the dimensions of an NCAA and NBA basketball court in yards, convert 94 feet and 50 feet by multiplying each to â…“. Hence, college and professional basketball courts measure 31.3 yards by 16.67 yards. If you want to know the dimensions of a basketball court in inches, then use the conversion 1 foot is equal to 12 inches.
Basketball courts come in different sizes based on the level of basketball being played. A professional NBA court is 94’ x 50’ | 28.65 x 15.24 m. Courts are comprised of several foundational components: the baskets, the three-point arcs, free-throw (foul) lines, and the half court line.
The test involves running a maximum sprint over 3/4 of a basketball court (75 feet or 22.86 meters). This test is used in the Basketball SPARQ testing and at the NBA combine. purpose: The aim of this test is to determine running speed over 3/4 court distance (75 feet, 22.86 meters).
We've handpicked 26 related questions for you, similar to «How many yards is a nba basketball court?» so you can surely find the answer!
How many players in basketball court?
#### five players
Basketball is a game played between two teams of five players each on a rectangular court, usually indoors. Each team tries to score by tossing the ball through the opponent's goal, an elevated horizontal hoop and net called a basket. How many square feet in a half court basketball court?
#### 2,350 square feet
A half-size court is 47 feet by 50 feet (2,350 square feet), which can run from \$8,600 to \$38,000. How many square feet is a half court basketball court?
#### 2,350 square feet
A half-size court is 47 feet by 50 feet (2,350 square feet), which can run from \$8,600 to \$38,000. Basketball court dimension?
Basketball courts come in different sizes based on the level of basketball being played. A professional NBA court is 94’ x 50’ | 28.65 x 15.24 m. Courts are comprised of several foundational components: the baskets, the three-point arcs, free-throw (foul) lines, and the half court line.
How many feet in a basketball court?
#### 91.86ft
A basketball court is 91.86ft long and 49.21ft wide.
### Video answer: Basketball court easy marking plan
How many feet is a basketball court?
#### 91.86ft
A basketball court is 91.86ft long and 49.21ft wide. How many inches is a basketball court?
In the National Basketball Association (NBA), the court is 94 by 50 feet (28.7 by 15.2 m). Under International Basketball Federation (FIBA) rules, the court is slightly smaller, measuring 28 by 15 meters (91.9 by 49.2 ft). In amateur basketball, court sizes vary widely.
### Video answer: Dominate the basketball court | vertimax training
How many meters is a basketball court?
The dimensions of a professional Basketball court are 28.65 meters long and 15.24 meters wide. High school and college courts are shorter in length measuring at 25.2 meters.
How many people on a basketball court?
#### five players
Basketball, game played between two teams of five players each on a rectangular court, usually indoors. Each team tries to score by tossing the ball through the opponent's goal, an elevated horizontal hoop and net called a basket. How many players on a basketball court?
#### five players
Basketball, game played between two teams of five players each on a rectangular court, usually indoors. Each team tries to score by tossing the ball through the opponent's goal, an elevated horizontal hoop and net called a basket.
### Video answer: Gareth bale smashes the nba half-court challenge
How many players on court for basketball?
#### five players
Basketball, game played between two teams of five players each on a rectangular court, usually indoors. Each team tries to score by tossing the ball through the opponent's goal, an elevated horizontal hoop and net called a basket. How many sections on a basketball court?
Dimensions. Basketball courts come in many different sizes. In the National Basketball Association (NBA), the court is 94 by 50 feet (28.7 by 15.2 m). Under International Basketball Federation (FIBA) rules, the court is slightly smaller, measuring 28 by 15 meters (91.9 by 49.2 ft).
How many steps is a basketball court?
#### two steps
A player who receives the ball while he is progressing or upon completion of a dribble, may take two steps in coming to a stop, passing or shooting the ball. A player who receives the ball while he is progressing must release the ball to start his dribble before his second step. Basketball: how many basketball players are there on the court?
How many players can be on the court at once? Professional and collegiate basketball (both men’s and women’s) limit teams to five players on the court at any one time. A team with more than five in play at once is assessed a technical foul for Too Many Players on the Court.
How many calories burned playing half court basketball?
Just shooting hoops helps burn around 300 calories an hour. Playing a half court game brings it to about 558 an hour, and stepping up to a full court game will burn around 747 calories an hour.
How many feet is a college basketball court?
#### 94 feet
Professional NBA and College Basketball court is 94 feet (29 m) by 50 feet (15 m). International Basketball the court 28 meters (92 ft) by 15 meters (49 ft). High school, and Junior High court 84 feet (26 m) by 50 feet (15 m). How many feet is a nba basketball court?
#### 50 feet
In the National Basketball Association (NBA), the court is 94 by 50 feet (28.7 by 15.2 m). Under International Basketball Federation (FIBA) rules, the court is slightly smaller, measuring 28 by 15 meters (91.9 by 49.2 ft). How many feet is half a basketball court?
Half-court dimensions are 47 feet long for the pros and 42 feet long for high school. What are the half-court dimensions for a backyard? Youth half court dimensions are usually 42 feet long by 37 feet wide. High school half courts are slightly larger, at 50 feet long by 42 feet wide.
### Video answer: 109' 9" world record basketball shot!!!
How many feet long is a basketball court?
#### 94 feet long
Basketball court dimensions vary depending on the age level of the athletes playing on them. The size of a standard basketball court used in the professional or college ranks measures 94 feet long x 50 feet wide while many high school courts measure 84 feet long x 50 feet wide. How many meters in a college basketball court?
#### 28.65 meters
The length of a college basketball court measures 94 feet or 28.65 meters from baseline to baseline. This is the same length as an NBA court with the width of the court being 50 feet or 15.24 meters. How many meters is a college basketball court?
How Long Is A Basketball Court In Meters? A regulation NBA or NCAA basketball court measures 94 feet long by 50 feet wide which is 28.7 meters long by 15.2 meters wide. The FIBA international basketball courts are a little smaller and were designed using meters as a measurement coming in at 28 meters long by 15 meters wide.
How many meters long is a basketball court?
The dimensions of a professional basketball court are 28.65 meters long and 15.24 meters wide. High school and college courts are shorter in length measuring at 25.2 meters. | 2,534 | 10,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-05 | latest | en | 0.960357 |
http://www.thesecretcode.co.uk/the-garden.html | 1,619,130,740,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039563095.86/warc/CC-MAIN-20210422221531-20210423011531-00297.warc.gz | 173,894,096 | 25,781 | # The Garden
"A garden in the east, in Eden . . ." (Genesis 2.8)
There exists a holographic matrix located at the head of the NIV Bible which I call The Garden. [1] The Garden is lit from above, landscaped, filled with abundant foliage, teeming with wildlife and, like the original Garden of Eden, inhabited by the first two human witnesses. Appropriately, as you will see, each encoding is a double witness to God's sovereignty and creative power. [2] This is perhaps the greatest wonder I have been led to in over fifteen years of work on the New Bible Code.
In this introductory page I show how the Garden 'double-witness' encodings have been constructed, give a few examples of typical encodings, compare them with randomly-chosen numbers and provide the results of a 'field' trial, displaying what I call the code signal, indisputable evidence that the double-witness codes are real.
The First-Day Verses
1 In the beginning God created the heavens and the earth. 2 Now the earth was formless and empty, darkness was over the surface of the deep and the spirit of God was hovering over the waters. 3 And God said "let there be light" and there was light. 4 God saw that the light was good and he separated the light from the darkness. 5 God called the light "day" and the darkness he called "night", and there was evening and there was morning, the first day. (Genesis 1.1-5)
In the First Day I show how the first five verses of Genesis have integrated geometric features that suggest it can be regarded as a self-contained unit. These verses describe the beginnings of Creation up to the end of the first day. [3] The ordinal values of each of the verses, individually and in combination with each other, also encode the geometry of the subset of numerical triangles known as G-triangles. [4]
Here is an overview of the geometric properties of those five verses.
The first row shows how the individual verses encode geometric figures derived from G-triangles. The second row shows that combinations of the five verses lead to three G-triangles, against odds of 600 to 1. [5] 'G-triangles' is short for 'generator triangles', a reference to their ability to generate fractal snowflakes and antisnowflakes - the first iterations of which are respectively the Star of David and the trefoil. Since the form of natural objects in our universe, from real snowflakes to spiral galaxies, can be modelled using fractal geometry, these G-triangle encodings are therefore a metaphor for God's creative power. Trefoils resemble sprigs of foliage and in fact the word 'trefoil' means 'three-leaved plant'. As you can see, the two trefoils are encoded in Genesis 1.1 and 1.5, the first and last of these first-day verses. The two 'plants' therefore grow at either end of the Garden, delineating it and marking it out for our attention.
The Double Witness
One witness is not enough to convict anyone accused of any crime or offense they may have committed. A matter must be established by the testimony of two or three witnesses. (Deuteronomy 19.15)
The New Bible Code in founded on a unique two-stage encryption-decryption process. Within the text of the NIV Bible numbers have been encrypted under the ordinal value system; these are then converted into words or phrases under the standard value system. The numbers 'planted' in the garden are no exception. However, in addition, each concept is itself encoded twice. These number-pairs act as a double witness to the reality of the code. They are usually relatively close within the Garden too. [6] As far as I am aware this double-witness phenomenon is another unique feature of the New Bible Code.
In many cases there were multiple instances of one or both numbers in the Garden. This is because different words and phrases can share the same standard value, and these 'extra' numbers appear to take part in other encodings. However, their presence is distracting if shown in the table, so they were left out to put the most closely-aligned pair in the limelight.
Versions and Translation
The Hebrew words and phrases came from the Biblia Hebraica Stuttgartensia, based on the Masoretic Text as preserved in the Leningrad Codex. This is the popular version of the Hebrew Bible. The online version used was the Study Bible. The English words and phrases came from the New International Version (NIV) Bible, 1984 Edition (2001 reprint), in British English. [7]
In the Hebrew Bible, words often come in several forms and spellings, giving a variety of possible numerical values. However, I have found that the first instance of the word or phrase to appear in the Hebrew Bible, along with its equivalent in the NIV, provide the two numbers for encoding, a restriction that defeats any accusations of cherry picking. [8]
Hebrew words can be prefixed with the definite article hey ('the'), but other modifiers (such as 'every', 'this') and conjunctions (such as'and', 'but') were not included. English nouns and noun phrases were treated the same way, although the indefinite article, 'a' (not found in Hebrew), was included if it was part of the translation, which was thought for thought, rather than word for word.
An Example: The Garden of Eden
To make things clear, I've shown below three selected pairs of encoded numbers from the opening words of Genesis. The Hebrew and English names of the objects they represent are also shown. All are from the first biblical references to these concepts, indicated in the lower right-hand box. Each word in the text itself has been replaced by its place value for compactness. Each pair of encodings is colour-coded for clarity.
Looking at the word pair 'ets/trees, these are the ordinal values of the portion of text coincident with the two word blocks. So the ordinal value of words 21 to 24, "the surface of the", is 160, which is the standard value of the Hebrew word 'ets (ayin, tzaddik - giving 70 + 90 = 160). The ordinal value of words 17 to 23, "empty, darkness was over the surface of", is 400, which is the standard value of 'trees' (200 + 90 + 5 + 5 + 100 = 400).
The word-strings shown here are the only examples of these numbers in the Garden. As you can see, each number-pair presents a different scenario.
1. For 'ets/'trees', there is only one instance of each number and the word strings overlap.
2. For B'Gan Eden M'Qedem/'A Garden in the east, in Eden', again there is one instance of each number, but the word strings touch, rather than overlap.
3. In the case of owph/'birds', there are multiple instances of each number. The closest Hebrew/English pair are the first two word strings in verses 1 and 2 (birds/birds overlaps in verse 4 but that is of course English/English, rather than the Hebrew/English double witness).
As I said earlier, for clarity I normally present only the closest matching pair, so the Garden of Eden table would be presented like this:
The reader will hopefully be impressed by the closeness of the pairs in each case. This is over and above the fact that both are found within the first five verses of Genesis, itself a measure of proximity. [9]
Control Experiment 1
As a check on the validity of these encodings, I duplicated the method with randomly-generated numbers. My calculator generates random numbers between 1 and 1000, which isn't too far from the spread of standard values to be expected from single words in English or Hebrew. I calculated them in pairs then looked for those pairs within the Garden. The first number in each pair represents a nominal Hebrew word, the second number represents a nominal English word.
As you can see, only six of the ten numbers can be found in the Garden and only the first and last of the five pairs are complete. This is in line with theoretical expectations, because only around 70% of numbers in the range 1 to 1000 are present, which would mean that only about 50% of randomly-generated pairs in this range would be found. [10] In the first pair (992/835) two of the word strings overlap. The other pair (786/100) is more widely separated. None of the other four pairs are complete and in the case of the second pair neither of the two numbers are found.
Showing only the closest matches once more, the table now looks like this:
Hopefully, it will now be easier to appreciate how unusual it is to consistently find, within this small piece of text, pairs of numbers derived from English and Hebrew words for the same concept.
Control Experiment 2: The Code Signal
Another test for the reality of the code, developed during discussions on the Biblewheel forum, detected what I call the 'code signal', empirical evidence for the reality of the double-witness encodings. The test was based on my claim in The Creation that everything God made during the Creation week is double-witnessed in the Garden and was designed to exclude the possibility of cherry picking and other forms of data manipulation in selecting items from Genesis 1 that might be encrypted in the Garden. The test applied the following selection criteria:
1. Every noun or noun phrase in the Creation narrative was included, so there could be no decision made about what to include and therefore no possibility of cherry picking. Words like 'beginning' and 'food', which are not created things, were therefore tested, so although this method would be expected to detect any code signal we should also expect some dilution from non-encrypted items.
2. The first biblical instance of each noun or noun phrase was chosen every time, again to preempt any possibility of cherry picking.
3. The shortest form of each noun or noun phrase necessary to identify the concept was selected. Adjectives were included if present, since they further defined a concept. So 'wild animals' gives more information than 'animals'. If a noun appeared in different places with and without an adjective, then they were treated as a separate items. So 'animals' and 'wild animals' are both found in Genesis 1 and both were included in the test. Plural forms were included too. However, all determiners, such as articles (eg. 'a', 'the'), demonstratives (eg. 'these', etc), possesives (eg. 'my'), quantifiers (eg 'some') and numbers (eg. 'one'), were excluded. [11]
4. The noun or noun phrase was isolated from the Hebrew Bible first, then the equivalent in the NIV Bible found. This was necessary, since for example three separate words in the Hebrew Bible were all translated as 'creatures' in the NIV Bible. So each Hebrew word was paired with the same word: 'creatures'.
Test Method
1. Every noun or noun phrase within the Creation narrative of the Hebrew Bible (Genesis 1.1 to Genesis 2.3) was isolated, using the stated criteria. The equivalent word or phrase from the NIV Bible was also isolated. This gave 80 word pairs, shown here:
beginning/Reshith 189/911; God/Elohim 71/86; heavens/shamayim 569/390; earth/Aretz 304/291; darkness/choshek 370/328; surface of the deep/paniy tehom 868/591; surface/paniy 505/140; deep/tehom 84/451; spirit of God/ruach Elohim 615/300; spirit/ruach 478/214; waters/mayim 896/90; light/aur 254/207; day/yom 705/56; night/layilah 274/75; evening/ereb 526/272; morning/boqer 306/302; expanse/raqiya 831/380; sky/shamayim 820/390; dry ground/yabbashah 1305/317; ground/yabbashah 511/317; one place/maqowm 'echad 224/199; place/maqowm 109/186; land/aretz 85/291; seas/Yam 206/50; vegetation/deshe' 937/305; seed-bearing plants/eseb zara zera 329/976; plants/eseb 451/372; seed/zera 114/277; trees on the land that bear fruit with seed in it/'ets periy 'asah periy l'miyn asher zeraw bw 'al ha'eretz 3019/2439; trees on the land that bear fruit/'ets periy 'asah periy 1920/1115; trees/'ets 400/160; fruit/periy 605/290; lights/ma'owr 354/641; signs/l'owth 266/831; seasons/mow'edim 416/170; days/yom 805/100; years/shanem 896/400; two great lights/sheni hama'owr hagadowl 1417/1098; greater light/ma'owr hagadol 652/289; lesser light/ma'owr haqatan 584/411; stars/kowkabim 491/98; expanse of the sky/raqiya hashamayim 1930/775; water/mayim 796/90; living creatures/sherets 1299/590; creatures/sherets 794/590; birds/owph 205/156; great creatures of the sea/hatanniym gadol 1482/642; living and moving thing/nephesh hachayah v'haremesht 1400/1403; winged bird/owph kanaph 680/306; living creatures/chay nephesh 1299/453; creatures/nephesh 794/430; livestock/behemah 827/52; creatures that move along the ground/remes 2580/540; creatures/remes 794/540; wild animals/chaiotu h'erets 774/715; animals/chaiotu 231/424; man in our image, in our likeness/'adam b'tselemenu b'demuwthenu 1490/789; likeness/demuwth 319/450; man in our image/'adam b'tselemenu 662/263; man/adam 91/45; image of God/tselem Elohim 199/246; image/tselem 62/160; male and female/zakar neqebah 218/390; male/zakar 76/227; female/neqebah 87/157; face of the whole earth/periy kol h'erets 1201/486; face/periy 15/140; whole earth/kol h'erets 907/346; fish of the sea/dagat h'yam 508/462; fish/dagah 123/407; sea/yam 106/50; birds of the air/'owph shamayim 584/551; air/shamayim 100/390; breath of life/nephesh chayah 422/453; breath/nephesh 306/430; life/chayah 50/23; green plant/yereq 'eseb 508/682; food/oklah 130/56; vast array/tsabam 1583/133; work/mela'kathu 670/497.
2. The standard value of each word pair was calculated, giving the 80 number pairs shown above with the word pairs.
3. Hundreds of sets of 80 randomised number-pairs in the same range as the biblical number-pairs were created. This was done by taking the bibiical list and randomly adding or subtracting numbers between 1 and 50. This method of generating the random numbers was useful here, because the percentage of random hits was critically dependent on the size of the number, with very low and very high numbers achieving less hits than numbers in between.
4. A search for each number was carried out in the Garden (Genesis 1.1 - 1.5).
Results and Discussion
Of the 80 biblical number pairs 58 complete pairs were found in the Garden, a 73% success rate. In hundreds of trials using sets of randomly-generated sets number pairs in the same range as the biblical pairs an average 54% success rate was achieved (this work was done by myself and Richard McGough during discussions with myself on his forum).
Since each pair tested gives a yes/no answer, like flipping a coin, and the expected success rate is close to 50%, the spread of results about the mean will produce a binomial distribution. Using the well-known binomial formula we can therefore calculate how likely we would be to achieve an average 73% success rate or better with 80 number pairs, when the average success rate should be 54%. [12]
For 80 trials and an expected success rate of 54%, the probability of acheiving a 73% success rate or better is 0.00053, or 1 in 1900.
If we increase the number of pairs tested the probability of achieving a success rate higher than the random average decreases. So the true improbability of the phenomenon depends on how much encoded material is present. In The Ark I show further encodings in the Garden, this time related to the Ark of the Testimony. Applying the same selection criteria as before gives a list of 25 items, 20 of which are found in the Garden. These are independent of the list of nouns in Genesis 1 and so can be added to them to give a larger list of 105 items, 78 of which are found in the Garden (74%).
For 105 pairs and an expected success rate of 54%, the probability of acheiving a 74% success rate or better is 0.000015, or 1 in 66000.
The results for the 80 pairs against hundreds of sets of 80 randomised pairs are presented in visual form below, showing the percentage of number-pairs in each set that are found in the Garden. The randomised sets produce a binomial distribution curve, which I have smoothed out. The success rate acheived in the test by the biblical pairs is an outlier of the main group, the diluted but still visible code signal.
As you can see, practically all of the random test results will fall within about fifteen percentage points either side of the mean. The Genesis test results are shown in comparison and they are well beyond the bulk of results to be expected from randomised pairs (roughly 3.5 standard deviations above the mean, for those with some understanding of probability). Further increasing the number of pairs in a set would lead to an even clearer code signal.
The actual encoded material displayed in the next two pages has a statistical significance well beyond four standard deviations. However, these encodings, like the rest of the New Bible Code, are more like works of art than encodings done by rote. Testing requires standardisation of the selection procedure to ensure there is no bias involved, which necessarily means that non-encoded material will be included and the success rate will be below 100%. However, even with dilution and with the likely presence of further as-yet-undiscovered material, the code signal shines through.
The next page in the Garden series is The Creation.
Bill Downie
3/7/17
Notes
1. The Genesis Watermark is the name I previously gave the initial few verses, but I now consider both this and the Signature of Christ, the first set of codes I found in that location, to be part of the Garden.
2. This part of the code was even discovered by a double witnessing, namely of myself and Kathryn LeCorre, who was given the revelations that led us to the Garden.
3. Other important numbers are the total ordinal value of 3764 and the total number of words, 83. If we bisect 3764 we obtain 37 and 64, geometrically-related factors of 2368, which is Ihsous Christos (Jesus Christ) in Greek. The geometric relationship between 64 and 37 is this: 64 is the cube of four and 37 is its 2D diagonal projection. In other words, if you observe a cube made of 64 smaller units along its diagonal axis, 37 of those units can be seen. The number of words, 83, is the reduced value of 'the Second Appearance'.
4. I am using the nomenclature of Vernon Jenkins, who first described these special triangular numbers in his own website.
5. There are 31 different combinations of the five verse totals and a pool of 3764 numbers to choose from. 3764 encompases the first 28 'G' triangles. By the binomial formula the odds against this happening by chance are 1 in 604.
6. The ordinal value of the five verses is 3764, therefore any number up to 3764 could theoretically be found in there too. The number of word strings of one or more words in length is 3486, so one might at first glance expect most numbers below 3764 to also be present. However, only 52% are actually are present, as many numbers are found multiple times, which 'pushes out' other possible numbers.
7. This version also contains the entire New Bible Code, of which the Garden is a small part. The US language version also contains the Garden and in fact will contain nearly all of the code, as will earlier editions of the NIV, the latest edition and even other versions: the code has been accumulating since the first Hebrew scribes put pen to parchment). But only the 1984 British-language version has it all.
8. Words and phrases are often found in different forms within the text and this often gives a choice of numbers to test for in the Garden. 'Cherry picking' means selecting individual test results that give the desired answer, and ignoring those that do not, thereby skewing the result in your favour. This was impossible here, because only the first biblical instance of a word or phrase was chosen and its gematria calculated.
9. In my eagerness to display the double-witness phenomenon I rushed it out in March 2017 without fully understanding it. However, after further reflection, including a lengthy discussion with Richard McGough on the Biblewheel forum, I have made some important changes to the page. If anyone has been misled I apologise, but I can see no alternative to presenting the code as I find it, since it will take many decades for a full exegesis.
10. If 70% of numbers in the range 1 to 1000 are present in the Garden we would expect (0.7 x 0.7) x 100, or 49% of randomly-generated pairs to be present. This is before we take account of the clustering phenomenon which also seems to be present. 52% of numbers between 1 and 3764 (the ordinal value of the entire Garden) are found. I am grateful to Richard McGough for his statistical analysis.
11. One could take issue with the selection criteria on other grounds, but it was virtually impossible to find a set of criteria that could be consistently applied. This set of criteria allowed a relatively simple way of extracting nouns and noun phrases from the text and were consistently applied across the board.
12. We are interested in the cumulative probability here, the probability of acheiving 58 hits or better, rather than exactly 58 hits. | 5,000 | 20,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | longest | en | 0.961559 |
https://hsm.stackexchange.com/questions/524/who-introduced-the-principle-of-mathematical-induction-for-the-first-time/527 | 1,620,586,172,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00495.warc.gz | 341,362,478 | 45,653 | # Who introduced the Principle of Mathematical Induction for the first time?
Can you tell me the name of the mathematician, who introduced the Principle of Mathematical Induction for the first time? (with reliable source).
Please don't say De Morgan because I have read the name of the mathematician somewhere, but I have forgotten.
The issue is thorny ...
According to Morris Kline, Mathematical Thought from Ancient to Modern Time. Volume I (1972), page 272 [only entry of the Subject Index regarding : mathematical Induction] :
The method was recognized explicitly by Maurolycus in his Arithmetica of 1575 and was used by him to prove, for example, that $1+3+5+ \ldots + (2n+1)=n^2$. Pascal in one of his letters acknowledged Maurolycus's introduction of the method and used it himself in his Traité du triangle arithmétique (1665), wherein he presents what we now call the Pascal triangle.
The modern source is Giovanni Vacca (1872 –1953) Italian mathematician, assistant to Giuseppe Peano and historian of science in his :
Acording to Kline :
the method [of mathematical induction] is implicit even in Euclid's proof of the infinitude of the number of primes [IX, 20].
This point is debatable.
Euclid, The Thirteen Books of the Elements, Vol. 2: Books III - IX (T.L.Heath editor), states IX.20 as follows [page 412] :
Prime numbers are more than any assigned multitude of prime numbers.
The proof is per impossibile. According to Heath's comment [page 413] :
We have here the important proposition that the number of prime numbers is infinite.
Neither in the statement of the proposition nor in its proof Euclid uses the work infinite [apeiros -on : adj, infinite].
In can be helpful to place it in the context of Grrek debate about the infinite : see Aristotle and Mathematics on the actual infinity.
According to the Aristotelian philosophy, we cannot legitimately "conceive" actual infinity; i.e. we have no experience of an infinite "collection" but only of an unlimited iterative process (the potential infinity).
Euclid's statement must be understood in this context : we never have a "complete" infinite set of prime numbers, but we have a "procedure" that, for a finite collection of prime numbers whatever, can "produce" a new prime which is not in the collection.
The only occurrence (not regarding straight lines) I've found of the word infinite (searching into : EUCLID’S ELEMENTS OF GEOMETRY, the Greek text of J.L. Heiberg (1883–1885), edited, and provided with a modern English translation, by Richard Fitzpatrick), is VII,31 [Heath ed, page 332] :
Any composite number is measured by some prime number.
The proof uses the fact that :
For, if it is not found [some prime number which will measure the number before it, which will also measure A], an infinite series of numbers [emphasis added] will measure the number A, each of which is less than the other: which is impossible in numbers.
It seems the only reasonable "candidate" for a precursor of the least number principle [see the comment of Ian Mueller, Philosophy of Mathematics and Deductive Structure in Euclid's Elements (1981, Dover reprint), page 77].
According to :
the Greeks [...] did not have mathematical induction [...].
Regarding Pappus, in
• +1, A very well thought out answer. One subtopic I would like to see added though is how geometric series figures into the story. I can't remember for sure, but I believe geometric sums were a very early result of Eudoxan geometry. – David H Nov 21 '14 at 14:08
• Induction doesn't have to apply to an actually infinite set, it is also needed for finite sets of indefinite length. So another precursor of an inductive argument is Euclid's proof of what is now called Euclidean algorithm aleph0.clarku.edu/~djoyce/java/elements/bookVII./propVII2.html. And that probably goes back to Eudoxus if not early Pythagoreans. – Conifold Nov 21 '14 at 19:57
• Related: Duhem's counter-argument, in his "The Nature of Mathematical Reasoning," to Poincaré's belief that "arithmetic frequently uses a reasoning which is not equivalent to a series of syllogisms of limited number; in reality, it condenses an infinity of successive syllogisms." – Geremia Nov 17 '16 at 2:47
Well, you could go with Plato. From this,
For a start, although the principle itself is not explicitly stated in any ancient Greek text, there are several places that contain precursors of it. Indeed, some historians see the following passage from Plato’s (427-347 BC) dialogue Parmenides (§147a7-c3) as the earliest use of an inductive argument.
The text quotes Parmenides:
Then they must be two, at least, if there is to be contact. - They must. - And if to the two terms a third be added in immediate succession, they will be three, while the contacts [will be] two. - Yes. - And thus, one [term] being continually added, one contact also is added, and it follows that the contacts are one less than the number of terms. For the whole successive number [of terms] exceeds the number of all the contacts as much as the first two exceed the contacts, for being greater in number than the contacts: for afterwards, when an additional term is added, also one contact to the contacts [is added]. - Right. - Then whatever the number of terms, the contacts are always one less. -True.
I attempted to find a second translation, and found this one, but I have been unable to identify this section, even though the section is supposedly listed. I found another transcript here, but it, too does not agree with the given text in the pdf. Still, the dialogue closely resembles the technique of mathematic induction, albeit used in a philosophical context.
The page goes on to add that Euclid and Pappus, too, played a part:
There are, however, several ancient mathematical texts that also contain quasi-inductive arguments. For instance Euclid (~330 - ~ 265 BC) in his Elements employs one to show that every integer is a product of primes. An argument closer to the modern version of induction is in Pappus' (~290-~350 AD) Collectio.
I have been unable to find a similar primary source for Euclid and Pappus. However, this appears to corroborate Euclid's role (though not as detailed as I would have likes) and also says that Wikipedia's claim about Pascal being important (mentioned below from another source) is accurate. I'm working on more primary sources, but at the moment no others are forthcoming.
This says that Pascal played a role in modern times:
Cantor in his Vorlesungen iuber Geschichte der Mathematik' says that Pascal was the originator of the method of complete induction
• I find it highly unsatisfying that you do not reproduce the relevant section of Plato's (and the other ancient Greeks') work in your answer. Please include them in your answer! – Danu Nov 19 '14 at 20:02
• @Danu I appreciate that you didn't merely downvote and not give any feedback (or downvote at all!). Yes, it would be good to get direct quotes from Plato et al.; I hadn't had time to get them earlier. I'll certainly add them it. – HDE 226868 Nov 19 '14 at 21:04
Victor Katz wrote in a discussion at the Historia mathematica discussion list, "as Barnabas Hughes notes, it is difficult to answer the question of the "first instances of proof by induction" unless one carefully defines what one means by "proof by induction." (The same general remark applies to lots of questions about firsts in the history of mathematics.) Noicomachus certainly has an argument which we would find very easy to convert into a modern formal "proof by induction." So do several Islamic authors around the 11th century, including al-Karaji and al-Samaw'al. (See my book, pp. 238-242.) Pascal may well be the first to state the modern principle of mathematical induction explicitly, but even he does not give proofs in the modern style - because he has no notation for a general "n". Thus he generally gives proofs by what I call the "method of generalizable example."
In another post Barnabus Hughes suggests yet an earlier "first use" of induction: If the essence of math induction lies in a process that begins at some small value, which process can be continued to larger values which regardless of their size maintain the pattern one wishes to accept, then I would hazard that Nicomachus of Geresa used the essence of math induction where he discussed figurate numbers (Arithmetica, Bk. 2, CC. 7ff.) In C. 7 he states, "Hence, the triangle is elementary among these
figures; for everything else is resolved into it, but it into nothing else." He then shows how, if the process established for the creation of each figurate number is followed, then that number is always seen. C. 12 shows how the other figures are resolved into triangles, including a table which lists the first five polygonal numbers to the tenth degree of each. I think that Nick established the pattern by induction. Comment?
Barnabas Hughes
According to Wikipedia this was Levi ben Gershon (a.k.a. Gershonides, RALBAG). 1288–1344
The work is notable for its early use of proof by mathematical induction, and pioneering work in combinatorics.
and
Gersonides was also the earliest known mathematician to have used the technique of mathematical induction in a systematic and self-conscious fashion
Remark. The word "induction" is used in a different sense in philosophy. One has to distinguish Mathematical induction from "induction" in philosophy. These are very different things. I am not aware of any ancient Greek use of Mathematical induction.
• About the "modern" sense of induction (as mathematical induction), see André Weil, Number Theory : An approach through history (1984), page 50 for Fermat's critique (1657) of the "inductive method" as an heuristical way to establish a "general" statement in mathematics (call it "incomplete induction") compared to a proof of the said "general" statement, and see page 77 for Fermat's enunciation (1659) of his newly discovered method of proof : the infinite descent. – Mauro ALLEGRANZA Nov 23 '14 at 11:36
writes:
The process of reasoning called "mathematical induction" has had several independent origins. It has been traced back to the Swiss Jakob (James) Bernoulli, the Frenchmen B. Pascal and P. Fermat, and the Italian F. Maurolycus.
It seems Fermat (1601-1665) might have been the first.
C. S. Peirce says "mathematical induction" is an improper term for "Fermatian inference" (CP 6.116):
In truth, of infinite collections there are but two grades of magnitude, the endless and the innumerable. Just as a finite collection is distinguished from an infinite one by the applicability to it of a special mode of reasoning, the syllogism of transposed quantity, so, as I showed in the paper last referred to, a numerable collection is distinguished from an innumerable one by the applicability to it of a certain mode of reasoning, the Fermatian inference, or, as it is sometimes improperly termed, “mathematical induction.”
He cites Fermat, Opera Omnia (Leipzig, 1911), vol. 1, §§340-351.
Also,
write, (ed. 2) xvii. 355:
In fact, mathematical induction, or reasoning by recurrence, sometimes is referred to as ‘Fermatian induction’, to distinguish it from scientific, or ‘Baconian,’ induction. | 2,603 | 11,239 | {"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} | 2.921875 | 3 | CC-MAIN-2021-21 | latest | en | 0.918591 |
http://mathhelpforum.com/pre-calculus/136993-equation.html | 1,481,056,605,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541995.74/warc/CC-MAIN-20161202170901-00357-ip-10-31-129-80.ec2.internal.warc.gz | 180,196,739 | 10,192 | 1. Equation
Hi people,
I want to show that $sin(Arctanx)=\frac{x}{\sqrt{1+x^2}}$, can you help me please???
2. let $tan^{-1}x=y$
$x=tan y$
$sec^2 y=1+tan^2 y$
$sec^2 y=1+x^2$
$sec y=\sqrt{1+x^2}$
$sin y=tan y cos y$
$sin y=\frac{tan y}{sec y}$
$sin y=\frac{x}{\sqrt{1+x^2}}$
$sin(tan^{-1}x)=\frac{x}{\sqrt{1+x^2}}$
3. Originally Posted by lehder
Hi people,
I want to show that $sin(Arctanx)=\frac{x}{\sqrt{1+x^2}}$, can you help me please???
For a (imho) somewhat more intuitive solution consider a right triangle with acute angle $\alpha$, opposite leg $x$ and adjacent leg 1: By construction we have $\arctan(x)=\alpha$. But by Pythagoras's theorem we also have that the length of the hypotenuse must be $\sqrt{1+x^2}$. By definition of $\sin$ in a right triangle we find that $\sin(\alpha)=\frac{x}{\sqrt{1+x^2}}$, i.e. the ratio of opposite leg to hypotenuse in that right triangle. | 317 | 899 | {"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": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-50 | longest | en | 0.72997 |
http://primes.utm.edu/curios/cpage/17117.html | 1,529,914,886,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867579.80/warc/CC-MAIN-20180625072642-20180625092642-00133.warc.gz | 247,414,952 | 2,436 | 7 (another Prime Pages' Curiosity)
Curios: Curios Search: Participate: Single Curio View: (Seek other curios for this number) Divisibility test for 7 (or 13): Combine the digits in order into groups of 3 (starting from the right) by alternating them with positive and negative signs. If the result is divisible by 7 (or 13 respectively), then so is the original number. For example, 7 divides 62540982 because 7 divides +(62)-(540)+(982). Submitted: 2008-09-05 11:19:07; Last Modified: 2008-09-05 11:22:02. Prime Curios! © 2000-2018 (all rights reserved) privacy statement | 171 | 585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.772706 |
https://ounces-to-grams.appspot.com/pl/1170-uncja-na-gram.html | 1,723,459,802,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00842.warc.gz | 335,785,435 | 6,301 | Ounces To Grams
# 1170 oz to g1170 Ounce to Grams
oz
=
g
## How to convert 1170 ounce to grams?
1170 oz * 28.349523125 g = 33168.9420563 g 1 oz
A common question is How many ounce in 1170 gram? And the answer is 41.270535481 oz in 1170 g. Likewise the question how many gram in 1170 ounce has the answer of 33168.9420563 g in 1170 oz.
## How much are 1170 ounces in grams?
1170 ounces equal 33168.9420563 grams (1170oz = 33168.9420563g). Converting 1170 oz to g is easy. Simply use our calculator above, or apply the formula to change the length 1170 oz to g.
## Convert 1170 oz to common mass
UnitMass
Microgram33168942056.2 µg
Milligram33168942.0562 mg
Gram33168.9420563 g
Ounce1170.0 oz
Pound73.125 lbs
Kilogram33.1689420563 kg
Stone5.2232142857 st
US ton0.0365625 ton
Tonne0.0331689421 t
Imperial ton0.0326450893 Long tons
## What is 1170 ounces in g?
To convert 1170 oz to g multiply the mass in ounces by 28.349523125. The 1170 oz in g formula is [g] = 1170 * 28.349523125. Thus, for 1170 ounces in gram we get 33168.9420563 g.
## Alternative spelling
1170 Ounces to Grams, 1170 oz to Grams, 1170 oz in Grams, 1170 Ounces to g, 1170 Ounces in g, 1170 Ounce to Grams, 1170 Ounce in Grams, 1170 Ounce to Gram, 1170 Ounce in Gram, 1170 Ounce in g, | 471 | 1,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.662383 |
kid.fiar.me | 1,679,653,069,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00316.warc.gz | 392,992,279 | 36,908 | # Graphs and Models
### Summary
• Sketch the graph of an equation
• Find the intercepts
• Test a graph for simetry with respect to an axis and the origin
• Find the points of intersection of two graphs
• Interpret mathematical models for real life-data
### The Graph of an Equation
In 1637 the French mathematician René Descartes revolutionized the study of mathematics by joining its two major fields: algebra and geometry. With Descarte’s coordinate plane, geometric concepts could be formulated analytically and algebraic concepts could be viewed graphically. The power of this approach was such that with within a century of its introduction, much of the calculus had been developed.
The same approach can be followed in your study of calculus. That is, by viewing calculus from multiple perspectives -graphically, analytically and numerically- you will increase your understanding of core concepts.
Consider the equation . The point is a solution point of the equation because the equation is satisfied (is true) when is substituted for and is substituted for . This equation has many other solutions, such as and . To find other solutions systematically, solve the original equation for .
Then construct a table of values by substituting several values of .
From the table, you can see that are solutions of the original equation . Like many equations, this equation has an infinite nuber of solutions. The sets of all solution point is the graph of the equation, as shown in Figure P1.
Even thought we refer the sketch shown in Figure P1 as the graph of , it really represents only a portion of the graph. The entire graph would extend beyond this page.
In this course, you will study many sketching techniques. The simplest is point plotting. That is, you plot points until the basic shape of the graph seems apparent.
#### EXAMPLE 1 – Sketching a Graph by Point Plotting
Sketch the graph of
#### Solution
First construct the table of values.
Then plot the points shown in the table
Finally, connect the points with a smooth curve as shown in Figure P2.
This graph is a parabola. It is one of the conics you will study later.
One disadvantage of plotting is that to get a good idea about the shape of a graph, you may need to plot many points. With only a few points, you could badly misrepresent the graph. For instance, suppose that to sketch the graph of
you plotted only five points: , as shown in Figure P3 (a). From these five points, you might conclude that the graph is line. This, however, is not correct. By plotting several more points, you can see that the graph is more complicated, as shown in Figure P3 (b).
[1, 2, 3, 4]
#### In exercises 5-14, sketch the graph of the equation by point plotting.
[5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
#### In exercises 17 and 18, use Geogebra to graph the equation. Find the unknown coordinate of each solution point accurate to two decimal places.
[17a, 17b, 18a, 18b]
### Intercepts of a Graph
Two types of solution points that are specially useful in graphing an equation are those having zero as – or -coordinate. Such points are called intercepts because they are the points at which the graph intersects the – or -axis.
The point is an x-intercept of the graph of an equation if it is a solution point of the equation. To find the – intercepts of a graph, let be zero and solve the equation for .
The point is a y-intercept of the graph of an equation if it is a solution point of the equation. To find the -intercepts of a graph, let be zero and solve the equation for .
It is possible for a graph to have no intercepts, or it might have several. For instance, consider the graphs shown in Figure P5.
#### EXAMPLE 2 – Finding x– and y-intercepts
Find the – and -intercepts of the graph of
#### Solution
To find the -intercepts, ley be zero and solve for .
Because this equation has three solutions, you can conclude that the graph has three -intercepts:
To find the -intercepts, let be zero.
Doing this proces . So the -intercept is
Example 2 uses an analytic approach to finding intercepts. When an analytic approach is not possible, you can use a graphical approach by finding the points at wich the graph intersects the axes. Use a graphing utility to approximate the intercepts.
#### In Exercises 19-28, find any intercepts.
[19, 20, 21, 22, 23, 24, 25, 26, 27, 28]
Los | 1,010 | 4,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2023-14 | longest | en | 0.940396 |
https://drmf-beta.wmflabs.org/wiki/Formula:KLS:14.10:13 | 1,721,720,947,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00226.warc.gz | 173,929,820 | 10,248 | # Formula:KLS:14.10:13
${\displaystyle{\displaystyle{\displaystyle P^{(\alpha,\beta)}_{n}\!\left(x|q% \right)=\frac{2^{n}q^{(\frac{1}{2}\alpha+\frac{1}{4})n}\left(q^{n+\alpha+\beta% +1};q\right)_{n}}{\left(q,-q^{\frac{1}{2}(\alpha+\beta+1)},-q^{\frac{1}{2}(% \alpha+\beta+2)};q\right)_{n}}{\widehat{P}}^{(\alpha,\beta)}_{n}\!\left(x% \right)}}}$
## Proof
We ask users to provide proof(s), reference(s) to proof(s), or further clarification on the proof(s) in this space. | 187 | 473 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.443445 |
http://stackoverflow.com/questions/13071485/matlab-slow-parallel-processing-with-distributed-arrays | 1,394,460,244,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010824553/warc/CC-MAIN-20140305091344-00050-ip-10-183-142-35.ec2.internal.warc.gz | 174,109,882 | 18,252 | # Matlab slow parallel processing with distributed arrays
I am new to using distributed and codistributed arrays in matlab. The parallel code I have produced works, but is much slower than the serial version and I have no idea why. The code examples below compute the eigenvalues of hessian matrices from volumetic data.
Serial version:
S = size(D);
Dsmt=imgaussian(D,2,20);
DHess = zeros([3 3 S(1) S(2) S(3)]);
d = zeros([3 S(1) S(2) S(3)]);
for i = 1 : S(1)
fprintf('Slice %d out of %d\n', i, S(1));
for ii = 1 : S(2)
for iii = 1 : S(3)
d(:,i,ii,iii) = eig(squeeze(DHess(:,:,i,ii,iii)));
end
end
end
Parallel version:
S = size(D);
Dsmt=imgaussian(D,2,20);
DHess = zeros([3 3 S(1) S(2) S(3)]);
CDHess = distributed(DHess);
spmd
d = zeros([3 S(1) S(2) S(3)], codistributor('1d',4));
for i = 1 : S(1)
fprintf('Slice %d out of %d\n', i, S(1));
for ii = 1 : S(2)
for iii = drange(1 : S(3))
d(:,i,ii,iii) = eig(squeeze(CDHess(:,:,i,ii,iii)));
end
end
end
end
If someone could shed some light on the issue I would be very grateful
-
How long does a single iteration take? – Jonas Oct 25 '12 at 19:21
are you opening your matlabpool? – Rasman Oct 26 '12 at 3:00
@Jonas A single iteration (over variable i) on the serial version takes around 1.7 seconds. A single iteration on the parallel version does not complete in over 5 minutes at which point I have terminated the execution. – Hampycalc Oct 26 '12 at 10:34
@Rasman Yes, I forgot to mention I am opening the matlabpool using the 'local' profile with 6 labs – Hampycalc Oct 26 '12 at 10:35
Here is a re-written version of your code. I have split the work over the outer-most loop, not as in your case - the inner-most loop. I have also explicitly allocated local parts of the d result vector, and the local part of the Hessian matrix.
In your code you rely on drange to split the work, and you access the distributed arrays directly to avoid extracting the local part. Admittedly, it should not result in such a great slowdown if MATLAB did everything correctly. The bottom line is, I don't know why your code is so slow - most likely because MATLAB does some remote data accessing despite the fact that you distributed your matrices.
Anyway, the below code runs and gives pretty good speedup on my computer using 4 labs. I have generated synthetic random input data to have something to work on. Have a look at the comments. If something is unclear, I can elaborate later.
clear all;
D = rand(512, 512, 3);
S = size(D);
% this part could also be parallelized - at least a bit.
tic;
DHess = zeros([3 3 S(1) S(2) S(3)]);
toc
d = zeros([3, S(1) S(2) S(3)]);
disp('sequential')
tic
for i = 1 : S(1)
for ii = 1 : S(2)
for iii = 1 : S(3)
d(:,i,ii,iii) = eig(squeeze(DHess(:,:,i,ii,iii)));
end
end
end
toc
% my parallel implementation
disp('parallel')
tic
spmd
% just for information
disp(['lab ' num2str(labindex)]);
% distribute the input data along the third dimension
% This is the dimension of the outer-most loop, hence this is where we
% want to parallelize!
DHess_dist = codistributed(DHess, codistributor1d(3));
DHess_local = getLocalPart(DHess_dist);
% create an output data distribution -
% note that this time we split along the second dimension
codist = codistributor1d(2, codistributor1d.unsetPartition, [3, S(1) S(2) S(3)]);
localSize = [3 codist.Partition(labindex) S(2) S(3)];
% allocate local part of the output array d
d_local = zeros(localSize);
% your ordinary loop, BUT! the outermost loop is split amongst the
% threads explicitly, using local indexing. In the loop only local parts
% of matrix d and DHess are accessed
for i = 1:size(d_local,2)
for ii = 1 : S(2)
for iii = 1 : S(3)
d_local(:,i,ii,iii) = eig(squeeze(DHess_local(:,:,i,ii,iii)));
end
end
end
% assemble local results to a codistributed matrix
d_dist = codistributed.build(d_local, codist);
end
toc
isequal(d, d_dist)
And the output
Elapsed time is 0.364255 seconds.
sequential
Elapsed time is 33.498985 seconds.
parallel
Lab 1:
lab 1
Lab 2:
lab 2
Lab 3:
lab 3
Lab 4:
lab 4
Elapsed time is 9.445856 seconds.
ans =
1
Edit I have checked the performance on a reshaped matrix DHess=[3x3xN]. The performance is not much better (10%), so it is not substantial. But maybe you can implement the eig a bit differently? After all, those are 3x3 matrices you are dealing with.
-
It's great that you have taken the time to provide this example as I will be able to use these ideas with many future projects. A couple of questions: I used 'drange' in my code, what is the purpose of this if you have to use 'getLocalPart'. Using your code I am getting an error: 'Error using distcompserialize Error during serialization' at the line DHess_dist = codistributed(DHess, codistributor1d(3)); The size of my input D is around 512x512x200, perhaps the size is an issue; although it seems it should not matter how big the array is, as this is a main purpose of parallel processing – Hampycalc Oct 26 '12 at 14:42
I should also state that D is a double. – Hampycalc Oct 26 '12 at 14:43
@Hampycalc I am sorry, I have obviously missed the fact that you used drange. I will edit my answer - you did divide your work after all. My bad. – angainor Oct 26 '12 at 14:55
@Hampycalc I have tried the code for other dimensions and it does not complain. D in my code is also of type double, so no problem there. I do not use drange and extract the local parts of my matrices to explicitly operate only on local data. In your case you relied on MATLAB to do the work splitting, which apparently went rather bad. – angainor Oct 26 '12 at 14:57
Yes, I will do as you have advised and extract local parts first in the future. I am still getting the error as before, but I shall search elsewhere for a solution as you have already answered my original question. Oddly I am using a x64 system with >50GB RAM and Matlab 2012a x64, so would be surprised if it is a memory issue. Anyway, thanks again for your help. – Hampycalc Oct 26 '12 at 15:12
If you are using some other scheduler with the workers on a separate machine then you might be able to get speedup, but that depends on what you're doing. There's an example here http://www.mathworks.com/products/parallel-computing/examples.html?file=/products/demos/shipping/distcomp/paralleldemo_backslash_bench.html which shows some benchmarks of MATLAB's \ operator. | 1,807 | 6,350 | {"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} | 2.796875 | 3 | CC-MAIN-2014-10 | latest | en | 0.811065 |
http://math.stackexchange.com/questions/61497/why-are-rings-called-rings?answertab=oldest | 1,462,426,338,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860125897.19/warc/CC-MAIN-20160428161525-00217-ip-10-239-7-51.ec2.internal.warc.gz | 182,873,035 | 20,235 | # Why are rings called rings?
I've done some search in Internet and other sources about this question. Why the name ring to this particular object? Just curiosity.
Thanks.
-
In German, "Ring" can also mean a (close) group of people with shared interests, an association. Maybe it is because of that: you've got some closely interacting elements but for all their interaction, they never leave the group. – Raphael Sep 2 '11 at 23:05
@Raphael Funny how you started with a ring and ended with a group :). – Srivatsan Sep 2 '11 at 23:09
Did nobody thought that maybe -- just maybe -- that perhaps the bloke who coined the name thought the name just sounds cool? I mean, if you had the chance to coin a name for something you discovered, wouldn't some of you want to name it Diamond or something like that? – Lie Ryan Sep 3 '11 at 1:43
@yoyo: yes, it is true, but for me in particular, is interesting to know the origin of this words. I think that all of us should be know a little about that. – leo Sep 3 '11 at 2:03
leo: I totally agree with your last comment. It goes hand in hand with the "read the masters" credo to know and try to find out a little about where the words we use on a daily basis come from. While in principle, as Hilbert allegedly put it, "One must be able to say at all times--instead of points, straight lines, and planes--tables, chairs, and beer mugs", I think knowing the etymology of words and the history of ideas is part of the general culture a mathematician should have. – t.b. Sep 3 '11 at 3:02
The name "ring" is derived from Hilbert's term "Zahlring" (number ring), introduced in his Zahlbericht for certain rings of algebraic integers. As for why Hilbert chose the name "ring", I recall reading speculations that it may have to do with cyclical (ring-shaped) behavior of powers of algebraic integers. Namely, if $\:\alpha\:$ is an algebraic integer of degree $\rm\:n\:$ then $\:\alpha^n\:$ is a $\rm\:\mathbb Z$-linear combination of lower powers of $\rm\:\alpha\:,\:$ thus so too are all higher powers of $\rm\:\alpha\:.\:$ Hence all powers cycle back onto $\rm\:1,\:\alpha,\:,\ldots,\alpha^{n-1}\:,\:$ i.e. $\rm\:\mathbb Z[\alpha]\:$ is a finitely generated $\:\mathbb Z$-module. Possibly also the motivation for the name had to do more specifically with rings of cyclotomic integers. However, as plausible as that may seem, I don't recall the existence of any historical documents that provide solid evidence in support of such speculations.
Beware that one has to be very careful when reading such older literature. Some authors mistakenly read modern notions into terms which have no such denotation in their original usage. To provide some context I recommend reading Lemmermeyer and Schappacher's Introduction to the English Edition of Hilbert’s Zahlbericht. Below is a pertinent excerpt.
Below is an excerpt from Leo Corry's Modern algebra and the rise of mathematical structures, p. 149.
Below are a couple typical examples of said speculative etymology of the term "ring" via the "circling back" nature of integral dependence, from Harvey Cohn's Advanced Number Theory, p. 49.
$\quad$The designation of the letter $\mathfrak D$ for the integral domain has some historical importance going back to Gauss's work on quadratic forms. Gauss $\left(1800\right)$ noted that for certain quadratic forms $Ax^2+Bxy+Cy^2$ the discriminant need not be square-free, although $A$, $B$, $C$ are relatively prime. For example, $x^2-45y^2$ has $D=4\cdot45$. The $4$ was ignored for the reason that $4|D$ necessarily by virtue of Gauss's requirement that $B$ be even, but the factor of $3^2$ in $D$ caused Gauss to refer to the form as one of "order $3$." Eventually, the forms corresponding to a value of $D$ were called an "order" (Ordnung). Dedekind retained this word for what is here called an "integral domain."
$\quad$The term "ring" is a contraction of "Zahlring" introduced by Hilbert $\left(1892\right)$ to denote (in our present context) the ring generated by the rational integers and a quadratic integer $\eta$ defined by $$\eta^2+B\eta+C=0.$$ It would seem that module $\left[1,\eta\right]$ is called a Zahlring because $\eta^2$ equals $-B\eta-C$ "circling directly back" to an element of $\left[1,\eta\right]$ . This word has been maintained today. Incidentally, every Zahlring is an integral domain and the converse is true for quadratic fields.
and from Rotman's Advanced Modern Algebra, p. 81.
-
"Beware that one has to be very careful when reading such older literature. Some authors mistakenly read modern notions into terms which have no such denotation in their original usage." deserves a +1 on its own. – J. M. Sep 3 '11 at 1:31
Thank you very much @Bill. – leo Sep 3 '11 at 2:05
Bill, I think it would be worth pointing readers to Kleiner's article From Numbers to Rings: The Early History of Ring Theory which you mentioned here. – t.b. Sep 3 '11 at 3:15
I don't think you missed anything. I just thought that people interested in the etymology of the word "ring" will likely be interested in the history of rings, and that's why I pointed to your other answer. – t.b. Sep 3 '11 at 3:57
Bravo for including 'original literature' – Simon S Jul 17 '15 at 11:50 | 1,337 | 5,219 | {"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} | 2.671875 | 3 | CC-MAIN-2016-18 | latest | en | 0.953575 |
https://physicsinventions.com/fourier-transforms-the-convolution-theorem/ | 1,606,702,435,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141204453.65/warc/CC-MAIN-20201130004748-20201130034748-00221.warc.gz | 421,894,972 | 9,385 | Fourier Transforms – The Convolution Theorem.
Ok so I’ve seen the convolution theorem written as:
F(h(x)$\otimes$g(x))=H(k)G(k)
(And this is how it appears when I have a quick google).
My book then does a problem in which is uses:
F(h(x)g(x))=H(k)$\otimes$G(k)
Where H(k)=F(h(x)) and similarly G(k)=F(g(x)),
and F represents a fourier transform
My question
– I can’t see how these are equivalent at all?
Many Thanks to anyone who can help shed some light !
http://ift.tt/1bSwkhd | 145 | 487 | {"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} | 2.796875 | 3 | CC-MAIN-2020-50 | longest | en | 0.869667 |
https://www.nagwa.com/en/videos/806185828161/ | 1,582,688,934,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146186.62/warc/CC-MAIN-20200226023658-20200226053658-00091.warc.gz | 815,863,009 | 5,390 | # Video: Identifying the Distributive Property
Identify the property demonstrated by the following equation: 4 × (14 + 9) = 4 × 14 + 4 × 9.
01:01
### Video Transcript
Identify the property demonstrated by the following equation: four times 14 plus nine equals four times 14 plus four times nine.
Let’s look closely at what is happening with the four on both sides of the equation. On the left side, four is being multiplied by one thing, whatever 14 plus nine is. On the right side of the equation, four is being multiplied by 14 and nine separately before you add them together. We call this distribution; we say that the four is being distributed to the 14 and the nine. The distributive property tells us that four times 14 plus nine will be the same thing as four times 14 plus four times nine. | 188 | 803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.922805 |
http://stackoverflow.com/questions/4129666/how-to-convert-hex-to-rgb-using-java?answertab=oldest | 1,438,081,048,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042981856.5/warc/CC-MAIN-20150728002301-00312-ip-10-236-191-2.ec2.internal.warc.gz | 228,745,410 | 21,895 | # How to convert hex to rgb using Java?
How can I convert hex color to RGB code in Java? Mostly in Google, samples are on how to convert from RGB to hex.
-
Can you give an example of what you're trying to convert from and what you're trying to convert to? Its not clear exactly what you're trying to do. – kkress Nov 9 '10 at 1:25
000000 will convert to black color rgb – user236501 Nov 9 '10 at 1:48
I guess this should do it:
``````/**
*
* @param colorStr e.g. "#FFFFFF"
* @return
*/
public static Color hex2Rgb(String colorStr) {
return new Color(
Integer.valueOf( colorStr.substring( 1, 3 ), 16 ),
Integer.valueOf( colorStr.substring( 3, 5 ), 16 ),
Integer.valueOf( colorStr.substring( 5, 7 ), 16 ) );
}
``````
-
I'd upvote, but I couldn't possibly break your 666 score. – Bob Jan 4 '13 at 22:32
Convert it to an integer, then divmod it twice by 16, 256, 4096, or 65536 depending on the length of the original hex string (3, 6, 9, or 12 respectively).
-
Hexidecimal color codes are already rgb. The format is #RRGGBB
-
Unless it's #RGB, #RRRGGGBBB, or #RRRRGGGGBBBB. – Ignacio Vazquez-Abrams Nov 9 '10 at 1:26
``````public static void main(String[] args) {
int hex = 0x123456;
int r = (hex & 0xFF0000) >> 16;
int g = (hex & 0xFF00) >> 8;
int b = (hex & 0xFF);
}
``````
-
A hex color code is #RRGGBB
RR, GG, BB are hex values ranging from 0-255
Let's call RR XY where X and Y are hex character 0-9A-F, A=10, F=15
The decimal value is X*16+Y
If RR = B7, the decimal for B is 11, so value is 11*16 + 7 = 183
``````public int[] getRGB(String rgb){
int[] ret = new int[3];
for(int i=0; i<3; i++){
ret[i] = hexToInt(rgb.charAt(i*2), rgb.charAt(i*2+1));
}
return ret;
}
public int hexToInt(char a, char b){
int x = a < 65 ? a-48 : a-55;
int y = b < 65 ? b-48 : b-55;
return x*16+y;
}
``````
-
I stumbled across post this when looking to do something else. Actually, there's an easier (built in) way of doing this:
``````Color.decode("#FFCCEE");
``````
-
unfortunately that is AWT :/ – wuppi Jan 4 '13 at 15:47
@wuppi I thought that was actually good news, as AWT is in JDK. What's so unfortunate about it? – chhh Mar 4 '14 at 21:22
Eclipse RCP is SWT, which I was working with. – wuppi May 27 '14 at 15:47
The accepted solution also uses AWT. AWT is not a problem for the original question asker. This should be the accepted solution. – axle123 Aug 6 '14 at 21:04
On android: Color.parseColor() – Dawid Drozd Apr 1 at 12:16
For Android development, I use:
``````int color = Color.parseColor("#123456");
``````
-
To elaborate on the answer @xhh provided, you can append the red, green, and blue to format your string as "rgb(0,0,0)" before returning it.
``````/**
*
* @param colorStr e.g. "#FFFFFF"
* @return String - formatted "rgb(0,0,0)"
*/
public static String hex2Rgb(String colorStr) {
Color c = new Color(
Integer.valueOf(hexString.substring(1, 3), 16),
Integer.valueOf(hexString.substring(3, 5), 16),
Integer.valueOf(hexString.substring(5, 7), 16));
StringBuffer sb = new StringBuffer();
sb.append("rgb(");
sb.append(c.getRed());
sb.append(",");
sb.append(c.getGreen());
sb.append(",");
sb.append(c.getBlue());
sb.append(")");
return sb.toString();
}
``````
-
you can do it simply as below:
`````` public static int[] getRGB(final String rgb)
{
final int[] ret = new int[3];
for (int i = 0; i < 3; i++)
{
ret[i] = Integer.parseInt(rgb.substring(i * 2, i * 2 + 2), 16);
}
return ret;
}
``````
For Example
``````getRGB("444444") = 68,68,68
getRGB("FFFFFF") = 255,255,255
``````
-
Lots of these solutions work, but this is an alternative.
``````String hex="#00FF00"; // green
long thisCol=Long.decode(hex)+4278190080L;
int useColour=(int)thisCol;
``````
If you don't add 4278190080 (#FF000000) the colour has an Alpha of 0 and won't show.
-
The other day I'd been solving the similar issue and found convenient to convert hex color string to int array [alpha, r, g, b]:
`````` /**
* Hex color string to int[] array converter
*
* @param hexARGB should be color hex string: #AARRGGBB or #RRGGBB
* @return int[] array: [alpha, r, g, b]
* @throws IllegalArgumentException
*/
public static int[] hexStringToARGB(String hexARGB) throws IllegalArgumentException {
if (!hexARGB.startsWith("#") || !(hexARGB.length() == 7 || hexARGB.length() == 9)) {
throw new IllegalArgumentException("Hex color string is incorrect!");
}
int[] intARGB = new int[4];
if (hexARGB.length() == 9) {
intARGB[0] = Integer.valueOf(hexARGB.substring(1, 3), 16); // alpha
intARGB[1] = Integer.valueOf(hexARGB.substring(3, 5), 16); // red
intARGB[2] = Integer.valueOf(hexARGB.substring(5, 7), 16); // green
intARGB[3] = Integer.valueOf(hexARGB.substring(7), 16); // blue
} else hexStringToARGB("#FF" + hexARGB.substring(1));
return intARGB;
}
``````
- | 1,513 | 4,765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2015-32 | latest | en | 0.685605 |
https://www.esaral.com/q/for-each-binary-operation-defined-below-determine-whether-is-commutative-or-associative-98553 | 1,718,680,439,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00051.warc.gz | 666,340,316 | 12,842 | # For each binary operation * defined below, determine whether * is commutative or associative.
Question:
For each binary operation * defined below, determine whether * is commutative or associative.
(i) On $\mathbf{Z}$, define $a^{*} b=a-b$
(ii) On Q, define $a^{*} b=a b+1$
(iii) On $\mathbf{Q}$, define $a^{*} b=\frac{a b}{2}$
(iv) On Z^ , define $a^{*} b=2^{a b}$
(v) On $\mathbf{Z}^{+}$, define $a^{*} b=a^{b}$
(vi) On $\mathbf{R}-\{-1\}$, define $a * b=\frac{a}{b+1}$
Solution:
(i) On $\mathbf{Z},{ }^{*}$ is defined by $a{ }^{*} b=a-b$.
It can be observed that $1^{*} 2=1-2=1$ and $2^{*} 1=2-1=1$.
$\therefore 1^{*} 2 \neq 2^{*} 1 ;$ where $1,2 \in Z$
Hence, the operation * is not commutative.
Also we have:
$\left(1^{*} 2\right)^{*} 3=(1-2)^{*} 3=-1 * 3=-1-3=-4$
$1^{*}\left(2^{*} 3\right)=1^{*}(2-3)=1^{*}-1=1-(-1)=2$
$\therefore\left(1^{*} 2\right) * 3 \neq 1^{*}(2 * 3)$; where $1,2,3 \in \mathbf{Z}$
Hence, the operation * is not associative.
(ii) On $\boldsymbol{O}^{*}$ is defined by $\boldsymbol{a}^{*} b=a b+1$
It is known that:
$a b=b a \& m n F o r E ; a, b \in \mathbf{Q}$
$\Rightarrow a b+1=b a+1 \& m n$ ForE; $a, b \in \mathbf{Q}$
$\Rightarrow a^{*} b=a^{*} b \& m n F o r E ; a, b \in \mathbf{Q}$
Therefore, the operation * is commutative.
It can be observed that:
$\left(1^{*} 2\right)^{*} 3=(1 \times 2+1)^{*} 3=3 * 3=3 \times 3+1=10$ $1^{*}\left(2^{*} 3\right)=1^{*}(2 \times 3+1)=1^{*} 7=1 \times 7+1=8$
$\therefore\left(1^{*} 2\right)^{*} 3 \neq 1^{*}\left(2^{*} 3\right) ;$ where $1,2,3 \in \mathbf{Q}$
Therefore, the operation * is not associative.
(iii) On $\mathbf{Q},{ }^{*}$ is defined by $a^{*} b=\frac{a b}{2}$.
It is known that:
$a b=b a \& m n F o r E ; a, b \in \mathbf{Q}$
⇒ &mnForE; a, b ∈ Q
⇒ * b = * a &mnForE; a, b ∈ Q
Therefore, the operation * is commutative.
For all $a, b, c \in \mathbf{Q}$, we have:
$(a * b) * c=\left(\frac{a b}{2}\right) * c=\frac{\left(\frac{a b}{2}\right) c}{2}=\frac{a b c}{4}$
$a *(b * c)=a *\left(\frac{b c}{2}\right)=\frac{a\left(\frac{b c}{2}\right)}{2}=\frac{a b c}{4}$
$\therefore(a * b) * c=a *(b * c)$
Therefore, the operation * is associative.
(iv) On $\mathbf{Z}^{+},{ }^{*}$ is defined by $a^{*} b=2^{a b}$.
It is known that:
$a b=b a \& m n F o r E ; a, b \in \mathbf{Z}^{+}$
$\Rightarrow 2^{a b}=2^{b a} \& m n F o r E ; a, b \in \mathbf{Z}^{+}$
$\Rightarrow a^{*} b=b^{*} a \& m n F o r E ; a, b \in \mathbf{Z}^{+}$
Therefore, the operation * is commutative.
It can be observed that:
$(1 * 2) * 3=2^{(1 \times 2)} * 3=4 * 3=2^{4 \times 3}=2^{12}$
$1 *(2 * 3)=1 * 2^{2 \times 3}=1 * 2^{6}=1 * 64=2^{64}$
$\therefore\left(1^{*} 2\right)^{*} 3 \neq 1^{*}\left(2^{*} 3\right) ;$ where $1,2,3 \in \mathbf{Z}^{+}$
(v) On $\mathbf{Z}^{+},{ }^{*}$ is defined by $a^{*} b=a^{b}$.
It can be observed that:
$1 * 2=1^{2}=1$ and $2 * 1=2^{1}=2$
$\therefore 1^{*} 2 \neq 2^{*} 1$; where $1,2 \in \mathbf{Z}^{+}$
It can also be observed that:
$(2 * 3) * 4=2^{3} * 4=8 * 4=8^{4}=\left(2^{3}\right)^{4}=2^{12}$
$2 *(3 * 4)=2 * 3^{4}=2 * 81=2^{81}$
$\therefore(2 * 3) * 4 \neq 2 *(3 * 4) ;$ where $2,3,4 \in \mathbf{Z}^{+}$
Therefore, the operation * is not associative.
(vi) On $\mathbf{R},{ }^{*}-\{-1\}$ is defined by $a * b=\frac{a}{b+1}$.
It can be observed that $1 * 2=\frac{1}{2+1}=\frac{1}{3}$ and $2 * 1=\frac{2}{1+1}=\frac{2}{2}=1$.
$\therefore 1^{*} 2 \neq 2^{*} 1$; where $1,2 \in \mathbf{R}-\{-1\}$
Therefore, the operation * is not commutative.
It can also be observed that:
$(1 * 2) * 3=\frac{1}{3} * 3=\frac{\frac{1}{3}}{3+1}=\frac{1}{12}$
$1 *(2 * 3)=1 * \frac{2}{3+1}=1 * \frac{2}{4}=1 * \frac{1}{2}=\frac{1}{\frac{1}{2}+1}=\frac{1}{\frac{3}{2}}=\frac{2}{3}$
$\therefore\left(1^{*} 2\right)^{*} 3 \neq 1^{*}\left(2^{*} 3\right) ;$ where $1,2,3 \in \mathbf{R}-\{-1\}$
Therefore, the operation * is not associative. | 1,825 | 3,866 | {"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} | 4.28125 | 4 | CC-MAIN-2024-26 | latest | en | 0.381106 |
https://www.perlmonks.org/?node_id=1014325 | 1,550,829,194,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247515149.92/warc/CC-MAIN-20190222094419-20190222120419-00427.warc.gz | 931,655,823 | 6,260 | Welcome to the Monastery PerlMonks
### Re^3: Prime Number Finder
by chith (Initiate)
on Jan 20, 2013 at 19:45 UTC ( #1014325=note: print w/replies, xml ) Need Help??
in reply to Re^2: Prime Number Finder
this is the actual code which works for windows
```#!usr/bin/perl -w
use strict;
use warnings;
my \$o = 2;
print "enter upto what number you wish to generate the primes: ";
my \$e = <STDIN>;
my (\$i,\$j,\$p);
my @prime_=();
print "prime numbers are:\n";
for(\$i=\$o; \$i<=\$e; \$i++)
{
\$p=0;
for(\$j=1; \$j<=\$i; \$j++)
{
if(\$i % \$j== 0)
{
\$prime_[\$p] = "\$j";
\$p++;
}
if (\$prime_[1] == \$i)
{
print "\$i\t";
}
}
}
print"\n";
Replies are listed 'Best First'.
Re^4: Prime Number Finder
by Anonymous Monk on Sep 17, 2013 at 08:46 UTC
This code is quite fast, since it skips a lot of unnecessary operations
```#!/usr/bin/perl
use strict;
use warnings;
use POSIX;
my (\$i,\$j,\$h,\$sentinel) = (0,0,0,0);
# i>=3
for(\$i=1000000000; \$i<=1000000500; \$i++){
# if \$i is an even number, it can't be a prime
if(\$i%2==0){}
else{
\$h=POSIX::floor(sqrt(\$i));
\$sentinel=0;
# since \$i can't be even -> only divide by odd numbers
for(\$j=3; \$j<=\$h; \$j+=2){
if(\$i%\$j==0){
\$sentinel++;
# \$i is not a prime, we can get out of the loop
\$j=\$h;
}
}
if(\$sentinel==0){
print "\$i \n";
}
}
}
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Notices? | 589 | 1,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-09 | longest | en | 0.646024 |
https://drawabox.com/community/submission/NR1JCXEH | 1,660,976,325,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573908.30/warc/CC-MAIN-20220820043108-20220820073108-00436.warc.gz | 217,440,944 | 12,352 | I'll be the TA handling your Lesson 2 critique.
You're making progress towards understanding the concepts introduced in this lesson and hopefully this critique will help you in your future attempts.
• Starting off in the arrows section your lines are looking smoothly and confidently drawn. You're doing a good job maintaining a consistent width as your arrows widen while moving closer to the viewer and with more mileage you'll become more consistent. There are spots where your arrows bulge/narrow suddenly, this is an issue because it gives the impression that your arrows are stretching which hurts their solidity. Remember that as our arrows move closer to the viewer we want them to widen consistently. This is a good exercise to experiment with line weight but when applying it we want to make sure we do subtly to key areas like overlaps to give clarity to our forms. Here are some things to look out for when applying line weight, and here are some reminders on how to apply it subtly. Great use of foreshortening so far, by utilizing it in both the arrows themselves as well as the negative space between their curves we can create a stronger illusion of an arrow moving through 3D space as you can see demonstrated here.
• Moving into the organic forms with contours exercise you're doing a good job keeping your forms simple, plenty of people tend to over-complicate them. You're keeping your line work confident here which is great, if you feel uncomfortable working with contours still don't stress with more mileage it'll become more natural. Speaking of contours I'd like you to try and shift the degree of your contours more. The degree of a contour line basically represents the orientation of that cross-section in space, relative to the viewer, and as we slide along the sausage form, the cross section is either going to open up (allowing us to see more of it) or turn away from the viewer (allowing us to see less), as shown here.
• In the texture exercises you're focusing largely on outlines, form shadows and negative space rather than cast shadows created by forms along the texture itself. This makes it difficult to create gradients with implied information which we could then use to create focal points in more complex pieces, by doing so we can prevent our viewers from being visually overwhelmed with too much detail. For more on the importance of focusing on cast shadows read here. I'd also like to quickly direct you to this image which shows that when we're working with thin line like textures if we outline and fill the shadow we will create a much more dynamic texture than simply drawing lines.
• It's quite common for people to feel like they don't fully grasp the form intersections exercise, if you feel like you may fall into this category try not to stress too much. This exercise is just meant to get students to start thinking about how their forms relate to one another in 3D space, and how to define those relationships on the page. We'll be going over them more in the upcoming lessons.Your forms are looking quite solid here and they believably appear to belong in the same cohesive 3D space, good work.
• While wrapping up your submission with the organic intersections exercise you do a great job demonstrating that your sense of 3D space is developing as your forms begin to wrap around each other believably. You're keeping your forms simple and easy to work with which is a good strategy to help produce good results. I'd like you to draw through all of your forms when trying this exercise again in the future. By drawing through our forms we'll develop a better understanding of the 3D space we're attempting to construct, just like when we did with our boxes. When it comes to your shadows you're pushing them enough so that they cast rather than just hugging the form that creates them which is a great start. Your shadows appear to be following a consistent light source, be sure to experiment with different angles and intensities when trying this exercise again in the future. I recommend pushing your light source to the top left or right corner of the page to start with, it's easier than working with a light directly above your form pile.
Overall this was a solid submission, while you may have some things to work on I have no doubt you will improve with more mileage. I'll be marking your submission as complete and moving you on to the next lesson.
Keep practicing previous exercises as warm ups and good luck in lesson 3! | 890 | 4,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-33 | longest | en | 0.969715 |
http://www.solving-math-problems.com/pre-algebra-fractions-parentheses.html | 1,653,361,923,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00390.warc.gz | 108,339,483 | 11,926 | # pre algebra - fractions - parentheses
by shannon
(pinehill)
how do you solve (2/3 + 1/6) = 2/3 + (1/6 + 1/3)
Is this equality true or false?
hint: switch the parentheses on the right side using the Associative Property of Addition.
### Comments for pre algebra - fractions - parentheses
Jan 30, 2011 pre algebra - fractions - parentheses by: Staff The question: by Shannon (Pine Hill, New Jersey, USA) how do you solve (2/3+1/6)=2/3+(1/6+1/3) The answer: The fastest way to see if this expression is a valid equality is to switch the parentheses on the right side using the Associative Property of Addition. Associative Property of Addition: Any series of addends can be grouped in any way and added in any order without changing the results. (2/3 + 1/6) = 2/3 + (1/6 + 1/3) (2/3 + 1/6) = (2/3 + 1/6) + 1/3 It is now obvious that the value on the right side of the equal sign is 1/3 larger than the value on the left side of the equal sign. Therefore, the expression should read: (2/3 + 1/6) ≠ (2/3 + 1/6) + 1/3 Or (2/3 + 1/6) < (2/3 + 1/6) + 1/3 Thanks for writing. Staff www.solving-math-problems.com | 354 | 1,111 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-21 | longest | en | 0.831537 |
https://studygolang.com/articles/34434 | 1,618,137,350,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061820.19/warc/CC-MAIN-20210411085610-20210411115610-00335.warc.gz | 646,180,741 | 11,675 | # Golang二维切片初始化
Dreamcats · · 132 次点击 · · 开始浏览
# Golang二维切片初始化
## 引言
``````var a = [5]int{1, 2, 3, 4, 5} // 用var
b := [5]int{1, 2, 3, 4, 5} // 用类型推断
var c = [...]int{1, 2, 3, 4, 5} // 不确定长度
d := [...]int{1, 2, 3, 4, 5}
``````
``````var a [5]int
b := [5]int{}
c := make([]int, 5)
``````
## 二维数组初始化
``````var a [3][4]int
``````
``````[[0 0 0 0] [0 0 0 0] [0 0 0 0]]
``````
``````a := [3][4]int{}
``````
``````[[0 0 0 0] [0 0 0 0] [0 0 0 0]]
``````
``````m, n := 3, 4
var a [m][n]int // 报错，必须是常量表达式
``````
``````func main() {
m, n := 3, 4
a := make([][]int, m) // 二维切片，3行
for i := range a {
a[i] = make([]int, n) // 每一行4列
}
fmt.Println(a)
}
``````
``````[[0 0 0 0] [0 0 0 0] [0 0 0 0]]
``````
``````var a []int // or a := []int{}
fmt.Println(a) // []
a[0] = 1 // 报错
``````
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# Golang二维切片初始化
## 引言
``````var a = [5]int{1, 2, 3, 4, 5} // 用var
b := [5]int{1, 2, 3, 4, 5} // 用类型推断
var c = [...]int{1, 2, 3, 4, 5} // 不确定长度
d := [...]int{1, 2, 3, 4, 5}
``````
``````var a [5]int
b := [5]int{}
c := make([]int, 5)
``````
## 二维数组初始化
``````var a [3][4]int
``````
``````[[0 0 0 0] [0 0 0 0] [0 0 0 0]]
``````
``````a := [3][4]int{}
``````
``````[[0 0 0 0] [0 0 0 0] [0 0 0 0]]
``````
``````m, n := 3, 4
var a [m][n]int // 报错，必须是常量表达式
``````
``````func main() {
m, n := 3, 4
a := make([][]int, m) // 二维切片，3行
for i := range a {
a[i] = make([]int, n) // 每一行4列
}
fmt.Println(a)
}
``````
``````[[0 0 0 0] [0 0 0 0] [0 0 0 0]]
``````
``````var a []int // or a := []int{}
fmt.Println(a) // []
a[0] = 1 // 报错
`````` | 949 | 1,660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-17 | latest | en | 0.32497 |
https://math.stackexchange.com/questions/1017759/graphing-in-polar-coordinates | 1,566,425,163,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316549.78/warc/CC-MAIN-20190821220456-20190822002456-00427.warc.gz | 539,361,660 | 30,512 | # Graphing in Polar Coordinates
I´m currently using polar coordinates to calculate some double and triple integrals. However, I have an small doubt; when you are want to express, lets say, a circle of radius $a$ centered in $(a,0)$ you can write it in polar coordinates and say that $x=r\cos\theta$ .and $y=r\sin \theta$ and so, as $(x-a)^2+y^2=a^2$, then $r=2a\cos\theta$.
My question is: Why is it that $-\pi/2<\theta<\pi/2$?
Please excuse me for such a dull question.
The graph of $r = 2a \cos \theta$ for $-\pi < \theta \leq \pi$ is a sideways figure 8, with the circle you want on the right, and the circle you don't want on the left. The circle on the left is obtained from $-\pi < \theta \ -\pi/2$ (lower half) and $\pi/2 \leq \theta \leq \pi$ (upper half). So to get just your one circle you have to restrict the range to not use those values. | 262 | 855 | {"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} | 3.359375 | 3 | CC-MAIN-2019-35 | latest | en | 0.880537 |
http://www.slideserve.com/kineta/flux-emergence-the-storage-of-magnetic-free-energy | 1,508,845,260,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828411.81/warc/CC-MAIN-20171024105736-20171024125736-00595.warc.gz | 566,884,205 | 20,564 | Flux Emergence & the Storage of Magnetic Free Energy
1 / 44
# Flux Emergence & the Storage of Magnetic Free Energy - PowerPoint PPT Presentation
Flux Emergence & the Storage of Magnetic Free Energy. Brian T. Welsch Space Sciences Lab, UC-Berkeley Flares and coronal mass ejections (CMEs) are driven by the release of free magnetic energy stored in the coronal magnetic field.
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Presentation Transcript
### Flux Emergence & the Storage of Magnetic Free Energy
Brian T. Welsch
Space Sciences Lab, UC-Berkeley
Flares and coronal mass ejections (CMEs) are driven by the release of free magnetic energy stored in the coronal magnetic field.
How does free magnetic energy enter the corona?
How is this energy stored?
What triggers its release?
How can ground-based solar observations address these questions?
Flares and CMEs are powered by energy in the coronal magnetic field.
From T.G. Forbes, “A Review on the Genesis of Coronal Mass Ejections”, JGR (2000)
The hypothetical coronal magnetic field with lowest energy is current-free, or “potential.”
• For a given coronal field Bcor, the coronal magnetic energy is:
• U dV (Bcor·Bcor)/8.
• The lowest energy coronal field would have current J = 0, and Ampére says 4πJ/c = x B, so x Bmin= 0.
• A curl-free vector field can be expressed as the gradient of a scalar potential, Bmin = -. (Since ·Bmin= -2 = 0, it’s easy to solve!)
• Umin dV (Bmin·Bmin)/8
• The difference U(F) = [U – Umin] is “free” energy stored in the corona, which can be suddenly released in flares or CMEs.
Coronal J cannot currently be observationally constrained; measurements of (vector) Bcor(x, y, z) haven’t been made.
In consequence, we must use indirect means to infer
properties of coronal currents.
When not flaring, Bphotosphere and Bchromosphereare coupled to Bcor, so provide valuable insights.
While Bcor can evolve on its own, changes in BphorBchwill induce changes in the coronal field Bcor.
In addition, following active region (AR) fields in time can provide information about their history and development.
Unfortunately, our ignorance regarding free magnetic energy in the corona is profound!
1. Physically, how does free energy enter the corona?
• Practically, how can we detect this buildup?
2. Physically, how is this energy stored?
• Practically, how can we quantify it once it’s there?
3. Physically, what triggers its release?
• Practically, how can we predict when release is imminent?
Common theme: we want to know vectorB (so we can infer J)from the photosphere upward into the corona.
Some Key Roles for Ground Based Efforts:
• Develop means to estimate vector B(x,y,z) in chromosphere and corona (radio, IR).
Short answer to #1: Energy comes from the interior!But how?
EUV image of ~1MK plasma
• Image credits: George Fisher, LMSAL/TRACE
What physical processes produce the electric currents that store energy in Bcor? Two options are:
(i) Currents could form in the interior, then emerge into the corona.
• Current-carrying magnetic fields have been observed to emerge (e.g., Leka et al. 1996, Okamoto et al. 2008).
(ii) Photospheric evolution could induce currents in already-emerged coronal magnetic fields.
• From simple scalings, McClymont & Fisher (1989) argued induced currents would be too weak to power large flares.
• Longcope et al. (2007), Kazachenko et al. (2009), and others argue that strong enough currents can be induced.
Both models involve slow buildup, then sudden release.
For (i), note that currents can emerge in two distinct ways!
b) vertical transport of cur-
rents in emerged flux
a) emergence of new flux
(Should be twisted, but I can’t draw
twist very well!)
Ishii et al., ApJ v.499, p.898 1998
NB: New flux only emerges along polarity inversion lines!
NB: This does not increase total unsigned photospheric flux.
For (ii), if coronal currents induced by post-emergence photospheric evolution drive flares and CMEs, then:
The evolving coronal magnetic field must be modeled!
NB: Induced currents close along or above the photosphere --- they are not driven from below.
==> All available energy in these currents can be released.
Longcope, Sol. Phys. v.169, p.91 1996
Back to the catalog of our ignorance regarding free magnetic energy in the corona:
1. Physically, how does free energy enter the corona?
• Practically, how can we detect this buildup?
2. Physically, how is this energy stored?
• Practically, how can we quantify it once it’s there?
3. Physically, what triggers its release?
• Practically, how can we predict when release is imminent?
Back to the catalog of our ignorance regarding free magnetic energy in the corona:
1. Physically, how does free energy enter the corona?
• Practically, how can we detect this buildup?
Statistically?
2. Physically, how is this energy stored?
• Practically, how can we quantify it once it’s there?
3. Physically, what triggers its release?
• Practically, how can we predict when release is imminent?
Statistical methods have been used to correlate observables with flare & CME activity, including:
• Total flux in active regions, vertical current (e.g., Leka et al. 2007)
• Flux near polarity inversion lines (PILs; e.g., Falconer et al. 2001-2009; Schrijver 2007)
• “Proxy” Poynting flux, vhBR2 (e.g., Welsch et al. 2009)
• Subsurface flows (e.g., Reinard et al. 2010, Komm et al. 2011)
• Magnetic power spectra (e.g., Abramenko & Yurchyshyn, 2010)
It’s challenging to infer physics from correlations, so I will emphasize more deterministic approaches here.
Back to the catalog of our ignorance regarding free magnetic energy in the corona:
1. Physically, how does free energy enter the corona?
• Practically, how can we detect this buildup?
Catch it in the act!
2. Physically, how is this energy stored?
• Practically, how can we quantify it once it’s there?
3. Physically, what triggers its release?
• Practically, how can we predict when release is imminent?
In principle, electric fields derived from magnetogram evol-ution can quantify energy or helicity fluxes into the corona.
• The fluxes of magnetic energy & helicityacross the magnetogram surface into the corona depend upon E:
dU/dt = ∫ dA (E x B)z /4π
dH/dt = 2 ∫ dA (E x A)z
U and H probably play central roles in flares / CMEs.
• Coupling of Bcor to B beneath the corona implies estimates of E there provide boundary conditions for data-driven,time-dependent simulations of Bcor.
• “Component methods” derive vor Eh from the normal component of the ideal induction equation,
Bz/t = -c[ hxEh ]z= [ x(vx B) ]z
• But the vectorinduction equation can place additional constraints on E:
B/t = -c(xE)= x(vx B),
where I assume the ideal Ohm’s Law,*so v<--->E:
E = -(vx B)/c==>E·B = 0
*One can instead use E = -(vx B)/c + R, if some model resistivity R is assumed.
(I assume R might be a function of B or J or ??, but is not a function of E.)
While tB provides more information about E than tBz alone, it still does not fully determine E.
• Faraday’s Law only relates tB to the curl of E, not E itself; a gauge electric fieldψ is unconstrained by tB.
(Ohm’s Law does not fully constrain E.)
• tBh also depends upon vertical derivativesin Eh, which single-heightmagnetograms do not fully constrain.
• Doppler data can provide additional info.
Some Key Roles for Ground Based Efforts:
• Develop means to estimate vector B(x,y,z) in chromosphere and corona (radio, IR).
• Routine magnetograms from atmospheric layers different than space-based observations. (Cadence c. 30 min?)
While tB provides more information about E than tBz alone, it still does not fully determine E.
• Faraday’s Law only relates tB to the curl of E, not E itself; a gauge electric fieldψ is unconstrained by tB.
(Ohm’s Law does not fully constrain E.)
• tBh also depends upon vertical derivativesin Eh, which single-heightmagnetograms do not fully constrain.
• Doppler data can provide additional info.
Doppler data helps because emerging flux might have little or no inductive signature at the emergence site.
Schematic illustration of flux emergence in a bipolar magnetic region, viewed in cross-section normal to the polarity inversion line (PIL).
Note the strong signature of the field change at the edges of the region, while the field change at the PIL is zero.
For instance, the “PTD” method (Fisher et al. 2010, 2011) can be used to estimate E:
• In addition to tBz, PTD uses information from tJz in the derivation of E.
• No tracking is used to derive E, but tracking methods (ILCT, DAVE4VM) can provide extra info!
• Using Doppler data improves PTD’s accuracy!
For more about PTD, see Fisher et al. 2010 (ApJ 715 242) and Fisher et al. 2011 (Sol. Phys. in press; arXiv:1101.4086).
Quantitative tests with simulated data show Doppler information improves recovery of E-field and Poynting flux Sz.
Upper left:
MHD Szvs.
PTD Sz.
Lower left:
MHD Szvs.
PTD +
FLCT Sz.
Upper right:
MHD Szvs.
PTD + Doppler Sz.
Lower right:
MHD Szvs.
PTD + Doppler +
FLCT Sz.
Poyntingflux units are in [105 G2 km s−1]
There’s a well-known intensity-blueshift correlation, because rising plasma (which is hotter) is brighter (see, e.g., Gray 2009; Hamilton and Lester 1999; or talk to P. Scherrer).
(Helioseismology uses time differences in Doppler shifts, so this issue isn’t a problem.)
Because magnetic fields suppress convection, lines are redshifted in magnetized regions.
Consequently, absolute calibration of Doppler shifts is essential.
From Gray (2009): Bisectors for 13 spectral lines on the Sun are shown on an absolute velocity scale. The dots indicate the lowest point on the bisectors. (The dashed bisector is for λ6256.)
Lines formed deeper in the atmosphere, where convective upflows are present, are blue-shifted.
HMI data clearly exhibit this effect. How can this bias best be corrected?
An automated method (from Welsch & Li 2008) identified PILs in a subregion of AR 11117. Note predominance of redshifts.
Some Key Roles for Ground Based Efforts:
• Develop means to estimate vector B(x,y,z) in chromosphere and corona (radio, IR).
• Routine magnetograms from atmospheric layers different than space-based observations. Cadence c. 30 min.?
• Minor: Estimates of absolute Doppler shifts.
Back to the catalog of our ignorance regarding free magnetic energy in the corona:
1. Physically, how does free energy enter the corona?
• Practically, how can we detect this buildup?
Catch it in the act!
2. Physically, how is this energy stored?
• Practically, how can we quantify it once it’s there?
Can infer existence of energy from coronal observations…
3. Physically, what triggers its release?
• Practically, how can we predict when release is imminent?
A cartoon model helps bias the mind…
A quasi-stable balance can exist between outward magnetic pressure and inward magnetic tension.
From Moore et al. (2001)
Non-potential fields are evinced by filaments / prominences, and sheared H-α fibrils & coronal loops.
Non-potential structures can persist for weeks, then flare or erupt suddenly.
Courtesy Tom Berger
Back to the catalog of our ignorance regarding free magnetic energy in the corona:
1. Physically, how does free energy enter the corona?
• Practically, how can we detect this buildup?
Catch it in the act!
2. Physically, how is this energy stored?
• Practically, how can we quantify it once it’s there?
Requires quantitative modeling of coronal field.
3. Physically, what triggers its release?
• Practically, how can we predict when release is imminent?
The Minimum Current Corona (MCC) approach can be used to estimate a lower-bound on coronal energy.
Method: Determine linkages from initial magnetogram, infer coronal currents (and therefore energy) based upon magnetogram evolution.
Separators with large currents have been related to flare sites.
Kazachenko et al. (2011)
Mark gets more free energy with an ad-hoc assumption for h·Eh -- estimates of E from observations would be better!
Some Key Roles for Ground Based Efforts:
• Develop means to estimate vector B(x,y,z) in chromosphere and corona (radio, IR).
• Routine magnetograms from atmospheric layers different than space-based observations. Cadence c. 30 min.?
• Minor: Estimates of absolute Doppler shifts.
• Modeling!
Back to the catalog of our ignorance regarding free magnetic energy in the corona:
1. Physically, how does free energy enter the corona?
• Practically, how can we detect this buildup?
Catch it in the act!
2. Physically, how is this energy stored?
• Practically, how can we quantify it once it’s there?
Requires quantitative modeling of coronal field.
3. Physically, what triggers its release?
• Practically, how can we predict when release is imminent?
Again, quantitative modeling of the coronal field is needed.
“Potentiality” should imply little free energy, and little likelihood of flaring.
Schrijver et al. (2005) found potential ARs were relatively unlikely to flare.
“Non-potentiality” should imply non-zero free energy, and increased likelihood of flaring.
Schrijver et al. (2005) found non-potential ARs were more likely to flare, with fields becoming more potential over 10-30 hours.
The hot, “chewynougat” in the core of thisnon-potential structure --- visible in SXT --- persists for months.
Evidently, the corona can store free energy for long times!
• What would cause this to erupt?
Inferring the presence of free energy is not enough to predict its release!
Non-potential structures can persist for weeks, then flare or erupt suddenly.
Hudson et al. (1999)
How can coronal “susceptibility” to destabilization (from emergence, or twisting, or reconnection) be inferred?
Schrijver includes several effects here; he argues only one (emergence) triggers eruptions. But this is speculative!
Schrijver ASR v. 43 p. 789 2009
Some Key Roles for Ground Based Efforts:
• Develop means to estimate vector B(x,y,z) in chromosphere and corona (radio, IR).
• Routine magnetograms from atmospheric layers different than space-based observations. Cadence c. 30 min.?
• Minor: Estimates of absolute Doppler shifts.
• Modeling!
• And more modeling!
Recommendations for Ground Based Efforts:
• Develop means to estimate vector B(x,y,z) in chromosphere and corona (radio, IR).
Facilities: ATST, FASR, COMP/COSMO; modeling!
• Routine magnetograms from atmospheric layers different than space-based observations. Cadence c. 30 min.?
• Global networks: GONG; SOLIS-like vect. m’graphs
• Modeling! And more modeling!
• Support both “practical” (for SWx) & “impractical”
(for curiosity) efforts, esp. data-driven modeling | 3,860 | 15,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.8492 |
http://math.stackexchange.com/questions/206342/exercise-on-fundamentals-of-divisibility?answertab=votes | 1,461,947,305,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111374.13/warc/CC-MAIN-20160428161511-00221-ip-10-239-7-51.ec2.internal.warc.gz | 182,913,133 | 16,846 | # Exercise on Fundamentals of Divisibility
(This question is on page 236 of Falko Lorenz's Algebra Volume 1: Fields and Galois Theory, exercise 4.5)
Prove that $Y^2 = X^3 - 2$ has exactly one solution in the natural numbers.
(Hint: use the fact that $\mathbb{Z}\sqrt{-2}$ is a Euclidean domain, and that in any unique factorization domain $R$, if $\alpha_1,\dots ,\alpha_n$ are pairwise rel. prime in $R$ and their product is an m-th power in $R$, each $\alpha_i$ is associated to an m-th power in $R$.
Here's what I know. Since $\mathbb{Z}\sqrt{-2}$ is a Euclidean domain, then it is also a unique factorization domain. As well, $X^3=Y^2+2=Y^2-\sqrt{-2}^2=(Y-\sqrt{-2})(Y+\sqrt{-2})$. Clues for the clueless?
-
An outline: The units are $\pm 1$. Show that $X$ and $Y$ must be odd. Then show $Y+\sqrt{-2}$ and $Y-\sqrt{-2}$ are relatively prime. So each is a cube.
Let $Y+\sqrt{-2}=(a+b\sqrt{-2})^3$. Expand. Compute in particular the coefficient of $\sqrt{-2}$. This must be $1$. That will tell you something very important about $b$. But then you will know what $a$ must be, more or less. That will give the only solutions, and the only positive one.
Ok I got that $1=3a^2b-2b^3$. I'm confused about what this should be telling me? – Chris Oct 3 '12 at 5:05
That $b$ divides $1$! – André Nicolas Oct 3 '12 at 5:06
Well I got it, $X=3$ and $Y=5$! – Chris Oct 3 '12 at 5:13
OK, but remember there are (deliberately) little missing steps in my outline. Little proofs to fill in. Like what are the units? (Norm argument) Any solutions must be odd (easy). $X+\sqrt{-2}$ and its conjugate are relatively prime in the ring. For that I used fact that solutions are odd. – André Nicolas Oct 3 '12 at 5:20 | 540 | 1,703 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2016-18 | latest | en | 0.888081 |
https://homework.cpm.org/category/CCI_CT/textbook/int1/chapter/11/lesson/11.1.3/problem/11-31 | 1,716,964,948,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059206.29/warc/CC-MAIN-20240529041915-20240529071915-00860.warc.gz | 248,347,747 | 15,351 | ### Home > INT1 > Chapter 11 > Lesson 11.1.3 > Problem11-31
11-31.
Over a three-year period, the value of a vintage VHS tape of “Charlene’s Greatest Hits” decreased from $20$ to $1.99$.
1. What were the annual multiplier and annual percent decrease?
$20(m)(m)(m) = 1.99$
The multiplier is $0.463$.
How can you use that to find the percent decrease?
2. Write the equation of a function that describes this situation.
Use the starting value and multiplier you found in part (a) to write the equation.
$y = 20(0.463)^{x}$
3. Make a sketch of the graph of the function and a table of values for the first 5 years. | 176 | 618 | {"found_math": true, "script_math_tex": 5, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-22 | latest | en | 0.898063 |
https://www.geegeez.co.uk/tag/individual-stall-bias/ | 1,653,734,316,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00412.warc.gz | 892,853,941 | 34,526 | ## Draw Bias 2022: Part 2
In the first article in this series I looked at how the draw can influence the market and how the market can change over time to compensate, writes Dave Renham.
Occasionally the market still gets it wrong regarding draw bias but that is increasingly rare. This is because horse racing betting markets are usually extremely efficient (by the time the race goes off, at least), not just taking the draw into account, but multiple other key factors. In this article I am going to share more draw-based research that I hope you will find interesting and ultimately useful for your own betting.
For those Gold members of Geegeez, the good news is that you are able to research the draw in two places: the Draw Analyser and the Query Tool. How you use each to study the draw is partly personal choice, but I would suggest that best insights are obtained when deploying both, not just one or the other; I use both tools for my research. Essentially, if I am just looking at the draw and nothing else I will use the Draw Analyser, but if I want to use the draw in conjunction with other factors then I’ll use the Query Tool.
When using the Geegeez Draw Analyser the stalls are split into three sections or ‘thirds’ – low, middle and high. What this means is that in a 12 runner race for example, draws 1 to 4 would be in the low third, 5 to 8 in the middle, and 9 to 12 high.
### TYPES OF DRAW BIAS
I want to start by talking about types of draw bias. I believe there are two types of bias. Firstly a bias that favours a particular section of the draw; secondly a bias against a particular section of the draw. Let me illustrate with a couple of examples using draw data from 2016 to 2021. Unless otherwise stated, in this article I am going to focus on 8+ runner handicaps during this six-year period.
#### Pontefract 1m 2f
It is rare to get effective draw biases at distances of 1m2f or more, but Pontefract is an exception. If we look at the track configuration we can perhaps see why this bias exists:
Low draws are positioned on the inside and with an early left turn this gives them the advantage of taking the shortest route assuming they break well. In contrast, higher drawn runners are either stuck out wide round the first turn or forced to tuck in mid pack or near the back, or they need to be rushed forward to get a position thus using energy very early in the race.
There is a second left hand turn after about another two furlongs cementing the early positional advantage for low drawn runners; and there is a third turn about a quarter mile from home which again favours those racing near to the inside rail. Let’s look at the most recent six-season data now:
The stats show a clear advantage to one section of the draw (LOW); there is a significant advantage in most areas. Low drawn runners win more often, place more often, have higher IV values and higher PRB figures, too. However, backing all such runners to SP would have made a small loss and the A/E index value is lower than the middle section’s A/E value. This factor was referenced in the first article: the market at Pontefract clearly appreciates there is a draw bias. Just because one section of the draw is clearly favoured, this not in itself a license to print money! For the record, however, you would have made a small profit of £11.98 during this period backing low draws to Betfair SP.
Pontefract over 1m 2f is an example of a bias strongly favouring a particular section. With middle draws out-performing higher draws, this is an example of a fairly linear relationship: the lower the draw the better. Draw 1 is better than draw 6; draw 6 is better than draw 10 etc.
Now for an example of a draw bias against a particular section of the draw.
#### Musselburgh 5f
The sprint 5f trip at Musselburgh is essentially a straight five but there is a slight kink to the left at the 3f pole which can slightly hinder wider drawn runners. With Musselburgh being a right handed course at longer distances, it means horses drawn next to the rail are the higher drawn runners. Here are the stats:
This is far from being a strong draw bias, but there is a bias against lower drawn runners compared with high and middle drawn runners. Low drawn runners come out comfortably bottom in all of the parameters as shown in the breakdown above. Looking at 2009 to 2015 we get a similar picture which gives further confidence that this is likely to continue this season and beyond.
It does seem that the kink to the left at the 3f pole is enough to make life more difficult for the wide (low)-drawn runners.
Indeed if we ignore 8- and 9-runner races (the smallest fields), and look at handicap races with ten or more runners we get the following results:
All of the low drawn variables deteriorate further, and such horses are winning only just above half of the races they statistically should (IV 0.53, an Impact Value of 1.00 being on par). Consequently, both middle and high draws are winning more races than they statistically should. One would expect to see those wider draws (low) struggling more over 5f at Musselburgh as the field size increases. However, it is always good to see results in black and white - as per the image above - to back up a theory.
### INDIVIDUAL DRAWS / STALLS
A question: when you look at draw biased course and distances, what do you focus in on? The so called favoured third of the draw only? The favoured half of the draw? Or do you go further and have a preference for specific draws / stalls?
There is an argument to back the horse that is in ‘pole position’ especially on a turning track. One would think that would be the horse housed closest to the inside (i.e. drawn 1). However, the stats I have uncovered suggest differently. The stats suggest the second closest horse to the inside (i.e. actual draw 2 - 'actual' draw being the real position a horse was drawn, after accounting for any non-runners) is generally most favoured.
To show this in more detail I have looked at all 8+ runner handicaps over 5f and 6f run around a bend (2016-2021). For the record there are 12 UK courses where 5f and/or 6f races occur round a bend (seven turf courses and five on the all-weather).
Firstly I want to compare win and placed strike rates (N.B. Place SR% includes winners with the placed runners).
The margins may look quite small but they are significant as the data set covers over 2400 handicap races over 5/6f. All other key stats also point in favour of 'actual' draw 2. Firstly A/E values:
Runners drawn 2 have been far better value than those drawn 1. This is a much bigger difference than I had expected.
Next a look at profit / loss figures. Firstly a comparison of traditional SP figures (to £1 level stakes):
Losses of nearly 26p in the £ if backing all horses drawn 1 are bankruptcy territory; a smaller 8p in the £ loss for all horses drawn 2 would see a far more protracted slide to the proverbial poorhouse. But, here's Betfair SP to save the day:
The flow of bleeding has been stemmed from stall 1 but there are still bank-destroying losses; whereas trap 2 is now in the black!
But... we already know that profit / loss figures can easily be skewed by big-priced outlier winners, especially using Betfair odds. So I thought it worth comparing stats for the two draws when the Betfair SP was no bigger than 16.0. Here is what I found:
We can now see that big priced winners are not skewing the stats. Draw 2 once again has a better strike rate (both win and placed), better returns and a much stronger A/E value.
So what is actually happening here to promote stall two above the notionally best-drawn box, stall one? That is something I have pondered for many years because I have seen this type of pattern repeating time and again.
One plausible theory is that it may simply be down to the fact that horses drawn right next to the rail have less room for manoeuvre. With a rail on their inside, if they break from the stalls poorly then they are very likely to be stuck behind one or more horses. Their options are compromised until they've completed the turn by which time it may be too late. Meanwhile, horses drawn 2 have a little more space either side of them and hence more options if they break slowly. Whether this theory is true or not I obviously cannot say, but there is logic there, and it is a pattern replicated in US dirt racing at sprint distances around a turn.
What is clear in terms of the stats: in 5-6f handicaps round a turn it is preferable to be drawn 2 rather than 1.
Before moving on, I mentioned that 12 courses were in that sample and, of those 12 courses, only Kempton saw a clear advantage to horses drawn 1 over those drawn 2. Two courses - Epsom (6f) and Wetherby 5½f - had limited data (just 16 and 15 races respectively), while the other nine courses all favoured horses drawn 2 over horses drawn 1, most of them fairly strongly.
### GOOD DRAWS WITH PRICE CONSIDERATIONS
As we have seen, backing a specific draw / stall under certain conditions could produce a profitable scenario. However, this idea is full of risks as we are pinning our hopes on one stall position and nothing else. So, how about combining a good draw with market factors? This is what we are going to look at next.
I have taken six of the strongest draw biases from the past six seasons (these are Chester over 5f and 7f; Goodwood over 7f and 1 mile; and Pontefract over 1 mile and 1 mile 2 furlongs). From there I have focused on the four stalls closest to the favoured inside rail: actual draws 1 to 4. Then I have ordered them depending on price. My idea is to compare price position of these good draws to see if there are patterns to be found.
By way of an example, let’s imagine the following scenario:
That would mean an order as follows:
Here are the actual results for the six course/distances (profit/loss has been calculated to Betfair SP and we are again focusing on handicaps with eight or more runners):
#### Pontefract 1 mile 2 furlongs
Combining the six courses we get the following results:
It seems therefore the best value lies at either end of the price position spectrum. The shortest priced runners drawn 1 to 4 have made the biggest profit. They have also had a decent strike rate of 28.6%. The biggest priced runner from draws 1 to 4 have also made good profits although it would have been a bit of a rollercoaster with just 13 wins from 258 runners (SR 5%).
So is this the way to go? I'm not sure, but I believe the idea is worthy of more digging in the future. I’ll add it to my rapidly expanding research list!
- DR | 2,431 | 10,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.955831 |
http://www.buenastareas.com/ensayos/Esfuerzo-Simple/5204322.html | 1,527,041,310,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865023.41/warc/CC-MAIN-20180523004548-20180523024548-00627.warc.gz | 359,992,032 | 21,469 | # Esfuerzo simple
Páginas: 3 (543 palabras) Publicado: 1 de septiembre de 2012
Teacher 2
Mechanics of Materials
1. The figure shows the union of a truss and the base of a wooden armor. Slighting the friction:
a) Determine the dimension b if the acceptable shear stress is900 KPa.
b) Calculate the dimension c if the bearing stress should not exceed 7MPa.
R: 0.32m, 0.041m
2. Two pieces of wood, with a thickness of 7/8in and 6in wide each of them, are pasted withglue as shown in
the figure. Knowing that the union will fail when the shear stress in the glue reaches 120psi. Find the
minimum permissible length (d) on the cuttings if the joint should support anaxial load of P=1200Lb.
R: 1.632in
FIME-UANL
1
Teacher 2
Mechanics of Materials
3. The shackle of the anchor cable supports the force of 600Lb. If the pin has a diameter of 0.25in.Determine
the shear stress at the pin.
R: 6111.549psi
4. Three steel plates, with 5/8in of thickness each, are joined by two rivets of ½in each:
a) If the load is P=10kips, which is the maximumbearing stress acting on the rivets?
b) If the maximum shear stress that the rivets can support is 32ksi, what is the force P required to fail by
shear?
R: 16ksi, 25.132kips
FIME-UANL
2 Teacher 2
Mechanics of Materials
5. The top wire of a circus is fixed to a vertical column AC and remains tense with a tensor wire BD. The
column AC is locked in C with a bolt of 10mm in diameterto the bracket shown in the detail a - a.
Calculate the average shear stress in the screw on C if the tension in the top wire is 5KN (horizontal).
R: 28.352MPa
6. In the figure, is shown theunion of a vertical column and a diagonal backup. The connection is composed
of three bolts of 5/8in that unite the two plates of 1/4in (base plate and connection plate), the load P
maintained by thebackup is of 5.5KN, determine:
a) The shear stress in the bolts.
b) The bearing stress between the plates and the bolts.
R: 5975.737psi, 11733.333psi
FIME-UANL
3
Teacher 2
Mechanics of...
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# Investigate and measure the speed of a ball rolling down a ramp.
Extracts from this document...
Introduction
Khalid Abu Rumman GCSE Physics Coursework Mr.Evison
11s
Introduction
In this piece of coursework I’m going to investigate and measure the speed of the ball rolling down a ramp. From the data that I’m going to collect I’m going to be able to work out the Gravitational potential energy when changing the height, the friction force acting on the ball whilst it rolling down, and finally the kinetic energy exerted by the ball.
Planning
Fair Testing
Before I would start the experiment I would have to devise a suitable method that would provide a safe test, which is fair and accurate. In this experiment I’m going to have to roll down a ball whilst changing the angle of elevation of the ramp, and the factors that would have an effect on the accuracy of the output results are;
• Size of the ball
• Heaviness of the ball
• Length of the ramp
• Height of the ramp
• Texture of the ball
• Smoothness of the ramp
• Angle of ramp from the ground surface
Here in this experiment the only factor that I'm going to change would be the height of the ramp which in turn would affect the angle of the ramp too, if the length of the ramp stays constant.
For the experiment to be a fair one I have to;
• Keep all the factors that affect the experiment constant except the ones that I'm testing.
• All the experiments have to be done by the same person on the same day, because if we changed the person that is doing the experiment it may lead to different reaction times making the experiment inaccurate
• All I have to change is the height of the of the ramp from the ground it’s resting on. And by keeping the length of the ramp constant that would in turn proportionally change the angle of the ramp.
Middle
5.16
2.75
2.89
2.75
2.79
Above are the preliminary results.
Selecting the most Suitable Apparatus
When looking at the preliminary results the golf ball traveled too fast for us to be able to stop the time accurately, the table tennis ball was too light so has gone off course most of the time, due to the air resistance, and that was enough to deflect it. The tennis ball was too big and I don’t think it would be a good idea to use it, as we would find it hard to place correctly on the starting point line, and it would take too much time which might lead to anomalies. But the squash ball to me was the most suitable ball to use, and it stays in its path all time which is in a straight line down the ramp. In the preliminary tests we used 3 cm think books to lift up the ramp, but I don’t think that that would be a good idea, because the books might be compressed a bit, which would decrease the height we are aiming to achieve. So I think I'm going to use building blocks that are 8.2cm in height each and that would cancel out the chance of any change in the height due to compression. When thinking about time, the most accurate stopwatch we can get would be a 2 d.p. stopwatch and that’s all we’ve got. There are lots of different types of ramps out there, with different types of surfaces but the only one that was available to me was a wooden ramp that has a groove going through it, to provide a path for the ball to roll through. Below is the apparatus that I'm going to use to do the experiment.
Apparatus
• Ramp that has been graved in a straight line to provide a course for the ball to roll through. (length of ramp is 200cm)
• Squash Ball (24.43 grams)
• Stop Watch (2.d.p accuracy)
• Building Blocks that are 8.2 cm in height each.
Prediction
I predict that the increase in the ramps angle is proportional to the speed of the ball. So if the angle increases the speed of the ball increases too. If the ramp was completely horizontal (angle 0o) the velocity of the ball would be zero as there would be no way of the gravitation pulling it through downwards, but if the ramp was put to a small gradient the ball would roll down slowly, and as you increase the gradient the speed of the ball increases too. But when the ramp is put vertically (900) the ball would free fall at the speed of gravity (9.81 m/s), due to the ramp not being there to put friction against the ball, and to deflect its course to another direction. As I would get different speeds, as I differentiate the steepness of the ramp, I will have to work out the mean speed using an equation which is the following;
Mean speed (m/s) = Distance Traveled (m) X Mean Time Taken (S)
Secondly I predict that as the Gravitational potential energy increases, the kinetic energy also increases proportionally
Thirdly I predict that as the height of the ramp increases the G.P.E also increases proportionally
Fourthly I predict that as the speed of the ball (squared) increases the kinetic energy exerted is also increased proportionally.
Fifthly I predict that as the kinetic energy increases the friction force also increases, and that the friction force is greater than the kinetic energy.
Number Predication 1 As the angle of the ramp increases the speed of the ball increases proportionally. 2 As the Gravitational potential energy increases, the kinetic energy also increases proportionally 3 As the height of the ramp increases the G.P.E also increases proportionally 4 As the speed of the ball (squared) increases the kinetic energy exerted is also increased proportionally. 5 As the kinetic energy increases the friction force also increases, and that the friction force is greater than the kinetic energy.
So to work out the G.P.E I would use this formula;
G.P.E (j) = Weight (N) X Change in Height (m)
But to work out the weight I would first have to use this formula;
Weight= Mass (kg) X Gravity (N/Kg) Where the gravity is constant at 9.81N/Kg
To work out Kinetic energy I would use this formula;
Ek (j) = ½ Mass (Kg) X Speed2 (m/s2)
And finally to work out the Friction force I would use;
Friction Force (j) = G.P.E (j) – K.E. (j)
Number & Range of Readings
As my main non constant value would be the height, I will have to choose the heights that I'm able to have using the building blocks. I will have six different heights which I think would give me a large range of results. The heights would be 8.2, 16.4, 24.6, 32.8, 41 and 49.2 centimeters. To work out how big the angles are I have to use this formula;
Angle = sin-1(Opp/Hyp)
Angle = sin-1(height of ramp/200)
So the angles of the heights are;
Height (cm) 8.2cm 16.4cm 24.6cm 32.8cm 41cm 49.2cm Angle (Ø0) 2.3490 4.7030 7.0650 9.4390 11.8290 14.2410
Conclusion
• Timing the stop watch incorrectly, for example, letting the ball roll first then pressing the start button on the stop watch a quarter of a second later may alter the tests accuracy dramatically.
• The grooves on the ramp might have parts that decrease the speed of the ball.
• The ball was not accelerating the whole time at the same acceleration, which that would make a difference when it came to do the calculations.
I don’t think repeating any more test for results would make it more accurate cause I’ve already had carried out 5 samples of each test.
I think I have carried out a suitable range of results but there was an area for improvement as I could have investigated larger heights, which would mean larger angles.
Improvements
Things that I would improve in this experiment are;
• Lager range of heights and angles
• Larger variety of ball to use
• A different surface for the ball to roll on, a smoother surface, such as plastic, or metal, with would reduce friction
• A stop watch that is connected to the computer, and uses lasers to start and stop the stopwatch when the ball rolls through them.
• Shorter ramp to keep the ball accelerating.
The whole experiment was a success in that I gained knowledge of things I didn’t know before, I have carried out the experiment safely, fairly and accurately.
Khalid Abu Rumman - -
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1. ## assembly hex global stiffness matrix
@tinh Thanks a lot for your advice. Yes, what you say is right. Before this I mess up the DOF of global matrix. Now it can work. Thank you indeed. (^_^)
2. ## assembly hex global stiffness matrix
@tinh Thanks a lot for your time. Is this means, I can use the order directly? Just input the order's node coordinate, then calculate the element stiffness matrix. And put the correspond DOF into global stiffness matrix? But I have tried that. It seems not work. Might you give me more detail? Thanks Donggua
3. ## assembly hex global stiffness matrix
Hello, everyone I have a model with hex elements, like below(model.png) : I want to get this model's global stiffness matrix. for element stiffness matrix, I use the local coordinate like the coordinate.png show: I ouput the ( node id, node coordinate, element id, node list of element) for get the global stiffness matrix. But it seems the local coordinate or the order of the node list of each element of hypermesh is different from mine. I don't know how to reorder the node list which I get from hypermesh. The node list order Which I get from hypermesh is show in nodeListHypermesh.png. It seems is anti-clockwise. Thank you so much for your time. Any advice will be appreciated. (^_^)
4. ## How to apply constrains to empty mass center of model
Hi, @tinh Thank you so much for your patient. But I don't learn a lot of FEA, I am a student of computer science. So... I don't know how to calculate it by theory expression. I really don't want to trouble you more. But en~~ may you give me some detail material about this problem? Thank you so much ~~(^_^) I don't know if the problem I want to solve has other method. I just want to add some random forces to model, and get the correspond strain energy. The directions and magtitude of force both random, so the model may fly away. Since this, I want to add a constraint to the mass center.
5. ## How to apply constrains to empty mass center of model
Hi, @tinh I have try to runing it without constraint. But It seems I get the wrong answer, I wonder if I miss something or make some mistakes...The result is shown in the following picture. As I search something from website, I set the inertial relief by using anlysis --> control cards --> param --> inrel set it to -2. When I solve this problem I get different stress distribution and strain energy(calculate by using derived result) from using spc constrain as shown in following picture. I have no idea about how to judge whether the result is correct or not? Thank you so much for your help~~(^_^)
6. ## How to apply constrains to empty mass center of model
Hi, @tinh Yes, yes. I mean this. May I get more detail about inertial relief support from you? Thank you so much for your time and this advice.. Thanks a lot~~(^_^)
7. ## How to apply constrains to empty mass center of model
Hi, @Prakash Pagadala Thank you so much for your reply. (^_^) I have tried the RBE2 element. I think I didn't make the problem clear enough. I can't constrain the node of the cubic, since every node may have force in x, y and z direction, because I add random force to this cubic. The direction and the magnitude are both random. So I can't fix any node on the cubic. I am not very sure if this problem is solvable. Thank you so much for your any advice. Waiting for your reply.(^_^)
8. ## How to apply constrains to empty mass center of model
Hello, everyone As shown in the picture, I add some forces to the model to do the linear static anlysis. I don't want to it move or rotation under that conditions, but the node of the cube can't be fixed in my problem. So I want to fixed the six DOFs in mass center which is the red point in the picture. But the mass center is empty i.e. don't have material and don't have connection with the model. So, the problem is : Is there any method to constrain the six DOFs of mass center and won't effect the result? Can the reb2 element use for solve this problem?(I am a newbie of FEA, I don't familiar with this element) Thank you so much for your time. And any advice will be appreciate. (^_^) Best wishes.
9. ## How can I get stress vs. time curve in hypermesh
Hi, @Prakash Pagadala Thank you so much for your time and patient. I am currently try another method to do my work. I will try your method if my current method can't work well. Thank you again. (^_^) Best regards, Dongying Fu
10. ## How to use force as response
@Prakash Pagadala Thank you for your time. When come to what type of optimizaiton, I think it's not shape/size/topology optimization. Since I have gotten the model, I just want to know how much froce when I apply, this model will get the specific stress. I am not very sure is this idea very strange. Thank you so much. (^_^)
11. ## How can I get stress vs. time curve in hypermesh
@Rahul R Thank you for your time. May you tell me what I can do to get this curve? Thanks a lot.
12. ## How to use force as response
Hi, everyone I am a newbie in hpermesh. As you can see in the picture, I have apply the load to the model. But now I want to know how much force I apply that the model will get the specific stress. So the problems I met are: 1. how to make force as response 2. how to add this specific stress (I guess maybe is the static stress=specific value) I don't know if this is possible. Thank you for you any advice. Thank you for your time. (^_^)
14. ## How can I get stress vs. time curve in hypermesh
Hello everyone, I want to get stress/strain vs. time curve in hyperview, but I have only learned how to do linear static analysis in hypermesh. I try it in hypermesh, but I only get one point in the hyperview 2D plot as the picture shown. I want to know if there are any video or may you tell me how to choose those cards can I get the stress vs. time curve in hyperview. Thank you so much for you time. ~~（^_^）
15. ## How to get strain energy of hyperview
Hi @Prakash Pagadala, Thank you for your reply. When I see your advice and I reopen this model in hyperview, the strain energy is in the result--> scalar. Maybe I made some mistakes at before. Surprising~~ Thank you so much~~(^_^)
16. ## How to get strain energy of hyperview
Hi, everyone I have get some problem when I try to get the strain energy in hyperview like this video show:http://www.altairhyperworks.com/training/new_feature_videos/11/videos/result_math_PartTotalStrainEnergy_final/result_math_PartTotalStrainEnergy_final.htm. And I get a screenshot --video.png-- for easy to understand. The GLOBAL_OUTPUT_REQUEST of control card have been set in hypermesh as the screenshot hm.png shown. But when I use Optistruct to analysis this model, it doesn't have the strain energy in result->scalar as hv.png shown. I am not very sure if I have miss some points. I want to get the strain energy of whole model to make sure my matlab code is correct. Thank you so much for your help, any advice will be appreciated. (^_^)
17. ## how to get the directions of principal stress
Hi, @Q.Nguyen-Dai I send the question to support@altair.com..They reply that hyperworks doesn't support to get this direction.. thank you so much..
18. ## how to get the directions of principal stress
hi, @Q.Nguyen-Dai Thank you so much for you reply. I am so sorry for I didn't express my question clearly. As your screenshot shows, different element have principal stress with different direction. I just want to get the direction/angle in numerical. Then I can calculate something based on this direction. I don't know if I have right expressed what I want to get. I want to a direction in numerical, such as l=(0,0,1) which means the major principal stress direction of one element parallel with Z-axis. Then the l=(0,0,1) is what I want to get. best wishes Donggua
19. ## how to get the directions of principal stress
hello,everyone I just want to know how could I get the direction or angle of principal stress, as it show in the tensor plot. I have been struggling with this problem since yesterday, but I didn't get anything. Thank you for you time..so..much.. Any advice will be appreciated.
20. ## how to get principal stress direction in hyperview by TCL
hello everyone, I want to get the principal stress direction in hyperview by TCL, as direction showing in the figure. I have get the principal stress(tensor value) of this element, but I don't find how to get the directions correspond with the principal stress. Thank you for your time..so..much..
21. ## How to get Elements density of each element in hyperview by TCL
I have solved this problem by polQueryCtrl and polQueryIterator, as code showing in the picture.
22. ## How to get Elements density of each element in hyperview by TCL
hello, everyone I want to get the topology optimization result in hyperview, such as displacement, stress state, elements desity and so on. But I don't know how to get those data. Hyperworks do not write out the hyperview command to command.tcl. And I try to get it throught hyperview hierarchy, as show in the figure. I do not find what I need. I have no idea if somethig wrong. Thank you for you any advice.
23. ## get node id of *createnodesbetweennodes
O.K..It worked... Thank you so much...
24. ## get node id of *createnodesbetweennodes
Hello everyone, I want to create node in TCL, and then I use the *createnodesbetweennodes command, but I can't get the node id of the node it's created, because the result is store in reject mark, I think, but I can't access this mark. I think the hypermesh's modify command is used in display. And the query command is used to query info. I don't know if it's correct. Thank you for you any help so much..
25. ## lattice analysis using pbeaml
Hi Prakash Pagadala Thank you for your help.. I have solve this problem by updating bar2 with all upadate options.. Thank you all the same..
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# Problem 4. Make a checkerboard matrix
Solution 188025
Submitted on 8 Jan 2013 by Michael A.
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 = 5; a = [1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1]; assert(isequal(a,checkerboard(n)))
2 Pass
%% n = 4; a = [1 0 1 0; 0 1 0 1; 1 0 1 0; 0 1 0 1]; assert(isequal(a,checkerboard(n)))
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Start Hunting! | 224 | 600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-04 | latest | en | 0.694252 |
http://fintechprofessor.com/completed-projects/fama-macbeth-1973-regressions-with-shanken-correction-in-stata/ | 1,653,379,362,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662570051.62/warc/CC-MAIN-20220524075341-20220524105341-00168.warc.gz | 21,690,824 | 18,181 | # Fama and MacBeth Procedure
The Fama and MacBeth (1973) procedure is a two-step process. It involves estimation of N cross-sectional regressions in the first step. And then in the second step, it requires calculation of T time-series averages of the coefficients of the N-cross-sectional regressions. The standard errors are adjusted for cross-sectional dependence. This is generally an acceptable solution when there is a large number of cross-sectional units and a relatively small time series for each cross-sectional unit. However, if both cross-sectional and time-series dependence are suspected in the data set, then adjusting standard errors merely for cross-sectional dependence will not be sufficient.
## Shanken Correction
In applying standard OLS formulas to a cross-sectional regression, we assume that the right-hand variables β are fixed. The β in the cross-sectional regression are not fixed, of course, but are estimated in the time-series regression. Therefore, there might be sampling error in the estimates of β. Shanken (1992) suggested a correction to the standard errors of the estimates. The code for Shanken correction is available for an additional fee of \$100
## Our Stata Code
We have developed easy to use yet robust codes for Fama and MacBeth procedure with Shanken correction. The codes need just a basic understanding of Stata. Further, our comments on each line of code will surely help you to not only apply the code but also understand the process more clearly.
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The code is available for \$ 100, plus a \$50 for raw data processing (in case the data is not in Stata format and variables are not already constructed). For further details, please contact us at:
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References
1. Fama, E. F., & MacBeth, J. D. (1973). Risk, return, and equilibrium: Empirical tests. Journal of political economy81(3), 607-636.
2. Shanken, J. (1992). On the estimation of beta-pricing models. The review of financial studies5(1), 1-33. | 464 | 2,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-21 | latest | en | 0.807771 |
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# Three numbers are given in which the second is triple the first and is also double the third. If the average of the three numbers is 66, find the first number.
Solution
Let the second number = x
Second number is triple the first.
Therefore
First nubmer = $$\frac{second number}{3}$$ =Â $$\frac{x}{3}$$
Second is double the third means,
Third number =$$\frac{second number}{2}$$ =Â $$\frac{x}{2}$$
Average of three numbers = 66
( $$\frac{x}{3}$$ + x +Â $$\frac{x}{2}$$ ) / 3= 66
Find the LCM of 2 and 3 = 6
x = 108
second number = 108
First number =$$\frac{x}{3}$$ = 36 | 195 | 596 | {"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} | 4.3125 | 4 | CC-MAIN-2022-21 | latest | en | 0.843813 |
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February 24, 2014, 16:26 Sphere in pointwise #1 New Member Join Date: Feb 2014 Posts: 10 Rep Power: 5 Hi, I am new to pointwise. I am trying to create a spherical geometry for the problem of flow over a sphere. I have been trying the 'rotate' tool along with 2D sketches to extrude it into a sphere but I keep getting an error that is something like - 'some mesh elements have negative Jacobian'. Can anyone suggest a simple way to do this? Thanks.
February 24, 2014, 23:29
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Chris Sideroff
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Quote:
Originally Posted by agv10 Hi, I am new to pointwise. I am trying to create a spherical geometry for the problem of flow over a sphere. I have been trying the 'rotate' tool along with 2D sketches to extrude it into a sphere but I keep getting an error that is something like - 'some mesh elements have negative Jacobian'. Can anyone suggest a simple way to do this? Thanks.
best
I'm a little bit confused how you're going about this but in general revolving a 2D mesh where there is mesh all the way to the axis will not result in a good 3D mesh.
The easiest way to do this and get a high quality structured mesh is with 6 blocks. It sounds complicated but it's not. Here's how ... start with a spherical geometry. If you don't have one, create a semi-circle db curve and revolve once 180 deg in one direction and revolve again 180 deg again in the other direction. It's generally not a good idea to do it in one step and revolve it 360 deg. Without the verbosity - trust me, do it in two halves. See first picture.
To mesh it create 12 connectors that form the sides of cude that enclose the sphere. My sphere has dia = 1 centered at (0,0,0) so I created a cube with sides L = 1 centered at (0,0,0). See second picture.
Project the 12 connectors onto the sphere by selecting all the connectors, Edit>Project (be sure to uncheck Interior Only). With all the connectors still selected set the dimension the connectors to your liking. I choose a dimension of 21 which results in a grid point spacing ~ 0.03. With all the connectors still selected create the domains with the Assemble Domains button on the toolbar. Select the 6 new domains and project them to the sphere (re-check Interior Only). With the 6 domains still selected smooth them 10-20 iterations with Grid>Solve. See third picture.
With the 6 domains still selected, create the volume mesh with Create>Extrude>Normal. In the attributes panel set the initial step size, growth rate and ensure the orientation normals point outward. I used a initial step size = 0.001, growth rate = 1.2. Go back to run tab and enter the number of steps, click Run. I choose 55 steps. The tricky part here is knowing how far to extrude to ensure the resulting farfield is far enough away. 55 steps resulted in farfield dia ~= 100. If you find it's too large or too small use Grid>Re-Extrude to add or remove steps. See the fourth image.
That should be it. Let me know if you have any questions.
Attached Images
sphere-01.jpg (9.5 KB, 166 views) sphere-02.jpg (12.5 KB, 152 views) sphere-03.jpg (11.3 KB, 145 views) sphere-04.jpg (11.7 KB, 132 views)
Last edited by cnsidero; February 25, 2014 at 14:19.
February 28, 2014, 19:22 #3 New Member Join Date: Feb 2014 Posts: 10 Rep Power: 5 Hi cnsidero, Thanks for your detailed response. I managed to create the sphere by revolving the db semi-circle curve 180 degrees in two directions (Figure 1) but I my projection is incorrect. The full geometry that I am trying to create is a sphere contained in a cube. Once I made the 12 connectors for the cube and projected them onto the sphere, only 4 domains were created and not 6 as you said. Two patches were left empty (Figure 2). I set projection control to the default parameter -'closest point' and set dimension to '20'. Have I missed something in the settings for the projection? Last edited by agv10; February 28, 2014 at 20:48. Reason: solved earlier problem
February 28, 2014, 21:01
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I have attached images of what I've done so far.
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March 1, 2014, 15:54
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Originally Posted by agv10 Hi cnsidero, Once I made the 12 connectors for the cube and projected them onto the sphere, only 4 domains were created and not 6 as you said. Two patches were left empty (Figure 2). I set projection control to the default parameter -'closest point' and set dimension to '20'. Have I missed something in the settings for the projection?
I see you're using an unstructured mesh. If that's the mesh type you want there's no need to use the approach I described. Instead simply select the two halves of the sphere and click the Create Domains on Database button in the tool bar.
My description was if you wanted a structured mesh.
March 4, 2014, 21:05 #6 New Member Join Date: Feb 2014 Posts: 10 Rep Power: 5 Hi, I tried doing that but once I select the two halves of the sphere (surface 1 + surface 2 under the tree on the left), the 'domains on database' option is disabled and I cannot select it. All my masks are checked at this point. I created a semi circle db curve and revolved it 180 degrees in two directions to create a sphere, so I now have 3 entities in the tree - curve-1, surface-1 and surface-2.
March 5, 2014, 09:04
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Originally Posted by agv10 Hi, I tried doing that but once I select the two halves of the sphere (surface 1 + surface 2 under the tree on the left), the 'domains on database' option is disabled and I cannot select it. All my masks are checked at this point.
Do you have a default dimension or average spacing set in the Defaults tab?
March 5, 2014, 14:27 #8 New Member Join Date: Feb 2014 Posts: 10 Rep Power: 5 No, they are disabled too.
March 5, 2014, 14:28
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Originally Posted by agv10 No, they are disabled too.
That's it. To automatically assemble an unstructured domains on a database entity you must have a non-zero dimension or average spacing set.
March 5, 2014, 14:45 #10 New Member Join Date: Feb 2014 Posts: 10 Rep Power: 5 But the dimension only gets enabled when I have connectors. I am starting from scratch by revolving the db semi-circle so I only have the sphere. At this point, can I enable the dimension or spacing without having connectors? If I click 'connectors on database entities', I see the dimension get enabled but after I set a dimension, the 'domains on database' is still not enabled.
March 5, 2014, 15:34
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Originally Posted by agv10 But the dimension only gets enabled when I have connectors. I am starting from scratch by revolving the db semi-circle so I only have the sphere. At this point, can I enable the dimension or spacing without having connectors? If I click 'connectors on database entities', I see the dimension get enabled but after I set a dimension, the 'domains on database' is still not enabled.
You're looking at the wrong dimension entry dialogue - not on the toolbar but in the Defaults panel. On the left side of Pointwise, is a panel with three tabs: List, Layers, Defaults. Click the Defaults tab. Within the Defaults panel, under Connectors are two entry entry dialogues: dimension and average ds. These are used to apply a non-zero dimension to newly created connectors including those created automatically by the Domains on Database command.
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Funnay Logic
Funnay Logic
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Contents : • Introduction • Boolean Vs Fuzzy • Fuzzy Sets • Fuzzy Set Operations • Time Dependent Fuzzy Logic • How does
fuzzy logic works? • Why use fuzzy logic? • Applications • Conclusion • References
Introduction
Many decision-making and problem-solving tasks are too complex to be understood quantitatively, however, people succeed by using knowledge that is imprecise rather than precise. Fuzzy set theory, originally introduced by Lotfi Zadeh in the 1960's, resembles human reasoning in its use of approximate information and uncertainty to generate decisions. It was specifically designed to mathematically represent uncertainty and vagueness and provide formalized tools for dealing with the imprecision intrinsic to many problems. By contrast, traditional computing demands precision down to each bit. Since knowledge can be expressed in a more natural by using fuzzy sets, many engineering and decision problems can be greatly simplified. Fuzzy set theory implements classes or groupings of data with boundaries that are not sharply defined (i.e., fuzzy). Any methodology or theory implementing "crisp" definitions such as classical set theory, arithmetic, and programming, may be "fuzzified" by generalizing the concept of a crisp set to a fuzzy set with blurred boundaries. The benefit of extending crisp theory and analysis methods to fuzzy techniques is the strength in solving real-world problems, which inevitably entail some degree of imprecision and noise in the variables and parameters measured and processed for the application. Accordingly, linguistic variables are a critical aspect of some fuzzy logic applications, where general terms such a "large," "medium," and "small" are each used to capture a range of numerical values. While similar to conventional quantization, fuzzy logic allows these stratified sets to overlap (e.g., a 85 kilogram man may be classified in both the "large" and "medium" categories, with varying degrees of belonging or membership to each group). Fuzzy set theory encompasses fuzzy logic, fuzzy arithmetic, fuzzy mathematical programming, fuzzy topology, fuzzy graph theory, and fuzzy data analysis, though the term fuzzy logic is often used to describe all of these. Fuzzy logic emerged into the mainstream of information technology in the late 1980's and early 1990's. Fuzzy logic is a departure from classical Boolean logic in that it implements soft linguistic variables on a continuous range of truth values which allows intermediate values to be defined between conventional binary. It can often be considered a superset of Boolean or "crisp logic" in the way fuzzy set theory is a superset of conventional set theory. Since fuzzy logic can handle approximate information in a systematic way, it is ideal for controlling nonlinear systems and for modeling complex systems where an inexact model exists or systems where ambiguity or vagueness is common. A typical fuzzy system consists of a rule base, membership functions, and an inference procedure. Today, fuzzy logic is found in a variety of control applications including chemical process control, manufacturing, and in such consumer products as washing machines, video cameras, and automobiles.
Boolean Vs Fuzzy
300 years B.C., the Greek philosopher, Aristotle came up with binary logic(0,1), which is now the principle foundation of Mathematics. It came down to one law: A or not-A, either this or not this. For example, a typical rose is either red or not red. It cannot be red and not red. Every statement or sentence is true or false or has the truth value 1 or 0. This is Aristotle's law of bivalence and was philosophically correct for over two thousand years. Two centuries before Aristotle, Buddha, had the belief which contradicted the black-andwhite world of worlds, which went beyond the bivalent cocoon and see the world as it is, filled with contradictions, with things and not things. He stated that a rose, could be to a certain degree completely red, but at the same time could also be at a certain degree not red. Meaning that it can be red and not red at the same time. Conventional(Boolean) logic states that a glass can be full or not full of water. However, suppose one were to fill the glass only halfway. Then the glass can be half-full and half-not-full. Clearly, this disprove's Aristotle's law of bivalence. This concept of certain degree or multivalence is the fundamental concept which propelled Zader Lofti of University Berkely in the 1960's to introduce fuzzy logic. The essential characteristics of fuzzy logic founded by him are as follows.
• • • • •
In fuzzy logic, exact reasoning is viewed as a limiting case of approximate reasoning. In fuzzy logic everything is a matter of degree. Any logical system can be fuzzified In fuzzy logic, knowledge is interpreted as a collection of elastic or, equivalently , fuzzy constraint on a collection of variables Inference is viewed as a process of propagation of elastic constraints.
The third statement hence, define Boolean logic as a subset of Fuzzy logic.
Fuzzy Sets
Fuzzy Set Theory was formalised by Professor Lofti Zadeh at the University of California in 1965. What Zadeh proposed is very much a paradigm shift that first gained acceptance in the Far East and its successful application has ensured its adoption around the world. A paradigm is a set of rules and regulations which defines boundaries and tells us what to do to be successful in solving problems within these boundaries. For example the use of transistors instead of vacuum tubes is a paradigm shift - likewise the development of Fuzzy Set Theory from conventional bivalent set theory is a paradigm shift. Bivalent Set Theory can be somewhat limiting if we wish to describe a 'humanistic' problem mathematically. For example, Fig 1 below illustrates bivalent sets to characterise the temperature of a room.
The most obvious limiting feature of bivalent sets that can be seen clearly from the
diagram is that they are mutually exclusive - it is not possible to have membership of more than one set ( opinion would widely vary as to whether 50 degrees Fahrenheit is 'cold' or 'cool' hence the expert knowledge we need to define our system is mathematically at odds with the humanistic world). Clearly, it is not accurate to define a transiton from a quantity such as 'warm' to 'hot' by the application of one degree Fahrenheit of heat. In the real world a smooth (unnoticeable) drift from warm to hot would occur. This natural phenomenon can be described more accurately by Fuzzy Set Theory. Fig.2 below shows how fuzzy sets quantifying the same information can describe this natural drift.
Fuzzy Set Operations.
Union
The membership function of the Union of two fuzzy sets A and B with membership functions and respectively is defined as the maximum of the two individual membership functions
The Union operation in Fuzzy set theory is the equivalent of the OR operation in Boolean algebra.
Complement
The membership function of the Complement of a Fuzzy set A with membership function is defined as
The following rules which are common in classical set theory also apply to Fuzzy set theory.
De Morgans law
,
Associativity
Commutativity
Distributivity
Time dependent fuzzy logic
• Crisp logics:
Traditional combinatorial logic is a static logic. It provides a logic output 1 or 0 based on the binary values at the inputs. Sequential logic takes combinatorial logic a step further; it considers events or states in sequential order. It is a process/state-based logic that, based on current states and input/output parameters, determines the next state. This may be encapsulated in an if….then, else statement: if {this happens}, then {do this}, else {do that}. The next stat is merely the next ‘step’of a process that takes place at a later, unspecified,time. The state diagram in figure (a) substantiates exactly this point: the next state {A ,B or C} in the state diagram depends on the current state{ A,B,or C} and input {1 pr 0}, whereby the dimension of time is not significant nor explicitly indicated. For instance, state A will change to state B only when the input will be 1; when the input will become 1, however, is not known. Combinatorial or sequential logic does not address many knowledge intensive and real-time processes where temporal reasoning plays an important role. The logic that extends the traditional logic and predicate calculus to include the notion of time is called temporal logic. However, combinatorial, sequential, and temporal logics are crisp and the parameter and variable values true/false, exactly 1 or 0.
• Fuzzy logics
Fuzzy logics may be considered a generalized combinatorial or sequential logic; however, the passage of time is not necessarily of the essence. In fuzzy control an if….then, else approach is also followed, where again the passage of time is not of the essence. As with combinatorial and sequential processes, there are real-time fuzzy processes where temporal reasoning is important. However, existing fuzzy control approaches are not result related, they are algorithmically oversimple, and they do not reflect real-time evaluation of the control objectives. To overcome this difficulty, different approaches have been proposed but again,, time is not explicit in these approaches.
How Does Fuzzy Logic Work?
In order to illustrate some basic concepts in Fuzzy Logic, consider a simplified example of a thermostat controlling a heater fan .The room temperature detected through a sensor is input to a controller which outputs a control force to adjust the heater fan speed. A conventional thermostat works like an on-off switch (Figure 2). If we set it at 78oF then the heater is activated only when the temperature falls below 75oF . When it reaches 81oF the heater is turned off. As a result the desired room temperature is either too warm or too hot. A fuzzy thermostat works in shades of gray where the temperature is treated as a series of overlapping ranges. For example, 78oF is 60% warm and 20% hot. The controller is programmed with simple if-then rules that tell the heater fan how fast to run. As a result, when the temperature changes the fan speed will continuously adjust to keep the temperature at the desired level. Our first step in designing such a fuzzy controller is to characterize the range of values for the input and output variables of the controller. Then we assign labels such as cool for the temperature and high for the fan speed, and we write a set of simple English-like rules to control the system. Inside the controller all temperature regulating actions will be based on how the current room temperature falls into these ranges and the rules describing the system behavior. The controller's output will vary continuously to adjust the fan speed.
The temperature controller described above can be defined in four simple rules: IF temperature IS cold THEN fan_speed IS high IF temperature IS cool THEN fan_speed IS medium IF temperature IS warm THEN fan_speed IS low IF temperature IS hot THEN fan_speed IS zero Here the linguistic variables cool, warm, high, etc. are labels which refer to the set of overlapping values shown in figure 2. These triangular shaped values are called membership functions. A fuzzy controller works similar to a conventional system: it accepts an input value, performs some calculations, and generates an output value. This process is called the Fuzzy Inference Process and works in three steps illustrated in Figure 3: (a) Fuzzification where a crisp input is translated into a fuzzy value, (b) Rule Evaluation, where the fuzzy output truth values are computed, and (c) Defuzzification where the fuzzy output is translated to a crisp value. During the fuzzification step the crisp temperature value of 78oF is input and translated into fuzzy truth values. For this example, 78oF is fuzzified into warm with truth value 0.6 (or 60%) and hot with truth value 0.2 (or 20%). During the rule evaluation step the entire set of rules is evaluated and some rules may fire up. For 78oF only the last two of the four rules will fire. Specifically, using rule three the fan_speed will be low with degree of truth 0.6. Similarly, using rule four the fan_speed will be zero with degree of truth 0.2. During the defuzzification step the 60% low and 20% zero labels are combined using a calculation method called the Center of Gravity (COG) in order to produce the crisp output value of 13.5 RPM for the fan speed.
Why Use Fuzzy Logic?
An Alternative Design Methodology Which Is Simpler, And Faster o Fuzzy Logic reduces the design development cycle o Fuzzy Logic simplifies design complexity o Fuzzy Logic improves time to market A Better Alternative Solution To Non-Linear Control o Fuzzy Logic improves control performance o Fuzzy Logic simplifies implementation o Fuzzy Logic reduces hardware costs
Fuzzy Logic is a paradigm for an alternative design methodology which can be applied in developing both linear and non-linear systems for embedded control. By using fuzzy logic, designers can realize lower development costs, superior features, and better end product performance. Furthermore, products can be brought to market faster and more cost-effectively. An Alternative Design Methodology Which Is Simpler, And Faster In order to appreciate why a fuzzy based design methodology is very attractive in embedded control applications let us examine a typical design flow. Figure 4 illustrates a sequence of design steps required to develop a controller using a conventional and a Fuzzy approach.
Using the conventional approach our first step is to understand the physical system and its control requirements. Based on this understanding, our second step is to develop a model which includes the plant, sensors and actuators. The third step is to use linear
control theory in order to determine a simplified version of the controller, such as the parameters of a PID controller. The fourth step is to develop an algorithm for the simplified controller. The last step is to simulate the design including the effects of nonlinearity, noise, and parameter variations. If the performance is not satisfactory we need to modify our system modeling, re-design the controller, re-write the algorithm and re-try. With Fuzzy Logic the first step is to understand and characterize the system behavior by using our knowledge and experience. The second step is to directly design the control algorithm using fuzzy rules, which describe the principles of the controller's regulation in terms of the relationship between its inputs and outputs. The last step is to simulate and debug the design. If the performance is not satisfactory we only need to modify some fuzzy rules and re-try. Although the two design methodologies are similar, the fuzzy-based methodology substantially simplifies the design loop. This results in some significant benefits, such as reduced development time, simpler design and faster time to market: Fuzzy Logic reduces the design development cycle With a fuzzy logic design methodology some time consuming steps are eliminated. Moreover, during the debugging and tuning cycle you can change your system by simply modifying rules, instead of redesigning the controller. In addition, since fuzzy is rule based, you do not need to be an expert in a high or low level language which helps you focus more on your application instead of programming. As a result, Fuzzy Logic substantially reduces the overall development cycle. Fuzzy Logic simplifies design complexity Fuzzy logic lets you describe complex systems using your knowledge and experience in simple English-like rules. It does not require any system modeling or complex math equations governing the relationship between inputs and outputs. Fuzzy rules are very easy to learn and use, even by non-experts. It typically takes only a few rules to describe systems that may require several of lines of conventional software. As a result, Fuzzy Logic significantly simplifies design complexity. Fuzzy Logic improves time to market Commercial applications in embedded control require a significant development effort a majority of which is spent on the software portion of the project. Development time is a function of design complexity, and the number of iterations required in a debugging and tuning cycle. As we explained above, a fuzzy based design methodology addresses both issues very effectively. Moreover, due to its simplicity the description of a fuzzy controller not only is transportable across design teams, but also provides a superior media to preserve, maintain, and upgrade intellectual property. As a result, Fuzzy Logic can dramatically improve time to market.
A Better Alternative Solution To Non-Linear Control Most real life physical systems are actually non-linear systems. Conventional design approaches use different approximation methods to handle non-linearity. Some typical choices are, linear, piecewise linear, and lookup table approximations to trade off factors of complexity, cost, and system performance. A linear approximation technique is relatively simple, however it tends to limit control performance and may be costly to implement in certain applications. A piecewise linear technique works better, although it is tedious to implement because it often requires the design of several linear controllers. A lookup table technique may help improve control performance, but it is difficult to debug and tune. Furthermore in complex systems where multiple inputs exist, a lookup table may be impractical or very costly to implement due to its large memory requirements. Fuzzy logic provides an alternative solution to non-linear control because it is closer to the real world. Non-linearity is handled by rules, membership functions, and the inference process which results in improved performance, simpler implementation, and reduced design costs: Fuzzy Logic improves control performance In many applications Fuzzy Logic can result in better control performance than linear, piecewise linear, or lookup table techniques. For instance, a typical problem associated with traditional techniques is trading-off the controller's response time versus overshoot. For the simple one-input temperature controller example this is illustrated in Figure 5:
The first linear approximation for the desired curve generates a slow output response with no overshoot, which implies that the room would be too cold for a while. The second linear approximation results in faster response with an overshoot and subsequent fluctuations, which implies that the temperature will be uncomfortable for a period of time.
With fuzzy logic we can use rules and membership functions to approximate any continuous function to any degree of precision. Figure 6 illustrates how we can approximate the desired control curve for our temperature controller using four points (or four rules). We can also add more rules to increase the accuracy of the approximation (similar to a Fourier transform), which yields an improved control performance. Rules are much simpler to implement and much easier to debug and tune than piecewise linear or lookup table techniques.
IF temperature IS cold THEN force IS high IF temperature IS cool THEN force IS medium IF temperayure IS warm THEN force IS low IF temperature IS hot THEN force IS zero Rules are not like a lookup table because the fuzzy arithmetic interpolates the shape of the non-linear function. The combined memory required for the labels and fuzzy inference is substantially less than a lookup table, especially for multiple input systems. As a result, processing speed can be improved as well. Another example of robust control that can be achieved with Fuzzy Logic is the classical problem of the inverted pendulum. A conventional controller for the pendulum depends on system parameters such as length, weight, and mass. If the parameters change, then we need to re-design our controller. With fuzzy control this is not necessary because a fuzzy system is robust. Aptronix has demonstrated an actual device where we can vary the weight or length of the pendulum and the system is still stable using the original set of rules.
By using a more natural rule-based approach which is closer to the real world, Fuzzy control can offer a superior performance and a better trade-off between system robustness and sensitivity, which results into handling non-linear control better than traditional methods.
Fuzzy Logic simplifies implementation
The one input temperature controller presented so far has helped us illustrate some fundamental concepts, however real life control is much more complex in nature. Most control applications have multiple inputs and require modeling and tuning of a large number of parameters which makes implementation very tedious and time consuming. Fuzzy rules can help you simplify implementation by combining multiple inputs into single if-then statements while still handling non-linearity. Fuzzy Logic reduces hardware costs Using a lookup table the two-input temperature controller requires 64Kb of memory, while the fuzzy approach is accomplished with less than 0.5Kb of memory for labels and object code combined. This difference in memory savings implies a cheaper hardware implementation. In addition, conventional techniques in most real life applications require complex mathematical analysis and modeling, floating point algorithms, and complex branching. This typically yields a substantial size of object code which requires a high end DSP chip to run. Fuzzy Logic enables you to use a simple rule based approach which offers significant cost savings, both in memory and processor class.
Applications
Fuzzy Logic - a powerful new technology
As we all know, Japanese products have higher standards in comparision with other countries. Because they use fuzzy logic technique in most of the products. The Japanese have a famous automatically operated train in Sendai that moves so smoothly you can hardly tell it's travelling. A load of clothes into fuzzy washer and press start, and the machine begins to churn , automatically choosing the best cycle. Place chili, potatoes, or etc in a fuzzy microwave and push single button, and it cooks for the right time at the proper temperature. Fuzzy Logic was being made over cars: cushioning their ride, enhancing safety, and cutting gas consumption by certain percentage. There's a prototype of a prototype of a fuzzy helicopter that hovers automatically, without adjustments by the pilot. It schedules elevators and traffic lights, and prevents tunnel cave-ins at constructions at constructions sites. If one were to visit Japan, one could see nation living slightly further in the future. Fuzzy logic also has its applications in the fields of information retrieval systems, a navigation system for automatic cars. A predicative fuzzy logic controller for automatic operation pf trains, laboratory water level controllers, controllers for robot arc-welders, feature definition controllers for robot vision, graphics controllers for automated police sketchers, and many more.
Conclusion
Fuzzy logic is a powerful problem-solving methodology with a myriad of applications in embedded control and information processing. Fuzzy provides a remarkably simple way to draw definite conclusions from vague, ambiguous or imprecise information. In a sense, fuzzy logic resembles human decision making with its ability to work from approximate data and find precise solutions. Unlike classical logic which requires a deep understanding of a system, exact equations, and precise numeric values, Fuzzy logic incorporates an alternative way of thinking, which allows modeling complex systems using a higher level of abstraction originating from our knowledge and experience. Fuzzy Logic allows expressing this knowledge with subjective concepts such as very hot, bright red, and a long time which are mapped into exact numeric ranges. Fuzzy Logic has been gaining increasing acceptance during the past few years. There are over two thousand commercially available products using Fuzzy Logic, ranging from washing machines to high speed trains. Nearly every application can potentially realize some of the benefits of Fuzzy Logic, such as performance, simplicity, lower cost, and productivity. Fuzzy Logic has been found to be very suitable for embedded control applications. Several manufacturers in the automotive industry are using fuzzy technology to improve quality and reduce development time. In aerospace, fuzzy enables very complex real time problems to be tackled using a simple approach. In consumer electronics, fuzzy improves time to market and helps reduce costs. In manufacturing, fuzzy is proven to be invaluable in increasing equipment efficiency and diagnosing malfunctions.
References:
Daniel Mcneil and Paul Freiberger " Fuzzy Logic". Fuzzy sets and fuzzy logic (Theory and applications) by George J. Klir/ Bo Yuan. Understanding neural networks and fuzzy logic by Stamatios V. Kartalopoulos. http ://www.quadralay.com “Fuzzy Fundamentals” by E.Cox (IEEE Spectrum, October 1992,pp.-58-61).
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https://relax-sakura.info/and-relationship/sine-wave-frequency-and-amplitude-relationship.php | 1,568,556,022,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571360.41/warc/CC-MAIN-20190915114318-20190915140318-00081.warc.gz | 587,915,739 | 9,111 | # Sine wave frequency and amplitude relationship
### Phase (waves) - Wikipedia
Calculate the amplitude and period of a sine or cosine curve. of sine and cosine waves involving changes in amplitude and period (frequency). .. you were given the frequency and asked to find the period using the following relationship. Where sine waves occur in nature - sound waves, mechanical motion, The amplitude of a sine wave is the maximum distance it ever reaches from zero. Since the In the bouncing weight above, the frequency is about one cycle per second. A seagull bobs up and down on a sinusoidal-shaped periodic ocean wave with a The Relationship between Wave Frequency, Period, Wavelength, and.
## Amplitude, Period, Phase Shift and Frequency
Time zones are also analogous to phase differences. A real-world example of a sonic phase difference occurs in the warble of a Native American flute.
The amplitude of different harmonic components of same long-held note on the flute come into dominance at different points in the phase cycle. The phase difference between the different harmonics can be observed on a spectrogram of the sound of a warbling flute. In time and frequency, the purpose of a phase comparison is generally to determine the frequency offset difference between wave cycles with respect to a reference.
The oscilloscope will display two sine waves, as shown in the graphic to the right. In the adjacent image, the top sine wave is the test frequencyand the bottom sine wave represents a signal from the reference.
### Phase, Frequency, Amplitude, and all that..
If the two frequencies were exactly the same, their phase relationship would not change and both would appear to be stationary on the oscilloscope display. Since the two frequencies are not exactly the same, the reference appears to be stationary and the test signal moves.
Wave Amplitude
By measuring the rate of motion of the test signal the offset between frequencies can be determined. Vertical lines have been drawn through the points where each sine wave passes through zero.
The bottom of the figure shows bars whose width represents the phase difference between the signals. In this case the phase difference is increasing, indicating that the test signal is lower in frequency than the reference.
We say that it has a period which we will denote by given by The height of the peaks and valleys in this function will be given by its amplitude. We are now ready to consider the effect of the phase-shift.
### | CK Foundation
In fact, we can make note of the fact that the graph of the function will cross the t axis when The first time that this happens is when which corresponds to a value of t given by Thus, the graph will be shifted so that it crosses the t axis at this value.
The shape of the curve does not change, only its position on the t axis. Superimposing sines and cosines Let us take a second look at the function we investigated above, and notice that when we apply the trigonometric identity we obtain is a constant, and therefore so isand assigning the names we have found that Thus, by using a trigonometric identity for the sums of angles, we have reduced a problem we needed to understand the question we started with, at the top of this page with a problem that we already know how to solve.
We have found that the sum of a sine and a cosine curve is actually equivalent to a sine with a phase shift. A bit of care is required, however, since in order for this conversion to work, it must be true that For your consideration: Describe the behaviour of the function Solution: We observe that the constants in front of the trigonometric functions have the values We would like to find the angle and the amplitude that fit with this pattern.
The ratio of the constants Thus, looking up the angle that has a value of we find that Thus the phase shift is.
We further calculate that which tells us that. | 785 | 3,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | longest | en | 0.925313 |
https://chem.libretexts.org/Courses/Heartland_Community_College/CHEM_120%3A_Fundamentals_of_Chemistry/04%3A_Quantities_in_Chemical_Reactions/4.03%3A__Avogadro's_Number | 1,726,158,642,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00623.warc.gz | 154,316,895 | 33,008 | # 4.3: Avogadro's Number: Equality Pattern and Conversions
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##### Learning Objectives
• Use information in a given phrase or word problem to write Avogadro's number equalities.
• Apply an Avogadro's number conversion factor to convert between a molar quantity and a chemical particle count.
Upon establishing that Avogadro's number is required to solve a problem, a corresponding Avogadro's number equality must be developed. Then, using dimensional analysis, the resultant equality can be applied as a conversion factor, in order to bring about a desired unit transformation.
As stated in Section 4.1, an equality pattern contains one number and two units on both sides of an equal sign. On the left side of the Avogadro's number equality pattern shown below, which contains a numerical value of "1," one of these units, "mol," is defined. Neither unit is specified on the right side of this equality pattern, which utilizes Avogadro's number as its numerical quantity. The remaining positions, which are indicated as "blanks" in the equality pattern shown below, should be occupied by units that are relevant to the identity of the specific chemical that is referenced in a given problem. As indicated below, the secondary unit on the left side and the first unit on the right side of an Avogadro's number equality should be the chemical formula of the chemical that is being considered. A chemical name should not be used in this, or any, equality. As the desired unit transformations in this chapter become increasingly complex, multiple chemicals will be referenced within a single problem. In these scenarios, the formula for the chemical that is written in closest physical proximity to the Avogadro's number indicator word should be selected and incorporated into the equality. The remaining unit on the right side of an Avogadro's number equality should be the indicator word that corresponds to the chemical formula that is written. Usually, this indicator word will also be explicitly-written within an Avogadro's number problem. Finally, note that the relative order of the two units on the right side of an Avogadro's number equality can be interchanged.
For example, consider the phrase "ions of Ca+2."
The word "ions" indicates that an Avogadro's number equality should be developed. Furthermore, since "ions" is an indicator word, this word is inserted as the second unit on the right side of the Avogadro's number equality that is being created. The chemical that is referenced in the given statement is "Ca+2." As this chemical information was given as a chemical formula, the symbol "Ca+2" is directly incorporated into the remaining unit positions in the equality. The resultant Avogadro's number equality is shown below.
1 mol Ca+2 = 6.02 × 1023 Ca+2 ions
Note that the given chemical formula is actually an ion symbol, as evidenced by the charge that is written as a superscript on the elemental symbol. Therefore, the indicator word "ions" appropriately corresponds to the chemical identified in the given statement.
##### Example $$\PageIndex{1}$$
Write an equality appropriate to the phrase "atoms of silicon."
Solution
The word "atoms" indicates that an Avogadro's number equality should be developed. Furthermore, since "atoms" is an indicator word, this word is inserted as the second unit on the right side of the Avogadro's number equality that is being created. The chemical that is referenced in the given statement is "silicon." However, this information was given in the form of a chemical name, which should not be used in this, or any, equality. Instead, the corresponding elemental symbol, "Si," is incorporated into the remaining unit positions in the equality that is being developed. The resultant Avogadro's number equality is shown below.
1 mol Si = 6.02 × 1023 Si atoms
Because the given chemical name is an elemental name, the indicator word "atoms" appropriately corresponds to the chemical identified in the given statement.
##### Exercise $$\PageIndex{1}$$
Write an equality appropriate to the phrase "molecules of magnesium acetate."
The word "molecules" indicates that an Avogadro's number equality should be developed. Furthermore, since "molecules" is an indicator word, this word is inserted as the second unit on the right side of the Avogadro's number equality that is being created. The chemical that is referenced in the given statement is "magnesium acetate." However, this information was given in the form of a chemical name, which should not be used in this, or any, equality. Instead, the corresponding chemical formula, "Mg(C2H3O2)2," which is derived by applying the Chapter 3 rules for determining ionic chemical formulas, is incorporated into the remaining unit positions in the equality that is being developed. The resultant Avogadro's number equality is shown below.
1 mol Mg(C2H3O2)2 = 6.02 × 1023 Mg(C2H3O2)2 molecules
Because the given chemical information is the name of an ionic compound, the indicator word "molecules" appropriately corresponds to the chemical identified in the given statement.
## Applying Avogadro's Number Equalities as Conversion Factors
Once an appropriate Avogadro's number equality has been developed, the information that it contains can be re-written in the form of a conversion factor, which can then be applied to bring about a desired unit transformation. Recall that the quantity containing the unit being canceled must be written in the denominator of a conversion factor. This will cause the given unit, which appears in a numerator, to be divided by itself, since the same unit appears in the denominator of the conversion factor. Since any quantity that is divided by itself "cancels," orienting the conversation factor in this way results in the elimination of the undesirable unit. However, remember that both components of the equalities that are developed in this chapter contain two units. Therefore, in order to achieve complete unit cancelation, a conversion factor that results in the simultaneous elimination of both units must be applied.
For example, use a conversion factor based on the equality developed above for Ca+2 to calculate how many ions of Ca+2 are present in 9.74 moles of Ca+2.
As stated above, the word "ions" indicates that an Avogadro's number equality pattern should be developed and applied to solve this problem. The equality that was generated to correspond with the phrase "ions of Ca+2" is replicated below.
1 mol Ca+2 = 6.02 × 1023 Ca+2 ions
To create a conversion factor from this equality, the quantity on the left side of the equal sign is written in the numerator of a fraction, and the other quantity is written in the denominator. A second conversion factor can be developed by interchanging where each quantity is written, relative to the fraction bar. Both of the resultant conversion factors are shown below.
$$\dfrac{ \text{1 mol Ca}^{+2}}{6.02 × 10^{23} {\text{ Ca}^{+2}} \text { ions }}$$ and $$\dfrac{6.02 × 10^{23} {\text{ Ca}^{+2}} \text { ions }}{ \text{1 mol Ca}^{+2}}$$
However, only one of these conversion factors will allow for the complete cancelation of the given unit, "moles of Ca+2," since both of the units that are being canceled must be written in the denominator of the conversion factor that should be applied to solve the given problem. Since the intent of this problem is to eliminate the unit "moles of Ca+2," the conversion factor on the right must be used. As stated above, the relative order of the two units on the right side of an Avogadro's number equality can be interchanged. While reversing the order of these units is not absolutely necessary, doing so more clearly illustrates that the answer will ultimately be expressed in the desired unit. Therefore,
$${9.74 \; \cancel{\rm{mol} \; \rm{Ca^{+2} }}} \times$$ $$\dfrac{6.02 \times 10^{23} \; \rm{ions} \; \rm{Ca}^{+2}}{1 \; \cancel{\rm{mol} \; \rm{Ca}^{+2}}}$$
The solution is calculated by multiplying the given number by the value in each numerator, and then dividing by the quantity in each denominator. Recall that, when using a calculator, each conversion factor should be entered in parentheses, or the "=" key should be used after each division. Furthermore, any quantity that is expressed in scientific notation should be offset by an additional set of parentheses when entered into a calculator. In this case,
9.74 × ((6.02 × 1023) ions Ca+2 ÷ 1) = 5.86348 × 1024 ions Ca+2 ≈ 5.86 × 1024 ions Ca+2
Finally, remember that the correct number of significant figures should be applied to any calculated quantity. Since the math involved in dimensional analysis is multiplication and division, the number of significant figures in each number being multiplied or divided must be counted, and the answer must be limited to the lesser count of significant figures. Neither the given number nor the rounded version of Avogadro's number that was utilized above are exact numbers. As each of these values contains three significant figures, the final answer should be rounded to three significant digits, as shown above.
##### Example $$\PageIndex{2}$$
Use a conversion factor based on the equality developed in Example $$\PageIndex{1}$$ to calculate how many moles of silicon contain 4.3 × 1025 atoms of silicon.
Solution
The word "atoms" indicates that an Avogadro's number equality should be developed and applied to solve this problem. The Avogadro's number equality that was generated to correspond with the phrase "atoms of silicon" is replicated below.
1 mol Si = 6.02 × 1023 Si atoms
In order to completely eliminate the unit "atoms of silicon," the quantity on the right side of this equality becomes the denominator in the conversion factor that is applied to solve the given problem. Again, the relative order of the two units on the right side of an Avogadro's number equality can be interchanged. While reversing the order of these units is not absolutely necessary, doing so more clearly illustrates the desired unit cancelation for this particular problem. The remaining portion of the Avogadro's number equality is written in the numerator in the resultant conversion factor, as shown below.
$${\text {4.3}} \times {\text{10}^{25}}$$ $${\cancel{\rm{atoms } \; \rm{Si}}} \times$$ $$\dfrac{1 \; \rm{mol} \; \rm{Si}}{6.02 \times 10^{23} \; \cancel{\rm{atoms} \; \rm{Si}}}$$ = $${\text {71.4285714...}}$$ $${\rm{mol} \; \rm{Si}}$$ ≈ $${\text {71}}$$ $${\rm{mol} \; \rm{Si}}$$
The solution is calculated by multiplying the given number by the value in each numerator, and then dividing by the quantity in each denominator. When using a calculator, each conversion factor should be entered in parentheses, or the "=" key should be used after each division, and any quantity that is expressed in scientific notation should be offset by an additional set of parentheses. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown above.
##### Exercise $$\PageIndex{2}$$
Use a conversion factor based on the equality developed in Exercise $$\PageIndex{1}$$ to calculate how many molecules of magnesium acetate are present in 0.5177 moles of magnesium acetate.
The word "molecules" indicates that an Avogadro's number equality pattern should be developed and applied to solve this problem. The Avogadro's number equality that was generated to correspond with the phrase "molecules of magnesium acetate" is replicated below.
1 mol Mg(C2H3O2)2 = 6.02 × 1023 Mg(C2H3O2)2 molecules
In order to completely eliminate the unit "moles of magnesium acetate," the quantity on the left side of this equality becomes the denominator in the conversion factor that is applied to solve the given problem. The remaining portion of the equality is written in the numerator in the resultant conversion factor, as shown below. While reversing the order of the two units on the right side of an Avogadro's number equality is not absolutely necessary, doing so more clearly illustrates that the answer will ultimately be expressed in the desired unit.
$${0.5177 \; \cancel{\rm{mol} \; \rm{Mg(C_2H_3O_2)_2 }}} \times$$ $$\dfrac{6.02 \times 10^{23} \; \rm{molecules} \; \rm{Mg(C_2H_3O_2)_2}}{1 \; \cancel{\rm{mol} \; \rm{Mg(C_2H_3O_2)_2}}}$$ = $${\text {3.116554}} \times {\text{10}^{23}}$$ $${\rm{molecules} \; \rm{Mg(C_2H_3O_2)_2}}$$
≈ $${\text {3.12}} \times {\text{10}^{23}}$$ $${\rm{molecules} \; \rm{Mg(C_2H_3O_2)_2}}$$
The solution is calculated by multiplying the given number by the value in each numerator, and then dividing by the quantity in each denominator. When using a calculator, each conversion factor should be entered in parentheses, or the "=" key should be used after each division, and any quantity that is expressed in scientific notation should be offset by an additional set of parentheses. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown above.
4.3: Avogadro's Number: Equality Pattern and Conversions is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by LibreTexts. | 4,797 | 17,477 | {"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} | 3.921875 | 4 | CC-MAIN-2024-38 | latest | en | 0.199424 |
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