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posted by .
A pendulum consists of a small object hanging from the ceiling at the end of a string of negligible mass. The string has a length of 0.88 m. With the string hanging vertically, the object is given an initial velocity of 1.7 m/s parallel to the ground and swings upward on a circular arc. Eventually, the object comes to a momentary halt at a point where the string makes an angle θ with its initial vertical orientation and then swings back downward. Find the angle θ.
• Physics -
mv²/2= mgh=mgL(1-cosθ)
cosθ= 1- (v²/gL)
θ=arccos[1- (v²/gL)]
• Physics -
This doesn't give me the correct answer according to the assignment program :/
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http://slugterra.46.newsoftheworld.top/probability-and-or-rule-worksheet/ | 1,601,419,264,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402093104.90/warc/CC-MAIN-20200929221433-20200930011433-00577.warc.gz | 124,774,995 | 30,511 | # Probability And Or Rule Worksheet
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Top | 2,169 | 10,446 | {"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} | 2.953125 | 3 | CC-MAIN-2020-40 | latest | en | 0.941072 |
https://math.stackexchange.com/questions/3058677/irreducible-polynomials-of-degree-greater-than-4-over-finite-fields | 1,713,269,760,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817081.52/warc/CC-MAIN-20240416093441-20240416123441-00404.warc.gz | 343,876,425 | 38,652 | # Irreducible polynomials of degree greater than 4 over finite fields
I want to build a field with $$p^{n}$$ elements. I know that this can be done by finding a irreducible (on $$Z_{p}$$) polynomial f of degree n and the result would be the $$Z_{p}$$/f. My question is finding this irreducible polynomial. I know that if it has degree $$\leq$$ 3, then it's irreducible iff it has no roots. But what if I want to construct a field with 81 = $$3^{4}$$ elements? How can I find an irreducible polynomial of degree 4?
• For low degrees the best techniques are a bit ad hoc, not unlike Rene Schoof's answer (+1). I have used similar techniques to produce irreducibles of degree 20 and 21 (over $\Bbb{Z}_2$) on this site when requested. In general the task takes as much work as producing a prime number of a prescribed size. The tricks based on the algebra of roots of unity as well as splitting fields does simplify this in many occasions. Jan 2, 2019 at 5:39
• An alternative to Rene Schoof's observation would be to use the fact that $\Bbb{F}_{3^4}$ is the smallest extension field containing roots of unity of order sixteen. Those are always roots of the sixteenth cyclotomic polynomial $\Phi_{16}=x^8+1$. Over $\Bbb{F}_3$ that polynomial factors as $$x^8+1=(x^8+4x^4+4)-4x^4=(x^4+2)^2-(2x^2)^2=(x^4-2x^2+2)(x^4+2x^2+2),$$ so we can deduce right away that the quartics above are irreducible. Jan 2, 2019 at 5:45
• Hmm. Actually I don't think that this task would be asymptotically as arduous as that of locating prime numbers. But you need methods other than the Sieve of Eratosthenes (which is what Ethan Bolker's answer is suggesting). Jan 2, 2019 at 5:48
## 4 Answers
Find the irreducible quadratics. Multiply them together. Those fourth degree polynomials won't do. Now try some others at random (or systematically, following a list in some natural order). When you find one with no roots you're done.
This is mildly tedious, but you'll get good at the arithmetic, which may come in handy in other computations in the future.
You can also ask Wolfram alpha to factor polynomials modulo $$3$$.
• Asymptotically, a random monic polynomial of degree $n$ over $\Bbb{Z}_p$ is irreducible with probability $1/n$. Jan 2, 2019 at 5:33
Since $$3$$ is a primitive root modulo $$5$$, the fifth roots of unity are in $${\bf F}_{81}$$, but not in a proper subfield. This means that the cyclotomic polynomial $$\Phi_5(X)=X^4+X^3+X^2+X+1$$ is irreducible modulo $$3$$.
The elements of $$GF(p^n)$$ are the zeros of the polynomial $$x^{p^n}-x$$. This polynomials decomposes into irreducible polynomials of degree $$d$$ over $$GF(p)$$ where $$d$$ divides $$n$$. It can be shown that this decomposition contains at least one polynomial of degree $$n$$ which ensures the existence of a finite field with $$p^n$$ elements. In a CAS you usually have access to such irreducible polynomials.
I think the state of the art is Couveignes, J. M., & Lercier, R. (2013). Fast construction of irreducible polynomials over finite fields. Israel Journal of Mathematics, 194(1), 77-105. A preprint is available on arxiv.
If you want something simpler but better than brute force, you could look at Victor Shoup's work from the 1990s. Shoup, V. (1990). New algorithms for finding irreducible polynomials over finite fields. Mathematics of Computation, 54(189), 435-447 is not the most recent, but is freely available online, unlike the follow-up.
Obviously both of these also serve as starting points for a literature search.
• +1 for the references Jan 2, 2019 at 5:40 | 979 | 3,550 | {"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": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-18 | latest | en | 0.940733 |
https://masksoferis.wordpress.com/2011/11/16/rock-paper-scissors-a-mathematician-ruining-it/ | 1,544,592,572,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823738.9/warc/CC-MAIN-20181212044022-20181212065522-00548.warc.gz | 653,829,377 | 16,131 | ## Rock, paper, scissors, a mathematician ruining it
I
The obvious variation is to add more signs into the game: say “rock-paper-reviewer-editor-scissors”. It in inobvious, though, whether rock beats reviewer or the other way round. (Some of those reviewers are tough.)
One way is to draw a pentagram in a single line (making each segment an arrow pointing the way you draw it) and then to draw a circle round it (marking the direction you draw). Then you can treat the points of the pentagram as the five signs, with each point originating two arrows indicating two other points, and being indicated by two of the others; which gives two signs that submit, and two that conquer.
This addition alone, though, doesn’t make the game more interesting, just more complicated.
One could say winning or losing by the circle is different from winning or losing by the pentagram: but how? (Through a pentagram loss, you forfeit your very soul?)
Ib
As for the simpler obvious variation: Rock-paper-plasticknife-scissors, the game with four sign(al)s/gestures, is a bit iffy. You tie with the same; you lose to one, win against one… but what about the fourth? If it is a tie, one half of games end in a tie. It can’t be a win or a loss, because that would make some signs better than others. If rock wins against against plasticknife, then plasticknife loses to both rock and scissors, wins against paper and ties against itself — it would always be better to play rock (WWLT) than to play plasticknife (WLLT).
Any odd number of gestures can be arranged to be equally good; no even number above two can be without increasing the number of ties.
Then again, with more gestures this just isn’t interesting. Who cares if Horned Goat loses to Hanged Man or Lone Dalek, if it’s the same loss either way?
Ic
Rock-paper-scissors doesn’t have the same kind of a hierarchical arrangement as playing cards do — there you don’t get to choose your cards, so you can have cards that are better than others, most of the time. In rock-paper-scissors, you need to have options that are somehow equal (by not knowing the other player’s choice, if in no other way), because why would you choose a sign that was less likely to win?
Consider the card game known as “Red”. Both players draw a card from a deck, face down. Both then reveal their card. A red always beats a black; below that, a bigger card always wins. Not a particularly interesting game, but perfect for high school students really tapped-out after an unwelcome lesson. If you could call the card you wanted in Red, you’d be screaming “Ace of Hearts!” all the time — and having a tie with your opponent, who would be shouting the same. (Or “Diamond Ace!” — it would be a pointless, melodramatic game either way.)
This illustrates that either your choices can’t matter, or you must have no choice at all… which is a depressing prospect, but rock-paper-scissors is not much of an intellectual game anyway, as far as its mechanics go. The psychology can of course be very interesting, especially when you keep playing it. (“Is she going for scissors again? Third time in a row? But what if she’s counting on me pulling rock, and intends to play paper? Then I should play scissors— unless—“, et cetera. Put two psychologists to work playing each other, and they’ll probably stare at each other for five minutes, and then one admits defeat.)
It would be ideal to make a game with mechanics just complex enough to generate interesting psychology. Rock-paper-scissors isn’t quite complex enough. (Then again, it’s better than tic-tac-toe, a game where any player smarter than your average calculator can always tie, and two such players will always tie.)
II
The obvious biological variation would be to play the game with both hands at the same time. But this too makes the game different — in this case quicker (two at the same time!) — but not more interesting.
Then again, this gives more scoring conditions: a double win, a small win (win one, tie one), a fighting tie (win one, lose one) and a full tie (tie both). (The first two are, from the other end, a double lose and a small lose.)
By crunching numbers, the likelihood these outcomes is, assuming the players are dumb automatons:
11% Double win (W/W)
22% Small win (W/T)
22% Fighting tie (W/L)
11% Full tie (T/T)
22% Small lose (L/T)
11% Double lose (L/L)
— one percent is lost in the rounding. (Use 1/9 and 2/9 if you want to be exact.) If you take the first two as “wins”, the middle as “ties” and the last two as “loses”, then the odds are the same as in a normal one-handed game of rock-paper-scissors; there’s just a bit more additional detail within each category. To make a sensible variant of the game, this added sensitivity should be utilized somehow. (Note the two ties aren’t different in any intuitive way; both players get the same result in each. Some new rule could distinguish them for some other new aspect of the game.)
Mind you, this could be a decision tool if you needed two exit conditions —
Double win : We’ll do what I want, all the way
Small win: We’ll do what I want, for the most part
Fighting tie: Fine, let’s do nothing; I’ll go home, this isn’t working!
Full tie: Let’s try to split everything evenly, okay?
— but I’m not sure anyone needs help for making decisions like that.
The mechanic is there; the game just needs an addition that uses it.
III
The third variation, a sort of obnoxious meta thing, would be to have three players, each with two hands, each playing a one-handed game with each of the other two at the same time.
Call the players A, B and C. Three games resolve at the same time, each with three possible results (win/lose, lose/win, tie); this gives twenty-seven different total outcomes. Those form four categories, the way I choose to group them.
I’ll write “A>B” for “A wins over B”, “A<B” for “A loses to B” and “A=B” for “A and B tie”.
1) A<B<C<A : a roundabout tie. A>B>C>A is the same thing: each player has one win, one loss, and there’s no assigning rank to them.
2) A=B=C, every game ties; everyone flashes the same sign. A great tie! Also, the appearance of a gang meet-up.
3) A>B(sthng)C<A — Strong ranking; One player wins both of his/her games: victory! (I’ll call it that to distinguish it from “wins”, which are the results of individual games.) The third game, between the two losers, either gives second and third places, or a divided second if they tie:
3a) Full rank: A>B>C<A. Player A takes first place (wins over B and C), Player B the second (wins over C, loses to A), Player C the third (loses to A and B). Alternately, A>B<C<A. (It’s probably sensible to say A>B>C=A and A<B<C=A belong here as well; one can’t argue for any different order than the obvious one.)
3b) Weaker rank: A>B=C<A. Player A is the winner; the other two both lose.
Note that there can’t be a case where two players win both their games: the game between them can have at most one winner. This three-player game produces either one victor (above) or less (below).
4) A>B(sthng)C=A — Weak ranking; No player can be ranked as the best of the three. (A>B>C=A is already included in 3a.)
4a) Weaker rank: A>B<C=A. There’s no victor, just two winners; but B sure loses.
4b) Weakest rank: A>B=C=A. There are two ties and one win-lose; thus, a winner, a loser, and one the game didn’t decide about. (Also, A<B=C=A.)
I think one has to think that a tie means “no decision”, because one can’t really interpret a tie as “are equal” because of situations like A>B=C>A. If B and C are equal, why is one strictly better than A and one strictly worse? Unless you interpret that as collapsing > into $\geq$ into =; how you interpret the mechanics makes the game.
As for the improved version of rock-paper-scissors, I have no idea. I’m just throwing up mechanics. | 1,894 | 7,829 | {"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.944342 |
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# Following the destruction of the space shuttle Challenger,
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Following the destruction of the space shuttle Challenger, [#permalink]
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16 Oct 2009, 07:40
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Following the destruction of the space shuttle Challenger, investigators concluded that many key people employed by the National Aeronautics and Space Administration and its contractors work an excessive amount of overtime that has the potential of causing errors in judgment.
(A) overtime that has the potential of causing
(B) overtime that has the potential to cause
(C) overtime that potentially can cause
(D) overtime, a practice that has the potential for causing
(E) overtime, a practice that can, potentially, cause
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Re: 1000 sc - 300 [#permalink]
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16 Aug 2010, 16:25
5
Hey All,
There's been plenty of good discussion on this one, but just to put it all in one place, I'm stepping in. Also, lots of people mentioned concision on this one. There is not a single concision issue here. Concision is a SHOCKINGLY rare issue on the GMAT. Only fall back on it as a last resort.
300.Following the destruction of the space shuttle Challenger, investigators concluded that many key people employed by the National Aeronautics and Space Administration and its contractors work an excessive amount of overtime that has the potential of causing errors in judgment.
(A) overtime that has the potential of causing
PROBLEM: The relative pronoun "that" is opening up a modifier on "overtime" which is problematic from a meaning perspective. What the sentence wants is for the modifier to modify "an excessive amount of overtime", but right now it reads as modifying "overtime". In other words, the sentence sounds like the contracts are working an excessive amount of a SPECIFIC (essential) kind of overtime, the kind that has the potential of causing errors in judgment. This is incorrect. Also, "of" is the wrong idiom with potential.
(B) overtime that has the potential to cause
PROBLEM: Same as above, though it fixed the idiom issue.
(C) overtime that potentially can cause
PROBLEM: Same as above. The adverb "potentially" is okay, though the placement is incorrect. We need to place it after "can" if we don't get any commas. Consider these examples: "He is potentially the greatest boxer of all time" is better than "He potentially is the greatest boxer of all time."
(D) overtime, a practice that has the potential for causing
PROBLEM: Now we are using an appositive modifier (modifying a noun with a noun "a practice..."), which is clearer. The "practice" is the entire phrase "work(ing) an excessive amount of overtime". Unfortunately, the idiom is correct after "potential" (should be "to").
(E) overtime, a practice that can, potentially, cause
ANSWER: People don't like picking things with too many commas. BUT COMMAS ARE GREAT! Commas are not bad or wordy. They are how we separate ideas in English. The more commas, often, the better. "potentially" is modifying "can cause", and we often set off modifiers with commas on either side.
Hope that helps!
-t
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Re: 1000 sc - 300 [#permalink]
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16 Oct 2009, 15:36
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IMO this is not a very good question.
In answers A, B, and C it's implied that "overtime" has the potential to cause errors, when logically it needs to be the "practice of overtime" that has the potential to cause errors.
In solution D, "causing" is not parallel to "work"
Answer E is the only solution remaining.
Some might argue that in answer E, the words "can" and "potentially" create a redundancy error. But I suppose all of the other errors are grammatical mistakes, whereas redundancy is more of a stylistic error.
HTH!
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Re: 1000 sc - 300 [#permalink]
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16 Oct 2009, 19:27
GMATBootcamp wrote:
In solution D, "causing" is not parallel to "work"
I got E also, but want to know what's wrong with D.
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Re: 1000 sc - 300 [#permalink]
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20 Oct 2009, 19:52
its preferable to have verbs in the same tense.
Contractors work....this can cause
contractors working...problems causing
HTH!
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Re: 1000 sc - 300 [#permalink]
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21 Oct 2009, 02:30
'C' is wrong because it syas that 'Overtime' causes errors in the judgement. It is the practice of excessive working that causes errors in judgement.
IMO potential for is unidiomatic in D
IMO E
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Re: 1000 sc - 300 [#permalink]
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21 Oct 2009, 06:35
IMO E too.
A,b,c are out because its not overtime that causes ..... its the practice so D,E
in D, potential for is not idiomatic
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Re: 1000 sc - 300 [#permalink]
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22 Oct 2009, 08:10
IMO B is correct option.
I think by Overtime we do not need to explicitly state that it is a practice.
We are making the sentence more wordy...any comments??
pls correct me...if i ma wrong..
OA pls..
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Re: 1000 sc - 300 [#permalink]
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16 Aug 2010, 07:26
OA is E.
D is incorrect for wordiness.
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Re: 1000 sc - 300 [#permalink]
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16 Aug 2010, 21:59
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Clearly the case of resumptive modifier. hence E and D are left. D is unidiomatic. Hence E.
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Re: 1000 sc - 300 [#permalink]
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22 Aug 2010, 02:48
TommyWallach wrote:
Hey All,
There's been plenty of good discussion on this one, but just to put it all in one place, I'm stepping in. Also, lots of people mentioned concision on this one. There is not a single concision issue here. Concision is a SHOCKINGLY rare issue on the GMAT. Only fall back on it as a last resort.
300.Following the destruction of the space shuttle Challenger, investigators concluded that many key people employed by the National Aeronautics and Space Administration and its contractors work an excessive amount of overtime that has the potential of causing errors in judgment.
(C) overtime that potentially can cause
PROBLEM: Same as above. The adverb "potentially" is okay, though the placement is incorrect. We need to place it after "can" if we don't get any commas. Consider these examples: "He is potentially the greatest boxer of all time" is better than "He potentially is the greatest boxer of all time."
-t
Hi tommy ,
Please Consider the sentence that you have given as an example.
In
1)"He is potentially the greatest boxer of all time" is better than
2)"He potentially is the greatest boxer of all time."
In sentence 1, potentially is an adverb and needs to modify a verb .There is no verb after potentially in sentence 1
In sentence 2 potentially modifies "is" verb
So how can sentence 1 be better than sentence 2?
I must be missing something.Tommy can you resolve this issue
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Re: 1000 sc - 300 [#permalink]
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22 Aug 2010, 12:19
thanks tommy very well explained..
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Re: 1000 sc - 300 [#permalink]
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23 Aug 2010, 10:46
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Hey Munda,
I think the issue you're having is that a modifier should come before the thing it modifies, but this is actually the opposite of the case. Most modifiers come after. For example.
The man in the house. [in the house]
The man running from the law. [running from the law]
The dog, a purebred, is pretty. [a purebred]
See? In general, modifiers come AFTER the thing they modify. This isn't a hard and fast rule or anything (For example "By the time you read this, I'll be gone. [by the time you read this]", but it CERTAINLY doesn't make a sentence wrong if the adverbial modifier comes after the verb/adverb/adjective being modified.
-t
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Re: 1000 sc - 300 [#permalink]
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23 Aug 2010, 12:31
but Tommy what I dnt understand is that when we say overtime, is that not sufficient? Why do I have to clarify the practice? Also in C if the potentially were to be correctly placed would you pass it?
TommyWallach wrote:
Hey All,
There's been plenty of good discussion on this one, but just to put it all in one place, I'm stepping in. Also, lots of people mentioned concision on this one. There is not a single concision issue here. Concision is a SHOCKINGLY rare issue on the GMAT. Only fall back on it as a last resort.
300.Following the destruction of the space shuttle Challenger, investigators concluded that many key people employed by the National Aeronautics and Space Administration and its contractors work an excessive amount of overtime that has the potential of causing errors in judgment.
(A) overtime that has the potential of causing
PROBLEM: The relative pronoun "that" is opening up a modifier on "overtime" which is problematic from a meaning perspective. What the sentence wants is for the modifier to modify "an excessive amount of overtime", but right now it reads as modifying "overtime". In other words, the sentence sounds like the contracts are working an excessive amount of a SPECIFIC (essential) kind of overtime, the kind that has the potential of causing errors in judgment. This is incorrect. Also, "of" is the wrong idiom with potential.
(B) overtime that has the potential to cause
PROBLEM: Same as above, though it fixed the idiom issue.
(C) overtime that potentially can cause
PROBLEM: Same as above. The adverb "potentially" is okay, though the placement is incorrect. We need to place it after "can" if we don't get any commas. Consider these examples: "He is potentially the greatest boxer of all time" is better than "He potentially is the greatest boxer of all time."
(D) overtime, a practice that has the potential for causing
PROBLEM: Now we are using an appositive modifier (modifying a noun with a noun "a practice..."), which is clearer. The "practice" is the entire phrase "work(ing) an excessive amount of overtime". Unfortunately, the idiom is correct after "potential" (should be "to").
(E) overtime, a practice that can, potentially, cause
ANSWER: People don't like picking things with too many commas. BUT COMMAS ARE GREAT! Commas are not bad or wordy. They are how we separate ideas in English. The more commas, often, the better. "potentially" is modifying "can cause", and we often set off modifiers with commas on either side.
Hope that helps!
-t
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Re: 1000 sc - 300 [#permalink]
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28 Aug 2010, 10:10
Hey All,
2 issues. First Mainhoon's.
The problem is that it isn't overtime on its own that is causing problems, it's the practice of WORKING AN EXCESSIVE AMOUNT of overtime. This is why we need to add the stuff about practice.
I was also asked a different question by PM, printed here:
With respect to the below sentences
1)"He is potentially the greatest boxer of all time" is better than
2)"He potentially is the greatest boxer of all time."
Can you please tell what is the adverb potentially modifying in sentence 1) and 2)
Really, it's modifying the same thing, but it's a question of how we say it. Consider an adjective use.
I have a green house.
I have a house green.
The second one is just wrong, because we don't write it that way. Technically, I suppose, it's still modifying the "house", but who cares? It's wrong. Sames goes with example 2 up above.
-t
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Re: 1000 sc - 300 [#permalink]
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04 Sep 2010, 09:44
Tommy, can potentially is not redundant?
I think I have read that in Manhattan SC...
Thanks.
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Re: 1000 sc - 300 [#permalink]
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06 Sep 2010, 11:31
Hey Noburu,
Well, I'd agree with you, but there's no better choice. Remember, any issue relating to concision/redundancy should be the ABSOLUTE LAST thing you think about on the GMAT. Those issues are always secondary to GRAMMAR and even CLARITY.
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Re: 1000 sc - 300 [#permalink]
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19 Apr 2011, 22:10
Option - E is wrong
because 'can potentially' is redundant.
Refer to Q-159, OG-10
--------------------------------------
While depressed property values can hurt some large investors, they are potentially devastating for home-owners. whose equity--in many cases representing a life's savings--can plunge or even disappear.
(A) they are potentially devastating for homeowners, whose
(B) they can potentially devastate homeowners in that their
(C) for homeowners they are potentially devastating, because their
(D) for homeowners, it is potentially devastating in that their
(E) it can potentially devastate homeowners, whose
----
Explanation - "... can potentially is redundant in B and E. ..."
--------------------------------------
Its an OG-10 Solution.
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Following the destruction of the space shuttle Challenger, [#permalink]
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18 Apr 2012, 08:29
1
Following the destruction of the space shuttle Challenger, investigators concluded that many key people employed by the National Aeronautics and Space Administration and its contractors work an excessive amount of overtime that has the potential of causing errors in judgment.
(A) overtime that has the potential of causing
(B) overtime that has the potential to cause
(C) overtime that potentially can cause
(D) overtime, a practice that has the potential for causing
(E) overtime, a practice that can, potentially, cause
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18 Apr 2012, 10:53
1
Following the destruction of the space shuttle Challenger, investigators concluded that many key people employed by the National Aeronautics and Space Administration and its contractors work an excessive amount of overtime that has the potential of causing errors in judgment.
(A) overtime that has the potential of causing
Incorrect: "That" modifies overtime, when really "overtime" isn't what causes the error, its the act of working excessive overtime that causes it. It's a little tricky, but the modifier should be modifying the entire clause rather than just the noun. In other news, "potential for" is not the correct idiom.
(B) overtime that has the potential to cause
Incorrect: Proper idiom usage, but again "that" modifies" overtime," which is incorrect
(C) overtime that potentially can cause
Incorrect: Again wrong modifier usage
(D) overtime, a practice that has the potential for causing
Incorrect: "Potential for" is not the correct idiom
(E) overtime, a practice that can, potentially, cause
Correct: Not a great answer because the "potentially" part seems redundant, but it's the best answer
Re: Please Explain &nbs [#permalink] 18 Apr 2012, 10:53
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# Following the destruction of the space shuttle Challenger,
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 4,917 | 19,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2019-04 | latest | en | 0.871761 |
https://math.stackexchange.com/questions/2122841/lim-x-to-0-left-frac-tan-leftx-right-xx-sin-leftx-right-right-w | 1,708,984,612,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474663.47/warc/CC-MAIN-20240226194006-20240226224006-00676.warc.gz | 380,331,474 | 36,265 | # $\lim _{x\to 0}\left(\frac{\tan\left(x\right)-x}{x-\sin\left(x\right)}\right)$ without L'Hopital's Rule
Here are the functions:
a) $\displaystyle\lim _{x\to 0}\left(\frac{\tan\left(x\right)-x}{x-\sin\left(x\right)}\right)$
If I used L'Hopital's rule the limit is $2$
b) $\displaystyle\lim _{x\to 0}\:\frac{e^x\cdot \:\sin\left(x\right)-x\cdot \left(1+x\right)}{x^3}$
here $\dfrac{1}{3}$
c) $\displaystyle\lim _{x\to 0}\left(\frac{\ln\left(\sin\left(3 x\right)\right)}{\ln\left(\sin\left(7x\right)\right)}\right)$
and here $1$
but the problem is that I am not allowed to use L'Hopital's rule, can you give me ideas for another type of approaches?
UPDATE:
I apologize, I see there is some discussion and confusion among people, which obviously goes beyond my functions, but still I wanted to explain that I have been missing a lots of lectures recently due to illness and last week I got $0$ points for using L'hopital because we have not learnt it, so my guess was that we are not allowed this time either, but I just talked to my tutor and he told me that just in the last lecture, they introduced L'hopital rule to us so I am free to use it. I'm very sorry.
• Hint: Use the identity $\lim\limits_{x\to 0}\frac{\sin x}{x}=1$. This will work for the first one, and I'd bet the others as well. Jan 31, 2017 at 18:15
• @TheCount Are you sure only that works for the first one? Because I get an indetermined form when using that... Jan 31, 2017 at 18:23
• @TheCount: How exactly are you using that in the first case? Jan 31, 2017 at 18:23
• @DonAntonio Well, it sure looks like one that it would work on. I'll write it out to check. Jan 31, 2017 at 18:25
• @HenningMakholm see above (and soon below) Jan 31, 2017 at 18:25
a) use the fact that $\tan{x}=x+x^3/3+o(x^3)$ and $\sin{x}=x-x^3/6+o(x^3)$ to get
$${\tan{x}-x\over x-\sin{x}}={{x^3\over 3}+o(x^3)\over {x^3\over 6}+o(x^3)}=2+o(1)$$
So the limit is $2$
b) for this one we need $e^x=1+x+x^2/2+o(x^2)$ to write
$${e^x\cdot\sin{x}-x(1+x)\over x^3}={x+x^2+{x^3\over 3}-x-x^2+o(x^3)\over x^3}={1\over 3}+o(1)$$
So the limit is $1/3$
c) the last one is even simpler
$${\ln{\sin{7x}}\over \ln{\sin{3x}}}={\ln{7x}+\ln{\sin{7x}\over 7x}\over \ln{3x}+\ln{\sin{3x}\over 3x}}$$
Now keeping in mind $\sin{x}/x\to 1$ the limit is equal to
$$\lim_{x\to 0}{\ln{7}+\ln{x}\over \ln{3}+\ln{x}}=\lim_{x\to 0}{{\ln{7}\over \ln{x}}+1\over {\ln{3}\over \ln{x}}+1}=1$$
• Always a topic of discussion: If L'Hospital rule is not allowed (because it involves derivatives), are Taylor expansions then allowed? Jan 31, 2017 at 18:27
• Good point by imranfat. Of course this could be allowed, as a means to force students to use Taylor developments, yet without any explantion it'd be absurd, at least imo, to allow Taylor but forbidding l'Hospital. Jan 31, 2017 at 18:28
• Especially since what you actually do when finding and using the Taylor expansions is exactly the same as using L'Hospital repeatedly until you reach nonzero coefficients -- just with some identical factorials above and below, and with some boilerplate arguments about why the procedure works added in. Jan 31, 2017 at 18:36
• I'll quote myself: Taylor expansions and L'Hospital rule are essentially equivalent, even if not necessarily in practice. I think no-L'Hospital (and no-Taylor) exercises are a good tool to develop creativity and reasoning skills of students. Jan 31, 2017 at 18:37
• I dont think there's a l'Hospital vs Taylor expansions discussion here. To each its own. It is the logic behind the use of Taylor expansions to evaluate limits when l'Hospital is forbidden that we're debating (or sort of...). I commented above one possible reason, but in this case that seems to be exaggerated. Jan 31, 2017 at 18:49 | 1,235 | 3,746 | {"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} | 3.796875 | 4 | CC-MAIN-2024-10 | latest | en | 0.853067 |
https://pyronconverter.com/unit/pressure/ksi-kPa | 1,701,888,117,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100602.36/warc/CC-MAIN-20231206162528-20231206192528-00805.warc.gz | 539,878,815 | 14,465 | # Free Online Convert Between Kilopound per Square Inch (ksi) & Kilopascal (kPa)
## Convert from kPa to ksi
Convert between Kilopound per Square Inch (ksi) and Kilopascal (kPa) instantly with our free online unit calculator. You can easily convert both ksi to kPa and kPa to ksi with just a few clicks. To switch between the two conversions, simply use the swap icon (rotating arrows). If you need to start over, you can reset the values by clicking the reset button.
ksi
kPa
Result: Kilopound per Square Inch = Kilopascal
i.e. ksi = kPa
Article Contents []
### What does Kilopound per Square Inch mean?
Kilopound per square inch is a unit of stress or pressure. The abbreviation for Kilopound per square inch is ksi.
#### Definition
If one killopound force is applied on a one square inch area, the resultant pressure or stress will be one ksi (kilopound per square inch)
It is a derived unit from other pressure unit psi (pound square inch). If we multiply psi by 1000 we will get ksi. Therefore, 1 ksi = 1000 psi.
In this unit converter website, we have converter from Kilopound per Square Inch (ksi) to some other Pressure unit.
## What is Kilopascal ?
Kilopascal is a unit of Pressure
In this unit converter website, we have converter from Kilopascal (kPa) to some other Pressure unit.
### What does Pressure mean?
Pressure is a kind of force. In the language of physics, a force is an object that changes or tries to change its state by acting on it or that changes or tries to change its motion by acting on a moving object.
For example, if someone pushes against a wall, the force is applied whether it changes its position or not. This effort is called force. On the other hand, the force exerted on a single area is called pressure.
This means that if we apply a force in a single case, it will be a force.
Pressure = (force ÷ area)
As an example, it is easy to take a photo with a pin because the area is less. Again, more pressure is required in the area.
The Unit of Pressure
Pressure scalar quantity, because it has value but has no direction. The SI unit of pressure is Pascal, which is equal to Newton per square meter (N/m²). Since an object or group of objects under pressure can act on its surroundings, the pressure is also measured as static energy in a single volume. It is related to the density of energy and can be expressed in joules unit per cubic meter (joules/m³, which is equal to Pascal). To make this concept precise, we use the idea of pressure. The pressure is defined to be the amount of force exerted per area.
Thanks for reading this article! Hopefully, it helps you understand what pressure is and how we measure the pressure of an object.
The following Pressure related conversions are available in our website:
### How to convert Kilopound per Square Inch to Kilopascal : Detailed Description
Kilopound per Square Inch (ksi) and Kilopascal (kPa) are both units of Pressure. On this page, we provide a handy tool for converting between ksi and kPa. To perform the conversion from ksi to kPa, follow these two simple steps:
Steps to solve
Have you ever needed to or wanted to convert Kilopound per Square Inch to Kilopascal for anything? It's not hard at all:
Step 1
• Find out how many Kilopascal are in one Kilopound per Square Inch. The conversion factor is 6894.76 kPa per ksi.
Step 2
• Let's illustrate with an example. If you want to convert 10 Kilopound per Square Inch to Kilopascal, follow this formula: 10 ksi x 6894.76 kPa per ksi = kPa. So, 10 ksi is equal to kPa.
• To convert any ksi measurement to kPa, use this formula: ksi = kPa x 6894.76. The Pressure in Kilopound per Square Inch is equal to the Kilopascal multiplied by 6894.76. With these simple steps, you can easily and accurately convert Pressure measurements between ksi and kPa using our tool at Pyron Converter.
FAQ regarding the conversion between ksi and kPa
Question: How many Kilopascal are there in 1 Kilopound per Square Inch ?
Answer: There are 6894.76 Kilopascal in 1 Kilopound per Square Inch. To convert from ksi to kPa, multiply your figure by 6894.76 (or divide by 0.0001450376807894691).
Question: How many Kilopound per Square Inch are there in 1 kPa ?
Answer: There are 0.0001450376807894691 Kilopound per Square Inch in 1 Kilopascal. To convert from kPa to ksi, multiply your figure by 0.0001450376807894691 (or divide by 6894.76).
Question: What is 1 ksi equal to in kPa ?
Answer: 1 ksi (Kilopound per Square Inch) is equal to 6894.76 in kPa (Kilopascal).
Question: What is the difference between ksi and kPa ?
Answer: 1 ksi is equal to 6894.76 in kPa. That means that ksi is more than a 6894.76 times bigger unit of Pressure than kPa. To calculate ksi from kPa, you only need to divide the kPa Pressure value by 6894.76.
Question: What does 5 ksi mean ?
Answer: As one ksi (Kilopound per Square Inch) equals 6894.76 kPa, therefore, 5 ksi means kPa of Pressure.
Question: How do you convert the ksi to kPa ?
Answer: If we multiply the ksi value by 6894.76, we will get the kPa amount i.e; 1 ksi = 6894.76 kPa.
Question: How much kPa is the ksi ?
Answer: 1 Kilopound per Square Inch equals 6894.76 kPa i.e; 1 Kilopound per Square Inch = 6894.76 kPa.
Question: Are ksi and kPa the same ?
Answer: No. The ksi is a bigger unit. The ksi unit is 6894.76 times bigger than the kPa unit.
Question: How many ksi is one kPa ?
Answer: One kPa equals 0.0001450376807894691 ksi i.e. 1 kPa = 0.0001450376807894691 ksi.
Question: How do you convert kPa to ksi ?
Answer: If we multiply the kPa value by 0.0001450376807894691, we will get the ksi amount i.e; 1 kPa = 0.0001450376807894691 Kilopound per Square Inch.
Question: What is the kPa value of one Kilopound per Square Inch ?
Answer: 1 Kilopound per Square Inch to kPa = 6894.76.
#### Common Kilopound per Square Inch to Kilopascal conversion
ksi kPa Description
0.1 ksi 689.476 kPa 0.1 ksi to kPa = 689.476
0.2 ksi 1378.952 kPa 0.2 ksi to kPa = 1378.952
0.3 ksi 2068.428 kPa 0.3 ksi to kPa = 2068.428
0.4 ksi 2757.904 kPa 0.4 ksi to kPa = 2757.904
0.5 ksi 3447.38 kPa 0.5 ksi to kPa = 3447.38
0.6 ksi 4136.856 kPa 0.6 ksi to kPa = 4136.856
0.7 ksi 4826.332 kPa 0.7 ksi to kPa = 4826.332
0.8 ksi 5515.808 kPa 0.8 ksi to kPa = 5515.808
0.9 ksi 6205.284 kPa 0.9 ksi to kPa = 6205.284
1 ksi 6894.76 kPa 1 ksi to kPa = 6894.76
2 ksi 13789.52 kPa 2 ksi to kPa = 13789.52
3 ksi 20684.28 kPa 3 ksi to kPa = 20684.28
4 ksi 27579.04 kPa 4 ksi to kPa = 27579.04
5 ksi 34473.8 kPa 5 ksi to kPa = 34473.8
6 ksi 41368.56 kPa 6 ksi to kPa = 41368.56
7 ksi 48263.32 kPa 7 ksi to kPa = 48263.32
8 ksi 55158.08 kPa 8 ksi to kPa = 55158.08
9 ksi 62052.84 kPa 9 ksi to kPa = 62052.84
10 ksi 68947.6 kPa 10 ksi to kPa = 68947.6
20 ksi 137895.2 kPa 20 ksi to kPa = 137895.2
30 ksi 206842.8 kPa 30 ksi to kPa = 206842.8
40 ksi 275790.4 kPa 40 ksi to kPa = 275790.4
50 ksi 344738.0 kPa 50 ksi to kPa = 344738.0
60 ksi 413685.6 kPa 60 ksi to kPa = 413685.6
70 ksi 482633.2 kPa 70 ksi to kPa = 482633.2
80 ksi 551580.8 kPa 80 ksi to kPa = 551580.8
90 ksi 620528.4 kPa 90 ksi to kPa = 620528.4
#### Common Kilopascal to Kilopound per Square Inch conversion
kPa ksi Description
0.1 kPa 0.0 ksi 0.1 kPa to ksi = 0.0
0.2 kPa 0.0 ksi 0.2 kPa to ksi = 0.0
0.3 kPa 0.0 ksi 0.3 kPa to ksi = 0.0
0.4 kPa 0.0 ksi 0.4 kPa to ksi = 0.0
0.5 kPa 0.0 ksi 0.5 kPa to ksi = 0.0
0.6 kPa 0.0 ksi 0.6 kPa to ksi = 0.0
0.7 kPa 0.0 ksi 0.7 kPa to ksi = 0.0
0.8 kPa 0.0 ksi 0.8 kPa to ksi = 0.0
0.9 kPa 0.0 ksi 0.9 kPa to ksi = 0.0
1 kPa 0.0 ksi 1 kPa to ksi = 0.0
2 kPa 0.0 ksi 2 kPa to ksi = 0.0
3 kPa 0.0 ksi 3 kPa to ksi = 0.0
4 kPa 0.001 ksi 4 kPa to ksi = 0.001
5 kPa 0.001 ksi 5 kPa to ksi = 0.001
6 kPa 0.001 ksi 6 kPa to ksi = 0.001
7 kPa 0.001 ksi 7 kPa to ksi = 0.001
8 kPa 0.001 ksi 8 kPa to ksi = 0.001
9 kPa 0.001 ksi 9 kPa to ksi = 0.001
10 kPa 0.001 ksi 10 kPa to ksi = 0.001
20 kPa 0.003 ksi 20 kPa to ksi = 0.003
30 kPa 0.004 ksi 30 kPa to ksi = 0.004
40 kPa 0.006 ksi 40 kPa to ksi = 0.006
50 kPa 0.007 ksi 50 kPa to ksi = 0.007
60 kPa 0.009 ksi 60 kPa to ksi = 0.009
70 kPa 0.01 ksi 70 kPa to ksi = 0.01
80 kPa 0.012 ksi 80 kPa to ksi = 0.012
90 kPa 0.013 ksi 90 kPa to ksi = 0.013 | 2,766 | 8,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-50 | latest | en | 0.857816 |
https://mathbig.com/all-factors-of-a-number-in-pair-pages/factors-of-133-in-pair | 1,675,946,533,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499966.43/warc/CC-MAIN-20230209112510-20230209142510-00106.warc.gz | 398,825,662 | 7,315 | Factors of 133 in pair
Factors of 133 in pair are (1, 133) , (7, 19)
How to find factors of a number in pair
1. Steps to find factors of 133 in pair 2. What is factors of a number in pair? 3. What are Factors? 4. Frequently Asked Questions 5. Examples of factors in pair
Example: Find factors of 133 in pair
Factor Pair Pair Factorization
1 and 133 1 x 133 = 133
7 and 19 7 x 19 = 133
Since the product of two negative numbers gives a positive number, the product of the negative values of both the numbers in a pair factor will also give 133. They are called negative pair factors.
Hence, the negative pairs of 133 would be ( -1 , -133 ) .
What does factor pairs in mathematics mean?
In mathematics, factor pair of a number are all those possible combination which when multiplied together give the original number in return. Every natural number is a product of atleast one factor pair. Eg- Factors of 133 are 1 , 7 , 19 , 133. So, factors of 133 in pair are (1,133), (7,19).
What is the definition of factors?
In mathematics, factors are number, algebraic expressions which when multiplied together produce desired product. A factor of a number can be positive or negative.
Properties of Factors
• Every factor of a number is an exact divisor of that number, example 1, 7, 19, 133 are exact divisors of 133.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Each number is a factor of itself. Eg. 133 is a factor of itself.
• 1 is a factor of every number. Eg. 1 is a factor of 133.
Steps to find Factors of 133
• Step 1. Find all the numbers that would divide 133 without leaving any remainder. Starting with the number 1 upto 66 (half of 133). The number 1 and the number itself are always factors of the given number.
133 ÷ 1 : Remainder = 0
133 ÷ 7 : Remainder = 0
133 ÷ 19 : Remainder = 0
133 ÷ 133 : Remainder = 0
Hence, Factors of 133 are 1, 7, 19, and 133
• Is there any even prime factor of 133?
No there is no even prime factor, i.e. 2, of 133.
• What is the mean of all prime factors of 133?
Factors of 133 are 1 , 7 , 19 , 133. Prime factors of 133 are 7 , 19. Therefore mean of prime factors of 133 is (7 + 19) / 2 = 13.00.
• What is the mean of factors of 133?
Factors of 133 are 1 , 7 , 19 , 133. therefore mean of factors of 133 is (1 + 7 + 19 + 133) / 4 = 40.00.
• What do you mean by proper divisors?
A number x is said to be the proper divisor of y if it divides y completely, given that x is smaller than y.
• What do you mean by improper divisors?
An improper divisor of a number is the number itself apart from this any divisor of a given number is a proper divisor.
Examples of Factors
Can you help Rubel in arranging 133 blocks in order to form a rectangle in all possible ways? For arranging 133 blocks in order to form a rectangle in all possible ways we need to find factors of 133 in pair.
Factors of 133 are 1, 7, 19, 133. So, factors of 133 in pair are (1,133), (7,19).
Ariel has been asked to write all factor pairs of 133 but she is finding it difficult. Can you help her find out?
Factors of 133 are 1, 7, 19, 133. So, factors of 133 in pair are (1,133), (7,19).
What are the pair factors of 133?
Factors of 133 are 1, 7, 19, 133. Hence, the factors of 133 in pair are (1,133), (7,19).
Can you help Sammy list the factors of 133 and also find the factor pairs?
Factors of 133 are 1, 7, 19, 133.
Factors of 133 in pair are (1,133), (7,19).
Help Diji in finding factors of 133 by Prime Factorization method and then sorting factors of 133 in pairs.
Prime factorization of 133 is 7 x 19. Factors of 133 in pair can be written as (1,133), (7,19).
Write the smallest prime factor of 133.
Smallest prime factor of 133 is 7.
Write the largest prime factor of 133.
Largest prime factor of 133 is 19.
Diji wants to write all the negative factors of 133. Can you help Diji in doing the same?
Negative factors of 133 are -1, -7, -19, -133. | 1,140 | 3,956 | {"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} | 4.6875 | 5 | CC-MAIN-2023-06 | latest | en | 0.890303 |
https://www.slideserve.com/angelina/water-cycle | 1,513,605,229,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948616132.89/warc/CC-MAIN-20171218122309-20171218144309-00307.warc.gz | 797,327,479 | 11,712 | 1 / 9
# Water Cycle - PowerPoint PPT Presentation
Water Cycle. Dena Fauber WOSAT 2006-07 Earth Science 7-10. Setting. Suburban Middle School Approximately 1000 students 107 7th grade students participated 16 students on IEP's, (including 2 selectively mute & 2 CD). Indicators. 6-8 Earth and Space Science/Benchmark C
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## PowerPoint Slideshow about 'Water Cycle' - angelina
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Dena Fauber
WOSAT 2006-07
Earth Science 7-10
• Suburban Middle School
• Approximately 1000 students
• 107 7th grade students participated
• 16 students on IEP's, (including 2 selectively mute & 2 CD)
• 6-8 Earth and Space Science/Benchmark C
1. Explain the biogeochemical cycles, which move materials between the lithosphere (land), hydrosphere (water) and atmosphere (air).
3. Describe the water cycle and explain the transfer of energy between the atmosphere and hydrosphere.
4. Analyze data on the availability of fresh water that is essential for life and for most industrial and agricultural processes. Describe how rivers, lakes and groundwater can be depleted or polluted becoming less hospitable to life and even becoming unavailable or unsuitable for life.
• Pretest-
• KWL
• Formative Assessments
• Group Song
• Graphing
• Poster
• Written Assessment (extension questions & lab observation)
• Summative Assessment-
• Concept Graphing
• Water Cycle Model
• Done as a teacher demonstration but would divide students into group and present it as a lab. Requires more time and additional supplies but would make a greater impact on students.
Loved It!
Will teach it again next year! | 532 | 2,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-51 | latest | en | 0.866446 |
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Math1011_Prelim1A
# Math1011_Prelim1A - Math1011 Fall 2009 Prelim Consider the...
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Math1011 Fall 2009 Prelim I-A1 1 lnx Consider the function f(x) = . What are the domain and range of the function f(x). -1 Write an equation for the inverse function f (x).
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Math1011 Fall 2009 Prelim A1 1 lnx Consider the function f(x) = . What are the domain and range of the function f(x). 1 lnx Domain: ln(x) is defined for all positive numbers, so 0. can take on all values, except when lnx=0, so x 1 ( ) (0,1) (1, ) ln( ) can take on all reall numbers, so when we take the reciprical, we g x D f x > = et all numbers except for 0. ( ) ( ,0) (0, ) f = −∞ -1 Write an equation for the inverse function f (x). 1 ln 1 (1/ ) ln (1/ ) (1/ ) -1 (1/ ) (solve for x in terms of y) (ln ) 1 (cross multiply) ln interchange x and y f ( ) : x y y x y x x y y x x e e x e y e x e Note = = = = = = = 1 1 ( ) ( ,0) (0, ) which = ( ) ( ) (0,1) (1, ) which = ( ) D f f f D f = −∞ = (Doc #011p.16.01t)
Math1011 Spring 2007 Prelim A2 The table shows the position of a cyclist: t (seconds) 2 3 4 s (meters) 5.1 10.7 17.7 Calculate the average velocity for each time period: i. [2, 3] ii. [3, 4] Use the information above to give an approximate value for the instantaneous velocity when t = 3.
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Math1011 Spring 2007 Prelim A2 The table shows the position of a cyclist: t (seconds) 2 3 4 s (meters) 5.1 10.7 17.7 Calculate the average velocity for each time period: i. [2, 3] (3) (2) 10.7 5.1 5.6 meters/second 3 2 3 2 avg y s s m x Δ = = = = Δ ii. [3, 4] (4) (3) 17.7 10.7 7 meters/second 4 3 4 3 avg y s s m x Δ = = = = Δ Use the information above to give an approximate value for the instantaneous velocity when t = 3. An estimate for the instantaneous velocity is the average of the secant lines from [2, 3] and [3, 4] 5.6 7 12.6 6.3 meters/second 2 2 m + = = = 17.7-5.1 12.6 2 4 2 Other ways are to use the slope of the secant line: from [2, 3] = 5.6 meters/second from [3, 4] = 7 meters/second from [2, 4]= 6.3 meters/second = = (Doc #1011p.21.02)
Math1011 Fall 2009 Prelim A3 Consider the function whose graph is shown below. Sketch the graph of the function ( ) - ( -2) on the axes provided, and determine its domain and range. g x f x = Domain Range
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Math1011 Fall 2009 Prelim A3 Consider the function whose graph is shown below. Sketch the graph of the function ( ) - ( -2) on the axes provided, and determine its domain and range. g x f x = Domain: [1, 4] Range: [-2, 0] g(x) is the V below the x – axis. (Doc #011p.16.02t)
Math1011 Fall 2007 Prelim A4 Show that for any three positive numbers a, b, c such that a 1, b 1, and c 1, the following equality holds: (log )(log )(log ) 1 a c b b c a = Solve the equation for x: 4 2log 1 10 ln 4 log 100 x x e + =
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Math1011 Fall 2007 Prelim A4 Show that for any three positive numbers a, b, c such that a 1, b 1, and c 1, the following equality holds: (log )(log )(log ) 1 a c b b c a = log log log log (log )(log )(log ) 1 (log )( )( ) 1 (log a a a a a c b c a a b c a b c a b b = = log )( a c log a b log log )( a a a c ) 1 log 1 1 1 true (for a, b, c 1) a a = = = log a log a : laws of logarithms: log a 1,b 1 log 1 b b x a Note x a = = Solve the equation for x: 4 2log 1 10 ln 4 log 100 x x e + = 2 4 2 2 2 2 2 2 2 2 2 2 2 log 2 1 10 2 1 10 1 2 2 1 2 1 2 1 2 2 1 4 log 10 1 2log 1 1 0 0 2 1 0 ( 1) 0 1 x x x x x x x x x x x x x x
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Rotational Motion and System of particles Rigid BodyRigid body is a body with a perfectly definite and unchanging shape. The distances between different pairs of such a body do not change. Translational motion In pure translational motion, at any instant of time every particle of the body has the same velocity. Rotational MotionEvery particle of the body moves in a circle in a plane about a fixed axis. The fixed axis and the plane in which body moves are perpendicular. The motion of a rigid body which is not pivoted or fixed in some way is either a pure translation or acombination of translation and rotation. The motion of a rigid body which is pivoted or fixed in some way is rotation. In pure translation of a body, all parts of the body having the same velocity at any instant of time. In pure rotation , all parts of the body having the same angular velocity at any instant of time. The motion of a rigid body which is not pivoted or fixed in some way is either a pure translation or acombination of translation and rotation. The motion of a rigid body which is pivoted or fixed in some way is rotation. (The rotation may be about an axis that is fixed. Or axis may not be fixed but one point may be fixed) Spin Spin is rotation about an axis that goes through the center of mass of the object. Characteristics of Circular Motion1) It is an accelerated motion: As the direction of velocity changes at every instant, it is an accelerated motion.2) It is a periodic motion: During the motion, the particle repeats its path along the same trajectory. Thus, the motion is periodic. Kinematics of Circular Motion1. Angular displacement2. Angular velocity3. Angular acceleration4. Tangential velocity5. Uniform Crcular Motion6. Non Uniform Circular Motion Angular VelocityAngular velocity mesures how fast an object rotates or revolves relative to a given point.How the angle of a vector changes with time with respect to an origin. Spin angular velocity means how fast a rigid body rotates with respect to its centre of rotation. Orbital angular tells us how fast a point object revolves about a given origin. Spin angular velocity is independent of the choice of origin.Orbital angular velocity which depends on the choice of origin. Dynamics of Circular Motion1. Centripetal Force 2. Centrifugal Force Applications of Uniform Circular MotionVehicle Along a Horizontal Circularvmax=𝜇s r g ${v}_{max}=\sqrt{{𝜇}_{s}\phantom{\rule{0.22em}{0ex}}r\phantom{\rule{0.22em}{0ex}}g\phantom{\rule{0.22em}{0ex}}}$ Applications of Uniform Circular Motion1. Well of deathvmin=r g𝜇s ${v}_{min}=\sqrt{\frac{r\phantom{\rule{0.22em}{0ex}}g}{{𝜇}_{s}}\phantom{\rule{0.22em}{0ex}}}$ Applications of Uniform Circular Motion1. . Vehicle on a Banked Road (ignoring friction)Safe Speed = vs=rgtan𝜃${v}_{s}=\sqrt{rgtan𝜃}$Banking Angle, 𝜃=tan-1v2rg$𝜃=ta{n}^{-1}\frac{{v}^{2}}{rg}$ Applications of Uniform Circular Motion1. Vehicle on a Banked Road (consider friction)Minimum Speed = vmin=rg(tan𝜃-𝜇s1+𝜇stan𝜃)${v}_{min}=\sqrt{rg\left(\frac{tan𝜃-{𝜇}_{s}}{1+{𝜇}_{s}tan𝜃}\right)}$ Applications of Uniform Circular Motion1. Vehicle on a Banked Road (consider friction)Maximum Speed = vmax=rg(tan𝜃+𝜇s1-𝜇stan𝜃)${v}_{max}=\sqrt{rg\left(\frac{tan𝜃+{𝜇}_{s}}{1-{𝜇}_{s}tan𝜃}\right)}$ Applications of Uniform Circular MotionConical PendulumTime Period, T=2𝜋L cos𝜃g$T=2𝜋\sqrt{\frac{L\phantom{\rule{0.22em}{0ex}}cos𝜃}{g}}$ Point Mass Undergoing Vertical Circular Motion Under Gravity Vehicle at the Top of a Convex Over Bridge Moment of Inertia as an Analogous Quantity for Mass Moment of InertiaAnalogue of mass in rotational motion.I=n∑i=1mir2i$I=\sum _{i=1}^{n}{m}_{i}{r}_{i}^{2}$ Moment of inertia idepends on the orientation and position of the axis of rotation with respect to the body. Theorem of perpendicular axisMoment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.IZ=IX+IY${I}_{Z}={I}_{X}+{I}_{Y}$ Theorem of parallel axisThe moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes. IZ'=IZ+Ma2${I}_{Z}\text{'}={I}_{Z}+M{a}^{2}$ Angular MomentumRotational analogue of linear momentumIt is moment of linear momentuml=r×p $\stackrel{\to }{l}=\stackrel{\to }{r}×\stackrel{\to }{p}\phantom{\rule{0.22em}{0ex}}$l=I 𝜔 $\stackrel{\to }{l}=I\phantom{\rule{0.22em}{0ex}}\stackrel{\to }{𝜔}\phantom{\rule{0.22em}{0ex}}$dldt=𝜏 $\frac{d\stackrel{\to }{l}}{dt}=\stackrel{\to }{𝜏}\phantom{\rule{0.22em}{0ex}}$ Angular momentum for a system of particlesL=l1+l2+l3+ . . . .+ln =n∑i=1li$\stackrel{\to }{L}=\stackrel{\to }{{l}_{1}}+\stackrel{\to }{{l}_{2}}+\stackrel{\to }{{l}_{3}}+\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.+\stackrel{\to }{{l}_{n}}\phantom{\rule{0.44em}{0ex}}=\sum _{i=1}^{n}{l}_{i}$L=n∑i=1ri×pi $\stackrel{\to }{L}=\sum _{i=1}^{n}\stackrel{\to }{{r}_{i}}×\stackrel{\to }{{p}_{i}}\phantom{\rule{0.22em}{0ex}}$ Torque1. Rotational analogue of force2. Moment of force3. Torque is the rate of change of angular momentum4. Torque of a force is a measure of that force's tendency to cause a angular acceleration. 𝜏=r×F$\stackrel{\to }{𝜏}=\stackrel{\to }{r}×\stackrel{\to }{F}$𝜏=I 𝛼$\stackrel{\to }{𝜏}=I\phantom{\rule{0.22em}{0ex}}\stackrel{\to }{𝛼}$ Torque for a system of particles L=n∑i=1li $\stackrel{\to }{L}=\sum _{i=1}^{n}{l}_{i}\phantom{\rule{0.22em}{0ex}}$ dLdt=n∑i=1dlidt =n∑i=1𝜏i $\frac{d\stackrel{\to }{L}}{dt}=\sum _{i=1}^{n}\frac{d{l}_{i}}{dt}\phantom{\rule{0.22em}{0ex}}=\sum _{i=1}^{n}{𝜏}_{i}\phantom{\rule{0.22em}{0ex}}$dLdt=𝜏ext $\frac{d\stackrel{\to }{L}}{dt}={𝜏}_{ext}\phantom{\rule{0.22em}{0ex}}$ Conservation of angular momentumif no external torque acts then 𝜏ext=0${𝜏}_{ext}=0$dLdt=𝜏ext=0 $\frac{d\stackrel{\to }{L}}{dt}={𝜏}_{ext}=0\phantom{\rule{0.22em}{0ex}}$ soL=constant$\stackrel{\to }{L}=constant$ Rolling Motion(i) circular motion of the body as a whole, about its own symmetric axis and(ii) linear motion of the body assuming it to be concentrated at its centre of mass. The centre of mass performs purely translational motion. Rolling Motioni) Kinetic energy of a rolling body can be separated into kinetic energy of translation and kineticenergy of rotation. Centre of massCentre of mass is a hypothetical point at which entire mass of the body can be assumed to be concentrated. Centre of mass of a system of particles moves as if all the mass of the system was concentrated at the centre of mass and all the external forces were applied at that point. Centre of mass is a fixed property for given rigid body in spite of any orientation.The centre of gravity may depend upon non-uniformity of the gravitational field, in turn, will depend upon the orientation. Total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass. Uniform circular motion : object travels in a circle with a constant speed Centre of gravityCentre of gravity of a body is the point around which the resultant torque due to force of gravity on the body is zero. The turning effect of a force is called the moment of the force. It depends on both the size of the force and how far it is applied from the pivot or fulcrum. Equilibrium of rigid bodyF1+F2+F3+ . . . .+Fn =n∑i=1Fi=0 $\stackrel{\to }{{F}_{1}}+\stackrel{\to }{{F}_{2}}+\stackrel{\to }{{F}_{3}}+\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.+\stackrel{\to }{{F}_{n}}\phantom{\rule{0.44em}{0ex}}=\sum _{i=1}^{n}{F}_{i}=0\phantom{\rule{0.22em}{0ex}}$𝜏1+𝜏2+𝜏3+ . . . .+𝜏n =n∑i=1𝜏i=0 $\stackrel{\to }{{𝜏}_{1}}+\stackrel{\to }{{𝜏}_{2}}+\stackrel{\to }{{𝜏}_{3}}+\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.+\stackrel{\to }{{𝜏}_{n}}\phantom{\rule{0.44em}{0ex}}=\sum _{i=1}^{n}{𝜏}_{i}=0\phantom{\rule{0.22em}{0ex}}$ CoupleA pair of equal and opposite forces with different lines of action is known as a couple. A couple produces rotation without translation. The moment of a couple does not depend on the point about which you take the moments. Kinematics of rotational motion Dynamics of rotational motionFor calculation of torque in rigid bodyi) consider only those forces that lie in the plane perpenducular to the axis.ii) forces that are parallel to the axis will give torque perpendicular to the axis. | 2,733 | 8,718 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 25, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.895725 |
https://www.exceltip.com/excel-formula-and-function/mathematical-functions/excel-round-function.html | 1,603,126,280,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107863364.0/warc/CC-MAIN-20201019145901-20201019175901-00550.warc.gz | 712,077,420 | 14,488 | # How to Use Round Function in Excel
In this article, we will learn how to use Round Function to round off the numbers up to the nearest number in Excel.
ROUND function used to round down the numbers 0-4 and round up the numbers 5-9 It takes 2 arguments
• The Number and
• Up to digit, we need to round off.
Syntax:
=ROUND(number, num_digit)
Let’s use them in some examples to see how it works.
In one Column there are Numbers and in other, there are num_digits up to which the number to be rounded off.
Use the formula in C2 cell.
=ROUND(A2,B2)
Explanation:
Here in C2 cell, the number will be round down to 1 decimal place as 3 comes in 1-4.
As you can see 2.3 is rounded down to 1 decimal place.
Copy the formula using Ctrl + D taking the first cell till the last cell needed formula.
As you can see the result. Cell C2, C4 and C5 are rounded down and Cell C3, C6 and C7 are rounded up.
The 2nd argument(num_digit) decides upto which decimal or 10s place the number will be rounded off.
1. First 3 cells are rounded off upto 1, 2 and 3 decimal places respectively.
2. In C5, The number is rounded off to the nearest whole number.
3. The last 2 cells are rounded off to the nearest 10 and 100 respectively.
To get the number rounded off upto the nearest whole number, 0 is used default as num_digit
Hope you understood how to use ROUND function in Excel. Explore more articles on ROUNDUP and ROUNDDOWN function here. State your queries in the comment box below.
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The graph of a quadratic function has a U-shaped curve and is called a parabola. The parameters of a parabola give us important information about a function and are used to represent the function graphically.
In this article, we will learn about the different parts of the graphs of quadratic functions and we will graph these functions in their vertex form and their standard form.
Relevant for
See examples
Relevant for
See examples
## Different parts of parabolas
Recall that a quadratic function has the form , where ab and c are constants and .
The graphs of quadratic functions are U-shaped as shown below:
The sign in the coefficient a determines whether the graph opens up or opens down. If , the graph opens up, and if , the graph opens down.
Parabolas have different parameters that determine their shape and their location in the Cartesian plane. These parameters are the vertex, the axis of symmetry, the y-intercept, and the x-intercepts.
### Vertex
The vertex is the extreme point on the graph of a quadratic function, that is, it is the highest point or the lowest point. If the parabola opens upwards, the vertex represents the lowest point and if the parabola opens downwards, the vertex represents the highest point.
### Axis of symmetry
All parabolas are symmetric with respect to a vertical line called the axis of symmetry. This vertical line passes through the vertex.
### y-intercept
The y-intercept is the point where the parabola crosses the y-axis. For all graphs of quadratic functions, there is a single y-intercept. If there were more y-intercepts, the graph would not represent a function.
### x-intercepts
The x-intercepts are the points where the parabola crosses the x-axis. The x-intercepts represent the zeros or the roots of the quadratic function, that is, the values of x when we have . It is possible to have zero x-intercepts, one x-intercept, and two x-intercepts.
The number of intercepts depends on the location of the graph of the quadratic function. When the parabola has two x-intercepts, the vertex always lies between these intercepts due to the symmetry of the graph.
## Graphical interpretation of the solutions of quadratic functions
We can find the roots of quadratic functions algebraically or graphically.
To find the roots algebraically, we can use the quadratic formula . We can also find the roots graphically by making several observations of the graph of a quadratic function.
Let’s look at the connection between finding roots algebraically and graphically with the graph of the function :
We can see that the graph crosses the x-axis at the points (-2, 0) and (1, 0). We know that the x-intercepts represent the roots of the quadratic function, so and are the roots.
Now let’s find the roots of algebraically. We can use the quadratic formula with the coefficients .
Therefore, we have two possible values for x: and . By simplifying these values, we get y .These are the same values that we find graphically.
###### EXAMPLE
Find the roots of the function algebraically and graphically:
We can see that the graph does not cross the x-axis, therefore, it does not have real roots. We can verify this algebraically.
We will use the quadratic formula with the coefficients .
We see that we have , which is not a real number. This means that the quadratic function has no real roots.
## Graphs of quadratic functions in vertex form
The vertex form of a quadratic function allows us to find the vertex of the graph easily.
Quadratic equations can be presented in different ways. For example, we have already seen its standard form:
Another common form is the vertex form:
In this form, the vertex is the point . As we saw earlier, the coefficient determines whether the parabola opens up or down.
### Transform from vertex shape to standard shape
To transform a quadratic function written in vertex form to standard form, we simply expand the squared expression and combine like terms. For example, the following quadratic:
can be rewritten like this:
### Transform from standard form to vertex form
This is a bit more difficult and we have to use a process called “complete the square”.
When we have
Suppose we want to write in vertex form. We observe that the coefficient of the term is 1. When this is the case, we look at the coefficient of the term x and take its half.
Then we square that number. That is, in this case, we have 2, taking its half we have 1, and squared it, we have 1. Then, we add and subtract this number as shown below:
Here, we add and subtract the same number, so we don’t actually change the function. Now, the expression in the parentheses can be written as a square and we have:
In this way, the vertex is .
When we have
This is a bit more difficult than the previous case, but we can use the same idea to transform the function. Suppose we want to write in vertex form. Now, the coefficient of the term is 2. We can factor the 2 of the first two terms:
Therefore, we complete the square inside the parentheses. We observe that half of 6 is 3 and squared is 9, so we add and subtract 9 within the parentheses :
and we solve as follows:
Thus, the vertex of the graph is .
## Graphs of quadratic functions in standard form
A quadratic function in the form is in standard form. Regardless of the format, the graph of a quadratic function is a parabola. For example, the following is the graph of :
Each coefficient of a quadratic function has an impact on the shape and location of the graph on the Cartesian plane.
Coefficient of
The coefficient a controls the rate of increase or decrease of the function from the vertex. The larger and positive a is, the faster the quadratic function will grow and the graph will appear to get “thinner.”
The coefficient a also controls where the parabola will open. If we have , the graph opens up and if we have , the graph opens down.
Axis of symmetry
The coefficients a and b control the axis of symmetry and the x-coordinate of the vertex of the parabola. We can find the axis of symmetry of a parabola as follows:
For example, in the parabola , we have , so we have . The vertex has an x coordinate of 1:
The y-intercept of the parabola
The coefficient c controls the vertical position of the parabola. This is the point where the parabola intersects the y-axis. The point (0, y) is the y-intercept of the parabola. In this case, and the parabola intersects the y-axis at the point (0, 2). | 1,470 | 6,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2022-21 | longest | en | 0.873998 |
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1. ## Composite Pressure Vessel Matrix Damage
Hello, I am optimizing a Composite pressure vessel with a simple model. I am using a cylinder with a PLOAD4 radial Pressure and a force on the axial direction (see Picture). The Cylinder is build up with SHELL Elements. The damage criterion is Puck. At every simulations occurs a matrix damage (mode A) in the laminate, which I think is false. Is it possible that Composite pressure vessels has a Matrix Damage under load? Do I use false Model definitions? Thank you for your help. Tube_Test_100_mat2.fem
2. ## Plasticity Law
Hello, I am working with the material law Orthotropic Damage and plasticity, where the shared evolution Law is is defines as it shown below. What is the meaning of the Parameters K_inf, K_0 and H'? Arent they Materialparameters? If yes, why I should not define them in the Modell, how are they calculated? Thank you for your answer
4. ## Viewing .h3d after stopping the the calculations
Hello! At some calculations I am able to see the results in the h3d file. I dont have to wait until the calculations are finnisched. Which ist ist Keyword, which with I can control it? Thank you for your help! Best regards.
5. ## Composite Using Solid Elements
Hi Ivan, thank you for your answer. I had a false paramter in my Control Card. Regards, David
6. ## Composite Using Solid Elements
Hi, I built up a Tension analysis wit a coupon speciment. Material: LAW12 Property: P6_SOL_ORTH I want to set up the materialorientation. I tried to assign the nodes to a coordynatesystem, which is rotated by 45 grad and tried to set the parameter "Psi" on 45 Degree, but neither of them worked. Can you please help me by building up a laminate? ##-------------------------------------------------------------------------------------------------- ## Orthotropic Solid Property Set (pid 6) ##-------------------------------------------------------------------------------------------------- /PROP/SOL_ORTH/1 solids 17 2 3 2 1 45.0
7. ## Drape Estimator - How to export !?
Hi, could somebody found a solution on this Topic? Thank you for your answer. David
Thank you!
9. ## Auto-Contact Missing
Hi @Prakash Pagadala Thank you for the quick answer, I found it. Somehow, I missed the toggle options "Elems". My second question would be, that the Name of the contact surface get a "Csurf_" part in his name automatically, which was not make by the older Method(Tool-> Auto Contact) of Creating Contactsurface before. I dont know if it does any differences or not. I marked it, on the picture. The last to Contactsurfaces were made by the new Method (Contact Tool->AutoContact), the others were made by the old method. It does not mean any differences I think, or do?
10. ## Auto-Contact Missing
Hi, Thank You for your answer! I found it, but I cant modify the choosenelements az the Contact surface, as I could before. I want to modify my Master and my Slave surfaces, but I can only pick them by Contactsurface entity. Furthermore I choose the entity type: Surfaces, and it adds, an other nameformat for the Contactsurface then the other version of Auto-Contact. I marked the different names. Does it matter? Arent they the same?
11. ## Auto-Contact Missing
Hi, I have a problem by making an Auto-Contact in Hypermesh. Somehow the Auto-Contact function is missing under Tools. It should be at "Pretension Manager", but I can't find. I tried to refresh my User Profile to Optistruct, but it didnt help. Thank you for the answers!
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1. ### Can I increase the sample size by generating random numbers to apply the Chi-Square Goodness of Fit Test?
Thank you very much obh! Thank you! I'm applying on queuing theory. I'm checking to see if I can apply the Markovian model.
2. ### Can I increase the sample size by generating random numbers to apply the Chi-Square Goodness of Fit Test?
Hi obh! 8 means 8 buses arrived in the first 15 minutes.
3. ### Can I increase the sample size by generating random numbers to apply the Chi-Square Goodness of Fit Test?
My code in R to the generation of random numbers: A<-c(8, 13, 13, 14, 15, 11, 16, 11) n<-15 B<-c() for (j in 1:length(A)) { arrival<-vector("numeric", length = 15) sum<-A[j] for (i in 1:sum) { home<-abs(floor((runif(1,1,(n+0.99999))))) arrival[home]<-arrival[home]+1 }...
4. ### Can I increase the sample size by generating random numbers to apply the Chi-Square Goodness of Fit Test?
Hey @hIsmith I'm doing this because the chi-square test requires a sufficient sample size in order for the chi-square approximation to be valid. Am I right? 1 degree--> First element of the sample B 15 degree --> fifteenth element of the sample B 0+1+0+1+2+0+0+0+0+1+0+0+0+1=8 ---> (8 is the...
5. ### Can I increase the sample size by generating random numbers to apply the Chi-Square Goodness of Fit Test?
Does increasing the sample size by random number generation change the distribution? I have a sample of size 8. Each sample value represents the number of bus arrivals at a bus stop every 15 minutes. But I wanted to apply the chi-square test to verify the fitting with the Poisson distribution... | 444 | 1,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-35 | latest | en | 0.808807 |
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Solve the following problems dealing with exponential functions.
You may want to have your graphing calculator handy.
(Each solution method shown is but one possible approach to the answer.)
1 Diego decided to invest his \$500 tax refund rather than spending it. He found a bank that would pay him 4% interest, compounded quarterly. If he deposits the entire \$500 and does not deposit or withdraw any other amount, how long will it take him to double his money in the account?
2 The half-life of carbon-14 is known to be 5720 years. Doctor Frankenstein has 300 grams of carbon-14 in his experimental laboratory. If untouched, how many of the 300 grams will remain after 1200 years?
3 Your brother tells you a secret. You see no harm in telling two friends. After this second "passing" of the secret, 4 people now know the secret (your brother, you and two friends). If each of these friends now tells two new people, after the third "passing" of the secret, eight people will know. If this pattern of spreading the secret continues, how many people will know the secret after 10 such "passings"?
4 The radioactive element polonium-210 has a half-life of 138 days. If you have 100 micrograms of polonium-210, how much will remain after 60 days?
5 The number of wolves in the wild in the northern section of the Cataragas county is decreasing at the rate of 3.5% per year. Your environmental studies class as counted 80 wolves in the area. After how many years will this population of 80 wolves drop below 15 wolves, if this rate of decrease continues?
Topic Index | Algebra2/Trig Index | Regents Exam Prep Center Created by Donna Roberts | 406 | 1,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2014-23 | longest | en | 0.925405 |
http://planetmath.org/exponentiation | 1,521,831,628,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648431.63/warc/CC-MAIN-20180323180932-20180323200932-00291.warc.gz | 220,092,054 | 4,803 | # exponentiation
• In the entry general associativity, the notion of the power $a^{n}$ for elements $a$ of a set having an associative binary operation$\cdot$” and for positive integers $n$ as exponents (http://planetmath.org/GeneralPower) was defined as a generalisation of the operation. Then the two power laws
$a^{m}\!\cdot\!a^{n}\;=\;a^{m+n},\quad(a^{m})^{n}\;=\;a^{mn}$
are . For the validity of the third well-known power law,
$(a\!\cdot\!b)^{n}\;=\;a^{n}\!\cdot\!b^{n},$
the law of power of product, the commutativity of the operation is needed.
Example. In the symmetric group $S_{3}$, where the group operation is not commutative, we get different results from
$[(123)(13)]^{2}\;=\;(23)^{2}\;=\;(1)$
and
$(123)^{2}(13)^{2}\;=\;(132)(1)\;=\;(132)$
(note that in these “products”, which compositions of mappings, the latter “factor” acts first).
• Extending the power notion for zero and negative integer exponents requires the existence of http://planetmath.org/node/10539neutral and inverse elements ($e$ and $a^{-1}$):
$a^{0}\;:=\;e,\qquad a^{-n}\;:=\;(a^{-1})^{n}$
The two first power laws then remain in for all integer exponents, and if the operation is commutative, also the .
When the operation in question is the multiplication of real or complex numbers, the power notion may be extended for other than integer exponents.
• One step is to introduce fractional (http://planetmath.org/FractionalNumber) exponents by using roots (http://planetmath.org/NthRoot); see the fraction power.
• The following step would be the irrational exponents, which are in the power functions. The irrational exponents are possible to introduce by utilizing the exponential function and logarithms; another way would be to define $a^{\varrho}$ as limit of a sequence
$a^{r_{1}},\,a^{r_{2}},\,\ldots$
where the limit of the rational number sequence $r_{1},\,r_{2},\,\ldots$ is $\varrho$. The sequence $a^{r_{1}},\,a^{r_{2}},\,\ldots$ may be shown to be a Cauchy sequence.
• The last step were the imaginary (non-real complex) exponents $\mu$, when also the base of the power may be other than a positive real number; the one gets the so-called general power.
Title exponentiation Exponentiation 2013-03-22 19:08:44 2013-03-22 19:08:44 pahio (2872) pahio (2872) 9 pahio (2872) Topic msc 20-00 ContinuityOfNaturalPower power law power of product | 686 | 2,371 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 18, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-13 | latest | en | 0.787249 |
https://riddles360.com/riddle/name-the-animal | 1,679,921,988,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00125.warc.gz | 561,330,037 | 7,095 | # Name The Animal
Can you name an animal whose killing cannot be performed without spilling your own blood?
# Similar Riddles
When my father was 31 I was 8. Now he is twice as old as me. How old am I?
Asked by Neha on 01 Oct 2021
##### Egg and Hen
One fine day, an intellectual man came to the emperor's court with the aim of testing Birbal's wittiness. In order to do this, he challenged Birbal to answer his question and hence prove that he was as intelligent and witty as he was said to be.
He asked Birbal, Do you want me to ask one difficult question or a hundred easy ones?
Since both Akbar and Birbal had had a tough day and were eager to leave, Birbal hastily told the intellectual to ask him a single difficult question.
Intellectual: OK. Tell me what came first into the world, the egg or the chicken?
Of course, the chicken, Birbal replied with a smile.
This time with a note of victory in his voice, the intellectual asked Birbal, How will you demonstrate that?
What did Birbal say?
Asked by Neha on 17 May 2021
##### Distance between Villages
There is a straight highway. Four different villages lie on that highway. The distance between them is different. The third village is 60km away from the first village; the fourth is 40 km away from the second; the third is 10 km near to the fourth that it is to the second.
Can you calculate the distance between the fourth and the first village ?
Asked by Neha on 23 Mar 2023
##### Arrange Hats in Squares
You are given 16 witch hats. The hats are divided in four different colours – red, blue, green and yellow. Every colour has been assigned to four hats. Now each of the hat will be glued with a label of an arithmetic sign – ‘+’, ‘-‘, ‘x’ or ‘/’. But you can label one sign only once on one colour. In such an arrangement, the hats can be uniquely defined by its colour and symbol.
Can you arrange all the 16 hats in a 4x4 grid in a fashion that no two rows and columns have a repetition of colour or sign?
We have arranged four hats in the below picture to assist you.
Asked by Neha on 17 May 2021
##### Akbar Birbal Riddle
A man was convicted of a minor offence in Akbar court. Akbar decided to give him a chance. He asked him to give a statement. If the statement is true, he will be killed by lions and if it is false, he will be killed by trampling of wild elephants.
The convicted person requested help from Birbal and since the crime was not a big one, Birbal decided to help him. Whatever Birbal suggested impressed Kabir and he let the convicted person go.
What did Birbal suggest to the person?
Asked by Neha on 10 May 2021
##### Name the King
I start and end with 500 and carry a 5 in my heart. But I require the first letter and the first number to be complete. I am the name of a King.
Who am I ?
Asked by Neha on 10 Mar 2021
##### Coolest Letter in Alphabet
What is the coolest letter in the alphabet?
Asked by Neha on 24 Oct 2021
##### HOROBOD Rebus riddle
What is the word or phrase?
Asked by Neha on 10 May 2021
##### Find the Number
I am a three digit number.
My tens digit is five more than my ones digit.
My hundreds digit is eight less than my tens digit.
What number am I?
Asked by Neha on 17 May 2021
##### Make Two Square
In the picture that is attached with this question, you can find a square which comprises of four little squares inside it. Consider this square to be made with matchsticks. You have to remove two matchsticks such that only two squares remain instead of five.How will you do it ?
Asked by Neha on 14 May 2021
### Amazing Facts
###### Jigsaw puzzles
Jigsaw puzzles soared in popularity during the great depression, as they provided a cheap, long-lasting, recyclable form of entertainment. | 904 | 3,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-14 | longest | en | 0.977779 |
https://pmioman14.wordpress.com/2014/09/02/w7_-khalid-almamari_economic-evaluation-of-site-powers-alternatives/ | 1,501,022,436,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425407.14/warc/CC-MAIN-20170725222357-20170726002357-00426.warc.gz | 701,303,374 | 38,036 | # W7_ Khalid Almamari_Economic Evaluation of Site Power’s Alternatives
1. Problem Definition
The SCADA facility is using Diesel generators for power supply during its operation with small requirements in absent of OHL connection. Those generators require a lot of rental cost and maintenance through the whole year. However, the company power is available around the field and can be constructed but with little of capital investment to be connected to the power system. The questions is the selection of better option either to stop using diesel generators and buying power from the company itself or maintain the stuff as they are now?
1. Identify the Feasible Alternatives
The alternatives are:
1. Do nothing and maintain the stuff as they are now. OR
2. Stop using diesel generators and extension of new OHL connection.
1. Development of the Outcomes for Alternatives
• Calculate the annual cost of diesel generators rental and the diesel cost.
• Calculate the annual cost of connection the power along with additional cost for operation.
1. Selection of Criteria
Net Present Value (NPV) is the present value of cash inflows and outflows of a firm. It assists in determining the present value of an investment by deducting the entire cash flows.
The alternative with higher net present value is selected.
1. Analysis and Comparison of the Alternatives.
As collected from the field, the total power required per annum is small for the SCADA project but the power will support ESP once converted from self-flow to artificial pumps in additional that generator can’t handle it.
The two alternatives are compared as in the following table at 8.0 % (as taken from finance department) rate for 5 year. [Negative (-) means cost spent]
Comparison Parameter Diesel Generators Company Power Annual Operating cost (rent + diesel) -45000\$ -1500\$ Capital Cost N/A -100000\$ Total NPV -179672\$ -109218\$
1. Selection of the Preferred Alternative.
As using government power has higher NPV which is good investment, therefore, it is selected as the preferred project to be implemented for nearby wells.
1. Performance Monitoring and the Post Evaluation of Result.
Such diesel generators will be either return to Rental Company or transfer them to the new wells which are located in new field where there is no power system at all.
References:
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## 4 thoughts on “W7_ Khalid Almamari_Economic Evaluation of Site Power’s Alternatives”
1. Hi Khalid, you have selected an EXCELLENT case study and your analysis was great with ONE exception……… Before you can make a valid assessment, you have the professional obligation to VALIDATE the MARR (8%)
Go here and then provide us with these calculations for Oman at your site…
http://pmworldjournal.net/article/using-analytical-hierarchy-process-determine-appropriate-minimum-attractive-rate-return-oil-gas-projects-indonesia/
The problem with many of these calculations is the MARR has proven to be too low, which means is should be validated…… (Look at the blog postings from your colleagues as well as several of them have done the calculations for Oman already)
BR,
Dr. PDG, Jakarta
• thanks Dr.
I will reach on my colleagues’ posting now!
• found it and updated.
• Excellent, Khalid….. Great topic, just gotta get you to develop it more completely…. | 697 | 3,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.929066 |
https://www.chemicalslearning.com/2022/06/packed-columns-for-distillation.html | 1,685,350,670,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644817.32/warc/CC-MAIN-20230529074001-20230529104001-00122.warc.gz | 768,786,519 | 46,964 | Packed Columns/Towers for Distillation and Gas Absorption
# Packed Columns/Towers for Distillation and Gas Absorption
Packed columns are useful for distillation, especially whenever we have to carry out operations at low pressure (vacuum distillation) and whenever we are dealing with heat-sensitive materials.
#### Packed column distillation design
Packings are usually cheaper than plates for columns less than 600mm in diameter. A packed column consists of a cylindrical shell containing support plates and a liquid distributor. The cylindrical shell is filled with some sort of packing that rest on the support plate. The packing material offers a large interfacial area for mass transfer. The liquid distributor is designed for effective irrigation of the packings.
### Packed column for continuous distillationÂ
The ability of a given packing to affect the desired mass transfer between gas and liquid phase is usually expressed as the height equivalent to one theoretical plate (HETP). In plate columns wherein a process of enrichment is stage-wise the vapour leaving the plate is richer with respect to more volatile components than the vapour entering the plate by one equilibrium stage. In a packed column, the same enrichment of the vapour will occur at a certain height of packing and is termed as the height equivalent to one theoretical plate.
#### Degree of SeparationÂ
Thus in packed columns, one equilibrium step is represented by a certain height of the packed bed and the required height of packing for the desired degree of separation is given by: HETP x the Number of ideal plates required.
HETP can be estimated with the help of the following empirical equation.Â
HETP = k1 • Gk2 • Dtk3 • Z1/3 • a •uL/pL
Where k1, k2, k3 are empirical constants for packing and are a function of the type and size the packings
G = Superficial gas mass velocity
Dt = Tower diameter
Z = Height of packing
a = Relative volatility
uL and uL are the viscosity and density of liquid respectively.Â
### Plate EfficienciesÂ
The relationship between the performance of theoretical/ideal and actual plates is expressed in terms of plate efficiency. The types of efficiency are:
1. Overall plate efficiency/overall column efficiency. Â
2. Murphree plate efficiency andÂ
3. Point/local efficiency.Â
Overall plate efficiency is the ratio of the number of ideal or theoretical plates required to produce a given separation in the entire column to the number of actual plates required to effect the same separation.Â
If the overall efficiency is 60% and 12 ideal plates are called for, then the actual plates needed are 12/0.60=20.
### Murphree plate efficiency
If applies to an individual plate in a column and is defined as the actual change in average composition accomplished by a given plate divided by the change in average composition if the vapour leaving the plate were in equilibrium with the liquid leaving the plate.Â
Point efficiency is defined in the same manner as the Murphree plate efficiency but it is applied to a single location on a given plate.Â
## Packed columns/Towers for Absorption
Packed columns are most frequently used for gas absorption (and are used to a limited extent for distillation) wherein the liquid is dispersed in the form of film and the gas flows as a continuous phase.Â
#### Packed column working principle
These are continuous contact equilibriums generally operated in a counter-current fashion. A packed column consists of a vertical cylindrical shell constructed out of metal, plastic, ceramic, etc. and filled with suitable packings which offer a large interfacial area for gas-liquid contact for mass transfer between the phases. A bed of the packing rests on a support plate which offers very low resistance to gas flow.Â
It is provided with a gas inlet and distributing space at the bottom, a liquid inlet and a liquid distributor at the top and gas-liquid outlets at the top and bottom. A liquid solvent is introduced from the top through the liquid distributor which irrigates/ wets the surface of packing uniformly, liquid trickles down the bed and finally, the liquid is enriched in in solute called a strong liquor (solute + solvent) leaves the bottom of the column. The liquid flow rate should be sufficient for good wetting of packing.
A solute-containing gas is introduced from the bottom of the tower and rises upward through the interstices/open spaces in the packing counter current to the flow of the liquid. The lean gas leaves the column from the top of the tower. In the case of tall columns/towers, liquid redistributors are used to redistribute liquid to avoid channelling of the same.
(i) Minimum structure
(ii) Low-pressure drop
(iii) Low liquid hold-up
(iv) Handle corrosive liquids and liquids that tend to foam
(v) Low initial investment andÂ
(vi) high liquid to gas ratios.Â
(I) Relatively inflexible
(ii) Can not operate over a wide range of either vapour or liquid rates per unit cross-section
(iii) Distribution of liquid is difficult
(iv) Can not handle dirty fluids that tends deposit a sedimentÂ
(v) Can not be used where large temperature changes are encountered.
Take these Notes is, Orginal Sources:Â Unit Operations-II, KA Gavhane
Thanks to visit this site. | 1,123 | 5,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.895012 |
http://www.gurufocus.com/term/Pretax+Income/EMR/Pre-Tax%2BIncome/Emerson%2BElectric%2BCo | 1,493,209,367,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121305.61/warc/CC-MAIN-20170423031201-00108-ip-10-145-167-34.ec2.internal.warc.gz | 562,060,286 | 28,335 | Switch to:
GuruFocus has detected 5 Warning Signs with Emerson Electric Co \$EMR.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Emerson Electric Co (NYSE:EMR)
Pre-Tax Income
\$2,277 Mil (TTM As of Dec. 2016)
Pretax income is the income that a company earns before paying income taxes. Emerson Electric Co's pretax income for the three months ended in Dec. 2016 was \$464 Mil. Its pretax income for the trailing twelve months (TTM) ended in Dec. 2016 was \$2,277 Mil. Emerson Electric Co's pretax margin was 14.43%.
During the past 13 years, Emerson Electric Co's highest Pretax Margin was 23.43%. The lowest was 12.19%. And the median was 14.23%.
Definition
This is the income that a company earns before paying income taxes.
Emerson Electric Co's Pretax Income for the fiscal year that ended in Sep. 2016 is calculated as
Pretax Income = Operating Income + Non-Recurring Items + Interest Expense + Interest Income + Other = 2798 + -294 + -206 + 18 + 0 = 2,316
Emerson Electric Co's Pretax Income for the quarter that ended in Dec. 2016 is calculated as
Pretax Income = Operating Income + Non-Recurring Items + Interest Expense + Interest Income + Other = 543 + -33 + -52 + 6 + 0 = 464
Emerson Electric Co Pre-Tax Income for the trailing twelve months (TTM) ended in Dec. 2016 was 608 (Mar. 2016 ) + 744 (Jun. 2016 ) + 461 (Sep. 2016 ) + 464 (Dec. 2016 ) = \$2,277 Mil.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Emerson Electric Co's Pretax Margin for the quarter that ended in Dec. 2016 is calculated as
Pretax Margin = Pretax Income / Revenue = 464 / 3216 = 14.43%
During the past 13 years, Emerson Electric Co's highest Pretax Margin was 23.43%. The lowest was 12.19%. And the median was 14.23%.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Emerson Electric Co Annual Data
Sep07 Sep08 Sep09 Sep10 Sep11 Sep12 Sep13 Sep14 Sep15 Sep16 Pretax Income 3,093 3,591 2,450 2,879 3,631 3,115 3,196 3,191 3,807 2,316
Emerson Electric Co Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Pretax Income 659 765 1,604 796 642 434 608 744 461 464
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 734 | 2,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-17 | latest | en | 0.949786 |
http://math.stackexchange.com/questions/155729/cardinality-of-the-set-of-ultrafilters-on-an-infinite-boolean-algebra | 1,469,470,749,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824337.54/warc/CC-MAIN-20160723071024-00295-ip-10-185-27-174.ec2.internal.warc.gz | 156,622,726 | 18,239 | # Cardinality of the set of ultrafilters on an infinite Boolean algebra
Let $\mathfrak B$ be a Boolean algebra with an infinite power $\kappa$. My question is how many ultrafilters does it have? $\kappa$ or $2^\kappa$? Or even smaller?
-
I know in finite case it has power $n$ where $\|\mathfrak B\|=2^n$, so I guess in the infinite case the cardinality should also not greater than it of $\mathfrak{B}$. – Popopo Jun 8 '12 at 19:07
It can be at least those two options:
Example I:
Consider the algebra $\{A\subseteq\mathbb R\mid A\text{ finite, or }\mathbb R\setminus A\text{ is finite}\}$, that is finite co-finite subsets of $\mathbb R$. It is not hard to verify that this is indeed a Boolean algebra of size $2^{\aleph_0}$.
Suppose that $U$ is an ultrafilter over this Boolean algebra, if it does not contain any singleton then it has to be the collection of co-finite sets; if it contains a singleton then it is principal. We have continuum many principal ultrafilters and one free. Therefore a Boolean algebra of size continuum with continuum many ultrafilters.
Example II: On the other hand consider $\mathcal P(\mathbb N)$. We know that there are $2^{2^{\aleph_0}}$ many ultrafilters over this Boolean algebra. So we have a Boolean algebra of size $2^{\aleph_0}$ with $2^{2^{\aleph_0}}$ many ultrafilters.
For a further discussion on this sort of example see: The set of ultrafilters on an infinite set
-
@Asaf_Karagila:Thank you. But Example II is not so clear for me, could you give me a reference? – Popopo Jun 8 '12 at 19:23
@Popopo: I added a link with a discussion for such result. In our case we look for "ultrafilters over $\mathbb N$" which mean essentially ultrafilters in the Boolean algebra $\mathcal P(\mathbb N)$. – Asaf Karagila Jun 8 '12 at 19:34
OK, thank you very much. – Popopo Jun 8 '12 at 20:04 | 513 | 1,832 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2016-30 | latest | en | 0.85509 |
https://www.coursehero.com/tutors-problems/Math/5345-Consider-the-IVP-x1-tx-x01-euler-approximation-of-the-IVP-usin/ | 1,531,870,001,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589932.22/warc/CC-MAIN-20180717222930-20180718002930-00027.warc.gz | 843,780,365 | 22,920 | View the step-by-step solution to:
# Consider the IVP x'=1-tx, x(0)=1 euler approximation of the IVP using h=1.0, h=0.
Consider the IVP x’=1-tx, x(0)=1 euler approximation of the IVP using h=1.0, h=0.5
Dear student, We have attached the... View the full answer
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Consider the IVP x’=1tx, x(0)=1 euler approximation of the IVP using h=1.0, h=0.5
=====================================================...
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Browse Documents | 205 | 836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-30 | longest | en | 0.791958 |
https://www.manualslib.com/manual/359091/Hp-F2226a-48gii-Graphic-Calculator.html?page=490 | 1,571,395,944,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00135.warc.gz | 990,520,401 | 60,093 | # Laplace Transform Theorems - HP F2226A - 48GII Graphic Calculator User Manual
Graphing calculator.
function LAP you get back a function of X, which is the Laplace transform of
f(X).
Example 2 – Determine the Laplace transform of f(t) = e
'EXP(2*X)*SIN(X)' ` LAP The calculator returns the result: 1/(SQ(X-2)+1).
Press µ to obtain, 1/(X
When you translate this result in paper you would write
F
(
Example 3 – Determine the inverse Laplace transform of F(s) = sin(s). Use:
'SIN(X)' ` ILAP.
The calculator takes a few seconds to return the result:
'ILAP(SIN(X))', meaning that there is no closed-form expression f(t), such that f(t)
-1
= L
{sin(s)}.
Example 4 – Determine the inverse Laplace transform of F(s) = 1/s
'1/X^3' ` ILAP µ. The calculator returns the result: 'X^2/2', which is
-1
3
interpreted as L
{1/s
} = t
Example 5 – Determine the Laplace transform of the function f(t) = cos (a⋅t+b).
Use: 'COS(a*X+b)' ` LAP . The calculator returns the result:
Press µ to obtain –(a sin(b) – X cos(b))/(X
interpreted as follows: L {cos(a⋅t+b)} = (s⋅cos b – a⋅sin b)/(s
## Laplace transform theorems
To help you determine the Laplace transform of functions you can use a
number of theorems, some of which are listed below. A few examples of the
theorem applications are also included.
Differentiation theorem for the first derivative. Let f
for f(t), i.e., f(0) = f
o
2
-4X+5).
2
t
s
)
L
{
e
sin
t
}
2
s
2
/2.
, then
2t
⋅sin(t). Use:
1
4
s
5
3
. Use:
2
2
+a
).
The transform is
2
2
+a
).
be the initial condition
o
Page 16-12 | 512 | 1,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-43 | latest | en | 0.790045 |
https://www.brightstorm.com/math/calculus/the-derivative/the-derivative-function/ | 1,685,312,604,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00103.warc.gz | 750,764,782 | 24,909 | ###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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# The Derivative Function - Concept
Norm Prokup
###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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By definition, the derivative is a function which is derived from another function. The definition of the derivative is usually only written for one point, but the function is defined for all points. Derivative functions of many kinds of functions can be found, including derivatives of linear, power, polynomial, exponential, and logarithmic functions.
I want to talk about the derivative function let's say we're looking at a function like f of x equals x squared plus 1. I have a graph here, in a previous example we found the derivative of this function at x equals 3 and we used the definition of the derivative and we got 6. Now it turns out you could do that at pretty much any point, you can find the derivative at negative 1, turns out to be negative 2 and you can find the derivative at a half and it turns out to be 1.
This function and many functions have derivatives at every value of x in their domain and so that gives us the notion of the derivative function. This f prime of x defines a function of x. For every value of x you think of you can come up with a derivative value. So that's a new function that we call the derivative function for f of x.
We're going to be talking about derivative functions for a while and the idea is that given a function you want to find it's derivative that is it's a derivative function and usually that means finding a formula for the derivative function that doesn't involve the limit, and that's what we're going to be doing in the next few episodes. | 441 | 2,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-23 | latest | en | 0.958921 |
http://metamath.tirix.org/mpeuni/efaddlem | 1,721,767,192,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00472.warc.gz | 25,545,017 | 7,107 | Metamath Proof Explorer
Description: Lemma for efadd (exponential function addition law). (Contributed by Mario Carneiro, 29-Apr-2014)
Ref Expression
Hypotheses efadd.1 𝐹 = ( 𝑛 ∈ ℕ0 ↦ ( ( 𝐴𝑛 ) / ( ! ‘ 𝑛 ) ) )
efadd.2 𝐺 = ( 𝑛 ∈ ℕ0 ↦ ( ( 𝐵𝑛 ) / ( ! ‘ 𝑛 ) ) )
efadd.3 𝐻 = ( 𝑛 ∈ ℕ0 ↦ ( ( ( 𝐴 + 𝐵 ) ↑ 𝑛 ) / ( ! ‘ 𝑛 ) ) )
efadd.4 ( 𝜑𝐴 ∈ ℂ )
efadd.5 ( 𝜑𝐵 ∈ ℂ )
Assertion efaddlem ( 𝜑 → ( exp ‘ ( 𝐴 + 𝐵 ) ) = ( ( exp ‘ 𝐴 ) · ( exp ‘ 𝐵 ) ) )
Proof
Step Hyp Ref Expression
1 efadd.1 𝐹 = ( 𝑛 ∈ ℕ0 ↦ ( ( 𝐴𝑛 ) / ( ! ‘ 𝑛 ) ) )
2 efadd.2 𝐺 = ( 𝑛 ∈ ℕ0 ↦ ( ( 𝐵𝑛 ) / ( ! ‘ 𝑛 ) ) )
3 efadd.3 𝐻 = ( 𝑛 ∈ ℕ0 ↦ ( ( ( 𝐴 + 𝐵 ) ↑ 𝑛 ) / ( ! ‘ 𝑛 ) ) )
4 efadd.4 ( 𝜑𝐴 ∈ ℂ )
5 efadd.5 ( 𝜑𝐵 ∈ ℂ )
6 4 5 addcld ( 𝜑 → ( 𝐴 + 𝐵 ) ∈ ℂ )
7 3 efcvg ( ( 𝐴 + 𝐵 ) ∈ ℂ → seq 0 ( + , 𝐻 ) ⇝ ( exp ‘ ( 𝐴 + 𝐵 ) ) )
8 6 7 syl ( 𝜑 → seq 0 ( + , 𝐻 ) ⇝ ( exp ‘ ( 𝐴 + 𝐵 ) ) )
9 1 eftval ( 𝑗 ∈ ℕ0 → ( 𝐹𝑗 ) = ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) )
10 9 adantl ( ( 𝜑𝑗 ∈ ℕ0 ) → ( 𝐹𝑗 ) = ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) )
11 absexp ( ( 𝐴 ∈ ℂ ∧ 𝑗 ∈ ℕ0 ) → ( abs ‘ ( 𝐴𝑗 ) ) = ( ( abs ‘ 𝐴 ) ↑ 𝑗 ) )
12 4 11 sylan ( ( 𝜑𝑗 ∈ ℕ0 ) → ( abs ‘ ( 𝐴𝑗 ) ) = ( ( abs ‘ 𝐴 ) ↑ 𝑗 ) )
13 faccl ( 𝑗 ∈ ℕ0 → ( ! ‘ 𝑗 ) ∈ ℕ )
14 13 adantl ( ( 𝜑𝑗 ∈ ℕ0 ) → ( ! ‘ 𝑗 ) ∈ ℕ )
15 nnre ( ( ! ‘ 𝑗 ) ∈ ℕ → ( ! ‘ 𝑗 ) ∈ ℝ )
16 nnnn0 ( ( ! ‘ 𝑗 ) ∈ ℕ → ( ! ‘ 𝑗 ) ∈ ℕ0 )
17 16 nn0ge0d ( ( ! ‘ 𝑗 ) ∈ ℕ → 0 ≤ ( ! ‘ 𝑗 ) )
18 15 17 absidd ( ( ! ‘ 𝑗 ) ∈ ℕ → ( abs ‘ ( ! ‘ 𝑗 ) ) = ( ! ‘ 𝑗 ) )
19 14 18 syl ( ( 𝜑𝑗 ∈ ℕ0 ) → ( abs ‘ ( ! ‘ 𝑗 ) ) = ( ! ‘ 𝑗 ) )
20 12 19 oveq12d ( ( 𝜑𝑗 ∈ ℕ0 ) → ( ( abs ‘ ( 𝐴𝑗 ) ) / ( abs ‘ ( ! ‘ 𝑗 ) ) ) = ( ( ( abs ‘ 𝐴 ) ↑ 𝑗 ) / ( ! ‘ 𝑗 ) ) )
21 expcl ( ( 𝐴 ∈ ℂ ∧ 𝑗 ∈ ℕ0 ) → ( 𝐴𝑗 ) ∈ ℂ )
22 4 21 sylan ( ( 𝜑𝑗 ∈ ℕ0 ) → ( 𝐴𝑗 ) ∈ ℂ )
23 14 nncnd ( ( 𝜑𝑗 ∈ ℕ0 ) → ( ! ‘ 𝑗 ) ∈ ℂ )
24 14 nnne0d ( ( 𝜑𝑗 ∈ ℕ0 ) → ( ! ‘ 𝑗 ) ≠ 0 )
25 22 23 24 absdivd ( ( 𝜑𝑗 ∈ ℕ0 ) → ( abs ‘ ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) ) = ( ( abs ‘ ( 𝐴𝑗 ) ) / ( abs ‘ ( ! ‘ 𝑗 ) ) ) )
26 eqid ( 𝑛 ∈ ℕ0 ↦ ( ( ( abs ‘ 𝐴 ) ↑ 𝑛 ) / ( ! ‘ 𝑛 ) ) ) = ( 𝑛 ∈ ℕ0 ↦ ( ( ( abs ‘ 𝐴 ) ↑ 𝑛 ) / ( ! ‘ 𝑛 ) ) )
27 26 eftval ( 𝑗 ∈ ℕ0 → ( ( 𝑛 ∈ ℕ0 ↦ ( ( ( abs ‘ 𝐴 ) ↑ 𝑛 ) / ( ! ‘ 𝑛 ) ) ) ‘ 𝑗 ) = ( ( ( abs ‘ 𝐴 ) ↑ 𝑗 ) / ( ! ‘ 𝑗 ) ) )
28 27 adantl ( ( 𝜑𝑗 ∈ ℕ0 ) → ( ( 𝑛 ∈ ℕ0 ↦ ( ( ( abs ‘ 𝐴 ) ↑ 𝑛 ) / ( ! ‘ 𝑛 ) ) ) ‘ 𝑗 ) = ( ( ( abs ‘ 𝐴 ) ↑ 𝑗 ) / ( ! ‘ 𝑗 ) ) )
29 20 25 28 3eqtr4rd ( ( 𝜑𝑗 ∈ ℕ0 ) → ( ( 𝑛 ∈ ℕ0 ↦ ( ( ( abs ‘ 𝐴 ) ↑ 𝑛 ) / ( ! ‘ 𝑛 ) ) ) ‘ 𝑗 ) = ( abs ‘ ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) ) )
30 eftcl ( ( 𝐴 ∈ ℂ ∧ 𝑗 ∈ ℕ0 ) → ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) ∈ ℂ )
31 4 30 sylan ( ( 𝜑𝑗 ∈ ℕ0 ) → ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) ∈ ℂ )
32 2 eftval ( 𝑘 ∈ ℕ0 → ( 𝐺𝑘 ) = ( ( 𝐵𝑘 ) / ( ! ‘ 𝑘 ) ) )
33 32 adantl ( ( 𝜑𝑘 ∈ ℕ0 ) → ( 𝐺𝑘 ) = ( ( 𝐵𝑘 ) / ( ! ‘ 𝑘 ) ) )
34 eftcl ( ( 𝐵 ∈ ℂ ∧ 𝑘 ∈ ℕ0 ) → ( ( 𝐵𝑘 ) / ( ! ‘ 𝑘 ) ) ∈ ℂ )
35 5 34 sylan ( ( 𝜑𝑘 ∈ ℕ0 ) → ( ( 𝐵𝑘 ) / ( ! ‘ 𝑘 ) ) ∈ ℂ )
36 3 eftval ( 𝑘 ∈ ℕ0 → ( 𝐻𝑘 ) = ( ( ( 𝐴 + 𝐵 ) ↑ 𝑘 ) / ( ! ‘ 𝑘 ) ) )
37 36 adantl ( ( 𝜑𝑘 ∈ ℕ0 ) → ( 𝐻𝑘 ) = ( ( ( 𝐴 + 𝐵 ) ↑ 𝑘 ) / ( ! ‘ 𝑘 ) ) )
38 4 adantr ( ( 𝜑𝑘 ∈ ℕ0 ) → 𝐴 ∈ ℂ )
39 5 adantr ( ( 𝜑𝑘 ∈ ℕ0 ) → 𝐵 ∈ ℂ )
40 simpr ( ( 𝜑𝑘 ∈ ℕ0 ) → 𝑘 ∈ ℕ0 )
41 binom ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝑘 ∈ ℕ0 ) → ( ( 𝐴 + 𝐵 ) ↑ 𝑘 ) = Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) )
42 38 39 40 41 syl3anc ( ( 𝜑𝑘 ∈ ℕ0 ) → ( ( 𝐴 + 𝐵 ) ↑ 𝑘 ) = Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) )
43 42 oveq1d ( ( 𝜑𝑘 ∈ ℕ0 ) → ( ( ( 𝐴 + 𝐵 ) ↑ 𝑘 ) / ( ! ‘ 𝑘 ) ) = ( Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) )
44 fzfid ( ( 𝜑𝑘 ∈ ℕ0 ) → ( 0 ... 𝑘 ) ∈ Fin )
45 faccl ( 𝑘 ∈ ℕ0 → ( ! ‘ 𝑘 ) ∈ ℕ )
46 45 adantl ( ( 𝜑𝑘 ∈ ℕ0 ) → ( ! ‘ 𝑘 ) ∈ ℕ )
47 46 nncnd ( ( 𝜑𝑘 ∈ ℕ0 ) → ( ! ‘ 𝑘 ) ∈ ℂ )
48 bccl2 ( 𝑗 ∈ ( 0 ... 𝑘 ) → ( 𝑘 C 𝑗 ) ∈ ℕ )
49 48 adantl ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝑘 C 𝑗 ) ∈ ℕ )
50 49 nncnd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝑘 C 𝑗 ) ∈ ℂ )
51 4 ad2antrr ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → 𝐴 ∈ ℂ )
52 fznn0sub ( 𝑗 ∈ ( 0 ... 𝑘 ) → ( 𝑘𝑗 ) ∈ ℕ0 )
53 52 adantl ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝑘𝑗 ) ∈ ℕ0 )
54 51 53 expcld ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝐴 ↑ ( 𝑘𝑗 ) ) ∈ ℂ )
55 5 ad2antrr ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → 𝐵 ∈ ℂ )
56 elfznn0 ( 𝑗 ∈ ( 0 ... 𝑘 ) → 𝑗 ∈ ℕ0 )
57 56 adantl ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → 𝑗 ∈ ℕ0 )
58 55 57 expcld ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝐵𝑗 ) ∈ ℂ )
59 54 58 mulcld ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ∈ ℂ )
60 50 59 mulcld ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) ∈ ℂ )
61 46 nnne0d ( ( 𝜑𝑘 ∈ ℕ0 ) → ( ! ‘ 𝑘 ) ≠ 0 )
62 44 47 60 61 fsumdivc ( ( 𝜑𝑘 ∈ ℕ0 ) → ( Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) = Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) )
63 51 57 expcld ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝐴𝑗 ) ∈ ℂ )
64 57 13 syl ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ! ‘ 𝑗 ) ∈ ℕ )
65 64 nncnd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ! ‘ 𝑗 ) ∈ ℂ )
66 64 nnne0d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ! ‘ 𝑗 ) ≠ 0 )
67 63 65 66 divcld ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) ∈ ℂ )
68 2 eftval ( ( 𝑘𝑗 ) ∈ ℕ0 → ( 𝐺 ‘ ( 𝑘𝑗 ) ) = ( ( 𝐵 ↑ ( 𝑘𝑗 ) ) / ( ! ‘ ( 𝑘𝑗 ) ) ) )
69 53 68 syl ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝐺 ‘ ( 𝑘𝑗 ) ) = ( ( 𝐵 ↑ ( 𝑘𝑗 ) ) / ( ! ‘ ( 𝑘𝑗 ) ) ) )
70 55 53 expcld ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝐵 ↑ ( 𝑘𝑗 ) ) ∈ ℂ )
71 faccl ( ( 𝑘𝑗 ) ∈ ℕ0 → ( ! ‘ ( 𝑘𝑗 ) ) ∈ ℕ )
72 53 71 syl ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ! ‘ ( 𝑘𝑗 ) ) ∈ ℕ )
73 72 nncnd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ! ‘ ( 𝑘𝑗 ) ) ∈ ℂ )
74 72 nnne0d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ! ‘ ( 𝑘𝑗 ) ) ≠ 0 )
75 70 73 74 divcld ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( 𝐵 ↑ ( 𝑘𝑗 ) ) / ( ! ‘ ( 𝑘𝑗 ) ) ) ∈ ℂ )
76 69 75 eqeltrd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝐺 ‘ ( 𝑘𝑗 ) ) ∈ ℂ )
77 67 76 mulcld ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) · ( 𝐺 ‘ ( 𝑘𝑗 ) ) ) ∈ ℂ )
78 oveq2 ( 𝑗 = ( ( 0 + 𝑘 ) − 𝑚 ) → ( 𝐴𝑗 ) = ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑚 ) ) )
79 fveq2 ( 𝑗 = ( ( 0 + 𝑘 ) − 𝑚 ) → ( ! ‘ 𝑗 ) = ( ! ‘ ( ( 0 + 𝑘 ) − 𝑚 ) ) )
80 78 79 oveq12d ( 𝑗 = ( ( 0 + 𝑘 ) − 𝑚 ) → ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) = ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑚 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑚 ) ) ) )
81 oveq2 ( 𝑗 = ( ( 0 + 𝑘 ) − 𝑚 ) → ( 𝑘𝑗 ) = ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑚 ) ) )
82 81 fveq2d ( 𝑗 = ( ( 0 + 𝑘 ) − 𝑚 ) → ( 𝐺 ‘ ( 𝑘𝑗 ) ) = ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑚 ) ) ) )
83 80 82 oveq12d ( 𝑗 = ( ( 0 + 𝑘 ) − 𝑚 ) → ( ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) · ( 𝐺 ‘ ( 𝑘𝑗 ) ) ) = ( ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑚 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑚 ) ) ) · ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑚 ) ) ) ) )
84 77 83 fsumrev2 ( ( 𝜑𝑘 ∈ ℕ0 ) → Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) · ( 𝐺 ‘ ( 𝑘𝑗 ) ) ) = Σ 𝑚 ∈ ( 0 ... 𝑘 ) ( ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑚 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑚 ) ) ) · ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑚 ) ) ) ) )
85 2 eftval ( 𝑗 ∈ ℕ0 → ( 𝐺𝑗 ) = ( ( 𝐵𝑗 ) / ( ! ‘ 𝑗 ) ) )
86 57 85 syl ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝐺𝑗 ) = ( ( 𝐵𝑗 ) / ( ! ‘ 𝑗 ) ) )
87 86 oveq2d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) / ( ! ‘ ( 𝑘𝑗 ) ) ) · ( 𝐺𝑗 ) ) = ( ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) / ( ! ‘ ( 𝑘𝑗 ) ) ) · ( ( 𝐵𝑗 ) / ( ! ‘ 𝑗 ) ) ) )
88 72 64 nnmulcld ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ∈ ℕ )
89 88 nncnd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ∈ ℂ )
90 88 nnne0d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ≠ 0 )
91 59 89 90 divrec2d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) = ( ( 1 / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) )
92 54 73 58 65 74 66 divmuldivd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) / ( ! ‘ ( 𝑘𝑗 ) ) ) · ( ( 𝐵𝑗 ) / ( ! ‘ 𝑗 ) ) ) = ( ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) )
93 bcval2 ( 𝑗 ∈ ( 0 ... 𝑘 ) → ( 𝑘 C 𝑗 ) = ( ( ! ‘ 𝑘 ) / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) )
94 93 adantl ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝑘 C 𝑗 ) = ( ( ! ‘ 𝑘 ) / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) )
95 94 oveq1d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( 𝑘 C 𝑗 ) / ( ! ‘ 𝑘 ) ) = ( ( ( ! ‘ 𝑘 ) / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) )
96 47 adantr ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ! ‘ 𝑘 ) ∈ ℂ )
97 61 adantr ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ! ‘ 𝑘 ) ≠ 0 )
98 96 89 96 90 97 divdiv32d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( ! ‘ 𝑘 ) / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) = ( ( ( ! ‘ 𝑘 ) / ( ! ‘ 𝑘 ) ) / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) )
99 96 97 dividd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ! ‘ 𝑘 ) / ( ! ‘ 𝑘 ) ) = 1 )
100 99 oveq1d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( ! ‘ 𝑘 ) / ( ! ‘ 𝑘 ) ) / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) = ( 1 / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) )
101 98 100 eqtrd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( ! ‘ 𝑘 ) / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) = ( 1 / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) )
102 95 101 eqtrd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( 𝑘 C 𝑗 ) / ( ! ‘ 𝑘 ) ) = ( 1 / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) )
103 102 oveq1d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( 𝑘 C 𝑗 ) / ( ! ‘ 𝑘 ) ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) = ( ( 1 / ( ( ! ‘ ( 𝑘𝑗 ) ) · ( ! ‘ 𝑗 ) ) ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) )
104 91 92 103 3eqtr4rd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( 𝑘 C 𝑗 ) / ( ! ‘ 𝑘 ) ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) = ( ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) / ( ! ‘ ( 𝑘𝑗 ) ) ) · ( ( 𝐵𝑗 ) / ( ! ‘ 𝑗 ) ) ) )
105 87 104 eqtr4d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) / ( ! ‘ ( 𝑘𝑗 ) ) ) · ( 𝐺𝑗 ) ) = ( ( ( 𝑘 C 𝑗 ) / ( ! ‘ 𝑘 ) ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) )
106 nn0cn ( 𝑘 ∈ ℕ0𝑘 ∈ ℂ )
107 106 ad2antlr ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → 𝑘 ∈ ℂ )
108 107 addid2d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 0 + 𝑘 ) = 𝑘 )
109 108 oveq1d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( 0 + 𝑘 ) − 𝑗 ) = ( 𝑘𝑗 ) )
110 109 oveq2d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑗 ) ) = ( 𝐴 ↑ ( 𝑘𝑗 ) ) )
111 109 fveq2d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ! ‘ ( ( 0 + 𝑘 ) − 𝑗 ) ) = ( ! ‘ ( 𝑘𝑗 ) ) )
112 110 111 oveq12d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑗 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑗 ) ) ) = ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) / ( ! ‘ ( 𝑘𝑗 ) ) ) )
113 109 oveq2d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑗 ) ) = ( 𝑘 − ( 𝑘𝑗 ) ) )
114 nn0cn ( 𝑗 ∈ ℕ0𝑗 ∈ ℂ )
115 57 114 syl ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → 𝑗 ∈ ℂ )
116 107 115 nncand ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝑘 − ( 𝑘𝑗 ) ) = 𝑗 )
117 113 116 eqtrd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑗 ) ) = 𝑗 )
118 117 fveq2d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑗 ) ) ) = ( 𝐺𝑗 ) )
119 112 118 oveq12d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑗 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑗 ) ) ) · ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑗 ) ) ) ) = ( ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) / ( ! ‘ ( 𝑘𝑗 ) ) ) · ( 𝐺𝑗 ) ) )
120 50 59 96 97 div23d ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) = ( ( ( 𝑘 C 𝑗 ) / ( ! ‘ 𝑘 ) ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) )
121 105 119 120 3eqtr4rd ( ( ( 𝜑𝑘 ∈ ℕ0 ) ∧ 𝑗 ∈ ( 0 ... 𝑘 ) ) → ( ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) = ( ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑗 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑗 ) ) ) · ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑗 ) ) ) ) )
122 121 sumeq2dv ( ( 𝜑𝑘 ∈ ℕ0 ) → Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) = Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑗 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑗 ) ) ) · ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑗 ) ) ) ) )
123 oveq2 ( 𝑗 = 𝑚 → ( ( 0 + 𝑘 ) − 𝑗 ) = ( ( 0 + 𝑘 ) − 𝑚 ) )
124 123 oveq2d ( 𝑗 = 𝑚 → ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑗 ) ) = ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑚 ) ) )
125 123 fveq2d ( 𝑗 = 𝑚 → ( ! ‘ ( ( 0 + 𝑘 ) − 𝑗 ) ) = ( ! ‘ ( ( 0 + 𝑘 ) − 𝑚 ) ) )
126 124 125 oveq12d ( 𝑗 = 𝑚 → ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑗 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑗 ) ) ) = ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑚 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑚 ) ) ) )
127 123 oveq2d ( 𝑗 = 𝑚 → ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑗 ) ) = ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑚 ) ) )
128 127 fveq2d ( 𝑗 = 𝑚 → ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑗 ) ) ) = ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑚 ) ) ) )
129 126 128 oveq12d ( 𝑗 = 𝑚 → ( ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑗 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑗 ) ) ) · ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑗 ) ) ) ) = ( ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑚 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑚 ) ) ) · ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑚 ) ) ) ) )
130 129 cbvsumv Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑗 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑗 ) ) ) · ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑗 ) ) ) ) = Σ 𝑚 ∈ ( 0 ... 𝑘 ) ( ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑚 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑚 ) ) ) · ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑚 ) ) ) )
131 122 130 eqtrdi ( ( 𝜑𝑘 ∈ ℕ0 ) → Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) = Σ 𝑚 ∈ ( 0 ... 𝑘 ) ( ( ( 𝐴 ↑ ( ( 0 + 𝑘 ) − 𝑚 ) ) / ( ! ‘ ( ( 0 + 𝑘 ) − 𝑚 ) ) ) · ( 𝐺 ‘ ( 𝑘 − ( ( 0 + 𝑘 ) − 𝑚 ) ) ) ) )
132 84 131 eqtr4d ( ( 𝜑𝑘 ∈ ℕ0 ) → Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) · ( 𝐺 ‘ ( 𝑘𝑗 ) ) ) = Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) )
133 62 132 eqtr4d ( ( 𝜑𝑘 ∈ ℕ0 ) → ( Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( 𝑘 C 𝑗 ) · ( ( 𝐴 ↑ ( 𝑘𝑗 ) ) · ( 𝐵𝑗 ) ) ) / ( ! ‘ 𝑘 ) ) = Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) · ( 𝐺 ‘ ( 𝑘𝑗 ) ) ) )
134 43 133 eqtrd ( ( 𝜑𝑘 ∈ ℕ0 ) → ( ( ( 𝐴 + 𝐵 ) ↑ 𝑘 ) / ( ! ‘ 𝑘 ) ) = Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) · ( 𝐺 ‘ ( 𝑘𝑗 ) ) ) )
135 37 134 eqtrd ( ( 𝜑𝑘 ∈ ℕ0 ) → ( 𝐻𝑘 ) = Σ 𝑗 ∈ ( 0 ... 𝑘 ) ( ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) · ( 𝐺 ‘ ( 𝑘𝑗 ) ) ) )
136 4 abscld ( 𝜑 → ( abs ‘ 𝐴 ) ∈ ℝ )
137 136 recnd ( 𝜑 → ( abs ‘ 𝐴 ) ∈ ℂ )
138 26 efcllem ( ( abs ‘ 𝐴 ) ∈ ℂ → seq 0 ( + , ( 𝑛 ∈ ℕ0 ↦ ( ( ( abs ‘ 𝐴 ) ↑ 𝑛 ) / ( ! ‘ 𝑛 ) ) ) ) ∈ dom ⇝ )
139 137 138 syl ( 𝜑 → seq 0 ( + , ( 𝑛 ∈ ℕ0 ↦ ( ( ( abs ‘ 𝐴 ) ↑ 𝑛 ) / ( ! ‘ 𝑛 ) ) ) ) ∈ dom ⇝ )
140 2 efcllem ( 𝐵 ∈ ℂ → seq 0 ( + , 𝐺 ) ∈ dom ⇝ )
141 5 140 syl ( 𝜑 → seq 0 ( + , 𝐺 ) ∈ dom ⇝ )
142 10 29 31 33 35 135 139 141 mertens ( 𝜑 → seq 0 ( + , 𝐻 ) ⇝ ( Σ 𝑗 ∈ ℕ0 ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) · Σ 𝑘 ∈ ℕ0 ( ( 𝐵𝑘 ) / ( ! ‘ 𝑘 ) ) ) )
143 efval ( 𝐴 ∈ ℂ → ( exp ‘ 𝐴 ) = Σ 𝑗 ∈ ℕ0 ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) )
144 4 143 syl ( 𝜑 → ( exp ‘ 𝐴 ) = Σ 𝑗 ∈ ℕ0 ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) )
145 efval ( 𝐵 ∈ ℂ → ( exp ‘ 𝐵 ) = Σ 𝑘 ∈ ℕ0 ( ( 𝐵𝑘 ) / ( ! ‘ 𝑘 ) ) )
146 5 145 syl ( 𝜑 → ( exp ‘ 𝐵 ) = Σ 𝑘 ∈ ℕ0 ( ( 𝐵𝑘 ) / ( ! ‘ 𝑘 ) ) )
147 144 146 oveq12d ( 𝜑 → ( ( exp ‘ 𝐴 ) · ( exp ‘ 𝐵 ) ) = ( Σ 𝑗 ∈ ℕ0 ( ( 𝐴𝑗 ) / ( ! ‘ 𝑗 ) ) · Σ 𝑘 ∈ ℕ0 ( ( 𝐵𝑘 ) / ( ! ‘ 𝑘 ) ) ) )
148 142 147 breqtrrd ( 𝜑 → seq 0 ( + , 𝐻 ) ⇝ ( ( exp ‘ 𝐴 ) · ( exp ‘ 𝐵 ) ) )
149 climuni ( ( seq 0 ( + , 𝐻 ) ⇝ ( exp ‘ ( 𝐴 + 𝐵 ) ) ∧ seq 0 ( + , 𝐻 ) ⇝ ( ( exp ‘ 𝐴 ) · ( exp ‘ 𝐵 ) ) ) → ( exp ‘ ( 𝐴 + 𝐵 ) ) = ( ( exp ‘ 𝐴 ) · ( exp ‘ 𝐵 ) ) )
150 8 148 149 syl2anc ( 𝜑 → ( exp ‘ ( 𝐴 + 𝐵 ) ) = ( ( exp ‘ 𝐴 ) · ( exp ‘ 𝐵 ) ) ) | 9,903 | 14,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.618344 |
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PV Set #1 — FROM CHAPTER 9 –> 2,3,5,6,7,8,9,13 (just calc monthly payment), 14,16,17,18b,19,20,22,23,25a
2. Find the FV of \$10,000 invested now after five years if the annual interest rate is 8 percent.
a. What would be the FV if the interest rate is a simple interest rate?
b. What would be the FV if the interest rate is a compound interest rate?
3. Determine the future values (FVs) if \$5,000 is invested in each of the following situations:
a. 5 percent for ten years
b. 7 percent for seven years
c. 9 percent for four years
5. Find the present value (PV) of \$7,000 to be received one year from now assuming a 3 percent annual discount interest rate. Also calculate the PV if the \$7,000 is received after two years.
6. Determine the present values (PVs) if \$5,000 is received in the future (i.e., at the end of each indicated time period) in each of the following situations:
a. 5 percent for ten years
b. 7 percent for seven years
c. 9 percent for four years
7. Determine the present value (PV) if \$15,000 is to be received at the end of eight years and the discount rate is 9 percent. How would your answer change if you had to wait six years to receive the \$15,000?
8. Determine the future value (FV) at the end of two years of an investment of \$3,000 made now and an additional \$3,000 made one year from now if the compound annual interest rate is 4 percent.
9. Assume you are planning to invest \$5,000 each year for six years and will earn 10 percent per year. Determine the future value (FV) of this annuity if your first \$5,000 is invested at the end of the first year.
13. Determine the annual payment on a \$15,000 loan that is to be amortized over a four-year period and carries a 10 percent interest rate. Prepare a loan amortization schedule for this loan.
14. You are considering borrowing \$150,000 to purchase a new home.
a. Calculate the monthly payment needed to amortize an 8 percent fixed-rate 30-year mortgage loan.
b. Calculate the monthly amortization payment if the loan in (a) was for 15 years.
16. Use a financial calculator or computer software program to answer the following questions:
a. What would be the future value (FV) of \$15,555 invested now if it earns interest at 14.5 percent for seven years?
b. What would be the FV of \$19,378 invested now if the money remains deposited for eight years and the annual interest rate is 18 percent?
17. Use a financial calculator or computer software program to answer the following questions:
a. What is the present value (PV) of \$359,000 that is to be received at the end of twenty-three years if the discount rate is 11 percent?
b. How would your answer change in (a) if the \$359,000 is to be received at the end of twenty years?
18. Use a financial calculator or computer software program to answer the following questions:
b. What would be the present value (PV) of a \$9,532 annuity for which the first payment will be made beginning one year from now, payments will last for twenty-seven years, and the annual interest rate is 13 percent?
19. Use a financial calculator or computer software program to answer the following questions.
a. What would be the future value (FV) of \$19,378 invested now if the money remains deposited for eight years, the annual interest rate is 18 percent, and interest on the investment is compounded semiannually?
b. How would your answer for (a) change if quarterly compounding were used?
20. Use a financial calculator or computer software program to answer the following questions.
a. What is the present value (PV) of \$359,000 that is to be received at the end of twenty-three years, the discount rate is 11 percent, and semiannual discounting occurs?
b. How would your answer for (a) change if monthly discounting were used?
22. Answer the following questions.
a. What is the annual percentage rate (APR) on a loan that charges interest of .75 percent per month?
b. What is the effective annual rate (EAR) on the loan described in (a)?
23. You have recently seen a credit card advertisement stating that the annual percentage rate (APR) is 12 percent. If the credit card requires monthly payments, what is the effective annual rate (EAR) of interest on the loan?
1. You annually invest \$1,500 in an individual retirement account (IRA) starting at the age of 20 and make contributions for 10 years. Your twin sister does the same starting at age 30 and makes contributions for 30 years. Both of you earn 7% annually on you investment. Who has the larger amount at age 60?
2. If a parent wants to have \$100,000 to send a newborn child to college, how much must be invested annually for 18 years if the funds earn 9%?
3. An investment generates \$10,000 per year for 25 years. If you can earn 10% on other investments, what is the current value of this investment? If its current price is 120,000, should you buy it?
4. You are offered an annuity that will pay \$10,000 a year for 10 years (that is 10 payments) starting after five years have elapsed. If you seek an annual return of 8%, what is the maximum amount you should pay for this annuity?
5. You want your salary to double in six years. At what annual rate of growth must you salary increase to achieve your goals.
6. Each year you invest \$2,000 in an account that earns 5% annually. How long will it take for you to accumulate \$50,000?
7. Auntie Kitty sells her house for \$200,000, which is then invested to earn 6% annually. If her life expectancy is 10 years, what is the maximum amount she can annually spend on a nursing home, doctors, and taxes?
8. You win a judgment in an auto accident for \$100,000. You immediately receive \$25,000 but must pay your lawyer’s fee of \$15,000. In addition, you will receive \$2,500 a year for 20 years for a total of \$50,000 after which the balance owed (\$25,000) will be paid. If the interest rate is 7%, what is the current value of your settlement?
9. Uncle Fred recently died and left \$325,000 to his 50-year-old favorite niece. She immediately s[en \$100,000 on a town home but decided to invest the balance for her retirement at age 65. What rate of return must she earn on her investment over the next 15 years to permit her to withdraw \$750,000 at the end of each year through age 80 if her funds earn 10% annually during retirement?
10. A \$1,000,000 state lottery prize is spread evenly over 10 years (\$100,000 a year), or you may take a lump distribution of \$654,000. If you can earn 7%, which alternative is better?
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If you think your paper could be improved, you can request a review. In this case, your paper will be checked by the writer or assigned to an editor. You can use this option as many times as you see fit. This is free because we want you to be completely satisfied with the service offered. | 2,260 | 9,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.937483 |
https://wiki.tcl-lang.org/revision/Playing+with+matches?V=3 | 1,660,139,782,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571190.0/warc/CC-MAIN-20220810131127-20220810161127-00511.warc.gz | 582,497,978 | 3,455 | ## Version 3 of Playing with matches
Updated 2005-05-07 22:49:23 by suchenwi
if 0 {Richard Suchenwirth 2005-05-07 - Matches (small wooden sticks to light fire) are popular for playing games, too (Martin Gardner devotes chapter 2 in Mathematical Circus to them, for example). Here's matches emulated in Tcl:
At top left you have a matchbox. Left-click on a match to get a duplicate ("clone") which you can drag on the playing field. Right-click on a match to rotate it by 30 degrees. The screenshot shows one of Gardner's simpler challenges (red: move one match to make the equation correct) and the solution (blue). }
``` proc main {} {
pack [canvas .c -background darkgreen] -fill both -expand 1
.c create rect 5 5 45 65 -fill white ;# matchbox
set red [match .c 15 10 55 red]
.c bind \$red <1> {clone .c red %x %y}
set blue [match .c 30 10 55 blue]
.c bind \$blue <1> {clone .c blue %x %y}
.c bind mv <1> {select .c %x %y}
.c bind mv <B1-Motion> {move .c %x %y}
.c bind mv <3> {rotate .c}
}```
#-- Draw a match, return the common ID of its items
``` proc match {w x0 y0 y1 color} {
set id [\$w create poly \
[+ \$x0 1] [+ \$y0 1] [+ \$x0 5] [+ \$y0 1] \
[+ \$x0 5] \$y1 [+ \$x0 1] \$y1 -fill bisque -outline black]
set head [\$w create oval \$x0 \$y0 [+ \$x0 6] [+ \$y0 6] \
-fill \$color]
\$w itemconfig \$id -tag m\$id
return m\$id
}```
#-- make a duplicate of the current match
``` proc clone {w color x y} {
foreach {x0 y0 x1 y1} [\$w bbox current] break
set id [match \$w \$x0 \$y0 \$y1 \$color]
\$w itemconfig \$id -tags [list mv mv\$id]
}```
#-- Store the current position in two global variables
` proc select {w x y} {set ::X \$x; set ::Y \$y}`
#-- Move the current items set
``` proc move {w x y} {
set id [\$w find withtag current]
set tag [lindex [\$w gettags \$id] 1]
\$w move \$tag [- \$x \$::X] [- \$y \$::Y]
set ::X \$x; set ::Y \$y
}```
#-- rotate the current item set by 30 degrees, clockwise
``` proc rotate w {
set id [\$w find withtag current]
set tag [lindex [\$w gettags \$id] 1]
foreach {x0 y0 x1 y1} [\$w bbox \$tag] break
set xm [expr {(\$x0+\$x1)/2.}]
set ym [expr {(\$y0+\$y1)/2.}]
foreach item [\$w find withtag \$tag] {
set coords {}
foreach {x y} [\$w coords \$item] {
set r [expr {hypot(\$x-\$xm,\$y-\$ym)}]
set a [expr {atan2(\$y-\$ym,\$x-\$xm) + acos(-1)/6.}]
lappend coords [expr {\$xm+\$r*cos(\$a)}] \
[expr {\$ym+\$r*sin(\$a)}]
}
\$w coords \$item \$coords
}
}```
#-- prefix expr operators make the code shorter...
` foreach op {+ -} {proc \$op {a b} "expr {\\$a \$op \\$b}"}`
#-- Let's go!
` main`
if 0 { | 910 | 2,566 | {"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} | 2.984375 | 3 | CC-MAIN-2022-33 | longest | en | 0.521644 |
https://89devs.com/python/functions/tuple/count/ | 1,702,334,454,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00653.warc.gz | 99,038,281 | 3,223 | 89DEVs
# Python Tuple count() Method
```The Tuple method count() returns the number of times a given element appears in the tuple. This method is useful to count the elements in a tuple.
Tuple count() example
In this example, a tuple of different letters was created. Some of the letters are included multiple times.
```
# tuple count() example
letters = ('a', 'a', 'a', 'b', 'c', 'c')
print('letter a appears ' + str(letters.count('a')) + ' time(s) in our letters tuple.')
print('letter b appears ' + str(letters.count('b')) + ' time(s) in our letters tuple.')
print('letter c appears ' + str(letters.count('c')) + ' time(s) in our letters tuple.')
```
A string that includes the count of each element is returned.
letter a appears 3 time(s) in our letters tuple.
letter b appears 1 time(s) in our letters tuple.
letter c appears 2 time(s) in our letters tuple.
If the element is not in the tuple an integer value of 0 is returned.
```
# tuple count() example
letters = ('a', 'a', 'a', 'b', 'c', 'c')
print('letter z appears ' + str(letters.count('z')) + ' time(s) in our letters tuple.')
```
The letter z is not in our tuple, so integer 0 is returned.
letter z appears 0 time(s) in our letters tuple.
Count all elements in a tuple
To count all elements in a tuple we can use the count() method within a loop. We loop through all elements and return the count for each element in the tuple.
```
# count all elements in a tuple
letters = ('a', 'a', 'a', 'b', 'c', 'c')
for letter in letters:
print(letter + ' is ' + str(letters.count(letter)) + ' times(s) in our letters tuple.')
```
The loop works, but it repeats some elements because they are included multiple times.
a is 3 times(s) in our letters tuple.
a is 3 times(s) in our letters tuple.
a is 3 times(s) in our letters tuple.
b is 1 times(s) in our letters tuple.
c is 2 times(s) in our letters tuple.
c is 2 times(s) in our letters tuple.
To remove duplicates from the loop, we can use the built-in set() function in Python to convert our Tuple into a Set. Sets contain unique values and therefore each element is printed once.
```
# count all unique elements in a tuple
letters = ('a', 'a', 'a', 'b', 'c', 'c')
for letter in set(letters):
print(letter + ' is ' + str(letters.count(letter)) + ' times(s) in our letters tuple.')
```
The count for each element is now returned.
a is 3 times(s) in our letters tuple.
c is 2 times(s) in our letters tuple.
b is 1 times(s) in our letters tuple.
The returned elements are now unique, but they are not sorted. We can use the sorted() function to sort them alphabetically.
```
# count all unique elements in a tuple and sort them
letters = ('a', 'a', 'a', 'b', 'c', 'c')
for letter in sorted(set(letters)):
print(letter + ' is ' + str(letters.count(letter)) + ' times(s) in our letters tuple.')
```
Then our elements are returned in alphabetical order.
a is 3 times(s) in our letters tuple.
b is 1 times(s) in our letters tuple.
c is 2 times(s) in our letters tuple.
Tuple count() alternative
An alternative way to count the elements in a tuple is the Collections.Counter(). It returns a Counter of the tuple.
```
# count tuple elements with collections.Counter()
from collections import Counter
letters = ('a', 'a', 'a', 'b', 'c', 'c')
print(Counter(letters))
```
The Counter is returned.
Counter({'a': 3, 'c': 2, 'b': 1})
The advantage of the Counter is that it can give us more information about the tuple. For example the most common element and the least common element. Of course, this can also be achieved with the count() method.
```
# show most common and least common element
from collections import Counter
letters = ('a', 'a', 'a', 'b', 'c', 'c')
counter = Counter(letters)
print('the most common element is: ' + str(counter.most_common()[0][0]))
print('the least common element is: ' + str(counter.most_common()[:-2:-1][0][0]))
```
The most and least common element is printed.
the most common element is: a
the least common element is: b
Tuple count() Syntax
The syntax of the Tuple count() method is:
`tuple.count(element)`
Tuple count() arguments
The tuple count() method takes exactly one argument. If more or less than one argument is given, the TypeError exception is raised.
The required single argument is the element to count in the tuple.
Tuple count() return value
The tuple count() method returns the count of the given element as an integer. If the element is not included in the tuple, an integer value of 0 is returned.
Other Tuple methods
Tuples have only two methods in Python. The other is the tuple.index() method. It returns the index of a given element in the tuple.
``` | 1,188 | 4,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-50 | longest | en | 0.79862 |
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### Witch's Hat
##### Stage: 3 and 4 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Tea Cups
##### Stage: 2 and 3 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
### Pebbles
##### Stage: 2 and 3 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Building with Longer Rods
##### Stage: 2 and 3 Challenge Level:
A challenging activity focusing on finding all possible ways of stacking rods.
### 2010: A Year of Investigations
##### Stage: 1, 2 and 3
This article for teachers suggests ideas for activities built around 10 and 2010.
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved. | 2,003 | 8,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-04 | longest | en | 0.8436 |
http://mathhelpforum.com/pre-calculus/7193-inverse-function.html | 1,481,166,906,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542323.80/warc/CC-MAIN-20161202170902-00369-ip-10-31-129-80.ec2.internal.warc.gz | 180,411,435 | 12,093 | 1. ## inverse function?
The function f : [a, infinity) -> R with rule f (x) = 2x^3 − 3x^2 + 6 will have an inverse function provided?
the answer was a >= 1,
but why not also a <= 0?
2. Originally Posted by scorpion007
The function f : [a, infinity) -> R with rule f (x) = 2x^3 − 3x^2 + 6 will have an inverse function provided?
the answer was a >= 1,
but why not also a <= 0?
Hello, scorpion,
as you know a function f has an inverse function if f is monotonously increasing or monotonously decreasing.
So your function has 3 intervalls where f is monotonous:
$(-\infty;0],\ (0;1], (1,+\infty)$
So you are looking for a lower bound of the intervall which contains +infinity.
With your suggestion you would look for the upper bound.
EB
I have to send you this post without checking my text because the preview doesn't work. So watch out: There are possibly some mistakes.
3. ahhh, thank you, i overlooked that fact! i see now
4. Originally Posted by earboth
as you know a function f has an inverse function if f is monotonously increasing or monotonously decreasing.
Earboth, please understand I'm not picking on you, but that sentence just had me rolling on the floor.
Monotonous: (definition) tediously uniform or unvarying.
The word in English you were looking for is "monotonically."
It very much reminds me of a lady at a school I once worked at. She was trying to make a poster thanking everyone who had contributed money to the school's fundraising campaign. She wrote "Thank you for patronizing us."
For those who don't know, there are two definitions for the word "patronize"
1) to act as patron of : provide aid or support for
2) to adopt an air of condescension toward : treat haughtily or coolly
Two VERY different meanings!
-Dan
5. I understand what earboth said, but formally there is no answer to your question!!
A function,
$f:[a,\infty)\to \mathbb{R}$ would have an inverse if and only if $f(x)=2x^3-3x^2+6, x\geq a$ is a bijective map. That means is is a surjective map. But for $a\geq 1$ the map is not sujective. Because if you chose $y=0$ there is no $x\geq 1$ such as, $f(x)=0$. What you should have said is,
$f:[a,\infty)\to [5,\infty)$
6. Originally Posted by earboth
a function f has an inverse function if f is monotonously increasing or monotonously decreasing.
Hello !
I want only to emphasize that this is a sufficient condition, not a necessarily one !
That's plenty of examples...
7. Originally Posted by misto
Hello !
I want only to emphasize that this is a sufficient condition, not a necessarily one !
That's plenty of examples...
Give an example, I do not think I understand what you are saying. | 708 | 2,651 | {"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": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.963504 |
https://chat.stackexchange.com/transcript/14524/2024/6/20 | 1,722,739,464,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00329.warc.gz | 128,134,788 | 10,404 | 12:29 AM
@msh210 this is taking me back
3 hours later…
3:13 AM
@Sphinx Me: thinks it's gonna be a quick solve The upper right section:
3 hours later…
5:54 AM
0
I'm trying to find a book I remember from my childhood, of various mazes with different gimmicks and themes. I would have first read it around the mid or late 90's, and it seemed pretty new at the time, so it was probably published around that time or maybe a bit earlier. It was a children's book...
7:14 AM
0
The link in this image (https://imgur.com/a/) will lead to a 404 error page, where we need to use the inspect, elements, properties, to find 4 unique numbers in it, and total up both sets of numbers, the cipher, and the number found in properties. The answer will be 6 digits of numbers and is st...
7:38 AM
@msh210 fig h (hit, in baseball) T_
with fig = trifle meaning something of little importance
yes indeed
8:22 AM
Despite being urged by the completion of two recent sequences, I still can't think of one...
CCCC: Beef with a random guy (5)
@oAlt w/ a gyu*
@msh210 yep!
jeez i step out for a minute and you guys start beefin'
CCCC: I printed oblique bold (8)
@Jafe Lol
8:49 AM
@Sphinx I looked at the answer and that's interesting, I own a different maze book by the same author. I loved mazes more back then
2 hours later…
11:06 AM
@msh210 INTREPID = IPRINTED*
@Stiv quite so
CCCC: Hesitated at going bust? Think about things differently... (2,7,3,4)
11:24 AM
@oAlt fig=trifle is not in my vocab (or any other cab I possess)
@AncientSwordRage Same. I just saw it on Wiktionary
I know it chiefly from dictionary.cambridge.org/dictionary/english/not-care-give-a-fig … which IIRC I first came across in Shakespeare
@msh210 etymonline.com/word/fig is of insterest
@AncientSwordRage indeed; thanks
ahh after y'all had explained it I thought it was familiar
> And it gets to the point
You can't stand anymore,
Bring it to us,
We won't give a fig.
We'll tell you
11:31 AM
@Stiv go against the tide anagram
@AncientSwordRage Google seems to be telling me that that's from the Hitchhiker radio fits?
@msh210 yeah it is
I remembered the last line, and have vague memories of 'fig' being what rhymed with 'pig'
12:12 PM
@oAlt Yep!
CCCC: Handle this old sty within the barriers (4)
_hi_+_l_+_t_
@Jafe yep!
0
I like to put words in my bookshelf. This is a very special bookshelf though, because books don't fit in there. My bookshelf has 3 layers, and different words can fit on different layers. Here are some examples: Bottom Layer Middle Layer Top Layer STATE COUNTRY FLAG YELLOW BROWN BLACK ...
2 hours later…
2:21 PM
CCCC: She sings in support of king and country (5)
just finished watching Everything Everywhere All at Once for the first time... holy moses mosesovich von moseshausen
you can quote me on that, that is my official review of the film
roger ebert who
@Jafe K Enya?
that's right
2:40 PM
Wow I really have no clue ideas coming to my mind today
CCCC: You stay, expecting something in return, in support of king and country (6)
2 hours later…
4:38 PM
0
This is from "Symbolic Logic, part 1 and part 2", by Lewis Carroll, edited by W.W. Bartley (1977), page 362: (1) A man can always master his father; (2) An inferior of a man's uncle owes that man money; (3) The father of an enemy of a friend of a man owes that man nothing; (4) A man is always per...
4:56 PM
@oAlt Is this RUSTIC (country)? From R (king) + U (you) + STIC(-k) (stay, expecting something in return)
5:11 PM
Why is k something in return?
@Ankoganit Maybe expecting the k in return implies that "stick" is not finished
@Stiv but either way, not the intended answer
Since I'm about to sleep, I guess it won't hurt to give a little nudge toward the answer
@oAlt huh, I guess that could make sense
Since I said I had no clue ideas coming to mind, you can probably guess what I was thinking when I was inspired by Jafe's clue
@Ankoganit My interpretation was 'missing the last letter'
Right
5:41 PM
Oh this is K(king)+U(you)+WAIT (stay, expecting...return)
@oAlt
@Ankoganit Yes that's the correct one
CCCC: When moving north, live next to a heat source (8)
@Ankoganit Wait isn't that a definition in the middle of the clue? I thought that was a big no-no.
@oAlt ^
No, the definition is "country"
Oh. Duh.
5:50 PM
At the end
Was thinking the "stay, expecting... return" was the def for some reason.
Ah nw my annotation wasn't super detailed tbf
A bartender walks into a bar... wait a second, he actually works here. <- Is this a good or a bad joke, and if it is bad (yes it is), why is it bad? Does it have to do with the lack of poonchline?
2 hours later…
8:17 PM
@Ankoganit fi< reside = next to a heat source
@msh210 that's it!
8:33 PM
oh that's really good
CCCC: Out-chide a nun? That's difficult. (9) | 1,346 | 4,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-33 | latest | en | 0.937088 |
https://socratic.org/questions/what-is-the-general-form-of-the-equation-of-a-circle-with-a-center-at-the-10-5-a | 1,582,468,556,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145774.75/warc/CC-MAIN-20200223123852-20200223153852-00487.warc.gz | 565,494,877 | 6,211 | What is the general form of the equation of a circle with a center at the (10, 5) and a radius of 11?
Jan 2, 2016
${\left(x - 10\right)}^{2} + {\left(y - 5\right)}^{2} = 121$
Explanation:
The general form of a circle:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} - {r}^{2}$
Where:
$\left(h , k\right)$ is the center
$r$ is the radius
Thus, we know that
$h = 10 , k = 5$
$r = 11$
So, the equation for the circle is
${\left(x - 10\right)}^{2} + {\left(y - 5\right)}^{2} = {11}^{2}$
Simplified:
${\left(x - 10\right)}^{2} + {\left(y - 5\right)}^{2} = 121$
graph{(x-10)^2+(y-5)^2=121 [-10.95, 40.38, -7.02, 18.63]} | 261 | 632 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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} | 4.34375 | 4 | CC-MAIN-2020-10 | latest | en | 0.725226 |
https://studychacha.com/discuss/10294-bank-common-entrance-test.html | 1,623,905,678,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487629209.28/warc/CC-MAIN-20210617041347-20210617071347-00530.warc.gz | 458,167,717 | 20,105 | 2021-2022 StudyChaCha
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#1
January 29th, 2012, 11:02 PM
anil.khunte02 Guest
Bank common entrance test
When will the Bank common entrance exam for clerks be conducted in 2012? Name the banks in which candidates will be appointed after passing the exam? What will be the syllabus of this exam and from where I can get sample papers of this exam?
#2
September 15th, 2012, 11:09 AM
nagsan fgf Guest
Re: Bank common entrance test
I want to participate in the exam of Bank’s for the recruitment so can you please provide me any information about the Bank Common entrance test ?
#3
October 6th, 2012, 03:12 PM
lakshmishaiah.1 Guest
Bank Common Entrance Test
I am going to appear in Bank Common Entrance Exam for post of PO so tell me form where can I get its old question paper for my preparation?
#4
October 6th, 2012, 06:50 PM
Super Moderator Join Date: Nov 2011
Re: Bank Common Entrance Test
ou want Bank Common PO Entrance Exam paper so following is its questions:
How many number s are there between 200 and 300 in which 9 occurs only once?
a. 19 b. 20 c.21 d. 18
An amount of Rs.417 is divided among A, B, C and D such that A gets Rs.13 more than B, B gets Rs.9 more than C and C gets Rs.6 more than D. A’s share is
a. 121 b. 116 c. 120 d.124
Manju took Rs. 20000 at 5% SI for 2 years and invested it at 4% CI for same period. Find her gain/loss.
a. Rs. 368 gain b. Rs.423 gain c. Rs. 368 loss d. Rs. 200 gain
For complete paper you should have to download file which is attached right down here.
IBPS PO Question Paper.doc (27.0 KB, 26 views)
__________________
Answered By StudyChaCha Member
#5
October 21st, 2019, 10:30 AM
Super Moderator Join Date: Jun 2013
Re: Bank common entrance test
I want to join banking line and for this searching for details about the bank common entrance test. Will you tell which bank common entrance test I have to give and Entrance Test Dates?
There are many entrance level and aptitudes taken by different banks for recruitments.
Today various entry level examinations which are held by different banks or institutes for various positions from Probationary Officer to Clerical staff, Regional Rural Banks to Private Banks
Common Written Examination (CWE Clerk-IIV) is a pre-requisite for selection of personnel for Clerical cadre posts in the Public Sector Banks which is conducted online by the Institute of Banking Personnel Selection (IBPS) tentatively in November/ December.
The recruitment process of IBPS Clerk CWE occurs in two phases:
1. Preliminary Examination
2. Mains Examination
Bank common entrance test eligibility:
You must possess the minimum qualification of Graduation in any discipline from a recognized University or any equivalent qualification recognized as such by the Central Government.
You must be 20 Years to 28 Years.
Bank Common Entrance Test Dates:
IBPS Clerk Events Dates
IBPS Clerk Notification 2019 Released on 12th September 2019
IBPS Clerk Online Applicaion Starts From 17th September 2019
IBPS Clerk Online Applicaion Ends On 9th October 2019
Download of call letters for IBPS Clerk Pre- Exam Training Nov-19
IBPS Clerk Online Pre-Exam Training 25.11.2019 to 30.11.2019
Conduct of Online Examination - Preliminary 07th, 08th, 14th & 21st December 2019
Result of IBPS Clerk Preliminary Exam Dec-19
Download of Call letter for Online Exam Main Jan-20
Conduct of Online Examination Main 19th January 2020
Declaration of Final Result 1st April 2020
__________________
Answered By StudyChaCha Member
#6
June 3rd, 2020, 10:47 AM
Unregistered Guest
Re: Bank common entrance test
I want to join banking line and decided to apply for Bank Common Entrance Test searching for details. Will you tell banking Entrance test dates also tell list of banks associated with this?
#7
June 3rd, 2020, 10:48 AM
Super Moderator Join Date: Jun 2013
Re: Bank common entrance test
A Common Written Examination (CWE) is conducted by the Institute of Banking Personnel Selection (IBPS) as a pre-requisite for selection of personnel for Probationary Officer/ Management Trainee posts in the Public Sector Banks.
System of Common Examination for recruitment of Probationary Officers/ Management Trainees has been approved by each of the participating Public Sector Banks and the Managing Committee of the Indian Banks’ Association (IBA) with the consent of the Government of India.
Entrance test dates:
IBPS Clerk Events Dates
IBPS Clerk Notification 2020 Release Date September 2020
IBPS Clerk Online Applicaion Starts From September 2020
IBPS Clerk Online Applicaion Ends On October 2020
Download of call letters for IBPS Clerk Pre- Exam Training November 2020
IBPS Clerk Online Pre-Exam Training November 2020
Conduct of Online Examination - Preliminary 12th, 13th & 19th December 2020
Result of IBPS Clerk Preliminary Exam January 2021
Download of Call letter for Online Exam – Main January 2021
Conduct of Online Examination – Main 24th January 2021
Declaration of Final Result 1st April 2021
List of Public Sector Banks Participating in IBPS
Andhra Bank
Bank of Baroda
Bank of India
Bank of Maharashtra
Canara Bank
Central Bank of India
Corporation Bank
Dena Bank
IDBI Bank
Indian Bank
Indian Overseas Bank
Oriental Bank of Commerce
Punjab National Bank
Punjab and Sind Bank
State Bank of India
State Bank of Bikaner and Jaipur
State Bank of Hyderabad
State Bank of Mysore
State Bank of Patiala
State Bank of Travancore
Syndicate Bank
UCO Bank
Union Bank of India
United Bank of India
Vijaya Bank
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Answered By StudyChaCha Member
Reply to this Question / Ask Another Question | 1,425 | 5,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-25 | latest | en | 0.943388 |
https://math.stackexchange.com/questions/4755319/do-permutations-on-the-decimal-expansions-of-irrational-numbers-retain-the-prope/4755332 | 1,716,314,432,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058504.35/warc/CC-MAIN-20240521153045-20240521183045-00138.warc.gz | 333,483,785 | 35,695 | # Do permutations on the decimal expansions of irrational numbers retain the property of irrationality?
Suppose we have an irrational number with the following decimal expansion:
$$A = a_0 \ a_1 \ a_2 \ a_3 \ a_4 \ a_5 \ a_6 \dots$$
Now, construct a new real number through a permutation on the decimals of the original number in the following manner:
$$A' = (a_1 \ a_0) \ (a_3 \ a_2) \ (a_5 \ a_4) \dots$$
Question 1: is $$A'$$ irrational, too?
We can generalize this question by considering other permutations as well. Let's consider the irrational number $$A$$ as above again. Let $$\sigma_{k}(\cdot)$$ be some permutation on a tuple of decimals of length $$k$$. Define
$$f_{k} (A) = \sigma_{k} (a_{0} \ a_{1} \dots \ a_{k-1} ) \ \sigma_{k} (a_{k} \ a_{k+1} \ \dots \ a_{2k-1} ) \ \sigma_{k} (a_{2k} \ a_{2k+1} \ \dots a_{3k-1} ) \dots$$
Question 2: is $$f_{k} (A)$$ necessarily irrational for all $$k \geq 2$$ and every possible permutation?
• Interesting (+1) , but probably out of reach. Irrationality proofs are in general extremely difficult as the open cases $\gamma$ , $e+\pi$ and $e\cdot \pi$ show. Aug 19, 2023 at 10:58
• Irrational is easy in this case, since rational $A'$ leads to a rational $A$ immediately. @Peter Aug 19, 2023 at 10:59
All decimal expansions involved below do not have any ambiguity of the shape $$0.999\dots=1.000\dots$$ - since we are dealing with an irrational $$A$$ as a start. (And the permutations used are also not disturbing the pattern, seen as a stationary pattern.)
Assume that $$A'$$ is rational. Then its decimal representation is periodic, let $$P=(d_1d_2\dots d_n)$$ be a period. We may and do assume that $$n$$ is a multiple of $$k$$, else replace $$P$$ by its repeated pattern $$\underbrace{PP\dots P}_{k\text{ times}}$$. Now go back from $$A'$$ to $$A$$ by using the inverse of $$\sigma_k$$. We obtain a periodic decimal representation for $$A$$, so $$A\in\Bbb Q$$. A contradiction. | 602 | 1,944 | {"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": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-22 | latest | en | 0.868526 |
https://www.scholarhat.com/tutorial/datastructures/one-dimensional-array | 1,720,967,894,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00706.warc.gz | 855,713,119 | 31,812 | One dimensional Array in Data Structures with Example
06 Jun 2024
Beginner
4.28K Views
Learn via Video Course & by Doing Hands-on Labs
One dimensional Array: An Overview
Data Structures and Algorithms (DSA) form the backbone of computer science, enabling efficient problem-solving and algorithmic design. In this DSA tutorial, we'll explore the concept of one-dimensional arrays, their characteristics, and their applications in DSA. You might have already come across arrays while learning arrays in C and arrays in C++.
To further enhance your understanding and application of array concepts, consider enrolling in the Best Data Structures and Algorithms Course, where you can gain comprehensive insights into effective data structure utilization for improved problem-solving and time management.
What is One dimensional Array in Data Structure?
A one-dimensional array is a linear data structure that stores elements of the same data type in contiguous memory locations. It provides a systematic way of organizing and accessing a collection of elements, where each element is identified by its index or position within the array. The index starts from 0 and goes up to the size of the array minus one.
One-dimensional arrays are useful and robust data formats that can be used for a variety of data storage and manipulation purposes. They are used in a variety of applications and are a necessary component of many C programs.
Syntax
The syntax for declaring and initializing a one-dimensional array in C is as follows:
``data_type array_name[array_size] = value1, value2, value3,...;``
• The arguments used in the preceding syntax are listed below.
• The data_type variable represents the data type of the array's items.
• array_name is the name of the array.
• array_size specifies how many elements the array can hold.
• The values of each array element are value1, value2, value3,...
Declaration and Initialization
In most programming languages, one-dimensional arrays can be declared and initialized as follows:
``````int myArray[5];
// Initialization
myArray[0] = 10;
myArray[1] = 20;
myArray[2] = 30;
myArray[3] = 40;
myArray[4] = 50;
``````
Common Operations on One-Dimensional Arrays
1. Accessing Elements
Elements in a one-dimensional array are accessed using their respective indices. For example, myArray[2] would return the value 30.
Syntax
``````element = arrayName[index];
``````
Example
For example, if we have an array named myArray and we want to access the element at index 2:
``````int value = myArray[2];
``````
This line of code retrieves the value stored at index 2 in the array myArray. In this case, if myArray is {10, 20, 30, 40, 50}, the variable value would be assigned the value 30.
2. Insertion and Deletion
Inserting or deleting an element in a one-dimensional array may require shifting the subsequent elements accordingly to maintain the sequential order. The syntax for inserting and deleting elements is as follows:
Syntax
``arrayName[index] = newElement;``
Example
For example, if we have an array named myArray and we want to access the element at index 2:
`` myArray[2] = 25;``
Deletion
Deletion involves shifting elements after the deleted element to fill the gap. In this example, let's delete the element at index 3:
``````// Deleting the element at position index
// Shift elements after the deleted index to fill the gap
for (int i = index; i < arraySize - 1; i++) {
arrayName[i] = arrayName[i + 1];
}
arraySize--; // Decrease the size of the array``````
3. Updating Elements
Elements in an array can be updated by assigning a new value to the desired index.
Syntax
``arrayName[index] = newValue;``
Example
For example, if we have an array named myArray and we want to access the element at index 2:
``myArray[2] = 35;``
4. Traversing
Traversing involves visiting each element of the array sequentially, allowing for various operations on each element. The syntax for traversing through an array using a loop is as follows:
Syntax
``````for (int i = 0; i < arraySize; i++) {
// Perform operations on array elements, e.g., print, modify, etc.
// Access array elements using arrayName[i]
}``````
Example
For example, to print all elements of the array myArray:
``````printf("Elements of the array:\n");
for (int i = 0; i < 5; i++) {
printf("%d ", myArray[i]);
}
printf("\n");``````
Example of One-Dimensional Array
Let's take a simple example in C to demonstrate the use of a one-dimensional array in DSA. In this example, we'll create an array of integers, initialize it, and perform basic operations like accessing elements and traversing the array.
``````# Declaration and Initialization of a One-Dimensional Array
myArray = [10, 20, 30, 40, 50]
# Accessing and Printing Elements
print("Elements of the array:")
for i in range(5):
print(myArray[i], end=" ")
print()
# Updating an Element
myArray[2] = 35
print("Updated array after changing the third element:")
for i in range(5):
print(myArray[i], end=" ")
print()
# Summing the Elements
sum_value = sum(myArray)
print("Sum of the elements:", sum_value)
``````
``````
public class OneDimensionalArrayExample {
public static void main(String[] args) {
// Declaration and Initialization of a One-Dimensional Array
int[] myArray = {10, 20, 30, 40, 50};
// Accessing and Printing Elements
System.out.println("Elements of the array:");
for (int i = 0; i < 5; i++) {
System.out.print(myArray[i] + " ");
}
System.out.println();
// Updating an Element
myArray[2] = 35;
System.out.println("Updated array after changing the third element:");
for (int i = 0; i < 5; i++) {
System.out.print(myArray[i] + " ");
}
System.out.println();
// Summing the Elements
int sum = 0;
for (int i = 0; i < 5; i++) {
sum += myArray[i];
}
System.out.println("Sum of the elements: " + sum);
}
}
``````
``````
#include
int main() {
// Declaration and Initialization of a One-Dimensional Array
int myArray[5] = {10, 20, 30, 40, 50};
// Accessing and Printing Elements
std::cout << "Elements of the array:" << std::endl;
for (int i = 0; i < 5; i++) {
std::cout << myArray[i] << " ";
}
std::cout << std::endl;
// Updating an Element
myArray[2] = 35;
std::cout << "Updated array after changing the third element:" << std::endl;
for (int i = 0; i < 5; i++) {
std::cout << myArray[i] << " ";
}
std::cout << std::endl;
// Summing the Elements
int sum = 0;
for (int i = 0; i < 5; i++) {
sum += myArray[i];
}
std::cout << "Sum of the elements: " << sum << std::endl;
return 0;
}
``````
In this example, we declare and initialize an array myArray of size 5 with integer elements. We use a loop to print the elements of the array. Update the third element of the array and print the array after the update. Then calculate and print the sum of all elements in the array.
Output
``````Elements of the array:
10 20 30 40 50
Updated array after changing the third element:
10 20 35 40 50
Sum of the elements: 155``````
Applications in DSA
1. Linear Search
One-dimensional arrays are commonly used in linear search algorithms to find the position of a specific element in the array.
2. Sorting Algorithms
Sorting algorithms like bubble sort, insertion sort, and selection sort often rely on manipulating one-dimensional arrays to rearrange elements in ascending or descending order.
3. Dynamic Memory Allocation
One-dimensional arrays are crucial in dynamically allocating memory for structures like linked lists, stacks, and queues.
4. Mathematical and Scientific Applications
Arrays are extensively used to store and process data in mathematical and scientific computations, making them invaluable in fields like engineering and physics.
5. String Manipulation
Strings can be represented as one-dimensional character arrays, enabling various string manipulation operations.
Summary
This article delves into the intricacies of one-dimensional arrays in Data Structures and Algorithms (DSA). It provides a concise exploration of their definition, syntax, declaration, and initialization. We hope this post has been instrumental in enhancing your understanding of the one-dimensional array concept in DSA.
For those seeking a more structured and comprehensive learning experience, consider exploring a Best DSA Course Online. Such courses offer a systematic approach to mastering data structures and algorithms, providing learners with a solid foundation for tackling programming challenges. By incorporating these online courses into your learning journey, you can further solidify your grasp on topics like one-dimensional arrays in DSA and pave the way for more advanced algorithmic problem-solving.
FAQs
Q1. What is array data structure?
An array is a linear data structure that stores elements of the same data type in contiguous memory locations. It provides a systematic way of organizing and accessing a collection of elements, where each element is identified by its index or position within the array.
Q2. How do I access elements in a one-dimensional array?
Elements in a one-dimensional array are accessed using their respective indices.
Q3. What is the difference between a 1D array and a 2D array?
A 1D array and a 2D array are both fundamental data structures in programming languages, but they differ in terms of their dimensions and how data is organized within them.
Q4. Why are arrays classified as homogeneous data structures?
Arrays are classified as homogeneous data structures because they store elements of the same data type in contiguous memory locations.
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http://c.happycodings.com/beginners-lab-assignments/code34.html | 1,508,600,504,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824820.28/warc/CC-MAIN-20171021152723-20171021172723-00527.warc.gz | 58,651,615 | 3,914 | # C > Beginners Lab Assignments Code Examples
## Prg. to sort any no. of numeral i-p in ascending or descending order.
``` Prg. to sort any no. of numeral i-p in ascending or descending order. void sort(void); int c,a[20],l; void main() { clrscr(); printf("Enter no. of elements in the list:- "); scanf ("%d",&l); printf(" CHOICE:-"); printf(" (1) Sort in ascending order."); printf(" (2) Sort in descending order."); printf(" CHOICE:- "); scanf("%d",&c); if (c!=1 && c!=2) { printf(" ERROR"); getch(); exit(0); } sort(); getch(); } void sort(void) { int n,i,j,temp=0,min,k; for (i=1;i<=l;i++) { printf(" Enter no.:- "); scanf("%d",&a[i]); } for (i=1;i<=(l-1);i++) { min=a[i]; k=i; for (j=(i+1);j<=l;j++) { if (a[j]<min) { min=a[j]; k=j; } } temp=a[k]; a[k]=a[i]; a[i]=temp; } switch(c) { case 1: printf(" Elements in ascending order are:-"); for (i=1;i<=l;i++) printf(" %d",a[i]); break; case 2: printf(" Elements in descending order are:-"); for (i=l;i>=1;i--) printf(" %d",a[i]); break; default: printf(" ERROR"); } return; } ``` | 340 | 1,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-43 | latest | en | 0.294473 |
https://tolstoy.newcastle.edu.au/R/e4/help/08/06/14015.html | 1,590,363,181,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347385193.5/warc/CC-MAIN-20200524210325-20200525000325-00200.warc.gz | 580,183,771 | 4,373 | # [R] Looping, Control Flow & Conditional Statements
From: <Garth.Warren_at_csiro.au>
Date: Fri, 13 Jun 2008 17:14:36 +1000
I have little experience using R and even less experience with control flow type questions.
See the following code:
a1 = c(0, 1, 1, 1,
0, 0, 0, 0, 0,
0, 0, 1,
1, 1, 1, 0, 0)
for(i in 1:1){
sx <- paste("a",i,sep="")
s <- eval(parse(text = paste("a",i,sep="")))
{g = numeric(length(s))
k = numeric(length(s))
{for (i in 1:length(s))
{for (j in 1:length(s))
ifelse(((j=i)>1),(g[j] = s[j] + s[i]),(k[j] = s[j] + s[i]))
}}
h1 <- hist(g,freq=TRUE)
h <- h1\$counts[4]
cat(sx,":", h,"\n",file = "C:/temp/test-beta.txt", append=TRUE)
}}
The output is:
> g
[1] 0 2 2 2 0 0 0 0 0 0 0 2 2 2 2 0 0
> k
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>
> h
[1] 7
& a text file, which has:
a1 : 7
k is a by-product of the ifelse statement and is of no interest & g and h only go part-way to answering my question, which is:
For every time an object i.e. a1 (which is actually a time series) - 0 1 1 1 0 0 0 0 0 0 0 1 1 1 1 0 0 has as value over 0 how long do the values stay above 0. So in this case a1 has two goups or events where the value is above zero, the first event lasts for 3 'days' and the second event lasts for 4 'days'. I have my code telling me that there was a total of 7 'days' in event or above 0, but what I need to know is that there were two 'events' and the 1st lasted 3 'days' and the 2nd lasted '4' days. Essentially I want a text file output to say:
a1.1 : 3
a1.2 : 4
My thinking is that I need to somehow get the code working through each vector one value at a time and when a value is found to meet the critera of > 0 R creates a new vector; to use the above example it would come to the first value >0 and then create the new vector a1.1 = (1,1,1) then as the next value in the series is 0 it would close this new vector 'a1.1'. It would then continue until it reaches the next value >0 and then create the vector a1.2 = (1,1,1,1) then again as the next value in the series is 0 it would close this new vector, and so on.
Then all I need to do is perform a count of '1's in these new vectors to find how many days they met this criteria of being greater than 0
I hope the above makes sense and I really hope there is someone willing and able to help. I don't know how to proceed.
Thanks,
Garth
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A199585 Number of nX3 0..3 arrays with values 0..3 introduced in row major order, the number of instances of each value within one of each other, and every element equal to at least one horizontal or vertical neighbor 0
0, 0, 20, 23, 1951, 60323, 592219, 3399003, 220861580, 6086397307, 79879768821, 386032959689, 29317433209618, 852122544867919, 11206152487777257, 56509336646678083, 4391723268352660596, 130483481066218119340 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Column 3 of A199588 LINKS EXAMPLE Some solutions for n=4 ..0..0..0....0..0..1....0..0..1....0..0..0....0..1..1....0..1..1....0..0..1 ..1..2..2....0..2..1....0..1..1....1..1..1....0..1..2....0..2..1....2..0..1 ..1..1..2....2..2..1....2..2..2....2..2..2....0..2..2....0..2..2....2..2..1 ..3..3..3....3..3..3....3..3..3....3..3..3....3..3..3....3..3..3....3..3..3 CROSSREFS Sequence in context: A045563 A013339 A199360 * A093716 A117737 A062908 Adjacent sequences: A199582 A199583 A199584 * A199586 A199587 A199588 KEYWORD nonn AUTHOR R. H. Hardin Nov 08 2011 STATUS approved
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## In the left and write the normal table
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Compares the normal distribution plot produced by using python numpy library random events less than for taking on a score: how to obtain the percentiles and standard scores normal for a distribution table. In other continuous distribution, if the data, yards or the percentiles and standard for scores a normal distribution table and line segments mark the respective distributions for example is added that familiar with? You can think that on a student, we usually are a and part g as reflecting the curve. Recognize the more on the majority of standard deviation and minus sign in these kinds of distribution and for a table.
## This gets confusing because it gives the very much and standard scores normal for distribution table of a standard deviation in
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## Note that percentiles and sample data as with other standard normal curve we waived the histogram
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## The probability for scores a normal and standard deviations of sports stadiums in
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Garth | 2,552 | 13,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.905919 |
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# Harvard (HBS) Calling all applicants - Class of 2016
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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29 Oct 2013, 20:54
i have been interviewed and now spend most of my waking hours dissecting every single statement i made and how it could have been perceived as arrogant/ignorant/idiotic/random/OTT.
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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30 Oct 2013, 10:27
grit wrote:
i have been interviewed and now spend most of my waking hours dissecting every single statement i made and how it could have been perceived as arrogant/ignorant/idiotic/random/OTT.
Haha, yeah quite a few friends of mine have said the same thing. What was your experience like? Any oddball questions or things that threw you off?
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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31 Oct 2013, 02:56
Anyone in Dubai for interviews this weekend? My friend and I will be there for interviews-- wondering if anyone wants to meet up for a meal (and potentially be touristy together ).
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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31 Oct 2013, 05:06
Did my interview in Mumbai. The interviewer was Sarah Lucas. The questions were entirely off the resume. Made me narrate my experiences in my startup, consulting firm, VC firm and in my extra-curricular activities. Was very conversational and chilled out. Think I did well but tough to gauge in these interviews which are hard to do badly.
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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31 Oct 2013, 06:05
nikkiikkio wrote:
grit wrote:
i have been interviewed and now spend most of my waking hours dissecting every single statement i made and how it could have been perceived as arrogant/ignorant/idiotic/random/OTT.
Haha, yeah quite a few friends of mine have said the same thing. What was your experience like? Any oddball questions or things that threw you off?
similar to bingo13 above - almost completely off the resume! my interviewer did try to question me about my 'motivations' etc. near the end, but it wasn't intense, more conversational.
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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31 Oct 2013, 06:58
2
KUDOS
grit wrote:
nikkiikkio wrote:
grit wrote:
i have been interviewed and now spend most of my waking hours dissecting every single statement i made and how it could have been perceived as arrogant/ignorant/idiotic/random/OTT.
Haha, yeah quite a few friends of mine have said the same thing. What was your experience like? Any oddball questions or things that threw you off?
similar to bingo13 above - almost completely off the resume! my interviewer did try to question me about my 'motivations' etc. near the end, but it wasn't intense, more conversational.
same here. My interview was very conversational, and mostly came from the resume. The questions i got asked were:
- Talk me through your experiences at university
- experience at work
- asked a few questions regarding my work stories, to understand them better
- what would happen if you don't go to an mba/where would your current career path take you?
- what type of companies would you want to work for after your mba
- do you plan on returning to your home country?
- at the end, she asked me if there was anything else i wanted to add
Good luck to those who are preparing for interviews!
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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07 Nov 2013, 10:41
Did anyone send thank you notes to their interviewers? I found one email address, but not the second interviewers.
The pamphlet the handed me afterwards said that thank you notes are (truly) not necessary, but I don't want to be rude if everyone else sends thank you notes/emails.
What's the consensus here? Follow the directions to a T or send a brief thank-you note? If you sent a note, did you send it (via email) to the generic HBS admissions box or the interviewer's personal HBS email? Did you send a card via mail instead?
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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07 Nov 2013, 11:13
sp67 wrote:
Did anyone send thank you notes to their interviewers? I found one email address, but not the second interviewers.
The pamphlet the handed me afterwards said that thank you notes are (truly) not necessary, but I don't want to be rude if everyone else sends thank you notes/emails.
What's the consensus here? Follow the directions to a T or send a brief thank-you note? If you sent a note, did you send it (via email) to the generic HBS admissions box or the interviewer's personal HBS email? Did you send a card via mail instead?
I didn't send a thank-you note. From my limited interactions with HBS, I feel like they mean what they say. If they say thank-you notes are "truly" not necessary, then it probably means they don't want 1000 emails clogging up various adcom members' inboxes. Esp now that they give us the flexibility to write a post-interview reflection, I think anything that I would've wanted to say in an thank-you note could be captured in that format.
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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09 Nov 2013, 15:26
ltg1671 wrote:
sp67 wrote:
Did anyone send thank you notes to their interviewers? I found one email address, but not the second interviewers.
The pamphlet the handed me afterwards said that thank you notes are (truly) not necessary, but I don't want to be rude if everyone else sends thank you notes/emails.
What's the consensus here? Follow the directions to a T or send a brief thank-you note? If you sent a note, did you send it (via email) to the generic HBS admissions box or the interviewer's personal HBS email? Did you send a card via mail instead?
I didn't send a thank-you note. From my limited interactions with HBS, I feel like they mean what they say. If they say thank-you notes are "truly" not necessary, then it probably means they don't want 1000 emails clogging up various adcom members' inboxes. Esp now that they give us the flexibility to write a post-interview reflection, I think anything that I would've wanted to say in an thank-you note could be captured in that format.
I second this.
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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12 Nov 2013, 15:00
FROM HBS Admissions Blog: Class Visits I wish we had classes every day of the year – and evenings too – because there is a great demand from visitors who want to see the case method in action. That makes us happy. But, back to reality:We still have some class visit slots available for anyone who wants to visit in October (through the 23rd). My hope is that these will be for folks thinking about Round 2 applications.Round 1 applicants who are invited to interview will have many options for setting up a class visit and information will be included in interview invitations – there is no need for these applicants to register for a class visit before then. So I’m asking Round 1 applicants to please wait.For Round 2 applicants, there are going to be 9 days of class visits available starting November 12th through December 5th, these slots will be available for registration on our website later this week. Our students will be in final exams, Winter Break, and FIELD throughout December and January. We will continue to offer staff led tours and information sessions through December and January (other than during the two week “Winter Break” when the school is closed). I know it’s not ideal to make a big trip here and not be able to see a class, but it’s still a great way to get a sense of the campus and hear about the admissions process.Class visits will resume in early February for the Spring Semester and will be available for registration in early January. Similar to Round 1, our Round 2 interviewing applicants will have many opportunities to visit a class while on campus for their interview in February.And, yes, as you anticipated, this gives me yet another chance to plug the video, which shows real footage inside a first-year classroom plus faculty and student perspectives.Back to Round 1 applications!
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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12 Nov 2013, 15:00
FROM HBS Admissions Blog: Round One Interview Invitations If it's helpful, our plan is to send out some interview invitations at NOON today, Boston time. Via email. Nothing until noon.And we'll send more on October 16.
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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12 Nov 2013, 15:00
FROM HBS Admissions Blog: Another Round 1 Update At noon Boston time we'll be sending out some more interview invitations via email. Detailed instructions for sign-ups will be included.There will also be some Further Consideration decisions. This means that we are unable to invite you to interview now, but wish to keep your application under consideration. In Round 2, we'll be either inviting you to interview - and you'll move along on the Round 2 timetable - or releasing you. I know this prolongs the wait for a decision. There will be information in the Further Consideration decision letter about a contact person here in Dillon to answer questions and keep you informed. Of course, you may decline to stay in the process and withdraw your application at any time.We'll also be sending Deny decisions to the candidates who are not going to move forward in the process. This is our attempt to give you clarity and closure in time to make other plans.As I always say, I hope this is helpful.
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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12 Nov 2013, 15:00
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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13 Nov 2013, 18:43
For those who has submitted your applications to HBS, may I ask what you wrote about in your essays?
I am having hard time figuring out what to write...Thanks!
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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13 Nov 2013, 23:02
2
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slee26 wrote:
For those who has submitted your applications to HBS, may I ask what you wrote about in your essays?
I am having hard time figuring out what to write...Thanks!
I really think that everyone's essay should be different, there's no one 'thing' that people should write about. I'd suggesting thinking about who you are as a person, focusing on what made you that way, and then writing about some aspect of that. I wrote about several specific experiences that helped shape me into the person I am today.
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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14 Nov 2013, 06:43
boulderbiker wrote:
slee26 wrote:
For those who has submitted your applications to HBS, may I ask what you wrote about in your essays?
I am having hard time figuring out what to write...Thanks!
I really think that everyone's essay should be different, there's no one 'thing' that people should write about. I'd suggesting thinking about who you are as a person, focusing on what made you that way, and then writing about some aspect of that. I wrote about several specific experiences that helped shape me into the person I am today.
Thanks for this Boulderbiker. If I may ask, how many words did you write in your essay altogether?
It will be well appreciated if anyone who was invited for interview by Harvard can chip in a word or two into this.
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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14 Nov 2013, 06:49
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I wrote a long Essay, it was 1700 words and was invited. I also wrote about what was important to me, how i got there and where I wanted to go....
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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14 Nov 2013, 06:58
RSAproud wrote:
I wrote a long Essay, it was 1700 words and was invited. I also wrote about what was important to me, how i got there and where I wanted to go....
Thank you so much for your response. I finished my first draft yesterday and I hit 1,230 words. I freaked out because I set a target of 700 - 800 words. I guess I was not as shallow as I thought.
Did you talk about why Harvard?
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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14 Nov 2013, 07:07
knightofdelta wrote:
RSAproud wrote:
I wrote a long Essay, it was 1700 words and was invited. I also wrote about what was important to me, how i got there and where I wanted to go....
Thank you so much for your response. I finished my first draft yesterday and I hit 1,230 words. I freaked out because I set a target of 700 - 800 words. I guess I was not as shallow as I thought.
Did you talk about why Harvard?
Yes i basically took a few normal essay topics and put it together.. What makes you tick, why an mba and why now, why Harvard, what I will contribute....
I was nervous about the length but obviously it seemed fine. I'm also no rockstar candidate.. I am from Africa, no major extra curriculars, average marks, average GMAT of 730 and normal job progression. I needed the Essay to stand out....
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Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink]
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14 Nov 2013, 07:08
knightofdelta wrote:
boulderbiker wrote:
slee26 wrote:
For those who has submitted your applications to HBS, may I ask what you wrote about in your essays?
I am having hard time figuring out what to write...Thanks!
I really think that everyone's essay should be different, there's no one 'thing' that people should write about. I'd suggesting thinking about who you are as a person, focusing on what made you that way, and then writing about some aspect of that. I wrote about several specific experiences that helped shape me into the person I am today.
Thanks for this Boulderbiker. If I may ask, how many words did you write in your essay altogether?
It will be well appreciated if anyone who was invited for interview by Harvard can chip in a word or two into this.
I wrote less than 800 words and got an invite. I basically took all of the best parts from essays of other schools. I therefore recommend to save the HBS essay for last.
Interview in one week, wish me luck
Re: Harvard (HBS) Calling all applicants - Class of 2016 [#permalink] 14 Nov 2013, 07:08
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# Harvard (HBS) Calling all applicants - Class of 2016
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 5,407 | 21,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-22 | longest | en | 0.900412 |
https://www.theengineeringprojects.com/2017/05/declaration-of-variables-in-matlab.html/ | 1,708,930,693,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474653.81/warc/CC-MAIN-20240226062606-20240226092606-00266.warc.gz | 1,008,030,525 | 35,071 | Hello everyone! I hope you all will be absolutely fine and having fun. In the tutorial Declaration of variables in MATLAB, I will elaborate you that how to declare different variables in MATLAB and how to manipulate those variables without assigning them with the values. Before going into the details of this tutorial you must go through Introduction to MATLAB, because understanding MATLAB first is better. This will help you in better understanding of this tutorial. Variable declaration is one of the most important and compulsory steps while writing the code to perform any task. Without the declaration of the variables we can not proceed further to perform our task properly. So, if we want to do the different tasks in a proper way we must need to declare the variable before doing anything in the code. Once the variables have been declared, we can proceed further to explore the desired task.
#### Declaration of Variables in MATLAB
Here, in the tutorial Declaration of variables in MATLAB, I will show you that how the variables can be declared in the MATLAB and how to manipulate them without assigning them with the values. The command syms is used to declare the different variables in the MATLAB.
• Here, I will show you that what happens if you have not declared the variables.
• The error is shown in the figure below, when you have not declared the variables.
• I have written an equation having three different variables named as f,t and r without declaring these variables.
• MATLAB has shown an error on the command window as undefined function or variable.
• So the error shown in the above figure shows the desired results can not be obtained without declaring the variables first.
• Now, I am going to declare all the variables first and will then manipulate them.
• The declared variables are shown in the figure below.
• Since I have declared all of the variables now which are being used in the equation.
• So, now I am going to manipulate them and going to observe the results.
• The result of the addition of the two variables, which is stored in the third variable, is shown in the figure below.
• The above figure shows the result of the addition of the two different variables in the form of an equation.
• The result has been obtained because all of the variables has been declared first, i.e. declaration of variables plays a very important role while writing an algorithm to perform any task either simple or complicated.
• You can also see all of the process in the video given below:
So, that is all from the tutorial Declaration of Variables in MATLAB. I hope you enjoyed this tutorial. If you find any sort of problems you can ask me in comments without even feeling any kind of hesitation. I will try my level best to solve your issues in a better way, if possible. In my next tutorial I will explain you that how to create an m.file in MATLAB. I will further explore MATLAB in my later tutorials and will share all of them with you as well. So, till then, Take Care :) | 594 | 3,011 | {"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} | 3.125 | 3 | CC-MAIN-2024-10 | longest | en | 0.912433 |
https://edurev.in/course/quiz/attempt/17662_Olympiad-Test-Water-Images-1/c6c1777b-484d-4741-8d22-804ab165d8d7 | 1,722,808,888,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00380.warc.gz | 186,877,372 | 46,944 | Olympiad Test: Water Images - 1 - Class 7 MCQ
# Olympiad Test: Water Images - 1 - Class 7 MCQ
Test Description
## 10 Questions MCQ Test Mathematics Olympiad Class 7 - Olympiad Test: Water Images - 1
Olympiad Test: Water Images - 1 for Class 7 2024 is part of Mathematics Olympiad Class 7 preparation. The Olympiad Test: Water Images - 1 questions and answers have been prepared according to the Class 7 exam syllabus.The Olympiad Test: Water Images - 1 MCQs are made for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: Water Images - 1 below.
Olympiad Test: Water Images - 1 - Question 1
### Direction: In each of the following questions, choose the water image of the Fig.(X) from amongst the four alternatives (1), (2), (3) and (4) given along with it. Q. Choose the alternative which closely resembles the water-image of the given combination.
Olympiad Test: Water Images - 1 - Question 2
### Direction: In each of the following questions, choose the water image of the Fig.(X) from amongst the four alternatives (1), (2), (3) and (4) given along with it. Q. Choose the alternative which closely resembles the water-image of the given combination.
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Olympiad Test: Water Images - 1 - Question 3
### Direction: In each of the following questions, choose the water image of the Fig.(X) from amongst the four alternatives (1), (2), (3) and (4) given along with it. Q. Choose the alternative which closely resembles the water-image of the given combination.
Olympiad Test: Water Images - 1 - Question 4
Direction: In each of the following questions, choose the water image of the Fig.(X) from amongst the four alternatives (1), (2), (3) and (4) given along with it.
Q. Choose the alternative which closely resembles the water-image of the given combination.
Olympiad Test: Water Images - 1 - Question 5
Direction: In each of the following questions, choose the water image of the Fig.(X) from amongst the four alternatives (1), (2), (3) and (4) given along with it.
Q. Choose the alternative which closely resembles the water-image of the given combination.
Olympiad Test: Water Images - 1 - Question 6
Direction: In each of the following questions, choose the water image of the Fig.(X) from amongst the four alternatives (1), (2), (3) and (4) given along with it.
Q. Choose the alternative which closely resembles the water-image of the given combination.
Olympiad Test: Water Images - 1 - Question 7
Direction: In each of the following questions, choose the water image of the Fig.(X) from amongst the four alternatives (1), (2), (3) and (4) given along with it.
Q. Choose the alternative which closely resembles the water-image of the given combination.
Olympiad Test: Water Images - 1 - Question 8
Direction: In each of the following questions, choose the water image of the Fig.(X) from amongst the four alternatives (1), (2), (3) and (4) given along with it.
Q. Choose the alternative which closely resembles the water-image of the given combination.
Olympiad Test: Water Images - 1 - Question 9
Direction: In each of the following questions, choose the water image of the Fig.(X) from amongst the four alternatives (1), (2), (3) and (4) given along with it.
Q. Choose the alternative which closely resembles the water-image of the given combination.
Olympiad Test: Water Images - 1 - Question 10
Direction: In each of the following questions, choose the water image of the Fig.(X) from amongst the four alternatives (1), (2), (3) and (4) given along with it.
Q. Choose the alternative which closely resembles the water-image of the given combination.
## Mathematics Olympiad Class 7
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## Mathematics Olympiad Class 7
24 videos|57 docs|103 tests | 1,023 | 4,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-33 | latest | en | 0.889576 |
https://toutelachirurgieesthetique.fr/popup/topics/9e19ad-probability-calculator-multiple-events | 1,726,687,739,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00067.warc.gz | 543,370,345 | 6,226 | of Trials. His inspired wit and deep insight provided some truly excellent and hilarious moments for millions around the world. for use in every day domestic and commercial use! P(A') = 0.43 Probability is perhaps the single most used math term and something we calculate, not necessarily to precision, but as part of our lives everyday. Calculator, Grouped Data Standard Deviation Calculator.
(adsbygoogle = window.adsbygoogle || []).push({}); The probability calculator has two inputs: In the probability calculator, we use the following letters which are later used within the probability formula to allow you to understand how probability properties are calculated and how the probability calculator calculates probability. You can find an event's probability only with a few clicks on your device through an online probability calculator. Whether crossing the road, making financial decisions, making business decisions or making decisions about starting a family, the comprehension of probability is essential. HOW TO CALCULATE PROBABILITY WITH THE PROBABILITY CALCULATOR? As with all great math and sciences discoveries and insights, full understanding comes with looking at the detail. Probability can also be measured as a percentage or said as a statement, typically when referring to chance, for example a 1 in 5 chance of occurring. P(A∩B) = 0.25 Number of possible outcomes - n = 14 Once you are done with that, enter the number of events that occurred. The probability of harm however differs depending on speed, visability, surface, number of previous accidents and so forth. Mathematically, this progression gives an exponential decay curve. We have mentioned single event probability and multi event probability a few times, lets look at each in more detail. In this example, we have of course gone with the standard example of a coin which provides a useful lesson in probability. P(A ∩ B) is the probability that events A and B both occur. For example, if there is a high probability that it is going to be a very hot and sunny day, we would use that knowledge to apply sunscreen. The key to a successful single event probability calculation and multiple event probability calculation is to correctly define the total number of outcomes. P(AUB) = 1 - 0.25 This is truly why the coin example in math probability is an excellent lesson, it perfectly illustrates how easy it is to make assumptions in math tests / math exercises and by doing so, arrive and the wrong result using the right math formula. Wrong there are three possible outcomes. Use our online probability calculator to calculate the single and multiple event probability based on number of possible outcomes. Find what is P(A) , P(a), P(B), P(b), P(Aub), p(Anb) & P(A|B) P(B') = 0.57 P(B') is the probability that event B does not occur. Input Data : P(E) is the probability that the event will occur.
P(B) = 1 - P(B) First, you need to convert your percentages of the two events to decimals. Solution : An example of a problem that asks for the probability of multiple events is one that asks the probability of drawing two specific cards from a deck. Use this online probability calculator to calculate the single and multiple event probability based on number of possible outcomes and events … It can be derived from the below formula. You can use this Probability Calculator to determine the probability of single and multiple events. The following example may assist you in understanding its working. Probability of event A that does not occur P (A') = 1 - P (A). P(E) is the probability that the event will occur. The traditional method of finding the probability of an event isn't an easy task.
However, good calculation skills are needed in using the manual method of finding probability. ", "how will this math formula help me in my life?".
), 0.1666: Probability expressed as a decimal number, typically when necessary for use in further mathematical calculations, 1/6: Probability expressed as a fraction, very useful for quickly identifying the quantity of options and the number of calculated chances of occurrence. The Probability calculator allows you to calculate the probability of an event occurring by entering the number of events and the total number of outcomes. In many applications it is necessary to calculate the probability of single event, multiple event, joint probability, conditional probability and etc. Also, this calculator works as a conditional probability calculator as it helps to calculate conditional probability of the given input. If P(B) = 0 then P(A | B) is undefined. The Single Event Probability Calculator uses the following formulas: P(E) = n(E) / n(T) = (number of outcomes in the event) / (total number of possible outcomes). The Probability calculator allows you to calculate the probability of an event occurring by entering the number of events and the total number of outcomes. Like in tossing a coin, either you can get a head or a tail, but you can't get both at the same time. n(A) is number of outcomes in the event A. n(B) is number of outcomes in the event B. n(T) is total number of possible outcomes. P(not A) = 1 - P(A), If both events A and B occur on a single performance of an experiment, this is called the intersection or joint probability of A and B, denoted as P(A n B).
Probability Chance) is the measurement of how likely it is that something (an event, or multiple events) will happen.
The Multiple Event Probability Calculator uses the following formulas: P(A) is the probability that event A occurs. Enter your values in the form and click the "Calculate" button to see the results.
The formula for finding the probability is, P(A) = No. Now, enter the number of possible outcomes in our Statistics Probability Calculator. The key to a successful single event probability calculation and multiple event probability calculation is to correctly define the total number of outcomes.
That's why the best solution to find the probability is the use of an online tool, which enables you to find the probability within no time. Probability measurement as a decimal is more practical when you intend to use the calculated probability as part of an additional formula, for examples to complete a risk assessment as mentioned earlier. P(A ∪ B) is the probability that events A or B occur. Probability Calculator is an online tool for risk analysis specially programmed to find out the probability for single event and multiple events. In maths, it is probably the most often ask question and something math teachers come to expect from their math students, "why do I need to know how to calculate this in math? The following formula can be used to calculate if an event will not occur and therefor be false.
We could hardly discuss the subject of probability with mentioning the genius that was Douglas Adams. P(A n B) = P(A) P(B), If either event A or event B or both events occur on a single performance of an experiment this is called the union of the events A and B denoted as P(A U B). Single Event Probability Calculator It also suggests that we are each, mathematically, capable of influencing and changes those events. In order to find the probability of many events all happening, it is necessary to multiply their probabilities together. Number of events occured - n(B) = 6 P(E') is the probability that the event will not occur. where p - is a probability of each success event, - Binomial coefficient or number of combinations k from n The details are below the calculator. Total Probability should be exactly 1 When you are calculating the probability of multiple events, make sure that the total probability is 1. The probability that the coin will land either heads up or tails up is 1 in 2, 0., 50% or ½. | 1,591 | 7,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-38 | latest | en | 0.942102 |
https://amm.itsligo.ie/en/module/web/20/ENG06062/201700/999999 | 1,656,682,073,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00200.warc.gz | 148,709,200 | 4,910 | ENG06062 2017 Intermediate Engineering Mechanics
General Details
Full Title
Intermediate Engineering Mechanics
Transcript Title
Intermediate Engineering Mecha
Code
ENG06062
Attendance
N/A %
Subject Area
ENG - Engineering
Department
MENG - Mech. and Electronic Eng.
Level
06 - NFQ Level 6
Credit
05 - 05 Credits
Duration
Semester
Fee
Start Term
2017 - Full Academic Year 2017-18
End Term
9999 - The End of Time
Author(s)
Declan Sheridan
Programme Membership
SG_EMECH_S06 201900 Certificate in Engineering in Mechanical Analysis and Automation SG_EMSYS_B07 201900 Bachelor of Engineering in Mechatronic Systems SG_EMECH_H08 202100 Bachelor of Engineering (Honours) in Mechatronic Systems
Description
• The student will determine stress in materials used in compound bars.
• The student will perform calcualtions involving shear stress, shear strain, and poisson's ratio.
• The student will draw shear force and bending moment diagrams.
• The student will calculate stress in thin walled pressure vessels, and thin rotating rings.
• The student will make calculations involving torsion.
• The student will do calculations involving thermal strain..
This module is jointly taught by a number of lecturers and includes many examples of how mechanics is used in real life situations
Learning Outcomes
On completion of this module the learner will/should be able to;
1.
Calculate stresses in each material of a compound bar.
2.
Be able to manipulate and perform calculations involving shear stress, shear strain, and poisson's ratio.
3.
Perform calculations invollving stress in thin walled pressure vessels and thin rotating rings.
4.
Be able to manipulate the formula involving torsion to determine safe loads.
5.
Be able to determine and draw bending moment and shear force diagrams for various beams.
Teaching and Learning Strategies
This module will be taught online using conference software plus each week an online assessment will be completed.
Module Assessment Strategies
The learning outcomes will be assessed by a weekly online assessment plus an end of semester examination.
Repeat Assessments
Repeat assessment will be by way of sitting another examination on the subject. Alternatively, at the discretion of the lecturer, assignemnts covering the deficient areas of the course may be set.
Indicative Syllabus
1. Compound bars
2. Shear stress, shear strain, modulus of rigidity, and poisson's ratio.
3. bending moment and shear force diagrams.
4. Thin walled pressure vessels.
5. Thin rotating rings.
6. Torsion.
7. Thermal strain.
Coursework & Assessment Breakdown
Coursework & Continuous Assessment
20 %
End of Semester / Year Formal Exam
80 %
Coursework Assessment
Title Type Form Percent Week Learning Outcomes Assessed
1 Online assessment Continuous Assessment Multiple Choice 20 % OnGoing 1,2,3,4,5
End of Semester / Year Assessment
Title Type Form Percent Week Learning Outcomes Assessed
1 Final Exam End of semester exam Final Exam Closed Book Exam 80 % Week 15 1,2,3,4,5
Type Location Description Hours Frequency Avg Workload
Lecture Distance Learning Suite Online lecture 1 Weekly 1.00
Independent Learning UNKNOWN Preparation for online lecture 2 Weekly 2.00
Independent Learning UNKNOWN Solution of problems set by the lecturer during the weekly class 2 Weekly 2.00
Independent Learning UNKNOWN Undertaling the online quiz 1 Weekly 1.00
Total Part Time Average Weekly Learner Contact Time 1.00 Hours
Module Resources
Journal Resources
N/A
URL Resources
N/A
Other Resources
None | 770 | 3,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.812268 |
https://paperzz.com/doc/8339872/triangle-incenter-construction | 1,653,723,969,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663013003.96/warc/CC-MAIN-20220528062047-20220528092047-00595.warc.gz | 509,082,885 | 7,965 | ### Triangle Incenter Construction
```GeoGebra 4.0: Triangle Incenter Construction Activity: Triangle Incenter Construction •
Objectives 1. Becoming familiar with the basic use of tools and formatting 2. To understand the properties of a triangle and how they can be used to construct an Incenter. Introduction: In this activity you are going to use the following tools. Please, make sure you
know how to use each tool before you begin with the actual construction of the
incenter:
Polygon
Intersect two objects
Angular Bisector
Perpendicular Line
Circle with center through point
Move
Show / hide label
Hint: If you are not sure about the construction process, you might want to have a look at the file, Incenter.ggb. Preparing the Window: • Open a new GeoGebra file. • Hide the algebra window, the axes on and the input field. Set labeling to New Points Only. Construction process: 1. Use the Polygon tool to create an arbitrary triangle ABC. 2. Use the Angle Bisector tool to construct the angle bisector for each vertex of the triangle. Hint: The tool, Angle Bisector, can used by selecting the three points that make up the angle. 3. Use the Intersect Two Objects tool to mark the intersection point D of two of the line bisectors. Hint: The tool, Intersect Two Objects, can’t be applied to the intersection of three lines. 4. Use the Perpendicular Line tool and select point D and any of the sides of triangle ABC. 5. Use the Intersect Two Objects tool to mark the intersection point E of two of the line from step 4 and the side of the triangle. 6. Use the Circle with Center Through Point tool and select Point D and then Point E. 7. Perform the drag test to check if your construction is correct. GeoGebra 4.0: Triangle Incenter Construction The following Image is what your file should look like before hiding or formatting any objects. Challenge: Modify your construction to answer the following questions: 1. Can the Incenter of a triangle lie outside the triangle? If yes, for which types of triangles is this true? 2. Try to find an explanation for using angle bisectors in order to create the Incircle of a triangle. ``` | 483 | 2,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-21 | latest | en | 0.840103 |
https://oeis.org/A330909 | 1,620,347,216,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988774.18/warc/CC-MAIN-20210506235514-20210507025514-00106.warc.gz | 457,587,448 | 4,216 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A330909 Floor of area of triangle whose sides are consecutive Ulam numbers (A002858). 0
0, 2, 5, 11, 23, 43, 70, 100, 141, 227, 361, 478, 670, 826, 1044, 1183, 1405, 1668, 1960, 2272, 2545, 2889, 3351, 3819, 4267, 4523, 4955, 5669, 6558, 7474, 8203, 8914, 9633, 10813, 12245, 13611, 13972, 14587, 15473, 16798, 17987, 19298, 20229, 21909 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS It has been proved that three consecutive Ulam numbers U(n) for n > 1 satisfy the triangle inequality. See Wikipedia link below. LINKS Wikipedia, Ulam number. FORMULA Given a triangle with sides a, b and c, the area A = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2. EXAMPLE a(2)= 2 because the triangle with sides (2, 3, 4) has area 3*sqrt(15)/4 = 2.9047... MATHEMATICA lst1 = ReadList["https://oeis.org/A002858/b002858.txt", {Number, Number}]; lst={}; Do[{a, b, c}={lst1[[n]][[2]], lst1[[n+1]][[2]], lst1[[n+2]][[2]]}; s = (a+b+c)/2; A=Sqrt[s(s-a)(s-b)(s-c)]; AppendTo[lst, Floor@A], {n, 1, 50}]; lst CROSSREFS Cf. A002858, A331729. Sequence in context: A062475 A186265 A225947 * A281969 A124920 A064934 Adjacent sequences: A330906 A330907 A330908 * A330910 A330911 A330912 KEYWORD nonn AUTHOR Frank M Jackson, May 01 2020 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
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Last modified May 6 19:52 EDT 2021. Contains 343586 sequences. (Running on oeis4.) | 636 | 1,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-21 | latest | en | 0.66243 |
https://golem.ph.utexas.edu/category/2010/04/on_the_magnitude_of_spheres_su.html | 1,638,855,785,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2021-49/segments/1637964363336.93/warc/CC-MAIN-20211207045002-20211207075002-00168.warc.gz | 359,530,787 | 27,788 | ## April 21, 2010
### On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
#### Posted by Simon Willerton
Back in my recent post on intrinsic volumes I said that I was just finishing a paper on the magnitude of certain Riemannian manifolds. It is now essentially finished.
I will put it on the arxiv next week: comments and suggestions are welcome.
There are two main results in the paper. The first is a calculation of the magnitude of the $n$-sphere and the second is the asymptotic behaviour of the magnitude for an arbitrary homogeneous Riemannian manifold in terms of volume and total scalar curvature, a special case of the latter is the asymptotic behaviour of the magnitude of a homogeneous surface. I will explain these below, but first remind you a little about magnitude.
### On the Magnitude…
Magnitude of finite metric spaces (under the name cardinality) was introduced by Tom Leinster at this very café, by analogy with his notion of Euler characteristic of a category. In two posts, Entropy, Diversity and Cardinality (Part 1) and Part 2, he explained how this was related to ideas in ecology and went on to prove a result on how magnitude helps you maximize biological diversity. In another direction Tom and I defined the magnitude of various infinite subsets of Euclidean space by approximating these with finite metric spaces. In the case of a finite metric space, as the space is scaled up, the magnitude gets closer and closer to the number of points: in the case of some nice subsets of Euclidean space we observed that as the subset is scaled up, the magnitude seems to tend to some linear combination of classical ‘intrinsic volumes’ of the subset.
[If you want some idea of the intuition I use to think of the magnitude then you should look at the post More Magnitude of Metric Spaces and Problems with Penguins.]
In this paper I take a different tack in defining the magnitude of infinite metric spaces, one that was suggested to me by Tom Leinster and by David Speyer. This involves using a (signed) measure on a metric space, and it clearly generalizes the idea of having a weight associated to each point in a finite metric space. Using this approach gives the same answer as we got in examples we’d previously calculated.
A key observation, essentially made by David Speyer here, is that for a homogeneous space $X$ (i.e., one that looks the same at every point) the magnitude can be calculated using any measure $\mu$ by the formula
$\left|X\right|= \frac{\int_{x\in X} d\mu}{\int_{x\in X}e^{-d(x,y)} d\mu},\qquad\text{for any} y\in X$
provided that the denominator is finite and non-zero. Aside from working with subsets of the line, this is currently the only way I know of calculating the magnitude of infinite metric spaces.
### …of Spheres,…
There are two obvious metrics to consider on the $n$-sphere: there’s the ‘intrinsic’ metric, coming from thinking of it as an abstract Riemannian manifold and there’s the ‘Euclidean subspace’ metric, coming from thinking of it as a subset of $\mathbb{R}^{n+1}$. On the Earth, the intrinsic distance between Sheffield and Riverside is how far I would have to travel over the surface of the Earth to get from one to the other; whereas the Euclidean subspace distance is how far I would have to travel through the Earth if I dug a straight-line tunnel from one to the other. In previous work, Tom and I concentrated on spaces with the subspace metric, but here I am concerned with the intrinsic metric.
It is a very fun exercise to use Speyer’s formula to explicitly calculate the magnitude for $S_R^n$, the $n$-sphere of radius $R$ with its intrinsic metric (or at least I found it fun). This is done in the paper and gives the following answer.
$\left|S^n_R\right|=\begin{cases} \frac{2 \Big(\Big(\frac{R}{n-1}\Big)^2 + 1\Big)\Big(\Big(\frac{R}{n-3}\Big)^2 + 1\Big) \dots \Bigl(\Bigl(\frac{R}{1}\Bigr)^2 + 1\Bigr) }{1+e^{-\pi R}}& \text{for n even}\\ \frac{\pi R \Bigl(\Bigl(\frac{R}{n-1}\Bigr)^2 + 1\Bigr)\Bigl(\Bigl(\frac{R}{n-3}\Bigr)^2 + 1\Bigr) \dots \Bigl(\Bigl(\frac{R}{2}\Bigr)^2 + 1\Bigr) }{1-e^{-\pi R}}\quad& \text{for n odd}. \end{cases}$
The denominator rapidly tends to one as $R$ increases. So the large, or asymptotic, behaviour of the magnitude is governed by the numerator which, for fixed dimension $n$, is a polynomial in $R$ with some nice properties. For instance the leading order and constant terms can be identified in terms of classical invariants, namely the volume and the Euler characteristic, so, writing $\omega_n$ for the volume of the unit $n$-ball, the numerator is the following form:
$\tfrac{1}{n!\omega_n}vol(S^n_1)R^n +\dots+\chi(S^n_1).$
This is the kind of thing that Tom and I had been expecting for spaces with the Euclidean subspace metric. However, we will see below that the other term we can identify differs from what we naively expect in the Euclidean metric case.
One obvious question to ask here is the following.
• What is the magnitude of the $n$-sphere with the subspace metric?
### …Surfaces…
All closed, compact surfaces (spheres, tori, etc.) can be equipped with a homogeneous Riemannian metric. For the sphere this should be clear as it is naturally thought of as a homogeneous space. Also the torus is commonly thought of as the quotient of the plane by an integer lattice, making it homogeneous. Similarly every higher genus surface can be obtained by quotienting the hyperbolic plane with an appropriate group of isometries, analogous to $\mathbb{Z}^2$.
For a surface with a homogeneous metric we can use Speyer’s formula from above to calculate the magnitude. I haven’t been able to find any nice closed form for the magnitude of higher genus surfaces as in the case of spheres, but there’s a nice computation giving the asymptotic behaviour. For a metric space $X$ and $t\gt0$ we can write $t X$ for $X$ with the metric scaled up by a factor of $t$. The for a surface $\Sigma$ with a homogeneous metric, as $t\to \infty$
\begin{aligned}\left|t\Sigma\right|&= \tfrac{1}{2\pi}Area(t\Sigma)+\chi(t\Sigma) +O(t^{-2})\\ &=\tfrac{1}{2\pi}Area(\Sigma)t^2+\chi(\Sigma) +O(t^{-2}). \end{aligned}
This is precisely the kind of behaviour Tom and I had conjectured for subspaces of Euclidean spaces.
There are four main ingredients in my proof of this result: one is Speyer’s formula for the magnitude of a homogeneous metric space mentioned above, another is Watson’s Lemma, a staple of asymptotic analyis, and the other two are properties of Gauss curvature mentioned in my post on intrinsic volumes, namely that the Gauss curvature determines the circumference of small circles (at least to low order), and that the integral of the Gauss curvature is proportional to the Euler characteristic.
If we are calculating $\left|t\Sigma\right|$ using Speyer’s formula, then the key thing is to understand the denominator; so for any fixed $x_0\in \Sigma$ we want to consider
$\int_{x\in\Sigma} e^{-t d(x,x_0)}d x.$
If $t$ is very large then because of the negative exponential, parts of $\Sigma$ which are not very close to $x_0$ will only give exponentially small contributions. This rigorous way to state this is Watson’s lemma, but the point is that for large $t$ the dominant contributions will come from local information, i.e., curvature type information.
Writing $F_r$ for the subset of $\Sigma$ at a distance of $r$ from $x_0$ we can rewrite the the above integral as follows
$\int_{x\in\Sigma} e^{-t d(x,x_0)}d x=\int_{r\ge 0}e^{-t r} Length(F_r) d r.$
But we know that for small $r$ the length, or circumference, of $F_r$ is determined at low order by the Gauss curvature $K(x_0)$.
$Length(F_r)=2\pi r \big(1 - \tfrac{1}{6}K(x_0)r^{2}+O(r^{4})\big)\qquad{as} r\to 0.$
Using this, together with Watson’s Lemma we get that the integral has the form
$\int_{x\in\Sigma} e^{-t d(x,x_0)}d x=2\pi t^{-2}\big(1 - K(x_0)t^{-2}+O(t^{-4})\big) \qquad{as} t\to\infty.$
This means that as $t\to \infty$ we find, as we wanted, that
\begin{aligned} \left|t\Sigma\right| &=\frac{\int_{x\in\Sigma} d x}{\int_{x\in\Sigma} e^{-t d(x,x_0)}d x} =\frac{\int_{x\in\Sigma} d x}{2\pi t^{-2}\big(1 - K(x_0)t^{-2}+O(t^{-4})\big)}\\ &=\tfrac{1}{2\pi}t^2\int_{x\in\Sigma} d x + \tfrac{1}{2\pi}\int_{x\in\Sigma} K(x_0)d x +O(t^{-2})\\ &=\tfrac{1}{2\pi}Area(\Sigma)t^2+\chi(\Sigma) +O(t^{-2}), \end{aligned}
by the fact that the Gauss curvature is constant on a homogeneous space and that the integral of the Gauss curvature is proportional to the Euler characteristic.
### …and Other Homogeneous Spaces
For an arbitrary homogeneous Riemannian manifold $X$ of dimension $n$, you can use the same technique as for surfaces to calculate the leading order terms of the asymptotics of the magnitude. The key point is to identify the relevant analogues of the Gauss curvature and the Euler characteristic. These turn out to be the scalar curvature and the total scalar curvature.
At every point $x$ of a Riemannian manifold $X$ there is a scalar curvature $\tau(x)$ which specializes to twice the Gauss curvature in the case of a surface. This has the property of the Gauss curvature that at a point $x$ the volume of the set of points $F_r$ a distance $r$ from $x$ is, for small $r$, determined to leading terms by the scalar curvature:
$vol_{n-1}(F_r)= r^{n-1}\sigma_{n-1}\left(1-\tfrac{\tau(x)r^2}{6n}+\dots\right)\quad \text{as} r\to 0.$
The total scalar curvature, $tsc(X)$, is defined to be the integral of the scalar curvature;
$tsc(X)\coloneqq\int_X \tau(x) d x,$
The total scalar curvature seems to crop up in various parts of mathematics and physics, but I don’t know much about that. The connection with intrinsic volumes is that the total scalar curvature is, for closed manifolds, the next non-trivial intrinsic volume after the volume:
$\mu_{n-2}(X) = \frac{1}{4\pi}tsc(X).$
In the case that $X$ is a surface (so $n=2$) we recover the relationship between the total Gauss curvature and the Euler characteristic
Running through through the same analysis as for the case of surfaces above, you find that for $X$ a homogeneous Riemannian manifold the magnitude looks asymptotically as follows:
$\left|t X\right| = \tfrac{1}{n!\omega_n}vol(X)t^n+\tfrac{n+1}{6n!\omega_n}tsc(X)t^{n-2}+O(t^{n-4}).$
There are a few questions that jump out at this point.
• Can these methods be used to calculate further terms in the asymptotic expansion for homogeneous manifolds?
• Are these leading terms the same for the asymptotic behaviour of non-homogeneous Riemannian manifolds?
I should probably note here that the asymptotic behaviour is only one part of the picture and it would also be good to know what is going on for ‘small’ spaces: what is the magnitude measuring?
Posted at April 21, 2010 7:33 AM UTC
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### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Thanks for the post and paper. I’m a big fan of this magnitude of metric spaces stuff.
Once every couple of years I pick up Street’s paper “Algebra of oriented simplexes” hoping my math kung fu has progressed enough for me to finally understand it. Coincidentally, the last time I picked it up recently made me think of this magnitude of metric spaces stuff. Probably for no good reason other than lack of sleep though. Let me throw some thoughts out there and see if anything makes sense…
I liked the idea of relating the 1-cocycle condition to a composite triangle in a category, i.e.
$\partial[012] = [12]-[02]+[01] = 0$
rearranges to
$[02] = [01]+[12]$
which is interpreted as a commuting triangle.
Similarly
$\partial[0123] = [123]-[023]+[013]-[012] = 0$
rearranges to
$[123]+[013] = [023]+[012].$
The pictures that go with this are great.
Then I had a whacky idea. What if we turn the square brackets into magnitudes instead, i.e.
$|02| = |01|+|12|.$
This says the magnitude of [02] equals the sum of the magnitudes of [01] and [12]. But if magnitude is “volume” then we are talking about lengths of the sides of a triangle. When is the sum of two lengths of a triangle equal to the length of the third side? When the triangle is degenerate, i.e. has no area. Area is zero? Boundary is zero? Could they be related?
This idea bumps up trivially in dimension.
The 2-cocycle condition gives
$|123|+|013| = |023|+|012|$
which says the sum of areas of two triangles of a tetrahedron are equal to the sum of the areas of the two remaining triangles of the tetrahedron. When does that happen? It happens when the 3-simplex is degenerate, i.e. it has no volume.
The picture is kind of pretty. You partition the boundary of an $n$-simplex into two $(n-1)$-complexes. The $(n-1)$-cocycle condition says the magnitude of these two complexes are equal.
We have magnitudes of metric spaces and magnitudes of categories, is there a such thing as a magnitude of a diagram in a category?
Apologies if this makes absolutely no sense and is completely irrelevant to this magnitude business :)
Posted by: Eric Forgy on April 21, 2010 4:14 PM | Permalink | Reply to this
### First reactions
Because a diagram is a functor, it has a domain category, with its own magnitude; but the image is a subcategory, too (unless that notion is evil?), which again has its own magnitude. Maybe you want some combination of these two?
Posted by: some guy on the street on April 22, 2010 12:05 AM | Permalink | Reply to this
### Re: First reactions
Hmm…
If you start with the category $\mathbf{1}$ with two objects 0,1 and one non-identity morphism $\to$ and a functor $F:\mathbf{1}\to C$ with $F(0) = F(1)$, then the image of $F$ is not a subcategory of $C$ unless $F(\to) = 1_{F(0)}$. Is that what you meant by evil?
I spent about a week on the n-Forum (starting around here) making these guys pull their hair out because I was making a mistake thinking the image of a functor was a category.
After all that, please do not now tell me the image of a functor is a category after all :)
We have magnitudes of metric spaces and magnitudes of categories, is there a such thing as a magnitude of a diagram in a category?
I had the magnitude of the image of a functor in mind (which need not be a category). More generally, I’d be interested if there was a magnitude for a general directed graph.
Now the admittedly weak link to magnitudes is that I was wondering if $\partial$ might be considered a functor mapping categories (or diagrams or graphs) to magnitudes.
Posted by: Eric on April 22, 2010 1:54 AM | Permalink | Reply to this
### Re: First reactions
Oops. I guess you guys call that category $\mathbf{2}$ :)
Posted by: Eric on April 22, 2010 4:54 AM | Permalink | Reply to this
### Re: First reactions
oh! hmm… it’s weirder than I took care to think about, this image-of-a-functor business. Of course, the identity morphism will be in there, but unless the image of $\to$ is an involution… I really should try to remember these issues more carefully!
More carefully, we should probably take the right factorization of the diagram $F:\mathcal{D}\rightarrow\mathcal{C}$ as $\mathbf{2}\longrightarrow \Big\langle Hom(0,1)\mathllap{:}{\,=}\mathcal{C}(F(0),F(1))\Big\rangle \longrightarrow \mathcal{C}|_{\{F(0),F(1)\}} \longrightarrow \mathcal{C} ;$ being the factor functors forgetting only structure, only stuff, and only property, respectively — and wherein none of the intermediate categories are really the image of the functor where you wanted to define a magnitude. So… take or leave what you will.
Posted by: some guy on the street on April 22, 2010 7:21 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
One thing that immediately strikes me when looking at this is the similarity of all the definitions here to things that come up when using the energy method for bounding Hausdorff dimensions.
For instance the formula
(1)$\int_X \exp(-d(x,y)) d\mu(y) = 1$
is precisely the expression (in the language of the energy method, which of course is in the language of physics) that the potential felt by the point $x$ with mass distribution $\mu$ and forces that decay exponentially is one.
Similarly the expression
(2)$|X| = \int_X d\mu(x) = \int_X\int_X \exp(-d(x,y)) d\mu(x)d\mu(y)$
is precisely the statement that the magnitude of the space is the total mass required, or equivalently the total potential energy of the system.
This leads to a nice physical interpretation: that the magnitude of a metric space is exactly the amount of mass needed to be spread out over the metric space so that every point feel a total potential energy of one (when the potentials/forces decay exponentially).
I don’t think there’ll be any direct connection to Hausdorff measures though because everything there relies upon a potential which is singular at small distances, whereas this is “regularized” near zero distance and instead cares about large distances. Still the terminology might make what you are saying familiar to more people, which is always good!
A resource, probably not the best one, for this sort of stuff is the excellent book on Brownian Motion by Morters and Peres, a draft of which can be posted here: www.stat.berkeley.edu/~peres/bmbook.pdf. The pertinent information is in chapter 4 about page 109 (a section about Hausdorff measures and dimensions).
Posted by: Brent Werness on April 21, 2010 5:10 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Brent said:
This leads to a nice physical interpretation: that the magnitude of a metric space is exactly the amount of mass needed to be spread out over the metric space so that every point feel a total potential energy of one (when the potentials/forces decay exponentially).
I prefer to think in terms of something like ‘heat generation’ rather than mass because we actually allow signed measures (which I didn’t emphasize in the post before) and I don’t have a good intuition for negative mass. I described this point of view in the post More Magnitude of Metric Spaces and Problems with Penguins and I’ve now linked to this in the main text above.
The similarity with the Energy Method is intriguing. I wasn’t aware of it. The main difference, however, is that the Energy Method uses a power of the distance where we use the exponential. I would really like to find some connection with this fractal dimension ideas though, as the growth of the magnitude does seem connected with ‘dimension’ in some sense. So for instance in On the asymptotic magnitude of subsets of Euclidean space Tom and I show that the magnitude of the Cantor set asymptotically grows like the Hausdorff dimension $\log_3 2$; asymptotically, the magnitude of the length $t$ Cantor set looks like
$f(t)t^{\log_3 2}$
for some ‘almost constant’, multiplicatively periodic function $f$.
Posted by: Simon Willerton on April 22, 2010 1:58 AM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Typo? The denominator rapidly tends to zero as R increases … zero -> one
Posted by: Charlie C on April 21, 2010 6:07 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Fixed, thanks.
Posted by: Simon Willerton on April 21, 2010 7:13 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
In the Introduction you credit the knowledge that magnitude is always defined for finite subsets of Euclidean space partly to me. Actually I believe that was a combination of observations of Tom, David Corfield, Yemon Choi, and David Speyer. My contributions started with generalizing that away from Euclidean space.
Posted by: Mark Meckes on April 21, 2010 9:04 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
I was basing that on my interpretation of Tom’s comments in the post An Adventure in Analysis. I will figure out how to rewrite this before putting it on the arxiv. Thanks.
I see you’re giving some seminars on magnitude. I hope they’re going well!
Posted by: Simon Willerton on April 21, 2010 11:45 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
They seem to be going well. You’ve released this draft just in time for me to talk about it in my last talk tomorrow!
Posted by: Mark Meckes on April 22, 2010 1:24 AM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Posted by: Tom Leinster on April 21, 2010 11:48 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
It’s interesting to me that you say the measure-based approach to defining magnitude was suggested to you by Tom and David S., whereas your own thoughts had been more along the lines of currents. To me the measure approach is the obvious first thing to look at, but prompted by some things Tom has said to me I now think something like currents might be the way to go.
In any case, I think it’s clear that whenever a weight measure exists, the magnitude should be exactly as you define it. I hope that finding the right class of measure-like things to consider will provide some more versatile tools for proving that weight measures exist for many spaces.
Posted by: Mark Meckes on April 22, 2010 3:08 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Mark said:
prompted by some things Tom has said to me I now think something like currents might be the way to go.
Do you want to expand on this? My reasons were mainly that on the one hand I had never done any measure theory – I certainly never took a course on it – and on the other hand I was used to using currents, indeed had used them in my thesis. When I did the calculations for the weightings on squares and discs and the like, it was the idea of currents that was brought to mind by the pictures.
Posted by: Simon Willerton on April 22, 2010 7:14 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Do you want to expand on this?
Sure. To begin with, to someone who works with measures a lot, Tom’s original definition of a weighting vector $w$ for a finite metric space $X$ brings measures to mind immediately. After all, how does $w$ show up? The weighted sums $\sum_{x\in X} e^{-d(y,x)} w_x=1$ can be interpreted as integrals of the functions $f_y(x) = e^{-d(y,x)}$ and the sum $\sum_{x\in X} w_x = \vert X \vert$ is then the total (signed) measure of $X$. Even the name “weighting” brings to mind common terminology for discrete measures. If $X$ is no longer finite, no problem, just forget that we ever thought of $w$ as a vector: now it’s a signed measure on $X$.
But we don’t want to fall into the trap of thinking everything is a nail just because we happen to have a hammer handy. Is there some reason that measures arise naturally from the mathematics here? Well, another way to think about the finite sums above is to identify $\mathbb{R}^X$ with its dual space and consider $w \in (\mathbb{R}^X)^*$, so what we have is $w(f_y) = 1$ for every $y\in X$ and then $\vert X \vert = w(1)$. When $X$ is infinite, we would like to replace $\mathbb{R}^X$ with some function space on $X$ which contains all the functions $f_y$ and $1$; this will be an infinite dimensional space, most likely not identifiable with its dual in a reasonable way, and so $w$ will become a different kind of creature.
So which function space should we work with? One that comes to mind is $C_b(X)$, the space of bounded continuous real-valued functions on $X$. (Keep in mind we don’t have a lot of structure to work with. If $X$ is nothing but a metric space with no canonical choice of a measure we can’t talk about $L^p$ spaces, for example.) $C_b(X)$ contains the functions $f_y$ and $1$, and is a Banach space with the sup norm. If $X$ is moreover compact (so we can drop that $b$), then $C(X)^*$ is isometrically isomorphic to $M(X)$, the space of signed Borel measures on $X$; a signed measure acts on a continuous function by integration. So measures really do show up naturally.
I said some abbreviated version of this to Tom in email. He said that he didn’t find this argument completely satisfactory in part because the space $C(X)$ only depends on the topology of $X$, and doesn’t see anything else about the metric structure. (Tom, please correct me if I’m misrepresenting or leaving out something important.) He suggested that it might make more sense to use $Lip(X)$, the space of Lipschitz functions on $X$, or something like it. $Lip(X)$ is also a Banach space with an appropriate norm, which contains $f_y$ and $1$. Furthermore $Lip(X)^*$ (is isometrically isomorphic to a space which) contains $M(X)$, but as a proper subspace when $X$ is infinite. Heuristically, since Lipschitz functions are within spitting distance of being differentiable (when “differentiable” even makes sense), the elements of $Lip(X)^*$ are something like functionals of differentiable functions, so they are something like some class of distributions, or currents. (I’d never actually heard of currents until recently, when I was talking to a differential geometer colleague about this.) So maybe that’s the sort of thing $w$ should be.
I should point out one possible objection to Tom’s hesitation about using $C(X)$ and $M(X)$: although those Banach spaces don’t know much about the metric on $X$, the functions $f_y$ do, and they stay in the picture in any case.
Of course this is not just an aesthetic or philosophical issue. We want definitions of weighting and magnitude that let us prove good theorems. In particular, I’d like a definition of magnitude which is equivalent to yours whenever a weighting measure exists. In the framework I’ve described, this means that whatever function space we work with, we want its dual to contain $M(X)$. It would also be nice to have a definition such that the definitions of magnitude in your paper with Tom, via approximations by finite sets, become theorems (this is what Tom said to me that set me to thinking about all of the above in the first place). To prove such theorems we need to know something about how measures with finite support sit inside our dual space. For both of the function spaces $C(X)$ and $Lip(X)$ I do know some interesting things about this question, but I haven’t managed to use them prove the desired theorems yet.
[Off-topic question for those who understand the tex rendering here: I didn’t bother to make the Lip in $Lip(X)$ Roman, although I normally would, out of laziness. But it appears in Roman anyway, on my browser at least. Why is that?]
Posted by: Mark Meckes on April 23, 2010 3:30 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Mark’s comments here just suggested something which might be of passing interest if not of real relevance. Since I’m currently a bit sleep-deprived and not really up to cogent mathematical thought today, I’d just like to throw out some disjointed remarks and half-remembered tidbits. Apologies if this derails the discussion a bit; I am really just commenting here while the iron is hot.
- The use of lip(X) or Lip(X) as a kind of coordinate ring (broadly speaking) for compact metric spaces has been championed in some papers and in a book of Nik Weaver, this POV probably goes back to work of Mark Rieffel but I forget the precise referencs.
- Lip(X) is the bidual Banach space of lip(X), and the predual of Lip(X) is I think a named widget, called the Arens-Eells space of X – it is like some linear combination of “molecules” or “atoms” - hopefully someone who’s seen some modern-era Hardy space theory can correct or expand on this bit? I don’t know about Lip(X)* though.
- I seem to remember that the forgetful functor from (Banach spaces, linear maps of norm $\leq 1$) to (metric spaces, distance-non-increasing maps) has a left adjoint. I first saw this in a talk of Gilles Godefroy on some work he did with Nigel Kalton on “Lipschitz-free spaces” - although he didn’t once use the words functor or adjunction if I recall correctly.
Posted by: Yemon Choi on April 23, 2010 10:22 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Let me make some general comments as to why Lipschitz functions might be appropriate in this context, or at least let me speculate on why Tom might think so.
The main point is that the idea of magnitude of metric spaces arose from thinking of metric space as enriched categories (following Lawvere). In that situation, the obvious notion of function (though probably not the only one) is that of Lipschitz function. Lipschitz functions play the role of ‘enriched presheaves’. I’ll try to explain what kind of properties can be deduced from that point of view.
Firstly, I should say what I assume you mean by Lipschitz functions, or rather what I would like you to mean. I will use $[0,\infty]$ to denote the extended non-negative reals $\mathbb{R}_{\ge 0}\cup\infty$. In the present context I would think of a Lipschitz function as being a function $h: X\to [0,\infty]$ such that $d(x,y)\ge h(x) - h(y)$. A good class of examples, indeed a key class of examples, of such functions is given in the following way. Take any $y\in X$ and define the function $h_y:X\to [0,\infty]$ by $h_y(x)\coloneqq d(x,y)$. The triangle inequality ensures that this satisfies the Lipschitz condition. More generally, for any subset $Y\subset X$ we can define $h_Y:X\to [0,\infty]$ by $h_Y(x)\coloneqq inf_{y\in Y} d(x,y)$. I’ll come back to the relevance of these in a second, but the sharp-eyed category theorists should be shouting “Yoneda!” at their screen by now.
Now this wasn’t quite what you were talking about. You were talking about functions like $f_y:X\to [0,1]$ where $f_y=e^{h_y}$, so I would expect that your functions $f:X\to [0,1]$ will satisfy a ‘multiplicative Lipschitz condition’ $f(x)/f(y)\ge e^{-d(x,y)}$ for all $x$ and $y$. So you might want to translate what I’m going to say over to your kind of functions.
Now $Lip(X)$ the set of Lipschitz functions (in my sense) carries a ‘generalized metric’ defined on function $h'$ and $h''$ as follows:
$d(h', h'')\coloneqq sup_{x\in X} [h''(x)- h'(x)]$
where my ad-hoc notation $[a]\coloneqq max(0, a)$ is the ‘non-negative part’. This generalized metric is not symmetric, might take the value $\infty$, and $d(h',h'')=0$ does not imply that $h'=h''$; however it does satisfy the triangle equality and $d(h,h)=0$.
The key point about the above generalized metric is that the function
$X\to Lip(X); \quad y\to h_y$
is an isometric embedding – it’s a nice exercise to prove that. [For the cognoscetti this is an enriched version of the Yoneda embedding, $Lip(X)$ is the $[0,\infty]$-category of $[0,\infty]$-presheaves on $X$.] As John B keeps telling us in a more general setting, we should think of $Lip(X)$ as some kind of cocompletion of $X$. (Okay, yes, when people bandy about words like cocompletion it causes some glazed expression to pass across my eyes.)
That the set of Lipschitz functions $Lip(X)$ is a cocompletion means that it will have all colimits (whatever that means), in particular it will have coproducts, so we will be able to add things together. In $X$ we can’t add two points together, but we can in $Lip(X)$: it makes sense in many circumstances to take the distance from a point $x$ to the union of two points $y$ and $y'$ to be the minimum of distances from $x$ to $y$ and from $x$ to $y'$. In this way we can think of the union of two points as being a Lipschitz function.
This leads to thinking of any subset of points in $X$ as giving an element of $Lip(X)$ as we did above. For $Y\subset X$ define the function $h_Y\in Lip(X)$ by $h_Y(x)\coloneqq inf_{y\in Y} d(x,y)$.
Things get even nicer when we restrict to $Comp(X)$, the set of compact subsets of $X$. Firstly, if $Y$ is compact then we can recover $Y$ from the function $h_Y$ by just taking the zero set of the function. Moreover, we can define a generalized metric on $Comp(X)$ by saying $d(Y',Y'')=\epsilon$ if $\epsilon$ is the least number such that $Y'$ is contained in a closed $\epsilon$-neighbourhood of $Y''$. In particular $d(Y',Y'')=0$ if and only if $Y'\subset Y''$. We obtain the Haussdorf metric (a genuine metric) $D$ on $Comp(X)$ by symmetrizing $d$ in the following way:
$D(Y',Y'')\coloneqq max\bigl( d(Y',Y''),d(Y'',Y')\bigr).$
Although I guess the Haussdorf metric has less information in it. None-the-less, we get an isometric embedding of generalized metric spaces:
$(Comp(X),d)\to (Lip(X),d);\quad Y\to h_Y.$
This can be made into an isometric embedding of genuine metric spaces by using the Haussdorf metric on the set of compact subsets and similarly symmetrizing the generalized metric on the set of Lipschitz functions.
So $Lip(X)$ contains not only the points of $X$ but also the unions of points of $X$. I guess there’s lots more stuff in $Lip(X)$, but I’m not imaginative enough at this point in time to think of anything else. What other colimits would be in there other than coproducts?
Anyway, this is all very natural from the point of view of enriched category theory and everything I’ve said has analogues in other enriched situations. As magnitude of metric spaces arose by thinking of metric spaces as being enriched categories, it makes sense to hope/think/guess that these sorts of functions are the right ones to consider in this context.
Posted by: Simon Willerton on April 24, 2010 3:38 AM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Thanks for the comments, Simon (and thanks also to Yemon - I’ve only recently started learning about $Lip(X)$ and $lip(X)$ and all that, so pointers to the literature are good to have [1]).
To me your comments actually support working with $C(X)$ instead of $Lip(X)$. Let me explain why. First of all, I’m going to work only with symmetric metrics, not because I disagree with Lawvere but because I don’t know how much classical stuff I know will still work without symmetry [2]. For me, a function $f:X\to \mathbb{R}$ is $L$-Lipschitz for $L \ge 0$ if $sup_{x,y\in X}|f(x)-f(y)| \le L$, so what you call Lipschitz is what I call $1$-Lipschitz; $f$ is said to be Lipschitz if it is $L$-Lipschitz for some (finite) $L$. The functions $f_y(x)=e^{-d(x,y)}$ are $1$-Lipschitz, since $t \mapsto e^{-t}$ is $1$-Lipschitz on $[0,\infty)$.
Since I hope to use Banach space methods here, for me $Lip(X)$ is the set of all Lipschitz functions $f:X\to \R$, since $1$-Lipschitz functions alone don’t form a vector space. Next I need a norm. The symmetric analogue to the distance you put on $1$-Lipschitz functions is the sup norm: $\|f\|_\infty = \sup_{x\in X} |f(x)|$. Now $Lip(X)$ is not complete in this norm. It’s a dense subspace of $C(X)$, and so $(Lip(X), \| \cdot \|_\infty)^* = M(X)$. So your comments seem to support what I meant by “working with $C(X)$”. By “working with $Lip(X)$” I mean instead equipping $Lip(X)$ with a different norm (the bounded Lipschitz norm) which makes it a Banach space, whose dual (and in fact predual) contain $M(X)$ as a proper subspace.
[1] One thing I will point out related to Yemon’s comments is that finitely supported signed measures are norm-dense in the predual of $Lip(X)$ (with the bounded Lipschitz norm). This is one thing I think could be useful in this context. Note that finitely supported signed measures are not norm-dense in $M(X)$, although they are dense in the weak-* topology.
[2] In case you ever meet a skeptical analyst or geometer who might think Lawvere lacks the metric-space-credibility to advocate dropping such a fundamental axiom, you can tell them that Gromov has also argued against the symmetry axiom as too restrictive for applications.
Posted by: Mark Meckes on April 24, 2010 6:48 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
To answer your parenthetical question about typesetting. In itex (which is what is used here) consecutive, abutting Roman letters in mathmode are interpreted as being a word, so are typeset upright.
• $L i p$ gives $L i p$
• $Lip$ gives $Lip$
Posted by: Simon Willerton on April 24, 2010 4:00 AM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
On page 3, you write “It does seem that the subspace distance metric and the intrinsic metric are infinitessimally the same. I would hope that the asymptotics of the magnitude only depend on metric infinitessimally.” I though I found a counter-example to this here (and you did as well, I think?)
I would not say that I hoped something was true if I knew a counter-example.
Also, on page 6, “we will note distinguish” should read “we will not distinguish”.
Posted by: David Speyer on April 22, 2010 3:37 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
I’m glad you’re there to keep me on my toes David.
I guess I had in my memory that there was some problem with your calculation for the subspace metric on $S^2$ (I don’t think I could quite figure it out or something, I don’t remember). But I hadn’t got round to going back to look at it. Anyway, I just sat down and calculated the thing on the nose today. It’s very nice. As you said, the thing to calculate is
$2\pi\int^\pi_{\theta=0} e^{-R 2\sin(\theta/2)} \sin(\theta) d \theta.$
The cunning observation is that
$\sin(\theta) d \theta = 2\sin(\theta/2)\cos(\theta/2)d\theta =2\sin(\theta/2)d(2\sin(\theta/2)).$
So substituting $s=2\sin(\theta/2)$ we get that the integral equals $2\pi\int_{s=0}^2 e^{-R s} s d s$ which evaluates to
$\frac{2\pi(1-e^{-2R}(2R+1))}{R^2}.$
The magnitude of the radius $R$ sphere with the subspace metric is the volume of $S^2$ divided by that integral, whence
$\left|S^2_{R,sub}\right| =\frac{2R^2}{1-e^{-2R}(2R+1)}= 2R^2+exponentially decaying terms.$
However, for the intrinsic metric you we have the magnitude is of the form $2R^2+2 +O(R^{-2})$. So asymptotically we get different answers for the magnitude.
You can do a similar calculation for higher dimensional spheres.
I’m sure I only put those questions there about the subspace metric so that you would goad me into actually doing the calculations!
This then leads to the question: in what sense are the subspace metric and the intrinsic metric infinitesimally the same? I still haven’t got my head round this.
Posted by: Simon Willerton on April 23, 2010 7:10 AM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
in what sense are the subspace metric and the intrinsic metric infinitesimally the same?
One sense in which this is true is that an $\varepsilon$-ball w.r.t. the intrinsic metric is the same as a $2R \sin (\varepsilon / 2R)$-ball w.r.t. the subspace metric, and $2R\sin (\varepsilon / 2R) \approx \varepsilon$ as $\varepsilon \to 0$.
Posted by: Mark Meckes on April 23, 2010 12:34 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
You say you haven’t calculated the magnitudes of spheres with the subspace metric yet. If you haven’t even checked for $S^1$ yet, I just did and it’s easy to see (the hardest part is using a trig identity) that it matches the expression derived in your paper with Tom.
Also, a typo in Theorem 1: $\mu$ is an invariant measure.
Posted by: Mark Meckes on April 22, 2010 3:43 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Oh, I see the calculation for $S^1$ is also in the discussion David just linked to.
Posted by: Mark Meckes on April 22, 2010 3:48 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Mark said
You say you haven’t calculated the magnitudes of spheres with the subspace metric yet. If you haven’t even checked for $S^1$ yet, I just did and it’s easy to see (the hardest part is using a trig identity) that it matches the expression derived in your paper with Tom.
I’m not sure which bit you mean you’ve checked!
I should say certainly say a little more in the paper about $S^1$, as we do know a reasonable amount about it. Of course what I meant to say was “I haven’t calculated it for the general $n$-sphere, but we know a fair bit about the circle, though I don’t really want to generalize from that.” However, that isn’t what I said. It was one of the bits of the paper where I needed to fill in the blank before finishing it and I rushed it. Thanks.
In the paper with Tom, it was shown that that the magnitude of a circle has the expected asymptotic behaviour – namely half the length – when it is thought of a subspace of any constant curvature surface. In particular this covers both the Euclidean metric and the intrinsic metric. The non-asymptotic magnitude will in general be different metrics on the circle.
Posted by: Simon Willerton on April 22, 2010 7:05 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
What I checked is that the magnitude of a circle of arbitary fixed circumference, as a subspace of $\mathbb{R}^2$, has the same magnitude whether defined as in the present paper, or as in section 4.1 of your paper with Tom.
Actually I think this has to be the case from measure-theoretic “first principles”, but there are some fiddly functional-analytic issues I want to think about more before I say I’m sure of that.
Posted by: Mark Meckes on April 22, 2010 7:14 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
I tried to post another comment on this earlier but it vanished. In fact the reason they have to be the same appeared in your paper with Tom: the sums for the finite approximations of the circle are just Riemann sums for the integral corresponding to the whole circle.
To make it sound fancier, the measures $\mu_n$, which give mass $1/n$ to each of $n$ equally spaced points on a circle, converge in the weak-* topology of $M(S^1)$ (induced by the action as the dual of $C(S^1)$) to the invariant probability measure on $S^1$.
Posted by: Mark Meckes on April 23, 2010 12:58 AM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
It seems I should have waited a while to collect my comments before posting, but it’s too late for that now.
Regarding Theorem 1 (Speyer’s Homogeneous Magnitude Theorem), it’s worth pointing out that if $X$ is a compact homogeneous metric space then (up to normalization) there exists a unique invariant measure $\mu$ on $X$.
Posted by: Mark Meckes on April 22, 2010 4:30 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Furthermore this is a positive measure, so $\int_{x\in X} e^{-d(x,y)} d\mu$ is automatically nonzero and so the magnitude of $X$ is indeed defined.
Folklore has it that a nonzero invariant measure on a noncompact space is (typically at least) infinite. Does anyone know an actual theorem to that effect?
Posted by: Mark Meckes on April 22, 2010 7:04 PM | Permalink | Reply to this
### Re: On the Magnitude of Spheres, Surfaces and Other Homogeneous Spaces
Is http://arxiv.org/abs/math-ph/0210033 “Volumes of Compact Manifolds” of some use, here?
Posted by: Alejandro Rivero on April 27, 2010 1:57 AM | Permalink | Reply to this
Read the post Magnitude of Metric Spaces: A Roundup
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Post a New Comment | 11,531 | 44,033 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 279, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-49 | latest | en | 0.943456 |
https://homeworkcorp.com/the-federal-pell-grant-program-gives-grants-to-low-income-undergraduate-students-according-to-the-national-postsecondary-student-aid-study-conducted-2/ | 1,611,091,858,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519784.35/warc/CC-MAIN-20210119201033-20210119231033-00314.warc.gz | 398,819,196 | 8,192 | The Federal Pell Grant Program gives grants to low-income undergraduate students. According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, the average Pell grant award for 2007-2008 was \$2,600. We wonder if the mean amount is different this year for Pell grant recipients at San Jose State University. Suppose that we randomly select 50 Pell grant recipients from San Jose State University. For these 50 students, the mean Pell grant award is \$2,450 with a standard deviation of \$600.
Let μ
= the mean amount of Pell grant awards received by San Jose State University Pell grant recipients this year. We test the following hypotheses.
• H
• 0
• : μ
• =
• 2600
• H
• a
• : μ
• 2600
The sample size is greater than 30, so a t-model is a good fit for the sampling distribution.
1. What is the t-test statistic? If necessary, round to two decimal places.
2. What is the P-value? If necessary, round to three decimal places.
Suppose we’re conducting a t-test for a population mean with the following hypotheses.
• H
• 0
• : μ
• =
• 50
• H
• a
• : μ
• 50
A sample of size 30 has a t-statistic of −
2.43
1. . Find the corresponding P-value. Do not round. | 322 | 1,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-04 | latest | en | 0.886312 |
https://datasciencewiki.net/k-means/ | 1,726,338,135,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00617.warc.gz | 176,895,437 | 10,564 | # K-Means
## K-Means :
K-means is a popular clustering algorithm that groups similar data points together into clusters. It is an iterative process that involves dividing a dataset into a specified number of clusters (K) and then assigning each data point to the cluster with the closest mean.
For example, let’s say we have a dataset of five 2D points, represented by the coordinates (x,y): (1,2), (3,4), (5,6), (7,8), and (9,10). If we want to group these points into two clusters, we can use the K-means algorithm to find the cluster centers and assign each point to the closest cluster.
First, we need to randomly select two points as the initial cluster centers. Let’s say we choose (1,2) and (9,10) as the initial cluster centers. We then calculate the distance between each point and the two cluster centers using the Euclidean distance formula. This will give us the following distances:
Point (1,2): distance to (1,2) = 0, distance to (9,10) = 13.4
Point (3,4): distance to (1,2) = 2.8, distance to (9,10) = 10.6
Point (5,6): distance to (1,2) = 5.7, distance to (9,10) = 8.6
Point (7,8): distance to (1,2) = 8.5, distance to (9,10) = 6.4
Point (9,10): distance to (1,2) = 13.4, distance to (9,10) = 0
Since the distances are different, each point will be assigned to the cluster with the closest center. In this case, (1,2) and (3,4) will be assigned to the first cluster, while (5,6), (7,8), and (9,10) will be assigned to the second cluster.
Once the initial assignment is made, the algorithm will then update the cluster centers by calculating the mean of all the points assigned to each cluster. In this example, the first cluster center will be updated to (2,3) and the second cluster center will be updated to (7,8).
The K-means algorithm then repeats this process of assignment and update until the cluster centers converge, meaning they no longer change with each iteration. This can be determined using a stopping criterion, such as a maximum number of iterations or a threshold for the change in the cluster centers.
Once the algorithm has converged, the final cluster centers and assignments will be the result of the K-means algorithm. In our example, the final cluster centers would be (2,3) and (7,8) and the final assignments would be (1,2) and (3,4) in the first cluster, and (5,6), (7,8), and (9,10) in the second cluster.
Another example of K-means clustering can be seen in the context of customer segmentation. Imagine a retail company with a large customer base that wants to group its customers into different segments based on their purchasing behavior. The company collects data on the products each customer has purchased, the amount they have spent, and the frequency of their purchases.
Using the K-means algorithm, the company can group its customers into different segments, such as high-value customers, frequent shoppers, and occasional buyers. The algorithm will use the data on customer purchases to find the cluster centers for each segment and assign each customer to the appropriate cluster.
The K-means algorithm is a simple and effective way to group similar data points together into clusters. It can be applied to a wide range of problems, from clustering 2D points to customer segmentation. By iteratively updating the cluster centers and assignments, K-means can quickly and accurately group data into meaningful clusters. | 823 | 3,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-38 | latest | en | 0.884584 |
http://townhouserehab.com/about-time-jgg/bond-order-of-be2-plus-7df1aa | 1,627,930,948,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154356.39/warc/CC-MAIN-20210802172339-20210802202339-00670.warc.gz | 44,395,375 | 9,943 | Answer. From the electronic configuration it is clear that there is no singly filled atomic orbital present in beryllium. This problem has been solved! Assertion: H2, Li2 and B2 each has a bond order of 1 and hence are equally stable. Question. similar to Be 2 2+ just remove 2 electrons from 2 s sigma anti bonding and gives bond order of 1 diamagnetic. 10 0. Differentiate between bonding and antibonding molecular orbitals. ... We're trying to predict whether we expect brilliant two plus and brilliant to minus to be stable and if we … Want to see the step-by-step answer? Since both have a bond order of $+\frac{1}{2},$ both $\mathrm{Be}_{2}^{+}$ and $\mathrm{Be}_{2}^{-}$ will exist in gas phase. Ask Study Doubts; Sample Papers; Past Year Papers; Textbook Solutions; Sign Up. Bond Order. Bond order 3 and isoelectronics 2) Bond order 2 and isoelectronics 3) Bond order 2, 5 and isoelectronics 4) Bond order1.5 and isoelectronics 38. Trending questions. Verify … From the electronic configuration it is clear that there is no singly filled atomic orbital present in beryllium. 10 valence electrons.b. What is the bond order of Li2−? For this, we need to do the following steps: Step 1: Calculate the total number of valence electrons present. Graphical Presentation of Bond-Order. asked Jun 8, 2018 in Chemistry by Nisa (59.6k points) chemical bonding and molecular structure; class-11; 0 votes. Darren Herweyer. for each electron in a bonding MO, it adds #0.5# to the bond order, because more bonding character strengthens the bond...; for each electron in an antibonding MO, it subtracts #0.5# from the bond order, because more … Topics. Enter your answer as a decimal. Use the molecular orbital diagram shown to determine which of the following is most stable.a. 4 in all, have to be accommodated in various molecular orbitals in the increasing order of their energies..The molecular orbital electronic … The bond order can be calculated in a simple manner. From the above MO diagram we can see that number of elctrons in the bonding and antibonding orbital is same and hence Be does not form Be2 molecule (for. Bond order is an estimate of the total number of chemical bonds present in a compound. The valence shell of each beryllium atom is 2s2 so there are a total of four valence shell electrons for which we Furthermore, does be2 2+ exist? A bond order greater than zero means that more electrons occupy bonding MOs (stable) than antibonding MOs(unstable). You can specify conditions of storing and accessing cookies in your browser. Favourite answer. 6 years ago. II. Performance & security by Cloudflare, Please complete the security check to access. We’re being asked to determine the bond order of He 2 +. What is the bond order of Be2 −? Also asked, does be2 2+ exist? 13 valence electrons.d. Which of the following gases will have the lowest rate of diffusion? Bond number gives an … Discuss the consequences of high enthalpy of H-H bond in terms of chemical reactivity of dihydrogen. 12 valence electrons.c. Previous Question Next Question. ONE PLZ..............plz solve this question in ATTACHMENT, public ans fast.......my questions r collecting....in line..,, Calculate the percentage of monochloro isomers producs during the chlorination of tertiary butane, ohh u like to eat skull's???? Is Li2− Paramagnetic Or Diamagnetic? Is n2 stable or unstable? Please enable Cookies and reload the page. How are the shapes of molecular orbitals determined? $Bond \ Order = \frac{2 (bonding\ electrons)-2(anti-bonding\ e-)}{2} = 0$ However, removing an electron from the antibonding level produces the molecule He 2 +, which is stable in the gas phase with a bond … Do you expect these molecules to exist in the. Add your answer and earn points. Question: What Is The Bond Order Of Li2−? How are the shapes of molecular orbitals determined? The bond number itself is the number of electron pairs (bonds) between a pair of atoms. Without the half filled orbital,the overlapping is not possible ,therefore Be2 molecule does not exist. So we could have B2. See the answer. It is important because it tells how stable a bond is. Answer to Draw an MO energy diagram and predict the bond order of Be2+ and Be2−. 0 15. Draw an MO energy diagram and predict the bond order of Be2+ and Be2-. (a) H2 (b) N2 (c) F2 (d) O2. Calculation of bond order. Bond order also indicates the stability of the bond. Even rather simple molecular orbital (MO) theory can be used to predict which we start reading from the bottom of the diagram because this is how MO diagrams are constructed, Diberyllium, Be2, has a bond order of zero and is unknown. Join. Out of H and H 2, which has higher first ionisation enthalpy? If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. 0.5. Is Li2− paramagnetic or diamagnetic? Kindly Sign up for a personalised experience. 1 decade ago. Find an answer to your question compare bond order of Be2, Be2+ dhyeysonani dhyeysonani 2 hours ago Chemistry Secondary School Compare bond order of Be2, Be2+ 1 See answer dhyeysonani is waiting for your help. Do you expect these molecules to exist in the. The nature of the chemical bond in Be2+, Be2, Be2−, and Be3 Apostolos Kalemos Citation: The Journal of Chemical Physics 145, 214302 (2016); doi: 10.1063/1.4967819 ... BeH, BeHe, BeLi, and BeB in order to isolate that “binding force” in Be containing molecules. What is the relationship between bond order and the dissociation energy of a molecule? Q. Answer to Draw an MO energy diagram and predict the bond order of Be2+ and Be2−. This problem has been solved! F22+b. Cloudflare Ray ID: 609f8c896d1f051f Enter Your Answer As A Decimal. is it 1???? Just take electrons that are in each MO, and . Bond order of Be2 is 1) 0 2) 1 3) 2 4) 3 18. Your IP: 64.34.219.21 Differentiate between bonding and antibonding molecular orbitals. Step 2: Draw the molecular orbital diagram. Draw an MO energy diagram and predict the bond order of Be2 + and Be2 -. Which of the following species has the shortest bond length? Without the half filled orbital,the overlapping is not possible ,therefore Be2 molecule does not exist. Update: no i mean a Li^2+ charge... Answer Save. THE Be2 SAGA Probably the very first ab initio work is the RHF calcula-tion by Fraga and Ransil in 1962.18 The ground X1 + g state is repulsive but … 14 valence electrons. Check out a sample Q&A here. Q. Answer and Explanation: The bond order for Be2 is 0 (zero). A bond order of zero is obtained by placing the available electrons in the bonding and antibonding levels, indicating that dihelium does not exist according to valence bond and bond order theory. All chemical compounds are made up of bonds. This site is using cookies under cookie policy. 1 … Relevance. What Is The Bond Order Of Be2+? Do you expect these molecules to exist in the gas phase? Expert Answer . Still have questions? Kindly Sign up for a personalised experience. What is the bond order of Be 2 +? asked Oct 5 in Gaseous State by Manish01 (47.4k points) gaseous state; class-11; 0 votes. The graphical representation presented in Fig. Is Li2− Paramagnetic Or Diamagnetic? krishna56789 krishna56789 Answer: Be=0. Verify … Recall that the formula for bond order is: See Answer. (i) Be2 molecule: The electronic configuration of Be(Z = 4) is:4 Be 1s2 2s1Be2 molecule is formed by the overlap of atomic orbitals of both beryllium atoms.Number of valence electrons in Be atom = 2Thus in the formation of Be2 molecule, two outer electrons of each Be atom i.e. Lv 4. In this video lecture, MOT has been applied to Beryllium, Boron, Nitrogen, and Oxygen. … The atomic number of Be is 4. Q:- If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. the theory, molecular orbitals … Bond order is the number of chemical bonds between a pair of atoms and indicates the stability of a bond. Determine the bond order in a molecule or ion with:a. Draw an MO energy diagram and predict the bond order of Be2 + and Be2 -. Popular Questions of Class Chemistry. Q:-Density of a gas is found to be 5.46 g/dm 3 at 27 °C at 2 bar pressure. Do you expect these molecules to exist in the gas phase? Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. what is the bond order of Li2+ ion? If you mean Li2 with a +1 charge, YES. 1 answer. • • What Is The Bond Order Of Li2−? Elemental nitrogen is commercially obtained by the distillation of liquid air because oxygen boils at … Be2 molecule has MO electronic configuration K K (σs)2(σ∗2s)2. Bond order and length are inversely proportional to … 2 Answers. For example, in diatomic nitrogen, N≡N, the bond order is 3; in acetylene, H−C≡C−H, the carbon-carbon bond order is also 3, and the C−H bond order is 1. What is the relationship between bond order and the dissociation energy of a molecule? Ask Study Doubts; Sample Papers; Past Year Papers; Textbook Solutions; Sign Up. check_circle Expert Answer. See the answer. Robert. Click hereto get an answer to your question ️ Bond order of Be2 is: The higher the bond order, the more electrons holding the atoms together, and therefore the greater the stability. Ne22+c. Boron doesn't actually form ions in solution so I doubt it's that and in order to have a bond order, there has to be a bond. We we draw the molecular orbital diagram for neutral di-beryllium we find a bond order of zero since there are four bonding electrons and four anti-bonding electrons. Bonds order means the number of bonds that exist in two bonded atoms in a molecule. Enter Your Answer As A Decimal. dhyeysonani is waiting for your help. Bond order, as introduced by Linus Pauling, is defined as the difference between the number of bonds and anti-bonds. Q. 1 shows that bond-order gradually increases to 1 in the range (0-2) electrons then falls to zero in the range (2-4) electrons then it further rises to 1 for (4-6) electrons and once again falls to zero for (6-8) electrons then again rises to 3 in the range (8-14) electrons and then finally falls to zero for (14-20) electrons. Here we consider the molecular orbital diagram (MO) of #"Li"_2#:. Bond order and bond length indicate the type and strength of covalent bonds between atoms. ?#Real Dpif yes ur welcomed....., any technique to study types of unit cell. Do you mean a molecule of B with a subscript of 2 or do you mean an a single boron ion with a charge of -2? Question: What Is The Bond Order Of Be2+? Join Yahoo Answers and get 100 points today. Do you expect these molecules to exist in the gas phase? F2. F22-d. O22+e. What will be its density at STP? As a result of the stability of the N 2 molecule, many nitrogen compounds are unstable, some explosively so. For example, in diatomic nitrogen N≡N the bond number is 3, in ethyne H−C≡C−H the bond number between the two carbon atoms is also 3, and the C−H bond order is 1. The bond order is 22−2 = 0. If so the bond order is 1. Ask question + 100. Bond order= [No of bonding electrons]- [No of anti bonding electrons]/ 2 B.O= 2–2/2 B.O= 0/2 = 0 The bond order for He2 is 0. Add your answer and earn points. Out of H and H 2, which has higher first ionisation enthalpy? See drawing. Step 3: Calculate the bond order of the molecule/ion. Get answers by asking now. In each MO, and therefore the greater the stability of the N molecule..., is defined as the difference between the number of electron pairs bonds! ( a ) H2 ( b ) N2 ( c ) F2 ( )! Cloudflare, Please complete the security check to access it is clear that is... 59.6K points bond order of be2 plus chemical bonding and molecular structure ; class-11 ; 0 votes of valence electrons present 2 bar.! Answer and Explanation: the bond order of He 2 + greater the stability the! 2018 in Chemistry by Nisa ( 59.6k points ) Gaseous State ; ;! No singly filled atomic orbital present in a compound d ) O2 to … What is the order. Web property order can be calculated in a molecule to access diagram shown determine. Determine which of the stability: 609f8c896d1f051f • Your IP: 64.34.219.21 • Performance & by. Half filled orbital, the overlapping is not possible, therefore Be2 has! K K ( σs ) 2 has the shortest bond length indicate the type strength. Class-11 ; 0 votes these molecules to exist in the Dpif YES ur welcomed..... , any technique Study. Dpif YES ur welcomed..... , any technique to Study types of unit cell is important it! By Nisa ( 59.6k points ) Gaseous State ; class-11 ; 0.... Simple manner bonded atoms in a simple manner molecule or ion with a. Bar pressure filled atomic orbital present in a simple manner the overlapping is not possible, therefore molecule. Order can be calculated in a simple manner estimate of the bond order and the dissociation energy a! The following is most stable.a ’ re being asked to determine the bond order bond. Bond in terms of chemical bonds present in beryllium conditions of storing and cookies! Mo energy diagram and predict the bond order greater than zero means more... , any technique to Study types of unit cell of electron (! Are unstable, some explosively so points ) chemical bonding and molecular structure ; class-11 0... By Manish01 ( 47.4k points ) Gaseous State by Manish01 ( 47.4k points ) chemical bonding and molecular structure class-11! • Your IP: 64.34.219.21 • Performance & security by cloudflare, Please complete the security check access. To Study types of unit cell type and strength of covalent bonds between atoms occupy bonding MOs ( )... Orbital diagram shown to determine which of the N 2 molecule, many compounds... Is defined as the difference between the number of electron pairs ( bonds ) between pair! Difference between the number of electron pairs ( bonds ) between a of... A Li^2+ charge... answer Save: - Draw an MO energy diagram and the! To exist in the gas phase is most stable.a gives an … answer to Draw an MO energy and... Because it tells bond order of be2 plus stable a bond is 5 in Gaseous State ; class-11 ; 0 votes you a! Molecules to exist in the has MO electronic configuration it is clear that there is singly... In beryllium bond is atomic orbital present in beryllium Year Papers ; Textbook Solutions ; Sign Up overlapping. Take electrons that are in each MO, and K K ( σs ) 2: a 5 Gaseous!: 609f8c896d1f051f • Your IP: 64.34.219.21 • Performance & security by cloudflare, Please complete the security to. ( stable ) than antibonding MOs ( stable ) than antibonding MOs ( unstable ) bonds that exist in gas! A pair of atoms without the half filled orbital, the overlapping not., therefore Be2 molecule has MO electronic configuration bond order of be2 plus is important because it tells how stable a is. Sign Up as a result of the stability of the stability of the total number of valence electrons...., the overlapping is not possible, therefore Be2 molecule does not exist ( b ) N2 ( c F2... Higher the bond order and the dissociation energy of a molecule in the gas phase,. Class-11 ; 0 votes order and the dissociation energy of a molecule without the bond order of be2 plus filled orbital the., any technique to Study types of unit cell °C at 2 bar.... And accessing cookies in Your browser is an estimate of the following is most stable.a bond is any... In a molecule or ion with: a rate of diffusion can specify conditions of and. To … What is the bond order of Be2+ and length are inversely proportional …. Together, and therefore the greater the stability of the N 2,... And therefore the greater the stability of the stability, as introduced Linus. Molecules to exist in the gas phase, 2018 in Chemistry by Nisa ( 59.6k points ) chemical and... Of a molecule or ion with: a can be calculated in a molecule or ion with a... ( a ) H2 ( b ) N2 ( c ) F2 ( d ) O2 stability. The total number of valence electrons present order is an estimate of the following gases have. ) F2 ( d ) O2: 609f8c896d1f051f • Your IP: 64.34.219.21 • &. 0 votes to do the following gases will have the lowest rate of diffusion a +1,! Be 5.46 g/dm 3 at 27 °C at 2 bar pressure gas is found to be 5.46 g/dm 3 27! Filled orbital, the overlapping is not possible, therefore Be2 molecule does exist... H 2, which has higher first ionisation enthalpy technique to Study types of unit cell a! A bond is K ( σs ) 2 ( σ∗2s ) 2 ( σ∗2s ) 2 2, which higher. As a result of the bond order, the overlapping is not possible, therefore molecule... The difference between the number of electron pairs ( bonds ) between a pair of atoms and structure! Question: What is the bond order of Li2− and Be2 - the of... Take electrons that are in each MO, and order and the dissociation energy of a molecule or with... And Be2− bond order of Be2 + and Be2 bond order of be2 plus nitrogen compounds are unstable, explosively! It is important because it tells how stable a bond order, as introduced by Linus Pauling is. Electrons that are in each MO, and the electronic configuration K (. Can be calculated in a simple manner means the number of bonds exist! To … What is the relationship between bond order and bond length indicate the type strength! By Nisa ( 59.6k points ) Gaseous State ; class-11 ; 0.. Question: What is the bond order of Be2+ and Be2- Chemistry by (. Has the shortest bond length indicate the type and strength of covalent bonds between.... Bond length indicate the type and strength of covalent bonds between atoms you temporary access the... Step 1: bond order of be2 plus the total number of chemical bonds present in.. Number of electron pairs ( bonds ) between a pair of atoms introduced by Linus Pauling, is as. Answer and Explanation: the bond order in a molecule Be2+ and Be2- is important because it tells how a! Higher the bond order in a molecule the molecule/ion , any technique to Study types unit. Most stable.a: Calculate the total number of chemical bonds present in beryllium and length are inversely proportional to What... The electronic configuration K K ( σs ) 2 order for Be2 is (! ( a ) H2 ( b ) N2 ( c ) F2 ( d bond order of be2 plus O2 to 5.46! Be 2 + Performance & security by bond order of be2 plus, Please complete the check... Result of the bond number gives an … answer to Draw an energy! D ) O2 enthalpy of H-H bond in terms of chemical reactivity of dihydrogen of?! Is an estimate of the stability • Performance & security by cloudflare, complete. Nisa ( 59.6k points ) Gaseous State ; class-11 ; 0 votes +1,... Do the following gases will have the lowest rate of diffusion without the half filled orbital the... Stability of the bond order of Be2+ and Be2− a simple manner possible, Be2... … What is the number of bonds that exist in the simple manner two bonded atoms in molecule. Difference between the number of bonds that exist in the gas phase )... Sample Papers ; Textbook Solutions ; Sign Up will have the lowest rate of diffusion ; Sample Papers ; Year. Be2+ and Be2− following is most stable.a exist in the you are a and. Order can be calculated in a compound clear that there is no filled... With: a zero ), we need to do the following species has the shortest bond?.: no i mean a Li^2+ charge... answer Save and strength of covalent bonds between atoms use molecular. Cloudflare, Please complete the security check to access human and gives you temporary access to the web property to! Is found to be 5.46 g/dm 3 at 27 °C at 2 bar pressure zero ) strength covalent! Strength of covalent bonds between atoms a simple manner this, we need to do the following species has shortest. 1: Calculate the bond order of Be2 + and Be2 - bonds present in molecule! Exist in the an … answer to Draw an MO energy diagram and predict the bond order greater than means. In Your browser you can specify conditions of storing and accessing cookies in browser... Be2 molecule does not exist σ∗2s ) 2 is most stable.a have the lowest rate of diffusion,. | 5,013 | 19,998 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-31 | latest | en | 0.865676 |
http://www.expertsmind.com/questions/what-is-the-electric-potential-30180621.aspx | 1,670,144,883,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00785.warc.gz | 61,570,753 | 12,578 | ## What is the electric potential, Physics
Assignment Help:
Suppose you have two parallel, metal plates that have an electric field between them of strength 3.0 x 104 N/C, and are 0.050 m apart.
Consider a point, P, located 0.030 m from plate A, the negatively charged plate, when answering the following question.
a. What is the electric potential at P relative to plate A?
#### Evaluate the skier''s height h, A water skier lets go of the tow rope upon ...
A water skier lets go of the tow rope upon leaving the end oaf jump ramp at a speed of 14.0 m/s. The skier has a speed of 13.0m/s at the highest point of the jump. Ignoring air res
#### Magnetism, Differentiate between teporary and permanent magnet
Differentiate between teporary and permanent magnet
#### Evaluate intensity in youngs double slit experiment, In a young's double sl...
In a young's double slit experiment a point on the screen is similar distance from both the slits if any one of the slit is closed the intensity at P due to the other slit is 0.02
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Illustrate the principle of carbon dioxide laser in short. Principle of carbon dioxide (CO 2 ) Laser: To get Laser transition between two levels this is necessary to acquire
#### Maxwells electromagnetic equation, Starting from Maxwell's electromagnetic ...
Starting from Maxwell's electromagnetic equation in free space, Get Poynting theorem.
#### Gravity, what is gravity caused by and what is its exact pull
what is gravity caused by and what is its exact pull
#### Find the x-component of the electric field, Presume that you are given an e...
Presume that you are given an expression for the electric potential V(x,y) as a function of x and y valid for points in the first quadrant of a Cartesian coordinate system and aske
calculate
#### Motion in a straight line, is average speed the magnitude of average veloci...
is average speed the magnitude of average velocity?justify your answer with a suitable example?
#### Determine the magnitude, Determine the Magnitude The cantilevered beam...
Determine the Magnitude The cantilevered beam shown below has two force couples acting on it as shown. The value of C1 in Newtons is given in P25. Determine the magnitude of
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https://brane-space.blogspot.com/2023/08/ | 1,713,381,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817171.53/warc/CC-MAIN-20240417173445-20240417203445-00686.warc.gz | 135,816,423 | 36,081 | ## Wednesday, August 30, 2023
### Solutions To Clairault DE Problems Revisited
1) Solve: p2 x - y = 0
Solution:
Solve for y:
y = p2 x
Differentiate:
dy/dx = p = p2 + 2px (dp/dx)
Re-arrange:
2 dp/(p -1) + dx/ x = 0
Solve:
(p - 1) 2 x = c
Eliminate p between previous two eqns.
y - 2Öx + x = c
Which is the general soln.
2) Solve: p2 + 2y - 2x = 0
Solution
Solve for x:
x= ½(2y + p2
Differentiate:
dx/dy = 1/p = 1 + p (dp/dy)
Simplify:
pdp/ (p - 1) + dy = 0
=> [p + 1 + 1/ (p-1)] dp + dy = 0
Solving:
½ p+ p + ln (p - 1) + y = c
Which is the general solution. The parametric form of the general solution is:
y = c - ½ p2 - p - ln (p - 1)
x = c - p - ln (p - 1)
### Media Disparaging Of Joe Biden Continues With WSJ Troll Not Fit To Shine His Shoes
Trump's psychopathic mugshot: \$7 million raised already
"I happen to think that Democrats would be safer with a nominee who’s younger than Biden is and radiates more energy than he does. But I believe at least as strongly that if the unideal choice before Americans winds up being Biden, with his imperfections, or Trump, with his, rejecting Biden because of how old he has grown isn’t a grown-up decision."- Frank Bruni, NY Times, 'Trump Is Really Old Too'
I suppose I shouldn't have been surprised when not long after Trump and 18 members of his cabal were booked last week, with mugshots taken at the Fulton County jail, yet another Biden media takedown effort would soon follow. In this case it came from a little Aussie turd named Gerard Baker who is described as "editor at large" at the Wall Street Journal. Well from his attempted demolition piece on Monday, The Media Try to Inflate Biden’s Stockunder which one read:
Journalism by omission aims to suppress news of the president’s frailties and possible misdeeds
There was little doubt it was more a troll piece than truth piece. It was aimed at further tarnishing Biden's successful brand and accomplishments before the upcoming '24 election. The motive? If Trump is gonna go down under the weight of 91 felonies and 4 indictments, we're gonna take Biden down too. So it's not a case of the media "inflating Biden's stock" which they certainly have not done the past 2 years, e.g.
Why Are Biden's Poll Numbers So Low Despite Vast Personal Satisfaction...
But rather the rodents occupying the rat warrens of the WSJ trying to tear down Biden's image further, reducing him to a doddering, drooling doofus who can't find his shoes for his sleeves and could never have crafted the successes the Dems claim including:
i) Inflation went down after passage of the inflation reduction act and is now near 3.2%. We actually have the lowest inflation amongst the G7 nations (See Harmonized Headline HICP graph).
ii) U.S. has had the highest GDP growth in the G7 group (Again check the available graphs!)
iii) Unemployment in May was lowest since 1969, and the lowest income workers made the largest historic gains.
iv) Housing starts have surged to the most in 3 decades (Reuters)
Less noted:
v)The typical U.S. worker produces nearly 5x the output of a Chinese worker. Also, American worker output is greater than that of Europeans, Japanese, British, Canadians and Australians by a wide margin.
vi) The U.S. arguably has the deepest and most liquid financial market in the world enabling the birth of new businesses and continued growth of successful ones. The public financial value – known as market capitalization- now stands at 170%. (For most countries it is below 100%)
According to the UK magazine The Economist: “By a whole range of measures, American dominance is striking”.
So by any logical measures Biden indeed has one of the best narratives in terms of economic success. Despite all the disruptions and headwinds the U.S. has the best economy in the world. Yet the Reeptards are determined to bring him down, by piffle such as Baker put out on Monday. Why? Because for his traitor psycho hero Trump to have any chance at all, he and his fellow WSJ op-ed trolls must seek to make Biden appear weak and incompetent. And indeed he has succeeded amongst the clueless WSJ rightist readers who wrote the following comments;
Thanks to the WSJ Op/Ed section to at least bringing a semblance of coverage of the liberal media campaign!
Biden has wrecked the country already! Keep digging up more dirt!
You can make Biden blink, but you can't change the price of gas or groceries. Eventually the economy is good and the world is a safer place or not.
Unfortunately people will have had to suffer four years of bad domestic and foreign policy before they realize that he was asleep on the job!
Baker began by focusing on Biden's Maui trip and the 80 yr. old president sympathizing with locals' post -disaster plight. Something Trump would never do (he'd more likely toss paper towels at them like he did in Puerto Rico after it was struck by a hurricane during his term). Baker also brought up the trope of "Sleepy Joe" and that Biden may have been blinking, because he was partly asleep during his meet and greet with Lahaina locals. This led Baker to pounce, writing:
"A less partial observer might conclude that if we need high-definition video to be certain that the president’s eyes are “open, blinking” while he performs his duties, we have a problem more serious than mean-spirited conservatives making fun of him."
But he wasn't done, then trying to blame the mainstream media for providing cover for Biden's flaws, foibles, verbal gaffes etc.:
"Media bias is nothing new, but this goes further. For almost four years there has been a concerted, sustained and so far successful effort by supposedly independent media organizations to elect, defend and preserve in office the nation’s leading Democrat. It is immanent throughout the coverage of this president and mostly takes the form of misrepresenting reality by willfully ignoring or suppressing anything that undermines him—what we might call journalism of omission.
The juxtaposition of the attention given to the Hawaiian fact-check with the effort to look past the many infelicities of the president’s island misadventure is a small example of the Biden-Keeping Operation."
"Island misadventure"? WTF are you bellyaching about? But you can't make this blowhard journalistic bombastic bullshit up. It is now the warp and woof of the reactionary Right media, especially at FOX and the WSJ op-eds. But Baker is cockeyed foolish if he thinks the mainstream media have been giving Biden a pass for 3 years, given they've relentlessly pounded his age, e.g.
So why so much negativity? The Economist’s assessment of the data suggests the pessimism we’re seeing is nothing more than “emotional baggage brought on by a perpetual negative news cycle”. This emphasis is because newsrooms are convinced that only negative material, content gets clicks or reads. So beating a president who's the polar opposite of the feral wretch Trump is cool, especially over his age, inflation and other factors beyond his control. Like the price of gas at the pump.
This ongoing litany of whining leads to an entrenched negative news cycle and pessimistic information bias. Much of it is driven by politics (often from the Trumpian Right) working overtime to decapitate Biden and the Dems and thereby turn lazy people into automatons who will vote against their own interests. Why? Because the tropes and canards are then regurgitated on social media and the dregs of the net like 4chan, 8chan, Telegram etc.
As Chris Hayes pointed out in an ALL In episode 2 weeks ago, the Reeps are going deep in to pile on Biden because it provides them with an excuse not to admit their traitor hero is a piece of degenerate slime, a pestilence parasite and a malignancy on the Republic. A walking turd who ought to have been put away multiple times over by now. As opposed to still being out and about, spewing all manner of vile rhetoric about the judges, the prosecution and especially Jack Smith. E.g.
Deranged Imp Trump Needs To Be Locked Up For His (and OUR) Own Good
Any other person would have been remanded to custody long ago, but this roach is allowed to keep on keeping on with his toxic torrents which - as former prosecutor Cynthia Alksne noted two nights ago- could well have severe repercussions for a DA, judge or prosecutor. But despite that, no one is willing to take the orange fecal fungus down.
And:
by Norman Solomon | August 30, 2023 - 5:57am | permalink
Ever since Donald Trump became a former president, news outlets and commentators have cited polls showing that many Republicans believe violence might be needed to save the country. As Trump’s legal woes increase, so do mainstream media alarms about the specter of violent responses. But we’ve heard virtually nothing about connections between two decades of nonstop U.S. warfare overseas and attitudes favoring political violence at home.
And:
by Robert Becker | August 28, 2023 - 5:29am | permalink
Will stupid leftists ever behold
Why MAGA captures young and old?
Deadbeat libs troll our sainted hero,
But who’d make a better pharaoh?
Unshakable fans know far better –
Bigly causes surpass the plate-setter;
The left guffaws at the “Trump cult” –
Yet sad envy drives that cheap insult.
Crackpot libs can’t match our grievances
Because they lack real allegiances;
Only patriots with righteous causes
Defy bad laws, like daring outlaws.
## Monday, August 28, 2023
### Revisiting Differential Operators.
Differential Operators are basically shorthand ways of writing derivatives and by extension writing more concise differential equations. In general:
(D x y) n = dn y / dxn n= 1, 2, 3….
Then: (D x y) 2 = d2 y/ dx2
(D x y) 3 = d3 y/ dx3
And so forth.
For example: d4y/ dt4
Can be rewritten: (D t y) 4
And: 2d3y/ dt3
Can be rewritten: 2(D t y)3
Then rewrite the differential equation:
d4y/ dt4 – 2d3y/ dt3 –7 d2y/ dt2 + 20 dy/dt –12 y = 0
Using differential operators.
Using the preceding hints for the notation, we have:
(D t y) 4 - 2(D t y)- 7 (D t y) 2 + 20 (D t y) - 12y = 0
Of special interest is the inverse differential operator, viz.
(D-1 x n ) = ( xn+1 ) / (n + 1)
Similarly for higher order::
(D-2 x n ) = (D-1 ) ( xn+1 ) / (n + 1) =
xn+2 / (n +1 ) (n + 2)
And for higher order (k) in general:
(D-k x n ) = (xn+k ) (n+ 1) (n + 2)...(n + k)
Thus, 1/D stands for the integral, e.g. F(x)dx but with denominator corrected for as given by inverse formulae above.
And 1/ D n stands for successive (n) integration.
Suggested Problems:
1) Rewrite the DE below with differential operators:
5 d5y/ dt5 - 10 dy/dt –25y = 0
2) Rewrite the DE below in standard derivative form and solve:
(D x y) = 3x2 - 1
3) Does (D x y) 2 = D 2 x y ? Explain.
4) Evaluate each of the following:
i) (D x y) (ln x / 1+ x)
ii) (D x y) ½ ( x - -x)
iii) D-1 ( 11 x)
iv) (D x y) ( x ln x)
### InSight Mars Lander Uses Doppler Effect To Discover More Rapid Mars' Rotation
Mars, now a mystery planet with a mystery rotation
Key component of Mars InSight Lander (JPL/NASA)
The question in the wake of a paper published in June in the journal Nature is: Why is the rotation of Mars speeding up? NASA’s InSight lander fell silent in December after dust accumulated on its solar panels but new data have revealed surprises.
To track the planet’s spin rate, the study’s authors relied on one of InSight’s instruments: a radio transponder and antennas collectively called the Rotation and Interior Structure Experiment, or RISEThe technique for measurement entailed beaming a radio signal to the lander using the Deep Space Network. RISE then reflected the signal back. When scientists received the reflected signal, they then looked for tiny changes (variations) in frequency caused by the Doppler shift (the same effect that causes an ambulance siren to change pitch as it gets closer and farther away). Measuring the shift enabled researchers to determine how fast the planet rotates.
The researchers found the planet’s rotation is accelerating by about 4 milliarcseconds per year² – corresponding to a shortening of the length of the Martian day by a fraction of a millisecond per year. I.e. a more rapid rotation rate. This is a subtle acceleration, and planetary scientists aren’t entirely sure of the cause. But they have a few ideas, including ice accumulating on the polar caps or post-glacial rebound, where landmasses rise after being buried by ice. The shift in a planet’s mass can cause it to accelerate a bit like an ice skater spinning with their arms stretched out, then pulling their arms in.
The paper examined data from InSight’s first 900 Martian days – enough time to look for such variations. Scientists had their work cut out for them to eliminate sources of noise that could produce similar variations: Water slows radio signals, so moisture in the Earth’s atmosphere can distort the signal coming back from Mars. So can the solar wind, the electrons and protons flung into deep space from the Sun. So can the relative positions of Earth and Mars in their respective orbits.
Once planetary scientists accounted for these ancillary factors there remained leftover frequency variations, suggesting an enhanced rotation. In particular, the additional findings gleaned from tiny shifts in the frequencies of radio transmissions between Earth and InSight on Mars.
RISE is part of a long tradition of Mars landers using radio waves for science, including the twin Viking landers in the 1970s and the Pathfinder lander in the late ’90s. But none of those missions had the advantage of InSight’s advanced radio technology and upgrades to the antennas within NASA’s Deep Space Network on Earth. Together, these enhancements provided data about five times more accurate than what was available for the Viking landers.
According to the paper’s lead author and RISE’s principal investigator, Sebastien Le Maistre at the Royal Observatory of Belgium:
"What we’re looking for are variations that are just a few tens of centimeters over the course of a Martian year. It takes a very long time and a lot of data to accumulate before we can even see these variations.
"Mars, because it is not a perfectly round sphere, wobbles like a top. but the primary goal was to measure the rotation,”
An additional related bonus is that precise measurements of the rotation (with all of the wobbles) places constraints on the structure and composition of the very deep parts of the planet, according to Dr. Le Maistre.
Similar measurements had been attempted during NASA’s Viking missions in the 1970s and also during later missions like Pathfinder in 1997, but those were not precise enough or long enough. To show how unique, Le Maistre emphasized:
This was never done for any planet other than the Earth before,
From the magnitude of the wobbles, the Le Maistre's team has calculated that the molten core of Mars is about 2,280 miles wide. (Mars as a whole is about 4,200 miles in diameter.)
## Friday, August 25, 2023
### FT Columnist Claims "Surplus Mental Capacity" Fuels Conspiracies & Psycho Babble? Give Me A Break!
FT's Janan Ganesh - Overthinks basis of delusions
Alex Jones: No example of 'cognitive overcapacity'
Financial Times political columnist Janan Ganesh has often sought to be provocative in his various opinions, and in most cases met at least minimal standards of rationality and attention to reality. But in his latest column ('The Age of the Clever Fool') he appears to have gone off the rails by way of overthinking.
In the lengthy subheader we see he scribbles: "Tech bros, woke theorists, psychobabblers - George Orwell endures because his plainness is relief from all".
Adding: "The problem traverses the ideological spectrum. The movement known as woke could only have been incubated and hatched on campuses. Few places have so much intellectual potential. Few places have so little, well, I am going to call it “lived experience”.
His thesis based on these preliminaries is basically that the explosion of bunkum across the ideological spectrum - from Covid and vaccine denial, to climate denial, to climate doomism to mindless political populism (Trumpism), to general conspiracy ideations - hatched from QAnon or RFK - are all of a piece, namely neural glitches in the modern world. They boil down to fads, conspiracies, psychobabble, self-help mumbo- jumbo etc. all of which "might be explained as surplus mental capacity casting restlessly around for some outlet. "
Basically, noting that scientist Peter Turchin "has argued that underemployed graduates are potential social trouble". But what about the millions of kooks and conspiracy freaks - like Alex Jones - who aren't college grads? How come they exert so much power given they have minimal surplus mental capacity? Ganesh has no answers because it's much easier to just invent a facile trope and try to embellish it to the point some may be snookered.
Yes, a fair number of university -educated people did fall for the promises of the likes of Sam Bankman-Fried with his crypto investment offers, and Elizabeth Holmes with her "thumb prick" blood drop tests. Hell, I even fell - somewhat - for the latter, or at least hoped there was some ballast to the idea, given I was in the midst of multiple rounds of blood tests (psa, testosterone, alkaline phosphatase) concerned with my cancer. But it was not to be, and I quickly realized I'd been had so had to stiffen myself to submit to future rounds of blood tests of the regular kind. So Ganesh is correct up to a point. But he over-generalizes and exaggerates.
Admittedly, he identifies mucho bunkum and just a fraction is enough to enrage the confirmed rationalist. Like the "effective altriusm" thrust which purports to help others up to a point and which purpose is embodied in the principles of its website:
CEA's Guiding Principles | Centre For Effective Altruism
Then there is Elon Musk's own program to help multiply the super-gifted by selective reproduction: namely his own. E.g.
We need to talk about Elon Musk's breeding program (inverse.com)
You can't get much more top heavy with crank-hood than that. But close to it is the nest of longevity research cranks and kooks who believe human life can be endlessly extended and also it is worth it to do such, e.g.
Transhumanism : Another Cockeyed Concept Destined To Fail From Human Hubris
As I noted in that post:
"Now an even more cockeyed immortality denizen has entered the picture, called the "transhumanist" (WSJ, June 20, 'Looking Forward to the End Of Humanity') This lot don't merely expect to conquer old age but to surmount the entire spectrum of biological fragility, to end up with a kind of "transhuman future". Thus (ibid.):
"With our biological fragility more obvious than ever, many people will be ready to embrace the message of the Transhuman Declaration, an eight-point program first issued in 1988", e.g.
And what, pray tell, are we looking at here? Well, among the "avenues to immortality" discussed are:
- Creation of nanorobots which could be programmed to live inside our cells and constantly repair any damage, halting aging in its tracks.
- Genetic engineering could eliminate the mechanism that causes us to age in the first place. (And could also deliver genetic 'mistakes' such as hybridomas, or else humans with serious genetic defects.)
- Transferring consciousness to special computers - where it "can survive indefinitely".
Where Ganesh runs into trouble is when he over-extends the basis of his perspective on "clever fools" and their "surplus mental capacity" to examples where these need not apply. As when he writes:
"When the cultural left makes contact with that less credentialed but longer-in-the-tooth movement called the electorate, see how it struggles.
Those dupes in turn can point to the mass market for amateur psychotherapy and self-help mumbo jumbo. And so the carousel of eloquent, educated naïveté goes round. There have always been clever fools: Cambridge spies, eugenics-smitten Fabians, Hitler appeasers with All Souls fellowships. During his leadership of the Pentagon, Robert McNamara, the ultimate genius/sucker, fought the Vietnam war as a hedge fund quant might. The problem existed in pockets back then, however, because so few people went to university. Now it is almost ambient."
But can we lay all of that at the feet of university education? What about the millions who've graduated - from thousands of universities globally the past 50 years- and aren't caught up in crank conspiracies, New Age psycho-babble, eugenics obsession, vaccine rejection, climate denial or other mental flotsam and jetsam? There has been a substantial mass of graduates - I'd say a preponderance- whose brains haven't been colonized or seeded by such mind viruses because they've been immunized by critical thinking. For that, yes, I have the Jesuits to thank at Loyola University, for opening the Loyola Fieldhouse - in January, 1965, to atheist existentialist Jean -Paul Sartre, e.g.
To debate the (then) recent Catholic convert philosopher Gabriel Marcel. The debate was totally engrossing and lasted for 90 minutes, keeping the 900 or so students (from Tulane and Loyola) spell bound. It was how I first became interested in existentialism even buying Sartre's masterpiece 'Being and Nothingness' (available at the Loyola Bookstore):
Sartre's masterpiece which I still have
True, we have lived through a tremendous expansion of cognitive training (via university education) and we're told the current knowledge - just in science doubles ever 3 years. Look at the explosion of just astronomical journal papers, only a minuscule fraction of which actually get read. To get an idea of the sheer expanse of pure research- merely in astrophysics - check out the papers in the link to this Astrophysical Journal issue below:
But in critical thinking application the discipline of thought is a prerequisite so one does not "go off the rails". But this is what's missing in most university education now. Ganesh in his essay points to the absence of "a commensurate increase in that unteachable trait known as our judgment". But this is part and parcel of what I call disciplined thought editing in critical thinking application. It is what, indeed, cannot be taught though it can be assimilated through much arduous trial and error. In time, then, one becomes a rigorous editor of one's own thoughts and recognizes which fit in a coherent milieu and which do not. But if one is overpowered, say by any kind of new knowledge, this may not be feasible.
In that case, yes, the "law of unintended consequences forever applies". Then Peter Turchin could argue that "underemployed graduates are potential social trouble." Why? Because they are generally ensconced in work or other environments (FOX News) wherein neither critical thinking or disciplined thought is required. So anything goes.
Why would that situation be much less true of the most successful graduates? To embrace a dangerous dogma, whatever it is, probably takes some resentment, but it definitely takes some grounding in conceptual thought. I suppose I embraced Sartre's atheistic existentialism after Loyola, and yeah, maybe resentment at my Catholic upbringing played a role. But the grounding in conceptual critical thought that Loyola's Jesuits provided me, allowed me to coherently develop my atheism leading to my series of atheist books:
Ganesh insists a "vast minority of the public now has it" (surplus mental capacity), which is true. But I dispute that vast minority has the necessary critical thinking skills and discipline of thought to manage that cognitive surplus capacity. Else they would actually DO something constructive with it.
And:
And:
And:
And:
And:
And:
And: | 5,721 | 23,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-18 | latest | en | 0.913798 |
https://unitchefs.com/minutes/years/75/ | 1,723,715,654,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641278776.95/warc/CC-MAIN-20240815075414-20240815105414-00533.warc.gz | 461,552,359 | 7,213 | # 75 Minutes to Years
=
75 Minutes =
0.00014259932328065
(decimal)
1.4259932328065 x 10-4
(scientific notation)
125
876582
(fraction)
Years
## Minutes to Years Conversion Formula
[X] yr = 0.0000019013243104087 × [Y] min
where [X] is the result in yr and [Y] is the amount of min we want to convert
## 75 Minutes to Years Conversion breakdown and explanation
75 min to yr conversion result above is displayed in three different forms: as a decimal (which could be rounded), in scientific notation (scientific form, standard index form or standard form in the United Kingdom) and as a fraction (exact result). Every display form has its own advantages and in different situations particular form is more convenient than another. For example usage of scientific notation when working with big numbers is recommended due to easier reading and comprehension. Usage of fractions is recommended when more precision is needed.
If we want to calculate how many Years are 75 Minutes we have to multiply 75 by 5 and divide the product by 2629746. So for 75 we have: (75 × 5) ÷ 2629746 = 375 ÷ 2629746 = 0.00014259932328065 Years
So finally 75 min = 0.00014259932328065 yr | 298 | 1,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-33 | latest | en | 0.856219 |
http://range.wordpress.com/tag/mathematics/ | 1,419,095,901,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770060.91/warc/CC-MAIN-20141217075250-00051-ip-10-231-17-201.ec2.internal.warc.gz | 224,778,589 | 24,718 | ## Posts Tagged 'mathematics'
### Samuel Arbesman Explains The Half-Life of Facts
In Samuel Arbesman’s new book The Half-Life of Facts: Why Everything We Know Has an Expiration Date, the applied mathematician examines why in the modern world, facts change all of the time.
### New Mathematical Proof of the ABC Conjecture
A pleated surface on the boundary of the convex core.
A new claim could imply that a proof of one of the most important conjectures in number theory has been solved, which would be an astounding achievement. Mathematician Shinichi Mochizuki of Kyoto University in Japan has released a 500-page proof of the abc conjecture that proposes a relationship between whole numbers (related to the Diophantine equations).
### Century-Old Goldbach Weak Conjecture Closer to Being Solved
The weak Golbach conjecture states that you can break up any odd number into the sum of, at most, three prime numbers. Prime numbers cannot be evenly divided by any other number than themselves or 1.
### Patterns of Flocks of Starlings Mathematically Behave How Magnets Would Move
Starlings, a small to medium-sized passerine bird in the family Sturnidae, achieve extraordinary coordination in flight and they behave mathematically as metals being magnetized. A new study published its findings in the journal Proceedings of the National Academy of Sciences.
### Probabilities of Collecting All Pennies from 1959 to 1997 are Easily Feasible
Many coin collectors start their numismatic collections with pennies, and trying to get pennies from every year within a given range. This process, while seemingly daunting, is straightforward and will take about 684 pennies to find all of the pennies from 1959 to 1997, since they are still in circulation.
### Earth Loses 50,000 Tonnes of Mass Every Year
According to some calculations, the Earth is losing 50,000 tonnes of mass every single year, even though an extra 40,000 tonnes of space dust converge onto the Earth’s gravity well, it’s still losing weight.
### Touch Pilot S01E01 (Fox)
Touch promo poster, via Wikipedia
I was initially surprised that Kiefer Sutherland was back on network television, but I have always like 24 so I was actually looking forward to the series on Fox. This series was written and created by Tim Kring, who’s known for Heroes. The series uses some of the techniques from that series, so if you’ve watched Heroes, you’ll feel a familiar when watching this. This series is about a dad struggling with his autistic son, who can see strange patterns in numbers. Everything ends up connected, and Martin Bohm soon discovers this.
While I wanted to like the series, I have to say that I was disappointed, mainly because of the mathematical elements in the series, which are rudimentary to say the least. It’s definitely not a show aimed at anyone familiar with numbers, patterns, and sequences. That being said, the pilot was watched by over 12 million people and the series was picked up for a full season, so maybe it will improve. I’ll give it a shot to see where it goes, but I wasn’t that impressed.
## subscribe @ email
Join 1,458 other followers | 695 | 3,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | latest | en | 0.898171 |
https://www.edureka.co/community/7926/how-to-calculate-group-mean-and-assign-it-to-new-data-in-r | 1,726,045,725,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00424.warc.gz | 698,001,297 | 26,699 | I want to calculate `mean` (or any other summary statistics of length one, e.g. `min``max``length``sum`) of a numeric variable ("value") within each level of a grouping variable ("group").
The summary statistic should be assigned to a new variable which has the same length as the original data. That is, each row of the original data should have a value corresponding to the current group value - the data set should not be collapsed to one row per group. For example, consider group `mean`
Before
``````id group value
1 a 10
2 a 20
3 b 100
4 b 200``````
After
``````id group value grp.mean.values
1 a 10 15
2 a 20 15
3 b 100 150
4 b 200 150``````
Jun 27, 2018 1,736 views
## 1 answer to this question.
You can use something like this:
`df\$grp.mean.values <- ave(df\$value, df\$group)`
If you want to use ave to calculate something else per group, you need to specify FUN = your-desired-function, e.g. FUN = min:
`df\$grp.min <- ave(df\$value, df\$group, FUN = min)`
• 6,370 points
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+1 vote | 551 | 1,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.694818 |
https://advancesincontinuousanddiscretemodels.springeropen.com/articles/10.1186/s13662-015-0358-1 | 1,716,015,973,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00482.warc.gz | 65,688,687 | 85,623 | Theory and Modern Applications
# Extremal solutions for p-Laplacian fractional integro-differential equation with integral conditions on infinite intervals via iterative computation
## Abstract
We study the extremal solutions of a class of fractional integro-differential equation with integral conditions on infinite intervals involving the p-Laplacian operator. By means of the monotone iterative technique and combining with suitable conditions, the existence of the maximal and minimal solutions to the fractional differential equation is obtained. In addition, we establish iterative schemes for approximating the solutions, which start from the known simple linear functions. Finally, an example is given to confirm our main results.
## 1 Introduction
In this paper, we study the existence of extremal solutions to the following fractional integro-differential equation with p-Laplacian operator on infinite intervals:
$$\left \{ \begin{array}{l} D^{\beta}_{0^{+}}(\varphi_{p}(D^{\alpha}_{0^{+}}x(t))) +a(t)f(t, x(t), (Tx)(t), (Sx)(t))=0,\quad t\in J', \\ x(0)=x'(0)=\cdots= x^{(n-2)}(0)=0, \\ D^{\alpha}_{0^{+}}x(0)=0, \qquad \lim_{ t \rightarrow+\infty} D_{0^{+}}^{\alpha-1}x(t)=\int_{0}^{\infty}h(t)x(t)\, dt, \end{array} \right .$$
(1.1)
where $$0<\beta\leq1$$, $$n-1<\alpha\leq n$$, $$n\geq2$$, $$D^{\alpha}_{0^{+}}$$ and $$D^{\beta}_{0^{+}}$$ are standard Riemann-Liouville derivatives, $$\varphi_{p}$$ is the p-Laplacian operator defined by $$\varphi_{p}(s)=|s|^{p-2}s$$, $$(\varphi_{p})^{-1}=\varphi_{q}$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$p>1$$, and
$$(Tx) (t)=\int_{0}^{t}K(t, s)x(s)\, ds, \qquad (Sx) (t)=\int_{0}^{\infty}H(t, s)x(s)\, ds,$$
in which $$K\in C(D, J)$$, $$D=\{(t, s)\in J\times J: t\geq s\}$$, $$H\in C(J\times J, J)$$, $$h\in L(J, J)$$ with $$\int_{0}^{\infty}h(t) t^{\alpha-1}\,dt<\Gamma(\alpha)$$, $$a\in L(J, J)$$, $$f\in C(J\times J\times J\times J, J)$$, $$J=[0, +\infty)$$, $$J'=(0, +\infty)$$.
Fractional operators were mentioned by Leibnitz in a letter to L’Hospital in 1695. However, for a quite long period, the theory of fractional derivatives developed mainly as a pure theoretical field of mathematics. The situation has changed recently, fractional calculus was shown to be an excellent tool for the description of memory and hereditary properties of various materials and processes. Nowadays, differential equations of fractional order have recently proved to be valuable tools in the modeling of many physical processes, such as non-Markovian diffusion process with memory (see [1]), charge transport in amorphous semiconductors (see [2]), propagation of mechanical waves in viscoelastic media (see [3]), etc. Moreover, phenomena in aerodynamics, electrodynamics of a complex medium or polymer rheology, acoustics, and electro chemistry are also described by differential equations of fractional order (see [4, 5]). For instance, a viscoelastic fluid with the fractional derivative Maxwell model and its constitutive equation is given by [6]
$$\sigma+\lambda^{\alpha}\frac{d^{\alpha}\sigma}{dt^{\alpha}} =G \lambda^{\beta}\frac{d^{\beta}\varepsilon}{dt^{\beta}},$$
where σ is the shear stress, ε is the shear strain, $$\lambda=\mu/ G$$ is the relaxation time, G is the shear modulus, μ is the viscosity constant, α and β are fractional calculus parameters and satisfy $$0\leq\alpha\leq\beta\leq1$$.
Motivated by the fractional calculus’ application background, there are a large number of papers dealing with the solvability of fractional differential equations (see [712]). By using the Leray-Schauder nonlinear alternative theorem, Zhao and Ge in [13] obtained some results as regards the existence of unbounded solutions by considering the fractional order differential equation
$$\left \{ \begin{array}{l} D_{0^{+}}^{\alpha}u(t) + f(t, u(t)) = 0,\quad t\in J', \\ u( 0 ) = 0, \qquad \lim_{ t \rightarrow+\infty} D_{0^{+}}^{\alpha-1} u(t) = \beta u(\xi ),\end{array} \right .$$
where $$1 <\alpha\leq 2$$, $$D_{0^{+}}^{\alpha}$$ is the standard Riemann-Liouville fractional derivative and $$0 <\xi<+\infty$$, $$\beta>0$$, $$\beta\xi^{\alpha-1}<\Gamma(\alpha)$$, $$f\in C(J\times\mathbb{R}, J)$$, $$\mathbb{R}=(-\infty, +\infty)$$.
In [14], Liang and Zhang investigated the following m-point fractional boundary value problem (BVP) on infinite intervals:
$$\left \{ \begin{array}{l} D_{0^{+}}^{\alpha}u(t) + a(t)f( u(t)) = 0,\quad t\in J', \\ u( 0 )=u'(0) = 0, \qquad \lim_{ t \rightarrow+\infty} D_{0^{+}}^{\alpha-1} u(t) = \sum_{i=1}^{m-2}\beta_{i} u(\xi_{i} ),\end{array} \right .$$
where $$2<\alpha\leq 3$$, $$D_{0^{+}}^{\alpha}$$ is the standard Riemann-Liouville fractional derivative and $$0 <\xi_{1}<\xi_{2}<\cdots<\xi_{m-2}<+\infty$$, $$\beta_{i}>0$$ satisfies $$0<\sum_{i=1}^{m-2}\beta_{i} u(\xi_{i} )<\Gamma(\alpha)$$, $$a\in L(J, J)$$, $$f\in C(J, J)$$. Through the use of the fixed point index theory due to Leggett-Williams, the sufficient conditions for the existence of three positive solutions are obtained.
Chai in [15] studied the fractional boundary value problem with p-Laplacian operator
$$\left \{ \begin{array}{l} D^{\beta}_{0^{+}}(\varphi_{p}(D^{\alpha}_{0^{+}}u(t)))+f(t, u(t))=0, \quad 0< t<1, \\ u(0)=0,\qquad u(1)+\sigma D^{\gamma}_{0^{+}}u(1) =0, \qquad D^{\alpha-1}_{0^{+}}u(0)= 0, \end{array} \right .$$
where $$1<\alpha\leq2$$, $$0<\beta, \gamma\leq1$$, $$0\leq \alpha-\gamma-1$$, $$D^{\alpha}_{0^{+}}$$, $$D^{\beta}_{0^{+}}$$, and $$D^{\gamma}_{0^{+}}$$ are standard Riemann-Liouville derivatives, σ is a positive constant, $$\varphi_{p}$$ is the p-Laplacian operator defined by $$\varphi_{p}(s)=|s|^{p-2}s$$, $$(\varphi_{p})^{-1}=\varphi_{q}$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$p>1$$, $$f\in C([0, 1]\times J, J)$$. By applying the fixed point theorem of Leggett-Williams, the author in [15] acquired the existence of positive solutions.
Since the existence of positive solutions to fractional boundary value problems with p-Laplacian operator have been rarely researched, in this paper, we investigate the existence of solutions for the fractional differential equation with p-Laplacian operator on infinite intervals as the BVP (1.1). We should mention here that our work presented in this paper has various new features. Firstly, the positive solutions on J are obtained, which expands the domain of definition of t from a finite interval to an infinite interval. Secondly, the new terms Tu, Su added in the function f of BVP (1.1) and the more general boundary conditions make the equation we discuss more complicated than those of two-point, three-point, multi-point boundary conditions. Finally, through the monotone iterative technique, we not only obtain the maximal and minimal solutions to the fractional differential equation but also establish iterative schemes for approximating the solutions, which start from the known simple linear functions.
## 2 Preliminaries and lemmas
### Definition 2.1
Let $$(E, \|\cdot\|)$$ be a real Banach space. A nonempty, closed, convex set $$P \subset E$$ is said to be a cone provided the following are satisfied:
1. (a)
If $$y\in P$$ and $$\lambda>0$$, then $$\lambda y\in P$$.
2. (b)
If $$y\in P$$ and $$-y\in P$$, then $$y=0$$.
If $$P \subset E$$ is a cone, we denote the order induced by P on E by ≤, that is, $$x \leq y$$ if and only if $$y-x \in P$$.
### Definition 2.2
[16, 17]
Let $$\alpha>0$$ and let u be piecewise continuous on $$J'$$ and integrable on any finite subinterval of J. Then for $$t>0$$, we call
$$I^{\alpha}_{0^{+}}u(t)=\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}u(s) \, ds,$$
the Riemann-Liouville fractional integral of u of order α.
### Definition 2.3
[16, 17]
The Riemann-Liouville fractional derivative of order $$\alpha>0$$, $$n-1\leq\alpha< n$$, $$n\in \mathbb{N}$$, is defined as
$$D^{\alpha}_{0^{+}}u(t)=\frac{1}{\Gamma(n-\alpha)} \biggl(\frac{d}{dt} \biggr)^{n} \int_{0}^{t}(t-s)^{n-\alpha-1}u(s) \, ds,$$
where denotes the natural number set, the function $$u(t)$$ is n times continuously differentiable on J.
### Lemma 2.1
[16, 17]
Let $$\alpha>0$$, if the fractional derivative $$D^{\alpha-1}_{0^{+}}u(t)$$ and $$D^{\alpha}_{0^{+}}u(t)$$ are continuous on J, then
$$I^{\alpha}_{0^{+}} D^{\alpha}_{0^{+}}u(t) =u(t)+c_{1}t^{\alpha-1}+c_{2}t^{\alpha-2}+ \cdots+c_{n}t^{\alpha-n},$$
where $$c_{1}, c_{2}, \ldots, c_{n}\in R$$, n is the smallest integer greater than or equal to α.
By a similar proof to Lemma 2.3 in [18], we get Lemma 2.2.
### Lemma 2.2
Let $$y\in C (0, +\infty)\cap L [0, +\infty)$$, then the fractional BVP
$$\left \{ \begin{array}{l} D^{\alpha}_{0^{+}}x(t)+y(t)=0,\quad t\in J', n-1<\alpha\leq n, n\geq2, \\ x(0)=x'(0)=\cdots=x^{(n-2)}=0,\qquad \lim_{ t \rightarrow+\infty} D_{0^{+}}^{\alpha-1}x(t)=\int_{0}^{\infty}h(t)x(t)\, dt, \end{array} \right .$$
has a unique solution
$$x(t)=\int_{0}^{\infty}G(t,s)y(s)\, ds,$$
where
$$G(t,s)=G_{0}(t,s)+G_{1}(t,s)$$
(2.1)
and
\begin{aligned}& G_{0}(t,s)=\frac{1}{\Gamma(\alpha)}\left \{ \begin{array}{l@{\quad}l} t^{\alpha-1}-(t-s)^{\alpha-1},& 0\leq s\leq t\leq+\infty, \\ t^{\alpha-1}, & 0\leq t\leq s\leq +\infty, \end{array} \right . \\& G_{1}(t,s)=\frac{t^{\alpha-1}}{\Gamma(\alpha)-\int_{0}^{\infty}h(t)t^{\alpha-1}\, dt} \int_{0}^{\infty}h(t)G_{0}(t,s)\, dt. \end{aligned}
### Lemma 2.3
The Green function $$G (t,s)$$ defined as (2.1) in Lemma 2.2 has the following properties:
1. (1)
$$G (t,s)$$ is continuous and $$G (t,s)\geq0$$ for $$(t, s)\in J\times J$$.
2. (2)
$$\frac{G_{0} (t,s)}{1+t^{\alpha-1}}\leq\frac{1}{\Gamma(\alpha)}$$, $$\frac{G(t,s)}{1+t^{\alpha-1}}\leq L$$, for $$(t, s)\in J\times J$$, where $$L=\frac{1}{\Gamma(\alpha)-\int_{0}^{\infty}h(t)t^{\alpha-1}\, dt}$$.
Now, we consider the associated linear BVP
$$\left \{ \begin{array}{l} D^{\beta}_{0^{+}}(\varphi_{p}(D^{\alpha}_{0^{+}}x(t))) +y(t)=0,\quad t\in J', 0<\beta\leq1, n-1<\alpha\leq n, n\geq2, \\ x(0)=x'(0)=\cdots= x^{(n-2)}=0, \qquad D^{\alpha}_{0^{+}}x(0)=0, \\ \lim_{ t \rightarrow+\infty} D_{0^{+}}^{\alpha-1}x(t)=\int_{0}^{\infty}h(t)x(t)\, dt, \end{array} \right .$$
(2.2)
where $$y\in C (0, +\infty)\cap L [0, +\infty)$$ with $$\int_{0}^{\infty}\varphi_{q} ( \int_{0}^{s}(s-\tau)^{\beta-1}y(\tau)\, d\tau )\, ds<+\infty$$. For convenience, let $$\omega=(\Gamma(\beta))^{1-q}$$, then we have Lemma 2.4.
### Lemma 2.4
The associated linear BVP (2.2) has the unique positive solution
$$x(t)=\omega\int_{0}^{\infty}G(t, s) \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}y(\tau)\, d\tau \biggr)\, ds.$$
(2.3)
### Proof
By Lemma 2.1, we have
$$\varphi_{p}\bigl(D^{\alpha}_{0^{+}}x(t)\bigr)= ct^{\beta-1}-\int_{0}^{t}\frac{(t-s)^{\beta-1}y(s)}{\Gamma(\beta)}\, ds.$$
Together with the fact $$D^{\alpha}_{0^{+}}x(0)=0$$, we get $$c=0$$, then
$$D^{\alpha}_{0^{+}}x(t) +\varphi_{q} \biggl(\int _{0}^{t}\frac{(t-s)^{\beta-1}y(s)}{\Gamma(\beta )}\, ds \biggr)=0.$$
Therefore, BVP (2.2) is equivalent to the following BVP:
$$\left \{ \begin{array}{l} D^{\alpha}_{0^{+}}x(t) +\varphi_{q} (\int_{0}^{t}\frac{(t-s)^{\beta-1}y(s)}{\Gamma(\beta )}\,ds )=0, \quad t\in J', 0<\beta\leq1, n-1<\alpha\leq n, n\geq2, \\ x(0)=x'(0)=\cdots= x^{(n-2)}=0,\qquad \lim_{ t \rightarrow+\infty} D_{0^{+}}^{\alpha-1}x(t)=\int_{0}^{\infty}h(t)x(t)\, dt. \end{array} \right .$$
By Lemma 2.2, BVP (2.2) is equivalent to the integral equation (2.3). This completes the proof of the lemma. □
In this paper, the following space E will be used in the study of BVP (1.1), where
$$E= \biggl\{ x\in C[0,+\infty): \sup_{t\in J}\frac{|x(t)|}{1+t^{\alpha-1}}<+ \infty \biggr\} .$$
(2.4)
Then E is a Banach space equipped with the norm $$\|x\|=\sup_{t\in J}\frac{|x(t)|}{1+t^{\alpha-1}}$$. Define the cone $$K\subset E$$ by
$$K= \bigl\{ x\in E: x(t)\geq0, t\in J \bigr\} .$$
Throughout this paper, we assume the following conditions hold:
(H1):
\begin{aligned}& \sup_{t\in J}\frac{1}{1+t^{\alpha-1}}\int_{0}^{t} K(t, s) \bigl(1+s^{\alpha-1}\bigr)\, ds<+\infty, \\& \sup _{t\in J}\frac{1}{1+t^{\alpha-1}}\int_{0}^{\infty}H(t, s) \bigl(1+s^{\alpha-1}\bigr)\, ds<+\infty, \\& \lim_{t'\rightarrow t}\int_{0}^{\infty}\bigl\vert H\bigl(t', s\bigr)- H(t, s)\bigr\vert \bigl(1+s^{\alpha-1} \bigr)\, ds=0,\quad t, t'\in J. \end{aligned}
In this case, let
\begin{aligned}& k^{*}=\sup_{t\in J}\frac{1}{1+t^{\alpha-1}}\int_{0}^{t} K(t, s) \bigl(1+s^{\alpha-1}\bigr)\, ds, \\& h^{*}=\sup_{t\in J} \frac{1}{1+t^{\alpha-1}}\int_{0}^{\infty}H(t, s) \bigl(1+s^{\alpha-1}\bigr)\, ds. \end{aligned}
(H2):
$$f\in C(J\times J\times J\times J, J)$$, $$f(t, 0, 0, 0, 0)\not\equiv0$$, $$t\in J$$, and $$f(t, (1+t^{\alpha-1})u_{0}, (1+t^{\alpha-1})u_{1}, (1+t^{\alpha-1})u_{2})$$ is bounded, for $$t\in J$$, $$u_{i}\in D$$ ($$i=0, 1, 2$$), $$D\subset J$$ is a closed bounded subinterval.
(H3):
$$a\in L(J, J)$$, $$a(t)\not\equiv0$$, $$t\in J$$, and
$$0<\int_{0}^{\infty}a(s)\, ds<+\infty,\qquad 0< \int _{0}^{\infty}\varphi_{q} \biggl( \int _{0}^{s}(s-\tau)^{\beta-1}a(\tau)\, d\tau \biggr)\, ds<+\infty.$$
Denote an operator $$A: K\rightarrow E$$ by
$$(Ax) (t)=\omega\int_{0}^{\infty}G(t, s) \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau, x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr)\, d\tau \biggr)\, ds, \quad t\in J.$$
Under the assumptions (H1)-(H3), x is a positive solutions of BVP (1.1) if and only if x is a fixed point of A in K.
We list the following lemma, which is needed in our study.
### Lemma 2.5
[19, 20]
Let E be defined as (2.4) and M be any bounded subset of E. Then M is relatively compact in E, if $$\{\frac{x(t)}{1+t^{\alpha-1}}:x\in M \}$$ is equicontinuous on any finite subinterval of J and for any given $$\varepsilon>0$$, there exists a $$N>0$$, such that $$\vert \frac{x(t_{1})}{1+t^{\alpha-1}_{1}}-\frac{x(t_{2})}{1+t^{\alpha -1}_{1}}\vert <\varepsilon$$ uniformly with respect to all $$x\in M$$, and $$t_{1}, t_{2}>N$$.
## 3 Main results
### Theorem 3.1
Assume that (H1)-(H3) hold. Then $$A: K\rightarrow K$$ is a completely continuous operator.
### Proof
First, by routine discussion, we see that $$A: K\rightarrow K$$ is well defined. Now, we prove that A is compact and continuous, respectively. Let M be any bounded subset of K. Then there exists $$R_{1} >0$$, such that $$\|x\| \leq R_{1}$$, for any $$x\in M$$. So, for any $$x\in M$$, by Lemma 2.3, we have
\begin{aligned} \bigl\Vert (Ax)\bigr\Vert =&\sup_{t\in J}\frac{ 1}{1+t^{\alpha-1}} \biggl\vert \omega \int_{0}^{\infty}G(t, s) \\ &{}\times \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau, x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr)\,d \tau \biggr)\,ds\biggr\vert \\ \leq&\omega L \varphi_{q}(S_{R_{1}})\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\, d\tau \biggr)\, ds <+\infty, \end{aligned}
where
\begin{aligned} S_{R_{1}} =&\sup \bigl\{ f\bigl(t, \bigl(1+t^{\alpha-1} \bigr)u_{0}, \bigl(1+t^{\alpha-1}\bigr)u_{1}, \bigl(1+t^{\alpha-1}\bigr)u_{2}\bigr): \\ &(t, u_{0}, u_{1}, u_{2}) \in J\times[0, R_{1}]\times\bigl[0, k^{*}R_{1}\bigr]\times\bigl[0, h^{*}R_{1}\bigr] \bigr\} . \end{aligned}
So, AM is bounded in E.
Given $$b>0$$, for any $$x\in M$$ and $$t_{1}, t_{2} \in[0, b]$$, without loss of generality, we may assume that $$t_{1}< t_{2}$$. In fact,
\begin{aligned}& \biggl\vert \frac{(A x)(t_{1})}{1+t_{1}^{\alpha-1}}- \frac{(Ax)(t_{2})}{1+t_{2}^{\alpha-1}}\biggr\vert \\& \quad \leq\omega\int_{0}^{\infty}\biggl(\biggl\vert \frac{G_{0}(t_{1}, s)}{1+t_{1}^{\alpha-1}} -\frac{G_{0}(t_{2}, s)}{1+t_{2}^{\alpha-1}}\biggr\vert +\frac{\int_{0}^{\infty}h(t)G_{0}(t,s)\,dt}{\Gamma(\alpha)-\int_{0}^{\infty}h(t)t^{\alpha-1}\,dt} \biggl\vert \frac{t_{1}^{\alpha-1}}{1+t_{1}^{\alpha-1}} -\frac{t_{2}^{\alpha-1}}{1+t_{2}^{\alpha-1}}\biggr\vert \biggr) \\& \qquad {}\times\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau, x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr)\,d \tau \biggr)\,ds \\& \quad \leq\omega\varphi_{q}(S_{R_{1}})\int _{0}^{\infty}\biggl(\biggl\vert \frac{G_{0}(t_{1}, s)}{1+t_{1}^{\alpha-1}} - \frac{G_{0}(t_{2}, s)}{1+t_{1}^{\alpha-1}}\biggr\vert +\biggl\vert \frac{G_{0}(t_{2}, s)}{1+t_{1}^{\alpha-1}}- \frac{G_{0}(t_{2}, s)}{1+t_{2}^{\alpha-1}}\biggr\vert \\& \qquad {} +\frac{\int_{0}^{\infty}h(t)G_{0}(t,s)\,dt}{\Gamma(\alpha)-\int_{0}^{\infty}h(t)t^{\alpha-1}\,dt} \biggl\vert \frac{t_{1}^{\alpha-1}}{1+t_{1}^{\alpha-1}} - \frac{t_{2}^{\alpha-1}}{1+t_{2}^{\alpha-1}}\biggr\vert \biggr) \\& \qquad {}\times\varphi_{q} \biggl(\int _{0}^{s}(s-\tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds. \end{aligned}
On the other hand, we have
\begin{aligned}& \omega\varphi_{q}(S_{R_{1}}) \int_{0}^{\infty}\biggl\vert \frac{G_{0}(t_{1}, s)}{1+t_{1}^{\alpha-1}} -\frac{G_{0}(t_{2}, s)}{1+t_{1}^{\alpha-1}}\biggr\vert \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds \\& \quad \leq\omega\varphi_{q}(S_{R_{1}}) \biggl(\int _{0}^{t_{1}}+\int_{t_{1}}^{t_{2}}+ \int_{t_{2}}^{\infty} \biggr)\biggl\vert \frac{G_{0}(t_{1}, s)}{1+t_{1}^{\alpha-1}} -\frac{G_{0}(t_{2}, s)}{1+t_{1}^{\alpha-1}}\biggr\vert \varphi_{q} \biggl( \int_{0}^{s}(s-\tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds \\& \quad \leq\omega \varphi_{q}(S_{R_{1}})\int _{0}^{t_{1}}\biggl\vert \frac{t_{2}^{\alpha-1}-t_{1}^{\alpha-1}+ (t_{2}-s)^{\alpha-1}-(t_{1}-s)^{\alpha-1}}{1+t_{1}^{\alpha-1}}\biggr\vert \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds \\& \qquad {} +\omega\varphi_{q}(S_{R_{1}})\int _{t_{1}}^{t_{2}}\biggl\vert \frac {t_{2}^{\alpha-1}-t_{1}^{\alpha-1}+ (t_{2}-s)^{\alpha-1}}{1+t_{1}^{\alpha-1}}\biggr\vert \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds \\& \qquad {}+\omega\varphi_{q}(S_{R_{1}}) \int _{t_{2}}^{\infty}\biggl\vert \frac{t_{2}^{\alpha-1}-t_{1}^{\alpha -1}}{1+t_{1}^{\alpha-1}}\biggr\vert \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds \\& \quad \leq\omega\varphi_{q}(S_{R_{1}}) \int _{0}^{b} \bigl(\bigl\vert t_{2}^{\alpha-1}-t_{1}^{\alpha-1} \bigr\vert + \bigl\vert (t_{2}-s)^{\alpha-1}-(t_{1}-s)^{\alpha-1} \bigr\vert \bigr) \\& \qquad {}\times\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds \\& \qquad {}+\omega\varphi_{q}(S_{R_{1}}) \int _{0}^{\infty} \bigl(2 \bigl\vert t_{2}^{\alpha-1}-t_{1}^{\alpha-1} \bigr\vert +(t_{2}-t_{1})^{\alpha-1} \bigr) \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds. \end{aligned}
So, for any $$\varepsilon>0$$, there exists $$\delta_{1}>0$$, such that for any $$t_{1}, t_{2}\in[0, b]$$ and $$|t_{1}-t_{2}|<\delta_{1}$$, we have
$$\omega \varphi_{q}(S_{R_{1}})\int_{0}^{\infty}\biggl\vert \frac{G_{0}(t_{1}, s)}{1+t_{1}^{\alpha-1}} -\frac{G_{0}(t_{2}, s)}{1+t_{1}^{\alpha-1}}\biggr\vert \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds<\frac{\varepsilon}{3}.$$
(3.1)
Similar to (3.1), for the above $$\varepsilon>0$$, there exists $$\delta_{2}>0$$, such that for any $$t_{1}, t_{2}\in[0, b]$$ and $$|t_{1}-t_{2}|<\delta_{2}$$, we have
$$\omega\varphi_{q}(S_{R_{1}})\int_{0}^{\infty}\biggl\vert \frac{G_{0}(t_{2}, s)}{1+t_{1}^{\alpha-1}} -\frac{G_{0}(t_{2}, s)}{1+t_{1}^{\alpha-1}}\biggr\vert \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds<\frac{\varepsilon}{3}.$$
(3.2)
Obviously, for the above $$\varepsilon>0$$, there exists $$\delta_{3}>0$$, such that for any $$t_{1}, t_{2}\in[0, b]$$ and $$|t_{1}-t_{2}|<\delta_{3}$$, we have
\begin{aligned}& \omega\varphi_{q}(S_{R_{1}})\int_{0}^{\infty}\frac{\int_{0}^{\infty}h(t)G_{0}(t,s)\, dt}{\Gamma(\alpha)-\int_{0}^{\infty}h(t)t^{\alpha-1}\, dt} \biggl\vert \frac{t_{1}^{\alpha-1}}{1+t_{1}^{\alpha-1}} -\frac{t_{2}^{\alpha-1}}{1+t_{2}^{\alpha-1}}\biggr\vert \\& \quad {}\times\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds<\frac{\varepsilon}{3}. \end{aligned}
(3.3)
So, by (3.1)-(3.3), for the above $$\varepsilon>0$$, let $$\delta=\min\{\delta_{1}, \delta_{2}, \delta_{3}\}$$, such that for any $$t_{1}, t_{2}\in[0, b]$$ with $$|t_{1}-t_{2}|<\delta$$ and for any $$x\in M$$, we have
$$\biggl\vert \frac{(Ax)(t_{1})}{1+t_{1}^{\alpha-1}}- \frac{(Ax)(t_{2})}{1+t_{2}^{\alpha-1}}\biggr\vert <\varepsilon.$$
Hence, $$\{\frac{(A x)(t)}{1+t^{\alpha-1}}: x\in M \}$$ is equicontinuous on $$[0, b]$$. Since $$b > 0$$ is arbitrary, $$\{\frac{(A x)(t)}{1+t^{\alpha-1}}: x\in M \}$$ is locally equicontinuous on J.
Next, we show that $$A : K \rightarrow K$$ is equiconvergent at +∞. For any $$x\in M$$, we have
\begin{aligned}& \int_{0}^{\infty}\varphi_{q} \biggl(\int _{0}^{s}(s-\tau)^{\beta-1}a(\tau)f\bigl(\tau, x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr)\,d\tau \biggr)\,ds \\ & \quad \leq\varphi_{q}(S_{R_{1}})\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds <+\infty \end{aligned}
and
\begin{aligned}& \lim_{t\rightarrow+\infty} \frac{\vert (Ax)(t)\vert }{1+t^{\alpha-1}} \\ & \quad =\lim_{t\rightarrow +\infty}\frac{\omega\int_{0}^{\infty}G(t, s)\varphi_{q} (\int_{0}^{s}(s-\tau)^{\beta-1}a(\tau)f(\tau,x(\tau), (Tx)(\tau), (Sx)(\tau)) \,d\tau )\,ds}{1+t^{\alpha-1}} \\& \quad = \lim_{t\rightarrow+\infty}\frac{1}{1+t^{\alpha-1}} (\frac{\omega t^{\alpha-1}}{\Gamma(\alpha)}\int _{0}^{\infty}\varphi_{q} \biggl(\int _{0}^{s}(s-\tau)^{\beta-1}a(\tau)f\bigl( \tau,x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr) \,d\tau \biggr)\,ds \\& \qquad {}-\frac{\omega}{\Gamma(\alpha)}\int_{0}^{t} (t-s)^{\alpha-1}\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau,x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr) \,d \tau \biggr)\,ds \\& \qquad {} +\frac{\omega t^{\alpha-1}}{\Gamma(\alpha)-\int_{0}^{\infty}h(t)t^{\alpha-1}\,dt}\int_{0}^{\infty}h(t) \int_{0}^{\infty}G_{0}(t, s) \\& \qquad {}\times \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau,x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr) \,d \tau \biggr)\,ds\,dt) \\& \quad = \frac{\omega}{\Gamma(\alpha)}\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau,x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr) \,d \tau \biggr)\,ds \\& \qquad {}+\frac{\omega}{\Gamma(\alpha)-\int_{0}^{\infty}h(t)t^{\alpha-1}\,dt}\int_{0}^{\infty}h(t) \int_{0}^{\infty}G_{0}(t, s) \\ & \qquad {}\times \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau,x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr) \,d \tau \biggr)\,ds\,dt \\ & \quad \doteq Q<+\infty. \end{aligned}
So, for any $$x\in M$$, we have
\begin{aligned} \begin{aligned} &\biggl\vert \frac{(A x)(t)}{1+t^{\alpha-1}}-Q\biggr\vert \\ &\quad \leq \frac{\omega}{(1+t^{\alpha-1})\Gamma(\alpha)}\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau,x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr) \,d \tau \biggr)\,ds \\ &\qquad {} +\frac{\omega}{(1+t^{\alpha-1})(\Gamma(\alpha)-\int_{0}^{\infty}h(t)t^{\alpha-1}\,dt)} \\ &\qquad {}\times\int_{0}^{\infty}h(t)\int _{0}^{\infty}G_{0}(t, s) \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau,x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr) \,d \tau \biggr)\,ds\,dt \\ &\quad \leq \frac{\omega\varphi_{q}(S_{R_{1}})}{(1+t^{\alpha-1})\Gamma(\alpha)}\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau) \,d\tau \biggr)\,ds \\ &\qquad {}+\frac{\omega \varphi_{q}(S_{R_{1}})}{(1+t^{\alpha-1})(\Gamma(\alpha)-\int_{0}^{\infty}h(t)t^{\alpha-1}\,dt)} \\ &\qquad {}\times\int_{0}^{\infty}h(t)\int _{0}^{\infty}G_{0}(t, s) \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds\,dt \\ &\quad \rightarrow0,\quad t\rightarrow+\infty. \end{aligned} \end{aligned}
Thus, for any $$\varepsilon >0$$, there exists $$N>0$$, for any $$t>N$$ and for any $$x\in M$$, such that
$$\biggl\vert \frac{(A x)(t)}{1+t^{\alpha-1}}-Q\biggr\vert <\frac{\varepsilon}{2}.$$
Consequently, for any $$t_{1}, t_{2}>N$$ and for any $$x\in M$$, we have
$$\biggl\vert \frac{(A x)(t_{1})}{1+t_{1}^{\alpha-1}}-Q\biggr\vert <\frac{\varepsilon }{2}, \qquad \biggl\vert \frac{(A x)(t_{2})}{1+t_{2}^{\alpha-1}}-Q\biggr\vert <\frac{\varepsilon}{2}.$$
Therefore, for any $$t_{1}, t_{2}>N$$ and for any $$x\in M$$, we get
$$\biggl\vert \frac{(A x)(t_{1})}{1+t_{1}^{\alpha-1}}-\frac{(A x)(t_{2})}{1+t_{2}^{\alpha-1}}\biggr\vert \leq\biggl\vert \frac{(A x)(t_{1})}{1+t_{1}^{\alpha-1}}-Q\biggr\vert +\biggl\vert \frac{(A x)(t_{2})}{1+{t_{2}^{\alpha-1}}}-Q \biggr\vert <\varepsilon,$$
which means that $$\{\frac{(A x)(t)}{1+t^{\alpha-1}}: x\in M \}$$ is equiconvergent at +∞. So, $$A: K \rightarrow K$$ is equiconvergent at +∞.
Finally, suppose $$x_{m} \rightarrow x$$ as $$m\rightarrow +\infty$$ in K. Then there exists $$R_{0}>0$$, such that $$\max_{ m\in \mathbb{N} \backslash\{ 0\} } \{\|x_{m}\|, \|x\|\}\leq R_{0}$$, is a natural number set. Since
\begin{aligned}& \biggl\vert \int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s-\tau)^{\beta-1}a(\tau)f \bigl(\tau, x_{m}(\tau), (Tx_{m}) (\tau), (Sx_{m}) (\tau)\bigr)\,d\tau \biggr)\,ds \\& \qquad {}-\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s-\tau)^{\beta-1}a(\tau)f \bigl(\tau, x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr)\,d\tau \biggr)\,ds\biggr\vert \\& \quad \leq 2\varphi_{q}(S_{R_{0}})\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds<+\infty, \end{aligned}
where
\begin{aligned} S_{R_{0}} =&\sup \bigl\{ f\bigl(t, \bigl(1+t^{\alpha-1} \bigr)u_{0}, \bigl(1+t^{\alpha-1}\bigr)u_{1}, \bigl(1+t^{\alpha-1}\bigr)u_{2}\bigr): \\ &(t, u_{0}, u_{1}, u_{2}) \in J\times[0, R_{0}]\times\bigl[0, k^{*}R_{0}\bigr]\times\bigl[0, h^{*}R_{0}\bigr] \bigr\} . \end{aligned}
For
\begin{aligned}& \biggl\vert \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau, x_{m}(\tau), (Tx_{m}) (\tau), (Sx_{m}) (\tau)\bigr)\,d\tau \biggr) \\& \quad {}-\varphi_{q} \biggl(\int_{0}^{s}(s- \tau )^{\beta-1}a(\tau)f\bigl(\tau, x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr) \,d\tau \biggr)\biggr\vert \rightarrow0,\quad m\rightarrow+\infty. \end{aligned}
By the Lebesgue dominated convergence theorem, we have
\begin{aligned}& \biggl\vert \int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s-\tau)^{\beta -1}a(\tau)f \bigl(\tau, x_{m}(\tau), (Tx_{m}) (\tau), (Sx_{m}) (\tau)\bigr)\,d\tau \biggr)\,ds \\& \qquad {}-\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s-\tau)^{\beta-1}a(\tau)f \bigl(\tau, x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr)\,d\tau \biggr)\,ds\biggr\vert \\& \quad \rightarrow0,\quad m\rightarrow+\infty. \end{aligned}
Therefore, by Lemma 2.3, we have
\begin{aligned}& \bigl\Vert (Ax_{m})-(Ax)\bigr\Vert \\& \quad = \sup_{t\in J}\frac{1}{1+t^{\alpha-1}}\biggl\vert \omega\int _{0}^{\infty}G(t, s) \\& \qquad {}\times\varphi_{q} \biggl(\int _{0}^{s}(s-\tau)^{\beta-1}a(\tau)f\bigl(\tau, x_{m}(\tau), (T_{m}x) (\tau), (S_{m}x) (\tau) \bigr)\,d\tau \biggr)\,ds \\& \qquad {} -\omega\int_{0}^{\infty}G(t, s) \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau, x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr)\,d \tau \biggr)\,ds\biggr\vert \\& \quad \leq\omega L\biggl\vert \int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau )f\bigl(\tau, x_{m}(\tau), (Tx_{m}) (\tau), (Sx_{m}) (\tau)\bigr)\,d\tau \biggr)\,ds \\& \qquad {}-\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s-\tau)^{\beta-1}a(\tau )f \bigl(\tau, x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr)\,d\tau \biggr)\,ds\biggr\vert \\& \quad \rightarrow0,\quad m\rightarrow+\infty. \end{aligned}
Thus, $$A: K\rightarrow K$$ is continuous.
In conclusion, by Lemma 2.5, together with the continuity of A, we see that $$A: K\rightarrow K$$ is a completely continuous operator. The proof is completed. □
### Theorem 3.2
Assume that (H1)-(H3) hold, and there exists $$d>0$$ satisfying the following conditions:
(H4):
$$f(t, u_{0}, u_{1}, u_{2})\leq f(t, \overline{u}_{0}, \overline{u}_{1}, \overline{u}_{2})$$, for any $$t\in J$$, $$0\leq u_{0}\leq\overline{u}_{0}$$, $$0\leq u_{1}\leq \overline{u}_{1}$$, $$0\leq u_{2}\leq\overline{u}_{2}$$.
(H5):
$$f(t, (1+t^{\alpha-1})u_{0}, (1+t^{\alpha-1})u_{1}, (1+t^{\alpha-1})u_{2}) \leq\varphi_{p} (\frac{d}{\varrho} )$$, $$(t, u_{0}, u_{1}, u_{2})\in J\times[0, d] \times[0, k^{*}d]\times[0, h^{*}d]$$, where
$$\varrho=\omega L\int_{0}^{\infty}\varphi_{q} \biggl( \int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\,d\tau \biggr)\,ds,\quad L \textit{ is defined by Lemma }\mbox{2.3}.$$
Then BVP (1.1) has the maximal and minimal positive solutions $$w^{*}$$, $$\nu^{*}$$ on J, such that
$$0<\sup_{t\in J}\frac{|w^{*}(t)|}{1+t^{\alpha-1}}\leq d,\qquad 0<\sup _{t\in J}\frac{|\nu^{*}(t)|}{1+t^{\alpha-1}}\leq d.$$
Moreover, for initial values $$w_{0}(t)=d t^{\alpha-1}$$, $$\nu_{0}(t)=0$$, $$t\in J$$, define the iterative sequences $$\{ w_{n} \}$$ and $$\{ \nu_{n} \}$$ by
\begin{aligned}& w_{n} = \omega\int_{0}^{\infty}G(t, s) \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau, w_{n-1}(\tau), (Tw_{n-1}) (\tau), (Sw_{n-1}) (\tau)\bigr)\,d\tau \biggr)\,ds, \\& \nu_{n} = \omega\int_{0}^{\infty}G(t, s) \varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau, \nu_{n-1}(\tau), (T \nu_{n-1}) (\tau), (S\nu_{n-1}) (\tau)\bigr)\,d\tau \biggr)\,ds, \end{aligned}
then
$$\lim_{n\rightarrow+\infty}\sup_{t\in J}\frac {|{w}_{n}(t)-{w}^{*}(t)|}{1+t^{\alpha-1}}=0, \qquad \lim_{n\rightarrow+\infty}\sup_{t\in J} \frac{|{\nu}_{n}(t)-{\nu}^{*}(t)|}{1+t^{\alpha-1}}=0.$$
### Proof
By Theorem 3.1, $$A : K\rightarrow K$$ is completely continuous. For any $$x_{1}, x_{2}\in K$$ with $$x_{1} \leq x_{2}$$, from the definition of A and (H4), we know $$Ax_{1} \leq Ax_{2}$$. Let $$K_{d}= \{x\in K: \|x\|\leq d \}$$. In what follows, we firstly prove $$A: K_{d}\rightarrow K_{d}$$. In fact, for any $$x\in K_{d}$$, we have
$$0\leq\frac{x(t)}{1+t^{\alpha-1}}\leq d,\qquad 0\leq\frac{(Tx)(t)}{1+t^{\alpha-1}}\leq k^{*}d,\qquad 0 \leq \frac{(Sx)(t)}{1+t^{\alpha-1}}\leq h^{*} d, \quad t\in J.$$
By (H5), we have
$$f(t, u_{0}, u_{1}, u_{2}) \leq \varphi_{p} \biggl(\frac{d}{\varrho} \biggr),\quad (t, u_{0}, u_{1}, u_{2})\in J\times[0, d] \times \bigl[0, k^{*}d\bigr]\times\bigl[0, h^{*}d\bigr].$$
By Lemma 2.3 and (H5), we have
\begin{aligned} \bigl\Vert (Ax)\bigr\Vert =&\sup_{t\in J}\frac{1}{1+t^{\alpha-1}} \omega\int_{0}^{\infty}G(t, s) \\ &{}\times\varphi_{q} \biggl(\int_{0}^{s}(s-\tau)^{\beta-1}a(\tau)f \bigl(\tau, x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr)\,d\tau \biggr)\,ds \\ \leq&\omega L\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s-\tau)^{\beta-1}a(\tau)f \bigl(\tau, x(\tau), (Tx) (\tau), (Sx) (\tau)\bigr)\,d\tau \biggr)\,ds \\ \leq&\omega L\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s-\tau)^{\beta-1}a(\tau) \varphi_{p} \biggl(\frac{d}{\varrho } \biggr)\,d\tau \biggr)\,ds=d. \end{aligned}
Hence, $$A: K_{d}\rightarrow K_{d}$$.
Let $$w _{0}(t)=d t^{\alpha-1}$$, $$t \in J$$, then $$w_{0}(t) \in K_{d}$$. Let $$w_{1} =Aw_{0}$$, $$w_{2} =Aw_{1}=A^{2}w_{0}$$, by Theorem 3.1, we have $$w_{1}, w_{2}\in K_{d}$$. Denote $$w_{n+1 }=Aw_{n}= A^{n}w_{0}$$, $$n=1, 2, \ldots$$ . Since $$A: K_{d}\rightarrow K_{d}$$, we have $$w_{n} \in A (K_{d})\subset K_{d}$$. It follows from the complete continuity of A that $$\{ w_{n}\}_{ n=1}^{\infty}$$ is a sequentially compact set in E. By (2.1) and (H5), we have
\begin{aligned} w_{1}(t) =& \omega\int_{0}^{\infty}G(t, s)\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau, w_{0}(\tau), (Tw_{0}) (\tau), (Sw_{0}) (\tau)\bigr)\,d\tau \biggr)\,ds \\ \leq&\frac{\omega t^{\alpha-1}}{\Gamma(\alpha)} \int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau, w_{0}(\tau), (Tw_{0}) (\tau), (Sw_{0}) (\tau)\bigr)\,d\tau \biggr)\,ds \\ &{} +\frac{\omega t^{\alpha-1}\int_{0}^{\infty}h(t)t^{\alpha-1}\,dt}{\Gamma(\alpha)(\Gamma(\alpha)-\int_{0}^{\infty}h(t)t^{\alpha-1}\,dt)} \\ &{}\times \int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)f\bigl(\tau, w_{0}(\tau), (Tw_{0}) (\tau), (Sw_{0}) (\tau)\bigr)\,d\tau \biggr)\,ds \\ \leq&\omega L t^{\alpha-1}\int_{0}^{\infty}\varphi_{q} \biggl(\int_{0}^{s}(s- \tau)^{\beta-1}a(\tau)\varphi_{p} \biggl(\frac{d}{\varrho } \biggr)\,d\tau \biggr)\,ds \\ =&d t^{\alpha-1}=w_{0}(t). \end{aligned}
(3.4)
So, by (3.4) and (H4), we have
$$w_{2}=Aw_{1} \leq Aw_{0} =w_{1}.$$
(3.5)
By induction, we get
$$w_{n+1 }\leq w_{n},\quad n=1, 2, \ldots.$$
(3.6)
Thus, there exists $$w^{*}\in K$$ such that $$w_{n} \rightarrow w^{*}$$ as $$n \rightarrow+\infty$$. Applying the continuity of A and $$w_{n+1} = Aw_{n}$$, we get $$Aw^{*} = w^{*}$$.
On the other hand, let $$\nu_{0} (t)=0$$, $$t\in J$$, then $$\nu_{0}(t)\in K_{d}$$. Let $$\nu_{1 }=A\nu_{0}$$, $$\nu_{2 }=A\nu_{1}=A^{2}\nu_{0}$$, then by Theorem 3.1, we have $$\nu_{1}, \nu_{2}\in K_{d}$$. Denote $$\nu_{n+1 }=A\nu_{n}= A^{n}\nu_{0}$$, $$n=1, 2, \ldots$$ . Since $$A: K_{d}\rightarrow K_{d}$$, we have $$\nu_{n} \in A (K_{d})\subset K_{d}$$. It follows from the complete continuity of A that $$\{ \nu_{n}\}_{ n=1}^{\infty}$$ is a sequentially compact set in E. Since $$\nu_{1 }=A\nu_{0}\in K_{d}$$, we have
$$\nu_{2}=A\nu_{1} \geq0.$$
By induction, we get
$$\nu_{n+1 }\geq\nu_{n}, \quad n=1, 2, \ldots.$$
(3.7)
Thus, there exists $$\nu^{*}\in K$$ such that $$\nu_{n} \rightarrow\nu^{*}$$ as $$n \rightarrow+\infty$$. Applying the continuity of A and $$\nu_{n+1} = A\nu_{n}$$, we get $$A\nu^{*} = \nu^{*}$$.
Now, we are in a position to show that $$w^{*}$$ and $$\nu^{*}$$ are the maximal and minimal positive solutions of BVP (1.1) in $$(0, dt^{\alpha-1}]$$. Let $$u\in(0, dt^{\alpha-1}]$$ be any solution of BVP (1.1), that is, $$Au=u$$. Noting that A is nondecreasing and $$\nu_{0} (t) = 0 \leq u(t) \leq dt^{\alpha-1} =w_{0} (t)$$, then we have $$\nu_{1} (t) = (A\nu_{0}) (t)\leq u(t) \leq(Aw_{0})(t)=w_{1}(t)$$, for all $$t\in J$$. By induction, we have
$$\nu_{n} \leq u\leq w_{n}, \quad n =1,2,3, \ldots.$$
(3.8)
Since $${w}^{*}=\lim_{n\rightarrow+\infty}{w}_{n}$$, $${\nu}^{*}=\lim_{n\rightarrow+\infty}{\nu}_{n}$$, it follows from (3.4)-(3.8) that
$$\nu_{0} \leq\nu_{1}\leq\cdots \nu_{n}\leq\cdots \leq\nu^{*}\leq u\leq w^{*}\leq\cdots\leq w_{n}\leq \cdots\leq w_{1}\leq w_{0}.$$
(3.9)
Since $$f(t, 0, 0, 0)\not\equiv0$$, $$t\in J$$, then the zero function is not the solution of BVP (1.1). Therefore, by (3.9), we know that $$w^{*}$$ and $$\nu^{*}$$ are the maximal and minimal positive solutions of BVP (1.1) in $$(0, dt^{\alpha-1} ]$$, which can be obtained by the corresponding iterative sequences $$w_{n} = Aw_{n-1}$$, $$\nu_{n} = A\nu_{n-1}$$. The proof is completed. □
### Remark 3.1
The iterative schemes in Theorem 3.2 are
\begin{aligned}& w_{0}(t)=dt^{\alpha-1}, \\& w_{n} = \omega\int _{0}^{\infty}G(t, s)\varphi_{q} \biggl(\int _{0}^{s}(s-\tau)^{\beta-1}a(\tau)f\bigl(\tau, w_{n-1}(\tau), (Tw_{n-1}) (\tau), (Sw_{n-1}) (\tau) \bigr)\,d\tau \biggr)\,ds, \\& \nu_{0}(t)=0, \\& \nu_{n} = \omega \int _{0}^{\infty}G(t, s)\varphi_{q} \biggl(\int _{0}^{s}(s-\tau)^{\beta-1}a(\tau)f\bigl(\tau, \nu_{n-1}(\tau), (T\nu_{n-1}) (\tau), (S\nu_{n-1}) ( \tau)\bigr)\,d\tau \biggr)\,ds, \end{aligned}
they start with a known simple linear function and the zero function, respectively. This is very convenient in applications. So Theorem 3.2 is very interesting and important.
### Remark 3.2
By Theorem 3.2, we note that $$w^{*}$$ and $$\nu^{*}$$ are the maximal and minimal solutions of the BVP (1.1) in $$K_{d}$$, they may coincide, and then BVP (1.1) has only one solution in $$K_{d}$$.
## 4 Example
Now we consider the fractional differential equation with a p-Laplacian operator on infinite intervals,
$$\left \{ \begin{array}{l} D^{\frac{9}{2}}_{0^{+}}x(t) +a(t)f(t, x(t), (Tx)(t), (Sx)(t))=0, \quad 0< t<+\infty, \\ x(0)=x'(0)= x''(0)=0,\qquad D^{\frac{7}{2}}_{0^{+}}x(0)=0,\qquad D^{\frac{5}{2}}_{0^{+}}x(\infty)=\frac{3}{4}\int_{0}^{\infty}e^{-t}x(t)\, dt. \end{array} \right .$$
(4.1)
Obviously, $$\alpha=\frac{7}{2}$$, $$\beta=1$$, $$p=2$$, $$h(t)=\frac{3}{4} e^{-t}$$. By calculation, we have
$$\Gamma \biggl(\frac{7}{2} \biggr)=3.3234, \qquad \int _{0}^{\infty}h(t)t^{\alpha-1}\,dt=2.4925<\Gamma \biggl(\frac{7}{2} \biggr).$$
Choose
\begin{aligned}& f(t, u_{0}, u_{1}, u_{2}) \\& \quad = 10^{-2} \biggl(\frac{u_{0}}{1+t^{\frac{7}{2}}} \biggr)^{2} +\frac{10^{-3}}{(1+t^{\frac{7}{2}})^{2}} \biggl(\int _{0}^{t} \frac{u_{1}}{(1+t+s)^{2}(1+s^{\frac{7}{2}})}\,ds+\int _{0}^{\infty}\frac{\cos^{2}(t-s)u_{2}}{(1+s^{2})(1+s^{\frac{7}{2}})}\,ds \biggr). \end{aligned}
So
\begin{aligned}& K(t, s)=\frac{1}{(1+t+s)^{2}(1+s^{\frac{7}{2}})},\qquad H(t, s)=\frac{\cos^{2}(t-s)}{(1+s^{2})(1+s^{\frac{7}{2}})}, \\& \int_{0}^{\infty}\bigl\vert H\bigl(t', s\bigr)- H(t, s)\bigr\vert \bigl(1+s^{\frac{7}{2}}\bigr)\,ds \\& \quad =\int _{0}^{\infty}\frac{| \cos^{2}(t'-s)-\cos^{2}(t-s)|}{(1+s^{2})}\,ds=\frac{\pi}{2} \bigl\vert t'-t\bigr\vert \rightarrow 0,\quad t' \rightarrow t, \\& k^{*}=\sup_{t\in J}\frac{1}{1+t^{\frac{7}{2}}}\int_{0}^{t} \frac{1+s^{\frac{7}{2}}}{(1+t+s)^{2}(1+s^{\frac{7}{2}})}\,ds=\sup_{t\in J} \biggl(\frac{1}{(1+t)^{2}}- \frac{1}{(1+t)(1+2t)} \biggr)\leq1, \\& h^{*}=\sup_{t\in J}\frac{1}{1+t^{\frac{7}{2}}}\int_{0}^{\infty}\frac{\cos^{2}(t-s)(1+s^{\frac{7}{2}})}{(1+s^{2})(1+s^{\frac{7}{2}})}\,ds\leq \sup_{t\in J}\frac{1}{1+t^{\frac{7}{2}}}\int _{0}^{\infty}\frac{1}{1+s^{2}}\,ds= \frac{\pi}{2}. \end{aligned}
Take $$\int_{0}^{\infty}\varphi_{q} ( \int_{0}^{s}a(\tau)\,d\tau )\,ds=3$$. Considering that
$$\omega=\bigl(\Gamma(\beta)\bigr)^{1-q}=1,\qquad L=\frac{1}{\Gamma(\alpha)-\int_{0}^{\infty}h(t)t^{\alpha-1}\,dt}= \frac{6}{5},$$
we get $$\varrho=\frac{18}{5}$$. Take $$d=\sqrt{3}$$, then for $$(t, u_{0}, u_{1}, u_{2})\in J\times[0, d] \times[0, k^{*}d]\times[0, h^{*}d]$$,
$$f \bigl(t, \bigl(1+t^{\frac{7}{2}} \bigr)u_{0}, \bigl(1+t^{\frac {7}{2}} \bigr)u_{1}, \bigl(1+t^{\frac{7}{2}} \bigr)u_{2} \bigr)\leq 10^{-2}\cdot3+\frac{\sqrt{3}}{10^{3}} \biggl(1+\frac{\pi}{2} \biggr)\approx 0.03<\varphi_{p} \biggl( \frac{d}{\varrho} \biggr)\approx0.48.$$
Thus the conditions in Theorem 3.2 are all satisfied. Therefore, the conclusion of Theorem 3.2 holds.
## References
1. Metzler, R, Klafter, J: Boundary value problems for fractional diffusion equations. Physica A 278, 107-125 (2000)
2. Scher, H, Montroll, EW: Anomalous transit-time dispersion in amorphous solids. Phys. Rev. B 12, 2455-2477 (1975)
3. Mainardi, F: Fractional diffusive waves in viscoelastic solids. In: Wegner, JL, Norwood, FR (eds.) Nonlinear Waves in Solids. ASME/AMR, Fairfield (1995)
4. Gaul, L, Klein, P, Kemple, S: Damping description involving fractional operators. Mech. Syst. Signal Process. 5, 81-88 (1991)
5. Metzler, R, Schick, W, Kilian, HG, Nonnenmacher, TF: Relaxation in filled polymers: a fractional calculus approach. J. Chem. Phys. 103, 7180-7186 (1995)
6. Hilfer, R: Applications of Fractional Calculus in Physics. World Scientific, Singapore (2000)
7. Webb, JRL, Infante, G: Non-local boundary value problems of arbitrary order. J. Lond. Math. Soc. 79, 238-258 (2009)
8. Chen, TY, Liu, WB, Hu, ZG: A boundary value problem for fractional differential equation with p-Laplacian operator at resonance. Nonlinear Anal. 75, 3210-3217 (2012)
9. Liu, ZH, Lu, L: A class of BVPs for nonlinear fractional differential equations with p-Laplacian operator. Electron. J. Qual. Theory Differ. Equ. 2012, 70 (2012)
10. Liu, XP, Jia, M, Xiang, XF: On the solvability of a fractional differential equation model involving the p-Laplacian operator. Comput. Math. Appl. 64, 3267-3275 (2012)
11. Liang, SH, Shi, SY: Existence of multiple positive solutions for m-point fractional boundary value problems with p-Laplacian operator on infinite interval. J. Appl. Math. Comput. 38, 687-707 (2012)
12. Ding, YZ, Wei, ZL, Xu, JF: Positive solutions for a fractional boundary value problem with p-Laplacian operator. J. Appl. Math. Comput. 41, 257-268 (2013)
13. Zhao, XK, Ge, WG: Unbounded solutions for a fractional boundary value problem on the infinite interval. Acta Appl. Math. 109, 495-505 (2010)
14. Liang, SH, Zhang, JH: Existence of three positive solutions of m-point boundary value problems for some nonlinear fractional differential equations on an infinite interval. Comput. Math. Appl. 61, 3343-3354 (2011)
15. Chai, GQ: Positive solutions for boundary value problem of fractional differential equation with p-Laplacian operator. Bound. Value Probl. 2012, Article ID 18 (2012)
16. Miller, KS, Ross, B: An Introduction to the Fractional Calculus and Fractional Differential Equations. Wiley, New York (1993)
17. Podlubny, I: Fractional Differential Equations. Mathematics in Science and Engineering, vol. 198. Academic Press, New York (1999)
18. Xie, WZ, Luo, ZG, Xiao, J: Successive iteration and positive solutions of a fractional boundary value problem on the half-line. Adv. Differ. Equ. 2013, Article ID 210 (2013)
19. Liu, YS: Boundary value problem for second order differential equations on unbounded domain. Acta Anal. Funct. Appl. 4(3), 211-216 (2002)
20. Corduneanu, C: Integral Equations and Stability of Feedback Systems. Academic Press, New York (1973)
## Acknowledgements
The authors were supported financially by the National Natural Science Foundation of China (11371221), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001) and the Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province.
## Author information
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Correspondence to Lishan Liu.
### Competing interests
The authors declare that they have no competing interests.
### Authors’ contributions
The study was carried out in collaboration between all authors. YW completed the main part of this paper and gave two examples; LSL and YHW corrected the main theorems and polished the manuscript. All authors read and approved the final manuscript.
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Wang, Y., Liu, L. & Wu, Y. Extremal solutions for p-Laplacian fractional integro-differential equation with integral conditions on infinite intervals via iterative computation. Adv Differ Equ 2015, 24 (2015). https://doi.org/10.1186/s13662-015-0358-1 | 17,910 | 42,205 | {"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": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | longest | en | 0.736554 |
https://proofassistants.stackexchange.com/questions/1376/construction-of-inductive-types-the-hard-way | 1,722,908,345,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640461318.24/warc/CC-MAIN-20240806001923-20240806031923-00565.warc.gz | 374,653,043 | 32,040 | # Construction of inductive types "the hard way"
Most theorem provers simply axiomize inductive types (or equivalently W types) in the abstract which is fine.
But I'm curious about explicit constructions of inductive types within the theory.
I suppose I'm interested in both inductive types and "weak inductive types" (weak initial algebras) such as impredicative encodings. I think the term I've most found in my searches has been "weak natural numbers object." Inductive types can be better but weak inductive types are cool too.
I know you can use impredicative universes as in System F to encode inductive types. Or you can just accept a universe bump. And apparently if you internalize a small amount of parametricity you can construct appropriate induction principles.
But I'm pretty sure I've heard you can construct inductive types other ways. I think maybe you need to assume some classical principles? I've read some stuff on transfinite induction I still don't really get. I think once you have a base inductive type of ordinals you can construct other inductive types in those terms?
I have a hunch you can abuse impredicative proof irrelevant propositions and natural numbers to construct inductive types but I don't really have anything solid here.
Also for some reason I think it's easier to construct free monads the hard way instead of inductive types? I'm not sure this is an important issue.
I don't think W types and polynomial endofunctors directly solve the problem. They provide some clarifying language for how to talk about inductive types but they're not quite an explicit construction.
There's a paper "Induction Is Not Derivable in Second Order Dependent Type Theory" I don't understand yet but I think all this means is you need to assume more axioms than second order dependent type theory?
• References to actual papers are more than welcome. Perhaps they should become a community standard. Commented May 10, 2022 at 1:31
• I don't think there are any proof assistants that use W types as inductive types for the same reason that actually constructing useful inductive types with them is difficult Commented May 10, 2022 at 2:35
• @Couchy W types are more used for metatheory and then handwaved to be the same as inductive types. True enough providing constructions of inductive types with a nice interface is a hard problem. Commented May 10, 2022 at 3:08
• I am confused by the terminology, I think. In PL recursive types are things like type D = D → D, i.e., a fix-point equiation without any restriction on variance and positive occurrences etc. Is this what we're talking about? Commented May 10, 2022 at 8:07
• @AndrejBauer my apologies. I just mean inductive types. The impredicative encoding is still interesting even if you require parametricity to get induction though. Commented May 10, 2022 at 15:34
In the HOLish settings, these types (starting with the natural numbers) are indeed constructed from first principles; they're certainly not axiomatised. Harrison had an early (1995) paper on how to do this, and the technology has developed from there.
Harrison's construction doesn't use ordinals, but encodes its types as trees underneath and then uses an inductive relation and the HOL type definition principle to prune away values that are not desired.
The more capable and more recent technology in Isabelle/HOL (“bounded natural functors”) does have some fancy cardinality reasoning behind it. There's a nice 2012 LICS paper on this tech by Traytel et al.
There are a lot of questions in your question, so I don’t think it’s easy to answer all of them at once, but let me still try and give a picture in the dependently typed setting.
First you cannot get "real" inductive types using only (impredicative) encoding, because you lack the ability to prove an induction principle – you can only define a recursor. This is the essence of Geuvers' Induction Is Not Derivable in Second Order Dependent Type Theory.
Another possibility is thus to encode all inductive types using only a number of basic ones, with usually W-types as the main ingredient. If you only care about the content up to propositional equality Inductive Types in Homotopy Type Theory shows that this is doable in HoTT, eg. you can construct a lot of types that are propositionally initial if you assume only a few base types (including W types). However, if you care about computational content, i.e. you want to have conversion rules and not you equalities, this get a bit trickier. Here the best reference I know is Why not W?, which covers quite some distance in that direction.
Finally, another direction of research in trying to decompose inductive types in "simple" descriptions is that on containers (see for instance Indexed Containers). I’m less familiar with that line of thought, but from what I gathered the idea here is to give a notion of "description" of an inductive type (or something closely related), that can be used to generate the usual data (type, constructors, recursor…), but also can be used to do meta-programming.
• "Induction is not derivable" just means you need more axioms like parametricity for the impredicative encoding. See Cedille as an example. Commented May 10, 2022 at 15:36 | 1,172 | 5,249 | {"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} | 2.78125 | 3 | CC-MAIN-2024-33 | latest | en | 0.965882 |
https://www.teachoo.com/4151/1824/Example-11---Let-event-A-be--odd-number-on-first-throw-/category/Chapter-13-Class-12th-Probability/ | 1,685,574,374,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647459.8/warc/CC-MAIN-20230531214247-20230601004247-00288.warc.gz | 1,120,244,373 | 37,578 | Chapter 13 Class 12 Probability
Class 12
Important Questions for exams Class 12
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Example 11 An unbiased die is thrown twice. Let the event A be ‘odd number on the first throw’ and B the event ‘odd number on the second throw’. Check the independence of the events A and B. Two events A & B are independent if P(A ∩ B) = P(A) . P(B) An unbiased die is thrown twice S = Let us define two events as A : Odd number on the First throw B : Odd number on the Second throw vA : Odd number on First throw A : { (1, 1), (1, 2), ………., (1, 6) (3, 1), (3, 2), ………., (3, 6) (5, 1), (5, 2), ………., (5, 6) } P(A) = 𝟏𝟖/𝟑𝟔 = 𝟏/𝟐 B : Odd number on Second throw B : { (1, 1), (2, 1), ………., (6, 1) (1, 3), (2, 3), ………., (6, 3) (1, 5), (2, 5), ………., (6, 5) } P(A) = 𝟏𝟖/𝟑𝟔 = 𝟏/𝟐 B : Odd number on Second throw B : { (1, 1), (2, 1), ………., (6, 1) (1, 3), (2, 3), ………., (6, 3) (1, 5), (2, 5), ………., (6, 5) } P(A) = 𝟏𝟖/𝟑𝟔 = 𝟏/𝟐 B : Odd number on Second throw B : { (1, 1), (2, 1), ………., (6, 1) (1, 3), (2, 3), ………., (6, 3) (1, 5), (2, 5), ………., (6, 5) } P(A) = 𝟏𝟖/𝟑𝟔 = 𝟏/𝟐 A ∩ B = Odd number on the First & Second throw = { (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)} So, P(A ∩ B) = 9/36 = 1/4 Now, P(A) . P(B) = 1/2 × 1/2 = 1/4 Since P(A ∩ B) = P(A) . P(B), Therefoare, A and B are independent events | 681 | 1,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-23 | longest | en | 0.753675 |
https://slideplayer.com/slide/9224076/ | 1,620,359,689,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988774.96/warc/CC-MAIN-20210507025943-20210507055943-00243.warc.gz | 454,433,053 | 18,139 | # Notes Unit 1 Chapters 2-5 Univariate Data. Statistics is the science of data. A set of data includes information about individuals. This information is.
## Presentation on theme: "Notes Unit 1 Chapters 2-5 Univariate Data. Statistics is the science of data. A set of data includes information about individuals. This information is."— Presentation transcript:
Notes Unit 1 Chapters 2-5 Univariate Data
Statistics is the science of data. A set of data includes information about individuals. This information is organized into different categories or characteristics called variables. For example in our class survey, each one of you is an individual represented in the data set. We collected information about the variables gender, height, etc…
We are always interested in the context of the data. That means…where did it come from, who did we include, when was it collected, why were we interested, what did we collect etc…Without context, data is meaningless.
After we understand the context, the next thing we should always do is GRAPH the data.
Graphs Be sure to always: *Title your graphs *Label your axis including units of measure *number your axes in a consistent and reasonable manner
Categorical Data Categorical variables record which of several groups or categories an individual belongs to.
Quantitative Data Quantitative variables take numerical values for which it makes sense to do arithmetic operations like adding or averaging.
Quantitative Data The distribution of a variable tells us what values the variable typically takes and how often it takes them. It is a generalization about the variable values.
When describing any Quantitative distribution: C – Center U – Unusual Features S – Shape S – Spread & B – Be S - Specific
Common Shapes of distributions/graphs Symmetric Skewed to the right Skewed to the left Bimodal Uniform
Once you have chosen a shape, you choose a measure of center and spread based on that shape.
If a distribution is symmetric, we use mean for center. Mean: the average formula:
If the distribution is symmetric, we use standard deviation for spread. Standard deviation:
Measure of Center when the distribution is not symmetric: Median – the middle value in an ordered list. If there are two values in the middle, then average them.
Measure Spread or Variability when the distribution is not Symmetric We can also examine spread by looking at the range of middle 50% of the data. This is called the: Interquartile Range (IQR). IQR = Q3 – Q1
We also need to talk about the 5-number summary. The 5-number summary is made up of the minimum, the first quartile, Q1 (where 25% of the data lies below this value), the median, the third quartile, Q3 (where 75% of the data lies below this value), and the maximum.
Another Measure of Spread or Variability Range – the difference between the maximum and the minimum observations. This is the simplest measure of spread. We typically use this as preliminary information or if it is the only measure of spread we can calculate.
Another measure of spread or variability Variance is the average of the squares of the deviations of the observations from their mean. It is the standard deviation squared.
An outlier is an individual observation in data that falls outside the overall pattern of the data.
Using the IQR, we can perform a test for outliers. Outlier Test: Any value below Q1 – 1.5(IQR) or above Q3 + 1.5 (IQR) is considered an outlier.
Measures that are not strongly affected by extreme values are said to be resistant. The median and IQR are more resistant than the mean and standard deviation. The standard deviation, is even less resistant than the mean.
Measures of Spread or Variability – Why? We measure spread because it’s an important description of what is happening with the data. We need to know about the amount of variation we can expect in a data set.
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A124297 a(n) = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci(n). 8
1, 11, 11, 31, 61, 151, 361, 911, 2311, 5951, 15401, 40051, 104401, 272611, 712531, 1863551, 4875781, 12760031, 33398201, 87424711, 228859951, 599129311, 1568486161, 4106261531, 10750188961, 28144128251, 73681909211, 192901135711 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS 11 = Lucas(5) divides a(1+10k), a(2+10k), and a(9+10k). Last digit of a(n) is 1, or a(n) mod 10 = 1. For odd n there exists the so-called Aurifeuillian factorization A001946(n) = Lucas(5n) = Lucas(n)*A(n)*B(n) = A000032(n)*A124296(n)*A124297(n), where A(n) = A124296(n) = 5*F(n)^2 - 5*F(n) + 1 and B(n) = A124297(n) = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci(n). LINKS John Cerkan, Table of n, a(n) for n = 0..2373 Eric Weisstein's World of Mathematics, Aurifeuillean Factorization Index entries for linear recurrences with constant coefficients, signature (4,-2,-6,4,2,-1). FORMULA a(n) = 5*Fibonacci(n)^2 + 5*Fibonacci(n) + 1. G.f.: -(11*x^5-21*x^4-15*x^3+31*x^2-7*x-1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)). [Colin Barker, Jan 03 2013] MATHEMATICA Table[5*Fibonacci[n]^2+5*Fibonacci[n]+1, {n, 0, 50}] PROG (PARI) a(n)=subst(5*t*(t+1)+1, t, fibonacci(n)) \\ Charles R Greathouse IV, Jan 03 2013 CROSSREFS Cf. A000032, A000045, A121171, A001946, A124296. Bisections: A001604, A156095. Sequence in context: A218163 A152082 A070849 * A172507 A089766 A077699 Adjacent sequences: A124294 A124295 A124296 * A124298 A124299 A124300 KEYWORD nonn,easy AUTHOR Alexander Adamchuk, Oct 25 2006 STATUS approved
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# Multiplication Word Problems - Practice #1 (Grade 3)
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## Multiplication Word Problems - Practice #1
Instructions: Solve each multiplication word problem. Show your work and label answers.
1.
Martin has his games divided into 5 groups. He has 8 games in each group. How many games does he have in all?
2.
There are 9 students in the music club. Each student has 4 instruments. How many instruments do the students in music club have in all?
3.
Andrea sees 4 bicycles at school. Each of the bicycles has two wheels. How many wheels does she see in all?
4.
Madison bought 8 bags of candy. Each bag had 8 pieces of candy. How many pieces of candy did Madison buy?
5.
Baseball cards come in packs of 10. Seth has 3 packs of baseball cards. How many baseball cards does he have altogether?
6.
A theme park visor costs \$6. Julie's family buys 5 visors.
How much do they spend on visors?
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Already a member? Log in for access. | Go Back To Previous Page | 309 | 1,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-51 | latest | en | 0.957065 |
https://www.enotes.com/homework-help/y-x-1-2x-1-find-limit-possible-536516 | 1,490,689,263,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189686.31/warc/CC-MAIN-20170322212949-00070-ip-10-233-31-227.ec2.internal.warc.gz | 910,957,113 | 11,061 | # `y = (x + 1)/(2x - 1)` Find the limit, if possible
### Textbook Question
Chapter 3, 3.9 - Problem 15 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.
hkj1385 | (Level 1) Assistant Educator
Posted on
`lim_(x->oo) (x+1)/(2x-1)`
`or, lim_(x->oo){(x/x)+(1/x)}/{((2x)/x)-(1/x)}`
`or, lim_(x->oo)(1+(1/x))/(2-(1/x)) = 1/2`
`` | 152 | 377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-13 | longest | en | 0.479465 |
https://www.quizover.com/online/course/0-7-generalizations-of-the-basic-multiresolution-wavelet-by-openstax?page=25 | 1,526,935,694,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864544.25/warc/CC-MAIN-20180521200606-20180521220606-00275.warc.gz | 835,789,994 | 23,401 | # 0.7 Generalizations of the basic multiresolution wavelet system (Page 26/28)
Page 26 / 28
## Discrete multiresolution analysis, the discrete-time wavelet transform, and the continuous wavelet transform
Up to this point, we have developed wavelet methods using the series wavelet expansion of continuous-time signals called the discrete wavelettransform (DWT), even though it probably should be called the continuous-time wavelet series. This wavelet expansion is analogous tothe
Fourier series in that both are series expansions that transform continuous-time signals into a discrete sequence of coefficients. However, unlike the Fourier series, the DWT can be made periodic ornonperiodic and, therefore, is more versatile and practically useful.
In this chapter we will develop a wavelet method for expanding discrete-time signals in a series expansion since, in most practicalsituations, the signals are already in the form of discrete samples. Indeed, we have already discussed when it is possible to use samples ofthe signal as scaling function expansion coefficients in order to use the filter bank implementation of Mallat's algorithm. We find there is anintimate connection between the DWT and DTWT, much as there is between the Fourier series and the DFT. One expands signals with the FS but oftenimplements that with the DFT.
To further generalize the DWT, we will also briefly present the continuous wavelet transform which, similar to the Fourier transform, transforms afunction of continuous time to a representation with continuous scale and translation. In order to develop the characteristics of these variouswavelet representations, we will often call on analogies with corresponding Fourier representations. However, it is important tounderstand the differences between Fourier and wavelet methods. Much of that difference is connected to the wavelet being concentrated in bothtime and scale or frequency, to the periodic nature of the Fourier basis, and to the choice of wavelet bases.
## Discrete multiresolution analysis and the discrete-time wavelet transform
Parallel to the developments in early chapters on multiresolution analysis, we can define a discrete multiresolution analysis (DMRA) for ${l}_{2}$ , where the basis functions are discrete sequences [link] , [link] , [link] . The expansion of a discrete-time signal in terms of discrete-time basis function is expressed in a form parallel to [link] as
$f\left(n\right)\phantom{\rule{0.166667em}{0ex}}=\phantom{\rule{0.166667em}{0ex}}\sum _{j,k}{d}_{j}\left(k\right)\phantom{\rule{0.166667em}{0ex}}\psi \left({2}^{j}n-k\right)$
where $\psi \left(m\right)$ is the basic expansion function of an integer variable $m$ . If these expansion functions are an orthogonal basis (or form a tight frame), the expansion coefficients (discrete-time wavelet transform) arefound from an inner product by
${d}_{j}\left(k\right)\phantom{\rule{0.166667em}{0ex}}=\phantom{\rule{0.166667em}{0ex}}⟨f\left(n\right),\psi \left({2}^{j}n-k\right)⟩\phantom{\rule{0.166667em}{0ex}}=\phantom{\rule{0.166667em}{0ex}}\sum _{n}f\left(n\right)\phantom{\rule{0.166667em}{0ex}}\psi \left({2}^{j}n-k\right)$
If the expansion functions are not orthogonal or even independent but do span ${\ell }^{2}$ , a biorthogonal system or a frame can be formed such that a transform and inverse can be defined.
Because there is no underlying continuous-time scaling function or wavelet, many of the questions, properties, and characteristics of the analysisusing the DWT in Chapter: Introduction to Wavelets , Chapter: A multiresolution formulation of Wavelet Systems , Chapter: Regularity, Moments, and Wavelet System Design , etc. do not arise. In fact, because of the filter bank structure for calculating the DTWT,the design is often done using multirate frequency domain techniques, e.g., the work by Smith and Barnwell and associates [link] . The questions of zero wavelet moments posed by Daubechies,which are related to ideas of convergence for iterations of filter banks, and Coifman's zero scalingfunction moments that were shown to help approximate inner products by samples, seem to have no DTWT interpretation.
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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Got questions? Join the online conversation and get instant answers! | 2,034 | 8,239 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-22 | longest | en | 0.943038 |
www.approvedscholars.com | 1,713,901,355,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818740.13/warc/CC-MAIN-20240423192952-20240423222952-00528.warc.gz | 584,706,928 | 17,982 | # ITS836 Assignment 6: Data Analysis in R – 100 points 1) Read the income dataset, “zipIncomeAssignment.csv”, into R.
ITS836 Assignment 6: Data Analysis in R – 100 points 1) Read the income dataset, “zipIncomeAssignment.csv”, into R.
2) Change the column names of your data frame so that zcta becomes zipCode and meanhouseholdincome becomes income.
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3) Analyze the summary of your data. What are the mean and median average incomes?
4) Plot a scatter plot of the data. Although this graph is not too informative, do you see any outlier values? If so, what are they?
5) In order to omit outliers, create a subset of the data so that: \$7,000 < income < \$200,000 (or in R syntax, income > 7000 & income < 200000)
7) Create a simple box plot of your data. Be sure to add a title and label the axes. HINT: Take a look at: https://www.tutorialspoint.com/r/r_boxplots.htm (specifically, Creating the Boxplot.) Instead of “mpg ~ cyl”, you want to use “income ~ zipCode”. In the box plot you created, notice that all of the income data is pushed towards the bottom of the graph because most average incomes tend to be low. Create a new box plot where the y-axis uses a log scale. Be sure to add a title and label the axes. For the next 2 questions, use the ggplot library in R, which enables you to create graphs with several different types of plots layered over each other. 8) Make a ggplot that consists of just a scatter plot using the function geom_point() with position = “jitter” so that the data points are grouped by zip code. Be sure to use ggplot’s function for taking the log10 of the y-axis data. (Hint: for geom_point, have alpha=0.2).
9) Create a new ggplot by adding a box plot layer to your previous graph. To do this, add the ggplot function geom_boxplot(). Also, add color to the scatter plot so that data points between different zip codes are different colors. Be sure to label the axes and add a title to the graph. (Hint: for geom_boxplot, have alpha=0.1 and outlier.size=0).
10) What can you conclude from this data analysis/visualization? Note: You are required to copy and paste all your R programming commands to your document and you should take the screenshot of your graphs. Your Word document should contain the report of the analysis of your data. Your Word document should have a distinct cover-page and reference-page as usual. Useful commands: install.packages(“ggplot2”) library(ggplot2)
## Sample Solution
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http://www.indiabix.com/aptitude/surds-and-indices/formulas | 1,432,376,153,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207927458.37/warc/CC-MAIN-20150521113207-00061-ip-10-180-206-219.ec2.internal.warc.gz | 518,816,298 | 6,547 | # Aptitude - Surds and Indices
@ : Home > Aptitude > Surds and Indices > Important Formulas
### Overview
• Important Formulas
### Exercise
"Do not wait for leaders; do it alone, person to person."
- Mother Teresa
1. Laws of Indices:
1. am x an = am + n
2. am = am - n an
3. (am)n = amn
4. (ab)n = anbn
5. a n = an b bn
6. a0 = 1
2. Surds:
Let a be rational number and n be a positive integer such that a(1/n) = a
Then, a is called a surd of order n.
3. Laws of Surds:
1. a = a(1/n)
2. ab = a x b
3. = a b
4. (a)n = a
5. (a)m = am | 201 | 552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | longest | en | 0.723633 |
http://priorart.ip.com/IPCOM/000102475 | 1,516,509,267,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890187.52/warc/CC-MAIN-20180121040927-20180121060927-00489.warc.gz | 271,716,573 | 8,229 | Dismiss
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# Optimized Drawing of Filled And Unfilled Circles in a Two-Dimensional Graphics System
IP.com Disclosure Number: IPCOM000102475D
Original Publication Date: 1990-Nov-01
Included in the Prior Art Database: 2005-Mar-17
Document File: 5 page(s) / 160K
IBM
## Related People
Fleischman, RL: AUTHOR [+2]
## Abstract
Most algorithms are quite efficient in drawing large unfilled circles. However, in most cases, the circles appearing on the screen are quite small and in each one of the algorithms the time to calculate the points on the circle can be as much as 90% of the total time taken to render the circle. To improve this situation a customized look-up table can be used to reduce the calculation time to a minimum.
This text was extracted from an ASCII text file.
This is the abbreviated version, containing approximately 38% of the total text.
Optimized Drawing of Filled And Unfilled Circles in a Two-Dimensional Graphics System
Most algorithms are quite efficient in drawing large
unfilled circles. However, in most cases, the circles appearing on
the screen are quite small and in each one of the algorithms the time
to calculate the points on the circle can be as much as 90% of the
total time taken to render the circle. To improve this situation a
customized look-up table can be used to reduce the calculation time
to a minimum.
To optimize the drawing of filled circles, a version of the
Bresenham algorithm is used.
Circle Algorithms In reviewing the literature [1,2,3,4] for the
implementation of drawing circles in graphics systems, it becomes
apparent that the only algorithm that minimizes errors in finding the
points on the perimeter of circle is the Bresenham algorithm. This
algorithm computes the points on the perimeter one by one starting
from some point and proceeding until the total circle is drawn. The
Bresenham algorithm is described in detail in [4].
In the case where the perimeter of the circle is to be drawn,
there is a need to compute a sequence of vectors that makes up the
perimeter of the circle, which are given to the hardware generator to
plot these vectors in the video pixel memory. These vectors should
be computed sequentially in a clockwise or counterclockwise fashion
to enable drawing the circle's perimeter with different line types.
In using the Bresenham algorithm, points are generated and
accumulated to create vectors which are then passed to the hardware
generator. The polar coordinate version also may be used to
calculate the points. Here a combination of both is implemented for
drawing perimeters.
Filled Circles Every point on the perimeter of the circle is
needed in order to fill a circle. The Bresenham algorithm, shown
below, is the most attractive algorithm since the Bresenham algorithm
computes all the perimeter points needed for the filled circle
procedure.
Bresenham Algorithm for a Single Octant
Parametric Variables: U, V, and E
Initially:
Let U(i) = 1
V(i) = E(i) - (2 * Radius) | 719 | 3,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-05 | latest | en | 0.912239 |
http://www.foxnews.com/food-drink/2013/10/17/mathematician-comes-up-with-equation-for-perfect-pizza.html | 1,524,721,885,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948089.47/warc/CC-MAIN-20180426051046-20180426071046-00171.warc.gz | 419,639,533 | 10,061 | Mathematician comes up with equation for perfect pizza
• A mathematician has come up with the first-ever formula for the "perfectly proportioned" pizza. (istock)
• Mathematician Eugenia Cheng was asked by chain restaurant PizzaExpress to work out why its 14 inch pizzas were so popular. (PizzaExpress)
Have you ever tried making pizza at home and ended up with a soggy mess? Well, pizza mistakes like this may be a thing of the past.
The Daily Mail reports Dr. Eugenia Cheng from the University of Sheffield created a mathematical formula that ensures a perfectly proportioned pizza every time.
While working with the pizza chain PizzaExpress to work out why its larger, thinner crust 14 inch pizza was proving so popular, compared to its classic 11inch, Cheng calculated a dough to topping ratio based on the thickness of the crust and size of the pizza.
She concluded that even if a person keeps the same amount of dough and topping, the ratio of topping to crust in an average bite changes with the size of the pizza. Smaller pizzas typically have more topping per bite than larger ones, but larger pizzas give diners a more even spread of toppings per bite.
She used d as the constant volume of dough and t for the constant volume of topping to come up with a mathematical formula for the ratio of topping to base in a median bite, reported the Mail.
More On This...
She explains the size of the pizza's crust is proportional to the thickness of the pizza - the larger the pizza, the thinner the base. For larger pizzas, it is important how toppings are arranged. The 14 inch pizza cooks just as evenly as the smaller version, but the topping is spread over a larger area on the 14 inch and goes close to the edge of the pizza.
We're not entirely convinced that you need a mathematical formula to figure this out. But, it's good to keep in mind that the next time you make pizza at home, make sure you don't overload it with too many toppings. And when in doubt, check out how Dominos does it.
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See all Trends | 462 | 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} | 3.5 | 4 | CC-MAIN-2018-17 | longest | en | 0.951817 |
https://cstheory.stackexchange.com/questions/32698/how-quickly-can-we-decompose-a-number-into-a-set-of-residues | 1,718,653,057,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861737.17/warc/CC-MAIN-20240617184943-20240617214943-00853.warc.gz | 164,750,559 | 38,859 | # How quickly can we decompose a number into a set of residues?
Basically, if we are given a natural $x$, in binary, with $x < m_0 \cdot m_1 \cdot m_2 \cdot \dots \cdot m_n$, how quickly can we find $x \bmod m_0$, $x \bmod m_1$,..., and $x \bmod m_n$? In other words, how quickly can we find $n$ residues in modular arithmetic? Note that the residues needn't be primes.
I suppose that this is a fairly broad question, and I'm curious if there are special cases where we can find these residues much faster (in other words, if there are special values for the modulus that allow faster calculations). My motivation is that I'm entertaining some ideas concerning integer multiplication. Along these lines, I'd like to break apart numbers into smaller residues. I'm interested in any results besides the trivial situation when some of the moduli are powers of 2.
If $m=2^k-t$ where $t$ is small, then you can reduce modulo $m$ faster. In particular, $a \cdot 2^k + b \equiv at+b \pmod{m}$, so reducing an $\alpha k$-bit number modulo $m$ typically requires $\alpha$ multiplications by $t$ and $\alpha$ additions. This can be significantly faster than the normal algorithms, if $t$ is small enough.
Something similar holds for $m=2^k+t$, too.
So, one plausible approach is to choose a modulus $m$ that is a product $m=m_1 m_2 \cdots m_n$ where each $m_i$ has the form $2^k \pm t$, where $k$ is the word size of your computer (e.g., $k=64$).
• I had actually considered this already. I considered finding values modulo products of these $m$'s, and successively breaking apart the products into smaller pieces. The idea was to reduce the amount of multiplications. Along slightly different lines, I figured that since the multiplications could possibly be repetitive, it may be worthwhile to look into multiplication by a constant - the constant being any intermediate value that is used multiple times. Unfortunately, none of these ideas seemed to really help much with efficiency. Commented Oct 4, 2015 at 3:43 | 508 | 2,009 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-26 | latest | en | 0.913333 |
https://stacks.math.columbia.edu/tag/0B39 | 1,674,928,479,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499646.23/warc/CC-MAIN-20230128153513-20230128183513-00847.warc.gz | 555,831,553 | 6,383 | ## 20.5 First cohomology and extensions
Lemma 20.5.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. There is a canonical bijection
$\mathop{\mathrm{Ext}}\nolimits ^1_{\textit{Mod}(\mathcal{O}_ X)}(\mathcal{O}_ X, \mathcal{F}) \longrightarrow H^1(X, \mathcal{F})$
which associates to the extension
$0 \to \mathcal{F} \to \mathcal{E} \to \mathcal{O}_ X \to 0$
the image of $1 \in \Gamma (X, \mathcal{O}_ X)$ in $H^1(X, \mathcal{F})$.
Proof. Let us construct the inverse of the map given in the lemma. Let $\xi \in H^1(X, \mathcal{F})$. Choose an injection $\mathcal{F} \subset \mathcal{I}$ with $\mathcal{I}$ injective in $\textit{Mod}(\mathcal{O}_ X)$. Set $\mathcal{Q} = \mathcal{I}/\mathcal{F}$. By the long exact sequence of cohomology, we see that $\xi$ is the image of a section $\tilde\xi \in \Gamma (X, \mathcal{Q}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ X, \mathcal{Q})$. Now, we just form the pullback
$\xymatrix{ 0 \ar[r] & \mathcal{F} \ar[r] \ar@{=}[d] & \mathcal{E} \ar[r] \ar[d] & \mathcal{O}_ X \ar[r] \ar[d]^{\tilde\xi } & 0 \\ 0 \ar[r] & \mathcal{F} \ar[r] & \mathcal{I} \ar[r] & \mathcal{Q} \ar[r] & 0 }$
see Homology, Section 12.6. $\square$
Comment #1606 by Keenan Kidwell on
The cohomology group at the end of the statement of the lemma should be $H^1(X,\mathcal{F})$.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 592 | 1,576 | {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 2, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-06 | latest | en | 0.532272 |
https://www.instructables.com/Rubyrint-rubik-cube-labyrinth/ | 1,653,799,282,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00458.warc.gz | 948,781,611 | 23,485 | # Rubyrint - Rubik Cube Labyrinth
10,142
65
16
## Introduction: Rubyrint - Rubik Cube Labyrinth
The purpose of this instructable is to build a fun game with and old rubik cube laying around.
Rubyrinth is a Rubik cube that has paths, forks and 2 gates on its faces instead of colors.
The goal of the game is to connect the 2 gates with the longest path
You can find more projects on my blog
## Step 1: Draw the Labyrinth
The first thing to do is draw the labyrinth. You can do it following these steps:
• cover the rubik cube faces with a white label
• draw the labyrinth with a pencil
• sketch the resulting faces on a paper
• design a cool looking labyrinth with an editor following the paper sketch
If you want to fast the things up, you can download my photoshop template and edit with your own labyrinth
NOTE: if you add a lot of forks, the game will become easier
## Step 2: Print & Cut
• Print the resulting image and cut each face
• Attach each face on the cube following the paper sketch produced in the first step
• Look at your amazing 3d labyrinth
## Step 3: Play
Now play with it! Challenge your friend to find the longest path that connects the two gates
Participated in the
Hack It! Challenge
## Recommendations
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• ### Chocolate Challenge
https://www.instructables.com/id/Carcassonne-Rubicks-Cube/
I will just use this print to make a maze in Minecraft lol.
Thanks for the awesome idea. i used your template and drew the maze onto a black, polished cube with a neon green Sharpie oil paint marker and then cleaned up the bubbly edges with a pocket knife. It actually looks slightly better than the one you made because the two colours contrast. Thanks a bunch!
nice idea the proton. I'll search for a 15 tiles game and try to do a copy
Am I correct that there are many solutions and not just one way to "connect the gstes" ?
pretty cool
Yes, actually the game is to fine the longest path among all the possible paths. But you can transform the tiles/faces and make only one path possible (hard to design..)
With this configuration is really easy to solve, but you can remove some forks and male it harder. | 530 | 2,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-21 | latest | en | 0.888126 |
https://www.mathworks.com/matlabcentral/cody/problems/2678-find-out-sum-and-carry-of-binary-adder/solutions/1978344 | 1,585,505,057,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370495413.19/warc/CC-MAIN-20200329171027-20200329201027-00412.warc.gz | 1,016,410,808 | 15,782 | Cody
# Problem 2678. Find out sum and carry of Binary adder
Solution 1978344
Submitted on 16 Oct 2019 by Alvin LAU
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y=1; pc=1; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,1)) assert(isequal(carry,1))
x = 1 1 sum = 1 carry = 1 sum = 1 carry = 1
2 Pass
x = 1; y=1; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,0)) assert(isequal(carry,1))
x = 1 0 sum = 0 carry = 1 sum = 0 carry = 1
3 Pass
x = 1; y=0; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,1)) assert(isequal(carry,0))
x = 0 1 sum = 1 carry = 0 sum = 1 carry = 0
4 Pass
x = 0; y=0; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,0)) assert(isequal(carry,0))
x = 0 0 sum = 0 carry = 0 sum = 0 carry = 0
5 Pass
x = 1; y=1; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,0)) assert(isequal(carry,1))
x = 1 0 sum = 0 carry = 1 sum = 0 carry = 1 | 423 | 1,055 | {"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} | 3.125 | 3 | CC-MAIN-2020-16 | latest | en | 0.630379 |
https://www.atlashunsingo.nl/20-07-49150-how-to-calculate-torque-of-roller-machine.html | 1,604,076,170,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911027.72/warc/CC-MAIN-20201030153002-20201030183002-00457.warc.gz | 603,474,551 | 7,932 | how to calculate torque of roller machine
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## Ways to Measure the Force Acting on a Rotating Shaft
Ways to Measure the Force Acting on a Rotating Shaft How can we be more efficient? ... In-line torque systems require cutting the shaft or lengthening the machine train to accommodate the inserted in-line transducer. ... The direct torque sensor measurement is preferred over the remote or indirect methods of calculating torque. Selecting a ...
## Torque and Equilibrium - HyperPhysics Concepts
Note that the torque is maximum when the angle is 90 degrees. A practical way to calculate the magnitude of the torque is to first determine the lever arm and then multiply it times the applied force. The lever arm is the perpendicular distance from the axis of rotation to the line of action of the force.
## Torque Calculator - Omni
Enter these values into our torque calculator. It uses the torque equation: τ = r * F * sinΘ = 0.5 * 120 * sin(90°) = 60 Nm. The torque calculator can also work in reverse, finding the force or lever arm if torque is given. If you want to learn more about the concept of force and Newton's second law, try the acceleration calculator.
## How To Calculate Torque Of Roller Machine
torque in rolling mills calculation formula with example. how to calculate torque of roller machine . torque calculation for roller use to crush material torque calculation for roller a …
## HOW TO CALCULATE TORQUE REQUIRED TO PRODUCE …
Aug 17, 2017· 1.HOW TO CALCULATE TORQUE REQUIRED TO PRODUCE POWER FOR A ROTATING MACHINE 2. HOW TO CALCULATE TORQUE FOR A ROTATING MACHINE 3. CALCULATE OR FIND TORQUE IF POWER SOURCE IS KNOWN HOW TO CALCULATE ...
## Torque and power requirement for roller | Physics Forums
Sep 14, 2014· Over the roller, a fabric in a roll form is loaded. The fabric roll sits over this roller by means of gravity. The roller has to drive the fabric roll on its circumference & feed the fabric end to another machine. Problem: 1. What is the torque required to drive the roller at the shaft end. 2.
## calculate torque of roller machine - riad-darailen.fr
how to calculate torque of roller machine - cict.co.in. Calculation of Torque for Drive Pulley of Roller Conveyor. May 06, 2017 · Calculation of Torque for Drive Pulley of Roller Conveyor How do we calculate how much is the Torque of the side the conveyor and tilt the machine. Get More Info.
## Motor Sizing Calculations - Alaa Khamis
Motor Sizing Calculations ... Calculate the value for load torque, load inertia, speed, etc. at the motor drive shaft of the mechanism. Refer to page 3 for calculating the speed, load torque and load inertia for various mechanisms. ... Total Inertia of the Machine [oz-in2]
## ActuatorsRotary Linear and Selection Calculations G ...
Calculate the values for load torque and load inertia at the motor drive shaft. Refer to the left column on page G-3 for the calculation of load torque for representative mechanisms. Refer to the right column on page G-3 for the calculation of inertia for representative shapes.
## how to calculate torque of roller machine - kesaias.eu
how to calculate torque of roller machine: ...ngineers Edge Forum | am looking for information on designing a belt driven roller conveyor. how would I calculate the torque The machine … How can I calculate the force required to bend a 10 mm We have a rolling machine with 3 How can I calculate the force required to bend a 10 mm thick plate How to calculate torque required in bottom ...
## How To Calculate Torque Of Roller Machine - rolvaplast.be
How To Calculate Torque Of Roller Machine - mbokodoinns.co.za. Torque, moment, or moment of force is rotational force. Just as a linear force is a push or a pull, a torque …
## how to calculate torque of roller machine | Solution for ...
A roller chain transfers power from one sprocket to another and … agricultural machines,… How to Calculate … torque in rolling mills calculation formula with example … Home»Grinding Machine»torque in rolling mills calculation formula with example. … How to Calculate the Torque of a Rotating Roll | eHow.com. Calculate torque.
## Power Screws Design Equation and Calculator | Engineers ...
In many applications, the load slides relative to a collar, thereby requiring an additional input torque T c: T c = F µ c d c / 2. Ball and tapered-roller thrust bearings can be used to reduce the collar torque. The starting torque is obtained by substituting the static coefficients of friction into the above equations.
## how to calculate torque of roller machine - pavages.be
We have a rolling machine with 3 rollers; the bottom has two rollers with diameters of 168 mm and one top roller with a diameter of 219 mm. I want to calculate the required force to bend a 10 mm thick plate which is 2,500 mm wide. The rolling machine is connected to a 10 HP motor with a gear box and the rollers have an rpm of 8.
## Setting techniques for tapered roller bearings | Machine ...
Tapered roller bearings can be set to any desired axial or radial clearance when installed in equipment being manufactured on the production line. This feature lets a designer control bearing ...
## Calculating Torque With Examples - ThoughtCo
Oct 18, 2018· To begin calculating the value of the torque, you have to realize that there's a slightly misleading point in the above set-up. (This is a common problem in these situations.) Note that the 15% mentioned above is the incline from the horizontal, but that's not the angle θ. The angle between r and F has to be calculated.
## How to Calculate Frictional Torque | Sciencing
Mar 13, 2018· Torque is described as a force acting a measured distance from a fixed axis, such as a door rotating on a hinge or a mass suspended from a rope that is hung across a pulley. The torque can be affected by an opposing force that results from a resistant surface. This opposing force is …
## Book - The Mechanics of Rollers - tappi.org
35 Cantilevered Machines 36 Cantilever Calculation Example 37 Roller Deflection Summary 38 Bibliography 4 Traction, Torque and Power 39 Why Traction Is Important 41 Which Mode Is In Effect 42 The Normal Entry Law 43 Static Coefficient of Friction 44 Friction Complications 45 Air Entrainment 46 Air Entrainment Model and Application
## How to Calculate Total Mass or Load of Conveyor for Load ...
May 13, 2015· I want to select a motor for a conveyor or machine, for Load torque calculation, i need the weight of the parts which has to move by motor. Now the problem is that,can i simply take full weight of all parts which has to move or carry? or is there is any specific method to take weight of all parts like chain, shaft, belts ? for example like: if there is shaft with pulleys on two pillow blocks ...
## Simple Torque Calculation for Belt Conveyor - bulk-online
Mar 09, 2006· Simple torque calculation for belt conveyor i need a simple (or estimate) formula to calculate torque (to turn)for belt conveyor application based on belt length, overall load, belt speed, pully drum diameter, and incline angle.
## How can I calculate the force required to bend a 10 mm ...
We have a rolling machine with 3 rollers; the bottom has two rollers with diameters of 168 mm and one top roller with a diameter of 219 mm. I want to calculate the required force to bend a 10 mm thick plate which is 2,500 mm wide. The rolling machine is connected to a 10 HP motor with a gear box and the rollers have an rpm of 8.
## Load Analysis and Driven Power Calculation (4 Roll Bending ...
The upper roller of 4 rolls bending machine is driven roller. The total drive torque acting on the upper roller is adding up the torque consumed on the deformation and overcome the friction. Friction torque including frictional resistance consumption to overcome shaft roller rolling on bending plate and torque consumption in the roller bearing ...
## Drive Wheel Motor Torque Calculations - mae.ufl.edu
The Wheel Torque calculated in Step Five is the total wheel torque. This quantity does not change with the number of drive wheels. The sum of the individual drive motor torques (see . Motor Specifications) must be greater than or equal to the computed Wheel Torque. The Maximum Tractive Torque represents the maximum amount of torque that can be ...
## Calculation of Torque for Drive Pulley of Roller Conveyor ...
May 06, 2017· To move the conveyor, we install Drive Pulley with diameter of 550mm, and connected to Gear Box (RPM = 95) + Motor (5,5 HP ; RPM 1,500) You just put these on the sides of an equation and put in what you know to what you want to find out. It is difficult to find the torque …
## Lead Screw Torque and Force Calculator - daycounter.com
Lead Screw Torque and Force Calculator. When designing machinery that uses lead screws, it's a common task to try and figure out the size of motor needed to drive a given force with a lead screw. This calculator will calculate torque given the lead screw parameters and the required force.
## Physics - Mechanics: Application of Moment of Inertia and ...
Mar 27, 2013· Physics - Mechanics: Application of Moment of Inertia and Angular Acceleration (2 of 2) Michel van Biezen. ... Torque, Moment of Inertia, ...
## How to Calculate Torque: 8 Steps (with Pictures) - wikiHow
Jun 28, 2019· How to Calculate Torque. Torque is best defined as the tendency of force to rotate an object about an axis, fulcrum, or pivot. You can calculate torque by using force and the moment arm (the perpendicular distance from an axis to the line...
## Bearing Torque | Machine Design
Bearing torque is the moment required to overcome internal friction to start or maintain rotation of one ring while the other is stationary. Torque or friction generally increases when: Bearing ...
## Motor Torque Calculations - NEPSI
Known variables: Weight (lbs), Linear Velocity (ft/min), Speed of Driving Motor (RPM), Change in Speed (RPM), and Time to Accelerate Total System (sec) In addition to the torque required to drive the load at a steady speed, torque is required to accelerate the load. | 2,266 | 10,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-45 | latest | en | 0.801192 |
http://motorsports-network.com/j2sbt/shear-modulus-formula-e69c6e | 1,627,729,405,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.6/warc/CC-MAIN-20210731105716-20210731135716-00189.warc.gz | 29,003,765 | 10,474 | The elastic modulus for tensile stress is called Young’s modulus; that for the bulk stress is called the bulk modulus; and that for shear stress is called the shear modulus. Let's explore a new modulus of elasticity called shear modulus (rigidity modulus). When a shear force is applied on a body which results in its lateral deformation, the elastic coefficient is called the shear modulus of elasticity. Calculate Shear Modulus from Young’s Modulus (1) Calculate Shear Modulus from the Bulk Modulus (2) Calculate Bulk Modulus from Young’s Modulus (3) Calculate Bulk Modulus from the Shear Modulus (4) Calculate Young’s Modulus from the Shear Modulus (5) E = Young Modulus of Elasticity. The empirical temperature dependence of the shear modulus in the SCG model is replaced with an equation based on Lindemann melting theory. 4.6.1 Shear and Bulk Moduli. It is given as:G=FlAΔxG=\frac{Fl}{A\Delta x}G=AΔxFl Where, SI unit of G isPascali.e. Shear modulus is the slope of the linear elastic region of the shear stress–strain curve and Poisson's ratio is defined as the ratio of the lateral and axial strain. G = 1.25 *10 6 N/m 2. L is the perpendicular distance (on a plane perpendicular to the force) to the layer that gets displaced by an extent x, from the fixed layer. This is because large shearing forces lead to permanent deformations (no longer elastic body). The shear modulus is defined as the ratio of shear stress to shear strain. The shear strain is defined as ∆x/L. Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree.... Help support true facts by becoming a member. To compute for shear modulus, two essential parameters are needed and these parameters are young’s modulus (E) and Poisson’s ratio (v). Shear strain. Because the denominator is a ratio and thus dimensionless, the dimensions of the shear modulus are those of … Shear Modulus Calculator. Unit of shear modulus is Nm–2 or pascals (Pa). It is denoted by G . It is the ratio of shear stress to shear strain, where shear strain is defined as displacement per unit sample length. {{#invoke:Citation/CS1|citation Conceptually, it is the ratio of shear stress to shear strain in a body. Shear modulus of the material of a body is given by Relation Between the Moduli of Elasticity: Numerical Problems: Example – 1: The area of the upper face of a rectangular block is 0.5 m x 0.5 m and the lower face is fixed. The plus sign leads to νâ¥0{\displaystyle \nu \geq 0}. The shear modulus S is defined as the ratio of the stress to the strain. In materials science, shear modulus or modulus of rigidity, denoted by G, or sometimes S or μ, is defined as the ratio of shear stress to the shear strain: The following chart gives typical values for the shear modulud of rigidity. the Steinberg-Cochran-Guinan (SCG) shear modulus model developed by, the Nadal and LePoac (NP) shear modulus model. It measures the rigidity of a b ody. shear modulus with increasing level of treatment, and, therefore, a correlation between the two could be derived. |CitationClass=book These are all most useful relations between all elastic constant which are used to solve any engineering problem related to them. The simplest formula is the ratio of Shear Force and the Area on which it is acting. 2) A cylindrical bar of width 10 mm is stretched from its original length to 10 mm using a force of 100 N. What is the Shear modulus of the system? shear modulus = (shear stress)/(shear strain) = (F/A)/(x/y) . This equation is a specific form of Hooke’s law of elasticity. The following equation is used to calculate a shear modulus of a material. Solution: Given. Common sense and the 2nd Law of Thermodynamics require that a positive shear stress leads to a positive shear strain. It is expressed in Pascals (Pa), gigapascals (GPa) or KSI. The height of the block is 1 cm. Other elastic moduli are Young’s modulus and bulk modulus. The basic difference between young’s modulus, bulk modulus, and shear modulus is that Young’s modulus is the ratio of tensile stress to tensile strain, the bulk modulus is the ratio of volumetric stress to volumetric strain and shear modulus is the ratio of shear stress to shear strain. Modulus of Rigidity calculation is made simple here. This page was last edited on 13 September 2014, at 19:24. Experiments have found one known Formula which calculates the shear modulus from the matrix and fibers young modulus multiplied with with the volumes fractions : see my papers. G = F * L / A * D. Where G is the shear modulus (pascals) F is the force (N) L is the initial length (m) A is the area being acted on (m^2) D is the transfer displacement (m) In order to do this, you need the modulus of elasticity and shear modulus to determine deflection. Relation between Young Modulus, Bulk Modulus and Modulus of Rigidity: Where. (224) in the case of shear modulus evolution is plotted in Fig. Pa. Shear Modulus is related to other Elastic Moduli of the Material. By signing up for this email, you are agreeing to news, offers, and information from Encyclopaedia Britannica. This valuable property tells us in advance how resistant a material is to shearing deformation. The image above represents shear modulus. G = Shear Modulus, also known as Modulus of Rigidity; K = Bulk Modulus = Poisson’s Ratio . The Nadal-Le Poac (NP) shear modulus model is a modified version of the SCG model. Answer: The shear modulus is calculated using the formula, G= σ / ϵ. G = (5*10 4 N/m 2)/(4*10 (-2)) = 1.25 *10 6 N/m 2. In English units, shear modulus is given in terms of pounds per square inch (PSI) or kilo (thousands) pounds per … Homogeneous isotropic linear elastic materials have their elastic properties uniquely determined by any two moduli among these; thus, given any two, any other of the elastic moduli can be calculated according to these formulas. It is defined as the ratio of shear stress and shear strain. For the shear modulus evolution, x 0 and x ∞ in Eq. Therefore, the shear modulus G is required to be nonnegative for all materials, Question 1: Compute the Shear modulus, if the stress experienced by a body is 5×10 4 Nm 2 and strain is 4×10-2. https://www.britannica.com/science/shear-modulus. }}, https://en.formulasearchengine.com/index.php?title=Shear_modulus&oldid=238966. ShearModulus (G) =Shear stress/Shear strain. Answer: The shear modulus is found from the equation: G= (F L) / (A Δx) If a material is very resistant to attempted shearing, then it will transmit the shear energy very quickly. Stay tuned with BYJU’S to learn more on other Physics related concepts. K = Bulk Modulus . In materials science, shear modulus or modulus of rigidity, denoted by G, or sometimes S or μ, is defined as the ratio of shear stress to the shear strain: Young's Modulus from shear modulus can be obtained via the Poisson's ratio and is represented as E=2*G* (1+) or Young's Modulus=2*Shear Modulus* (1+Poisson's ratio). Young’s modulus. What is Shear Modulus? Shear-modulus (G): Where ρ is the density of the material and V s is the pulse velocity of the S-wave. Shear Modulus Formula. Stress = 5×10 4 Nm 2. The compression spring is a basic standard part used in a wide variety of machine design applications and mechanisms. G = Modulus of Rigidity. The formula for calculating the shear modulus: G = E / 2(1 + v) Where: G = Shear Modulus E = Young’s Modulus v = Poisson’s Ratio. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. There are two valid solutions. Shear modulus can be represented as; $$Shear.Modulus=frac{Shear.Stress}{Shear.Strain}$$ ¨ $$G=frac{f_{s}}{e_{s}}$$ Shear modulus is also known as modulus of elasticity of modulus of rigidity and it is the ratio of shear stress to shear strain. Mokarram Hossain, Paul Steinmann, in Advances in Applied Mechanics, 2015. The shear modulus itself may be expressed mathematically as. The shear modulus of material gives us the ratio of shear stress to shear strain in a body. a shearing force applied to the top face produces a displacement of 0.015 mm. Strain = 4×10-2. This relationship is given as below: E=2G(1+μ)E= 2G ( 1+\mu )E=2G(1+μ) And E=3K(1–2μ)E = 3K ( 1 – 2 \mu )E=3K(1–2μ) Where, The minus sign leads to νâ¤0{\displaystyle \nu \leq 0}. Young’s Modulus or Elastic Modulus or Tensile Modulus, is the measurement of mechanical properties of linear elastic solids like rods, wires, etc. Gain in Dynamic Shear Modulus Gains in dynamic shear modulus with treatment level for the sand, silty clay and the benton ite clay are shown in Figs. It is also known as the modulus of rigidity and may be denoted by G or less commonly by S or μ. S=±E2+9â¢M2â10â¢Eâ¢M{\displaystyle S=\pm {\sqrt {E^{2}+9M^{2}-10EM}}}. = 1), p is the pressure, and T is the temperature. This will also explain why our bones are strong and yet can be fractured easily. 9. Measured using the SI unit pascal or Pa. I know you can determine the shear modulus using Poissons ratio but doing testing to determine poissons seems a little excessive. Using the equations above we can determine Poisson’s Ratio (ν): So Poisson’s ratio can be determined simply by measuring the P-wave velocity and the S-wave Young’s modulus … Some of these are Bulk modulus and Shear modulus etc. The dimensional formula of Shear modulus is M 1 L-1 T-2. The SI unit of shear modulus is the Pascal (Pa), but values are usually expressed in gigapascals (GPa). It can be used to explain how a material resists transverse deformations but this is practical for small deformations only, following which they are able to return to the original state. Shear Modulus of elasticity is one of the measures of mechanical properties of solids. ShearModulus (G) = (5×10 4)/ (4×10-2) ShearModulus (G) = 1.25×10 6 Nm 2. Then, shear modulus: G = s h e a r s t r e s s s h e a r s t r a i n = F / A x / L = F L A x. The ratio of tensile stress to tensile strain is called young’s modulus. Let’s solve an example; Here is the Shear Modulus Calculator to calculate the Shear modulus or modulus of rigidity. The shear modulus G is also known as the rigidity modulus, and is equivalent to the 2nd Lamé constant m mentioned in books on continuum theory. In materials science, shear modulus or modulus of rigidity, denoted by G, or sometimes S or μ, is defined as the ratio of shear stress to the shear strain: = = / / = where = / = shear stress; is the force which acts Example 1. |CitationClass=journal Influences of selected glass component additions on the shear modulus of a specific base glass. Note that the relation between stress and strain is an observed relation, measured in the laboratory. There are some other numbers exists which provide us a measure of elastic properties of a material. Shear modulus tells how effectively a body will resist the forces applied to change its shape. Is this comparable for concrete as well? For masonry, they advise using a shear modulus of 0.4 X modulus of elasticity. Be on the lookout for your Britannica newsletter to get trusted stories delivered right to your inbox. All data can be recalculated and the is a … The NP shear modulus model has the form: and µ0 is the shear modulus at 0 K and ambient pressure, ζ is a material parameter, kb is the Boltzmann constant, m is the atomic mass, and f is the Lindemann constant. The dimensional formula of Shear modulus is M 1 L-1 T-2. The shear modulus is the earth’s material response to the shear deformation. In the Compression Spring Design article, we presented the basic formula for any spring constant: F = kΔH = k(Hfree-Hdef) where Hfree is uncompressed spring length and Hdef is spring length as a result of force applied, and the basic formula for a compression coil spring constant k= (Gd4) / 8D3Na where G is the S… (224) are replaced by initial and final shear moduli μ in and μ ∞, respectively, as well as the curvature parameter κ p by κ μ.An illustration of Eq. }}, {{#invoke:citation/CS1|citation I need to calculate shear modulus … ( NP ) shear modulus or modulus of rigidity is M 1 L-1 T-2 an equation based Lindemann! Why our bones are strong and yet can be fractured easily is used to solve any problem. But values are usually expressed in gigapascals ( GPa ) defined as the modulus of gives... ) shear modulus or modulus of 0.4 x modulus of rigidity and may be denoted by or. Determine the shear modulus evolution, x 0 and x ∞ in Eq two be! Which provide us a measure of elastic properties of solids Steinmann, Advances! Britannica newsletter to get trusted stories delivered right to your inbox advance how resistant a material to... Elastic properties of a material and LePoac ( NP ) shear modulus if! Shear-Modulus ( G ) = ( 5×10 4 Nm 2 and strain is called young ’ law. A wide variety of machine design applications and mechanisms will resist the forces to... Positive shear stress to shear strain is 4×10-2 Moduli are young ’ s modulus … = )... A body is 5×10 4 Nm 2, the Nadal and LePoac ( NP ) shear modulus evolution is in. ( G ): Where gives us the ratio of the SCG model is a specific base.... The empirical temperature dependence of the material observed relation, measured in the case of shear stress shear! Density of the shear modulus, Bulk modulus and modulus of rigidity and be... Is 5×10 4 ) / ( x/y ) or modulus of rigidity and may denoted... Per unit sample length order to do this, you are agreeing to news, offers, and from! Calculate a shear modulus using Poissons ratio but doing testing to determine deflection G ) Where. = 1.25×10 6 Nm 2 and strain is an observed relation, measured in the laboratory, unit. The SCG model is replaced with an equation based on Lindemann melting theory, you need the modulus of material... Melting theory ( shear strain in a wide variety of machine design applications and.! Are Bulk modulus and Bulk modulus less commonly by s or μ attempted shearing, then it will the... Pa. Mokarram Hossain, Paul Steinmann, in Advances in applied Mechanics,.! Material response to the top face produces a displacement of 0.015 mm -10EM } } delivered. Offers, and T is the pulse velocity of the S-wave the shear modulus is related to them tuned BYJU... 1 L-1 T-2 to shearing deformation of machine design applications and mechanisms usually expressed in (! ( Pa ), but values are usually expressed in pascals ( Pa ) body.! Displacement of 0.015 mm component additions on the shear energy very quickly,... { E^ { 2 } +9M^ { 2 } +9M^ { 2 } +9M^ { }! { 2 } +9M^ { 2 } +9M^ { 2 } +9M^ { 2 } -10EM } } } our. To attempted shearing, then it will transmit the shear modulus of rigidity: Where very resistant to attempted,... Commonly by s or μ \geq 0 } to determine deflection ) shear modulus of rigidity may. And the 2nd law of elasticity and shear modulus with increasing level of treatment, and T the! Attempted shearing, then it will transmit the shear modulus etc on lookout... Solve any engineering problem related to other elastic Moduli of the measures of mechanical properties of solids modulus with level. Applied Mechanics, 2015 increasing level of treatment, and, therefore, shear modulus formula between! Modulus to determine Poissons seems a little excessive is defined as the ratio of modulus... Your Britannica newsletter to get trusted stories delivered right to your inbox which it is expressed in (... Measured using the SI unit Pascal or pa. Mokarram Hossain, Paul Steinmann, in Advances applied. 6 Nm 2 are used to solve any engineering problem related to them can fractured. In pascals ( Pa ), p is the density of the S-wave question:. Minus sign leads to νâ¥0 { \displaystyle \nu \geq 0 } commonly by or! A shearing force applied to the top face produces a displacement of 0.015 mm,. Other elastic Moduli of the material your Britannica newsletter to get trusted stories delivered right your! Shearing, then it will transmit the shear modulus … = 1,. Be fractured easily the pressure, shear modulus formula, therefore, a correlation the! Strain in a body 6 Nm 2 and strain is an observed,... Base glass is replaced with an equation based on Lindemann melting theory to..., at 19:24 basic standard part used in a wide variety of machine applications. Elasticity and shear modulus is related to them is also known as ratio! Is very resistant to attempted shearing, then it will transmit the shear modulus is the ratio shear... Require that a positive shear stress to shear strain, Where shear.... Material and V s is defined as the ratio of shear modulus or modulus of a is. Modulus, Bulk modulus more on other Physics related concepts ): Where tuned with ’! ( 224 ) in the laboratory property tells us in advance how resistant a material any! Stress and strain is an observed relation, measured in the laboratory: Compute the shear modulus developed. Material gives us the ratio of shear modulus evolution, x 0 and ∞... You are agreeing to news, offers, and information from Encyclopaedia Britannica attempted,... Trusted stories delivered right to your inbox developed by, the Nadal and LePoac ( NP ) shear?. 13 September 2014, at 19:24 is to shearing deformation shearing forces lead to permanent deformations ( no longer body... Poac ( NP ) shear modulus of 0.4 x modulus of elasticity of solids the modulus of a material to! Stress and shear modulus is M 1 L-1 T-2 … What is modulus... The modulus of a material here is the pressure, and T is the density of the S-wave { x... 2Nd law of elasticity the two could be derived, in Advances in applied Mechanics 2015. On which it is defined as the ratio of tensile stress to shear strain be on the shear etc! Constant which are used to calculate the shear modulus model is replaced with an based. In the laboratory example ; shear strain ) = ( 5×10 4 ) / ( x/y ) the temperature. 1: Compute the shear modulus of elasticity variety of machine design applications and mechanisms in a wide variety machine. { E^ { 2 } +9M^ { 2 } +9M^ { 2 } +9M^ { 2 +9M^. Of mechanical properties of solids is very resistant to attempted shearing, then will... The ratio of shear stress to the strain edited on 13 September 2014, at 19:24 ( ). The SCG model ) in the case of shear stress to the top face produces a of. One of the measures of mechanical properties of solids shear deformation F/A ) / ( x/y ) an relation! Np ) shear modulus evolution, x 0 and x ∞ in Eq are shear modulus formula. S modulus and shear modulus tells how effectively a body are used to solve any engineering problem to... Positive shear stress leads to a positive shear strain Steinmann, in Advances in applied Mechanics,.! Ratio of shear stress to shear strain the compression spring is a specific form of Hooke ’ shear modulus formula response. Modulus with increasing level of treatment, and information from Encyclopaedia Britannica equation... Denoted by G or less commonly by s or μ newsletter to trusted. Resistant to attempted shearing, then it will transmit the shear modulus model is a specific base glass = )... Advances in applied Mechanics, 2015 the S-wave and modulus of rigidity and may be denoted G! The shear modulus, Bulk modulus and modulus of material gives us the ratio of shear stress the... | 4,762 | 19,325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2021-31 | latest | en | 0.887145 |
http://mp3pretraga.com/find-sample-size-desired-margin-error/ | 1,526,817,032,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863410.22/warc/CC-MAIN-20180520112233-20180520132233-00473.warc.gz | 193,067,440 | 5,981 | # Find Sample Size Desired Margin Error
RECOMMENDED: If you have Windows errors then we strongly recommend that you download and run this (Windows) Repair Tool.
is the population standard deviation, n is the sample size, and z* is the appropriate z*-value for your desired level of confidence (which you can find in the.
Overall, I’ll take what the rotation has shown as a whole for the entire season as a better sample size of what to expect in October. a best of five – or even a one.
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A tutorial on computing the sampling size for the desired margin of error of population proportion estimate at given confidence level. Tags: Elementary Statistics.
The critical value is a factor used to compute the margin of error. As a rough guide, many statisticians say that a sample size of 30 is large enough when the.
Confidence Interval Margin of Error for a Population Proportion. Calculating n for Estimating a Mean. Example. Suppose that you were interested in the.
How Large of a Sample Size Do Is Needed for a Certain Margin of Error. The following is an example of how we can use the formula to calculate the desired sample size.
. the sample size is based on how detailed a comparison is desired. if the sample size is 100, the margin of error is. Sampling and Sample Size Author: user
Survey Sample Sizes and Margin of Error. Written by Robert Niles. The most accurate survey of a group of people is a vote: Just ask everyone to make a decision and.
La Marge – If you read polls in the news, you’re probably familiar with the term "margin of error. Based on the sample size (and some other factors) and utilizing statistics, the pollster can calculate the margin of sampling error. This describes how.
Finding Sample Size with Predetermined Margin of Error and. – We show you how to calculate a desired sample size given a margin of error and confidence level. Finding Sample Size with Predetermined Margin of Error and Level.
Results of the full poll of 569 farmers across Ireland indicate that the biggest medium term problem for Irish dairy farmers is not being able to find enough. The.
Among the red flags you should watch for are surveys that fail to disclose the survey methodology, sample size, and margin of error. These details may be. to.
Given margins of error, the number could be as high as 23,000 — about as many.
Learn how to determine the sample size necessary for correctly representing your. to produce results accurate to a specified confidence and margin of error.
Explanation. The margin of error is usually defined as the "radius" (or half the width) of a confidence interval for a particular statistic from a survey.
Required Sample Size n to Estimate a Population Proportion p.95. The desired Margin of Error is ME =.03; Curdle wants to be 90% confident, so z*=1.645; the.
How to determine population and survey sample size? – CheckMarket – In addition, it is desired to calculate the average score obtained by age. You can calculate margin of error for your population and sample size using our. | 734 | 3,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-22 | latest | en | 0.894508 |
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Helper I
## Moving/rolling avarage (R12) of a cummulative measure
Hi,
I need help to build a measure to calculate a rolling avarage of a measure
These are the measures:
``WIP Month = [Poured]-SUM(Delivery[Peso])-SUM(Scrap[Peso])``
``````WIP cummulated =
CALCULATE(
[WIP month],
FILTER(ALL(Calendario),Calendario[Date]<=MAX(Calendario[Date])&&
Calendario[Date]>=DATE(2021,01,01)))``````
I need a measure for rolling 12 average "WIP cummulated"
Something like this (R12M):
(this was done in manually in excel)
Thiago Duarte
1 ACCEPTED SOLUTION
Community Support
You can create a measure as below and check if it can return the correct result...
``````WIP R12M =
VAR NumOfMonths = 12
VAR _selyear =
SELECTEDVALUE ( 'Calendario'[Ano] )
VAR _selmonth =
SELECTEDVALUE ( 'Calendario'[Monthno] )
VAR _selym =
VALUE ( _selyear & IF ( _selmonth >= 10, _selmonth, "0" & _selmonth ) )
VAR _date =
DATE ( 2021, 1, 1 ) + 365
VAR _date2 =
IF ( DAY ( _date ) > 1, _date + 1, _date )
VAR _ym =
VALUE ( YEAR ( _date2 ) & FORMAT ( _date2, "mm" ) )
VAR LastCurrentDate =
MAX ( 'Calendario'[Date] )
VAR Period =
DATESINPERIOD ( 'Calendario'[Date], LastCurrentDate, - NumOfMonths, MONTH )
VAR Result =
CALCULATE (
AVERAGEX (
GROUPBY ( 'Calendario', 'Calendario'[Ano], 'Calendario'[Mes] ),
[WIP cummulated]
),
Period
)
RETURN
IF ( _selym >= _ym, Result )``````
Rolling 12 Months Average in DAX
Rolling Averages In Power BI
If the above one can't help you get the desired result, please provide some sample data in the table Delivery, Scrap and other fact table(exclude sensitive data) with Text format and your expected result with backend logic and special examples. And share the visual field settings as below screenshot. It is better if you can share a simplified pbix file. You can refer the following link to upload the file to the community. Thank you.
How to upload PBI in Community
Best Regards
Community Support Team _ Rena
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
3 REPLIES 3
New Member
Hello there ,
The same moving or rolling avarage I wanted for Hourly basis, is this possible with above solution?
Community Support
You can create a measure as below and check if it can return the correct result...
``````WIP R12M =
VAR NumOfMonths = 12
VAR _selyear =
SELECTEDVALUE ( 'Calendario'[Ano] )
VAR _selmonth =
SELECTEDVALUE ( 'Calendario'[Monthno] )
VAR _selym =
VALUE ( _selyear & IF ( _selmonth >= 10, _selmonth, "0" & _selmonth ) )
VAR _date =
DATE ( 2021, 1, 1 ) + 365
VAR _date2 =
IF ( DAY ( _date ) > 1, _date + 1, _date )
VAR _ym =
VALUE ( YEAR ( _date2 ) & FORMAT ( _date2, "mm" ) )
VAR LastCurrentDate =
MAX ( 'Calendario'[Date] )
VAR Period =
DATESINPERIOD ( 'Calendario'[Date], LastCurrentDate, - NumOfMonths, MONTH )
VAR Result =
CALCULATE (
AVERAGEX (
GROUPBY ( 'Calendario', 'Calendario'[Ano], 'Calendario'[Mes] ),
[WIP cummulated]
),
Period
)
RETURN
IF ( _selym >= _ym, Result )``````
Rolling 12 Months Average in DAX
Rolling Averages In Power BI
If the above one can't help you get the desired result, please provide some sample data in the table Delivery, Scrap and other fact table(exclude sensitive data) with Text format and your expected result with backend logic and special examples. And share the visual field settings as below screenshot. It is better if you can share a simplified pbix file. You can refer the following link to upload the file to the community. Thank you.
How to upload PBI in Community
Best Regards
Community Support Team _ Rena
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Helper I
Thank you @v-yiruan-msft!
That worked beautifully. I just had to make a few adjustments in how de dates are declared
``````VAR _selyear =
SELECTEDVALUE ( 'Calendario'[Ano] )
Became
VAR _selyear =
SELECTEDVALUE ( Calendario[Date].[Ano] )``````
and so on
Now i'll try and understand your code so i can learn from it
Thanks again!
Best Regards,
Thiago Duarte
Announcements | 1,240 | 4,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-33 | latest | en | 0.712249 |
https://holooly.com/solutions-v20/a-three-story-office-building-located-on-main-street-in-nyack-ny-in-close-proximity-to-the-hudson-river-figure-1-8-has-an-upwind-direction-from-the-river-in-order-to-calculate-the-at-roof-snow-lo/ | 1,659,953,473,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570793.14/warc/CC-MAIN-20220808092125-20220808122125-00265.warc.gz | 296,417,603 | 27,058 | Products
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## Q. 1.2
A three-story office building located on Main Street in Nyack, NY, in close proximity to the Hudson River (Figure 1.8) has an upwind direction from the river. In order to calculate the at roof snow load for the building, we need to determine the variables in Equation 1.2.
$p_{f}=0.7C_{e}C_{t}I_{s}p_{g}$ (1.2)
## Verified Solution
Step 1: Obtain the ground snow load ($p_{g}$ ) from Figure 7.1 in ASCE 7-10. The ground snow load for Nyack, NY is 30 psf, ($p_{g}=30 psf$).
Step 2: Determine the roof exposure. The exposure factor, $C_{e}$ , is based on the wind exposure of the building and the surface roughness. The building, as shown in Figure 1.8, is 40 ft tall, and has a surface roughness in compliance with that of surface roughness B. The surface roughness prevails for a distance of approximately 2200 ft, which is less than the 2600 ft required to satisfy the condition for exposure category B. Additionally, exposure category D is not satisfied and consequently exposure category C prevails. Hence, use terrain category C in Table 1.3, and for a fully exposed roof,$C_{e}$ =.9.
Step 3: Determine the thermal factor, $C_{t}$ , which is based on the thermal condition of the building described in Table 1.4, hence $C_{t}$=1.0
Step 4: Determine the importance factor, $I_s$, which is based on the risk category assignment of the building in Table 1.5-1, in ASCE 7-10. Risk categories I, II, III, and IV are based on the potential loss of life during a catastrophic failure. According to Table 1.5-1, an office building has a risk category of II. From Table 1.5, a risk category II has an importance factor, $I_{s}$= 1.0.
Step 5: Finally, the at roof snow load for the building in Figure 1.8 is found from Equation 1.2
$p_{f}=0.7C_{e}C_{t}I_{s}p_{g} \\p_{f}=0.7(.9)(1.0)(1.0)(30 psf)=18.9 Ib/ft^2$
Table 1.3 Exposure factor, $C_e$ Terrain category Exposure of roof Fully exposed Partially exposed Sheltered B 0.9 1.0 1.2 C 0.9 1.0 1.1 D 0.8 0.9 1.0 Above tree line in mountainous areas 0.7 0.8 N/A
Table 1.4 Thermal factor, $C_t$ Thermal condition $C_{t}$ All structures except as indicated below 1.0 Structures keep just above freezing and others with cold, ventilated roofs in which the thermal resistance (R-value) between the ventilated space and the heated space exceeds 25°F × h × ft/Btu 1.1 Unheated and open air structures 1.1 Structures intentionally kept below freezing 1.2 Continuously heated greenhouses with a roof having a thermal resistance (R-value) 1.3 ess than 2.0°F × h × $ft^2$/Bt 0.85
Table 1.5 Important factors by risk category of buildings for snow, ice, and earthquake loads Risk category from Table 1.5-1 Snow importance factor, $I_{s}$ Ice importance factor–thickness, $I_{i}$ Ice importance factor–wind, $I_{w}$ Seismic importance factor, $I_{e}$ I 0.80 0.80 1.00 1.00 II 1.00 1.00 1.00 1.00 III 1.10 1.25 1.00 1.25 IV 1.20 1.25 1.00 1.50 | 1,025 | 3,423 | {"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": 0, "wp-katex-eq": 18, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2022-33 | latest | en | 0.904528 |
https://www.shaalaa.com/question-bank-solutions/show-that-every-positive-integer-form-2q-that-every-positive-odd-integer-2q-1-where-q-some-intege-euclid-s-division-lemma_5559 | 1,638,407,628,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.58/warc/CC-MAIN-20211201234046-20211202024046-00342.warc.gz | 1,052,757,978 | 9,016 | # Show that every positive integer is of the form 2q and that every positive odd integer is of the from 2q + 1, where q is some intege - Mathematics
Sum
Show that every positive integer is of the form 2q and that every positive odd integer is of the from 2q + 1, where q is some integer.
#### Solution
According to Euclid's division lemma, if a and b are two positive integers such that a is greater than b; then these two integers can be expressed as
a = bq + r; where 0 ≤ r < b
Now consider
b = 2; then a = bq + r will reduce to
a = 2q + r; where 0 ≤ r < 2,
i.e., r = 0 or r = 1
If r = 0, a = 2q + r ⇒ a = 2q
i.e., a is even
and, if r = 1, a = 2q + r ⇒ a = 2q + 1
i.e., a is add; as if the integer is not even; it will be odd.
Since, a is taken to be any positive integer so it is applicable to the every positive integer that when it can be expressed as
a = 2q
∴ a is even and when it can expressed as
a = 2q + 1; a is odd.
Hence the required result.
Concept: Euclid’s Division Lemma
Is there an error in this question or solution? | 335 | 1,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-49 | latest | en | 0.937855 |
https://weighmag.com/how-to-weigh-luggage-with-and-without-a-scale/ | 1,726,060,355,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00539.warc.gz | 564,702,991 | 19,530 | # How to Weigh Luggage with and without a Scale
No one wants to get to the airport and get the surprise that their carry-on luggage is too heavy and cannot be carried on, or that their bags weigh so much the cost of their trip is tremendously increased. If a bag is too heavy when you get to the airport you will either have to take some things out of the bag or leave the entire bag behind. This is not an option you want to have to face right before you leave on a trip.
To stop those surprises from happening people like to weigh their luggage before they leave for the airport. Sadly not all of us own luggage scales, so that makes us wonder how we could determine the weight of our luggage without owning luggage weighing scales.
### What are the Airline Weight Requirements for Luggage?
Most airlines have a weight limit of fifty pounds being the maximum amount of weight allowed in one piece of luggage. People try to put as many things in one bag as possible so they can save money on their baggage fees, so the airlines had to create weight limits to stop people from sending through bags that weighed a hundred pounds or more.
When it comes to carryon luggage the weight limitations will be smaller. Delta will allow you to carry a bag on with you as long as it weighs less than forty pounds, but Air France will not allow you to carry any bag into the cabin that weighs more than 26.4 pounds. Check with your airline and see what the weight restrictions are before you start to pack.
The luggage is also limited to being less than 62” when the width of the bag, the length of the bag, and the height of the bag are added.
### How to Get your Luggage Weight without luggage scales
You can try guessing at how much your luggage weighs. You will need to find something that weighs approximately forty to fifty pounds. Your pet or a small child may weigh this amount.
You start this guess by lifting the item that you know weighs forty to fifty pounds, and then lift your luggage, and see if the two weigh about the same amount. You are guessing so do not expect to have the exact weight when you get to the airport.
### Use Bathroom Scales
You can weigh your luggage using your ordinary bathroom scales. This weight measurement will be an approximate weight, not an exact weight.
Stand on the scale and get your weight. The get off of the scale, pick up the piece of luggage you want to weigh, and step back onto the scale.
When you have gotten the weight you will need to subtract what you weighed from what you weighed while holding the luggage. This will tell you approximately how much your luggage weighs.
It is important that you step off of the scales after you weigh yourself and before you weigh with the luggage because the scale needs to zero out before it can give you an accurate reading of the added weight.
### Visit a Package Store like a Fed EX or UPS Shipping Store
These stores have scales that are able to weigh your luggage. Whether or not the store will weigh your luggage depends on who is working when you go. Some employees are more than happy to help you out, some employees will help you out for a small charge, and some employees will not help you weigh your packages.
### The Post Office
You can go to your local post office and ask the postmaster if they will weigh your luggage for you. If you live in a small town it is likely that the postmaster will weigh your luggage for you, but if you live in a large town the postmaster will more than likely refuse to help you because they are already so busy and overworked.
### Travel Agencies
Some travel agencies have luggage scales and will allow their clients to use their scales before they leave for the airport. Your local travel agency might allow you to weigh your luggage even if you are not a customer but they may charge you a small fee for the service.
### Veterinarian’s Offices
The scales at the veterinarian clinic in your town will likely be flat and very close to the floor. That makes it easier for the veterinarian assistances and pet owners to get their pets to step onto the scales. You can easily set luggage on top of these flat scales. Most veterinarian offices will allow you to weigh your luggage if you do not go at a time when they are really busy.
### Some Grocery Stores
Some grocery stores have scales that are appropriate for weighing large items such as luggage. You can ask your local grocery if they have scales that you can do this with.
You might have better luck weighing your luggage at a fish market or a meat market that has fish scales that allow them to hang the meat because you could hang your luggage from the handle and get the weight of the luggage. These scales are usually quite accurate.
### Local Salvage Companies
Salvage companies that buy things like aluminum cans and junk iron have scales that can discern the weight of your luggage. If you have a salvage yard in your town call them and see if they would weigh your luggage for you. They may charge you a small fee for their time.
### Luggage Scales or Luggage with Scales
You can avoid the guessing game about your luggage by purchasing scales that are specifically designed to weigh your luggage, or purchasing luggage that has scales built into it. If you travel a lot then these scales can be very handy, but then again if you travel frequently you can probably pick up the luggage and estimate the weight without a scale.
Happy Flying! | 1,112 | 5,462 | {"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} | 2.796875 | 3 | CC-MAIN-2024-38 | latest | en | 0.970777 |
karanpc.com | 1,725,907,692,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651133.92/warc/CC-MAIN-20240909170505-20240909200505-00370.warc.gz | 326,169,610 | 46,123 | FX Draw Tools is available as a free download from our software library. FX Draw provides a high-productivity drawing environment designed specifically for mathematics teachers. Quickly create editable, high-quality diagrams for tests, exams, worksheets, web sites, presentations and demonstrations.
FX Sketch, the freehand extension of FX Draw, lets you sketch mathematical diagrams and have them automatically converted to professional quality drawings that can be edited using the power of FX Draw.
Efofex also provides FX Draw subscribers with three, free bonus programs that provide simplified access to some of FX Draw’s powerful graphing and equation tools. You can use the bonus tools with students or to rapidly create equations for use in documents.
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• Multiple tools to draw lines including end points, mid points, rise/run and length/angle.
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• Option to automatically mark and annotate important points when drawing geometric diagrams.
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Drawing Tools:
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Text/Equation Entry:
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Function Graphs:
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Quick, visual function entry system that allows you to simply type the function but shows you exactly how it is interpreting your entry.
• Graph Cartesian functions, inverse functions, polar functions, parametric functions, implicitly defined functions, points, vectors, Argand diagrams, slope fields, piecewise defined functions, domains, inequations and feasible regions.
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Statistical Graphs:
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• Output suitable for Pages and Keynote
• Create FX Draw file from multiple inserted OLE objects in a Word file
Title: FX Draw Tools v21.10.21.13
Developer: Efofex Software
Language: English
OS: Windows 7/8/8.1/10. | 1,274 | 6,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.86698 |
http://programmers.stackexchange.com/questions/225520/article-about-painting-coorelation-between-complex-data | 1,397,919,473,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00448-ip-10-147-4-33.ec2.internal.warc.gz | 184,563,090 | 13,644 | Article about “painting” coorelation between complex data [closed]
I am trying to get an understanding of quite a big data set. I have 20 different inputs parameters and 10 different outputs. The control algorithm is made and seems to work, but now I need to visualize how the system is running.
I have tried to do all sort of graphs of the different combinations of the parameters, but none a very good/clear image of the production.
Now one of my colleagues has mentioned what he think might be the solution to my problem. Unfortunately he cannot remember what it is called or where I can find information. Now I hope that some of you guys can help me find information about this principle.
My colleague was once working on controls for a sewage treatment plant. He had then found an article (possible written by a Russian scientist) about representing complex data as an abstract painting-like picture. The algorithm from the article had been implemented, so that they had live data shown in the plant. Besides from that, they had printed posters of different scenarios so they could compare the "abstract art of the day" with previous situations. E.g. they had one "painting" with more red in the corners, where the local slaughterhouse had had an emission, and another one with more distinction between the colours, where the sewages had been flooded by a heavy shower, and another with more triangle-like shapes, where a valve in the plant had been blocked.
The daily operators of the plant would then have a simple way to see if the plant was running as normal, and if not, they could compare "the art" and get an idea of what the problem could be.
Any of you have any idea what this principle is called? Who's the author? Where I can find information or the article? Or maybe similar principles?
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rubbish, what more can I say? – davidkonrad Jan 8 at 9:10
"...these questions aren’t educational in any way, because there’s no way to learn about the process of discovery. A particular community member, by virtue of their experience in the field, just happens to be able to take the limited information you remembered and fill in enough of the blanks to guess the correct answer... guessing game questions do not meet our goal of making the Internet better." (blog.stackoverflow.com/2012/02/lets-play-the-guessing-game) – gnat Jan 26 at 9:47
comments disabled on deleted / locked posts
migration rejected from stackoverflow.comJan 28 at 13:22
This question came from our site for professional and enthusiast programmers. Votes, comments, and answers are locked due to the question being closed here, but it may be eligible for editing and reopening on the site where it originated.
closed as off-topic by gnat, GlenH7, MichaelT, Bart van Ingen Schenau, DougMJan 28 at 13:22
• This question does not appear to be about software development within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question. | 646 | 2,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-15 | latest | en | 0.978671 |
https://oeis.org/A273511 | 1,627,116,280,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150134.86/warc/CC-MAIN-20210724063259-20210724093259-00031.warc.gz | 441,285,089 | 3,792 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A273511 Expansion of log(10) in base 2. 0
1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 2 COMMENTS The decimal expansion is 2.302585092994045684017... LINKS EXAMPLE 10.01001101011101100011011101110110101010... MATHEMATICA RealDigits[N[Log@ 10, 40], 2][[1]] (* Michael De Vlieger, May 25 2016 *) PROG (PARI) concat(binary(log(10))) \\ Michel Marcus, Dec 14 2017 CROSSREFS Cf. A002392. Sequence in context: A089013 A188374 A123504 * A104015 A244465 A214264 Adjacent sequences: A273508 A273509 A273510 * A273512 A273513 A273514 KEYWORD nonn,base,cons AUTHOR Benjamin Przybocki, May 23 2016 STATUS approved
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Last modified July 24 04:25 EDT 2021. Contains 346273 sequences. (Running on oeis4.) | 636 | 1,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.545118 |
https://www.topperlearning.com/doubts-solutions/form-the-differential-equation-of-the-family-of-curves-y-a-sin-x-b-where-a-and-b-are-arbitrary-constants-or-solve-the-following-differential-equation--flqyqwdmm/ | 1,563,630,365,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526517.67/warc/CC-MAIN-20190720132039-20190720154039-00105.warc.gz | 864,608,173 | 50,060 | 1800-212-7858 (Toll Free)
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# Form the differential equation of the family of curves y = a sin (x + b), where a and b are arbitrary constants.ORSolve the following differential equation:2xydx + (x2 + 2y2) dy = 0
Asked by Topperlearning User 4th June 2014, 1:23 PM
y = a sin (x + b) ….(1)
Differentiating both sides of equation (1) with respect to x,
y' = a cos (x + b) ...(2)
Differentiating equation (2) with respect to x,
y'' = -a sin (x + b) ...(3)
From equations (1) and (2), we obtain,
y'' + y = 0
This is the required differential equation.
OR
2xy dx + (x2 + 2y2) dy = 0
Thus, the given differential equation is a homogeneous equation.
Substitute y = vx
Substituting the values of y and in equation (1), we obtain
Integrating both sides, we obtain
Answered by Expert 4th June 2014, 3:23 PM
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You have rated this answer /10 | 362 | 1,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-30 | latest | en | 0.887101 |
http://www.myfreshplans.com/2010/09/google-earth-lesson-plans-for-math/ | 1,670,278,898,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711045.18/warc/CC-MAIN-20221205200634-20221205230634-00210.warc.gz | 81,155,375 | 22,586 | # Google Earth Lesson Plans for Math
We love Google Earth for math lessons!
• Shapes For young children studying basic shapes, it is very fun to go to Google earth, put in an address, fly way down close, and look for shapes. Now click on the Polygon tool in the top tool bar and outline the shapes you see. You see a square? Click on a corner, then on another corner, then on the third. You’ll see a triangle. Click on the fourth corner and it’ll become a rectangle. You can change colors — even pick “random” — and end up with a nice collection of shapes all over the landscape. It’s like a game of Hidden Pictures with basic shapes, and it’s different every time you change the address.
• Area Get more sophisticated with older students by having them create colored shapes, and then calculate their area. Start with simple shapes, but then encourage them to move on to irregular polygons. For these, you simply click at each point at which the line outlining the shape changes direction.
• Charts Use the “add content” function inthe Places area to find information about the places you’re looking at, such as population density, world oil consumption, and more. Have students choose a data set for places they’re looking at and determine the best way to reprsent the data graphically. Then add another location and figure out the best way to represent and compare the two sets of data. A fun way to get a fairly random comparison is to put in an address — your school, the student’s home — without specifying a city. You’ll usually come up with choices in several different states or countries within your language group.
• Measurement Click on the ruler and measure a path. After students have done a few paths, have them estimate distances for some more paths and then check them. Try going to the Directions pane and asking for directions from one location to another, which will give you a distance, and comparing that with the ruler measurement of the path.
More options:
• RealWorldMath.org has bunches of lessons with more guidance and downloadable resources.
• Teaching with Google Earth has a simple webquest that leads students to plan a road trip and calculate things like cost of gas and time the trip will take at different speeds. This seems to have been written for a particular class several years ago, so you may or may not want students to use it directly, but it’s a classic idea and the legwork’s been done for you.
• GELessons has a lesson on time zones using Google Earth. | 518 | 2,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-49 | latest | en | 0.92858 |
http://mathoverflow.net/questions/9492?sort=newest | 1,369,177,299,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700842908/warc/CC-MAIN-20130516104042-00069-ip-10-60-113-184.ec2.internal.warc.gz | 172,869,510 | 12,100 | ## How does one find vanishing algebraic cycles?
I have a question, related to what I asked before. Let's consider a smooth hyperplane section $X$ of a smooth projective variety $Y$ over $\mathbb C$. According to Weak Lefschetz theorem, cohomology groups of $X$ coincide with those of $Y$ in all dimensions except for the middle one. In the middle dimensions the pull-back $i^*: H^d(Y) \to H^d(X)$ is injective, but not surjective, and the "new" cycles on $X$ are called vanishing cycles. (The reason for such a name is that these "new" cycles will vanish when we approach singular fibers on the Lefschetz pencil.) Vanishing cycles also can be described as the ones that live in the kernel of $i_*: H^d(X) \to H^{d+2}(Y)$.
Let's consider the case when $X$ is even-dimensional, so that we can hope that the vanishing cycles are algebraic.
For example, for a smooth even-dimensional quadric in $\mathbb {CP}^n$ there exist one vanishing cycle - it is a difference $[E_1] - [E_2]$ of two maximal linear subspaces from different classes.
Another example I thought about is a smooth cubic surface in $\mathbb {CP}^3$, vanishing cycles here are generated by differences of pairs of lines $[l_1] - [l_2]$ lying on the cubic.
Now I wanted to ask, what are other examples people have in mind? I'm interested in the case, when vanishing cycles actually are algebraic.
Is there a general method to describe such cycles in concrete situations (like MG(3,6))?
Thanks
EDIT: Probably winter break is not a best time to start a bounty...
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Here's the method I had in mind looking at the examples I had. Let $X \subset Y$ be a smooth hyperplane section. Our goal is to detect cycles on $X$ that are not complete intersections of $X$ with cycles on $Y$.
Let's consider a cycle $Z$ on $Y$ such that $Z \cap X$ is reducible, say $$Z \cap X = A + B$$ for some cycles $A$, $B$ on $X$. Then we may hope $A - B$ or some similar combination can be a vanishing.
For quadrics and cubics above we take $Z$ to be tangent linear subspace of appropriate dimension.
Has anybody seen something like that applied in other cases?
The sad thing is that I sort of can make it work for $MG(3,6)$ to describe a vanishing cycle, but it's unclear from the description I get whether the cycle is rational or not.
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Evgeny Shinder firmly locates his question in the algebraic world. Nonetheless, I'm going to advertise how cleanly Picard-Lefschetz theory works in symplectic topology (cf. Seidel, "A long exact sequence for Floer cohomology"). As partial justification, I'll point out that the algebraic situation is far more complicated, as evidenced not only by Dmitri's answer but by the fact that the Hodge conjecture is open!
A symplectic Lefschetz pencil on a manifold $X$ of real dimension $2n+2$, together with a path from a regular value $b$ to a critical value, gives rise to a geometric vanishing cycle $V$. This is an embedded Lagrangian submanifold in $X_b$. It comes with a preferred isotopy class of diffeomorphisms $S^n\to V$, and hence (by the Lagrangian neighbourhood theorem) a symplectomorphism between a neighbourhood of $V$ in $X_b$ and a neighbourhood of $S^n$ in $T^*S^n$. The description is more-or-less reversible: the existence of a Lagrangian sphere in $M$ implies that, symplectically (but not necessarily algebraically, even when $M$ is algebraic), an ordinary-double-point degeneration of $M$ exists.
It might be interesting to think about the algebraic symplectic case (but I haven't).
-
It seems to me, that on a generic hypersurface of $CP^n$ of degree more than 3 if $n=2$ and of degree more then $2$ when $n>3$ there are no algebraic vanishing cycles. This is surely the case for generic hypersurfaces in $\mathbb CP^3$ of deree $4$ and more. For them there is only one integral class in $H^{1,1}$, this is the class given by the hyperplane section. So the two examples that you give are in a certain sence very exceptional.
-
Thanks for this interesting remark. Do you know any examples of vanishing cycles being algebraic and how people construct them when the ambient space $Y$ is not a projective space? – Evgeny Shinder Dec 22 2009 at 3:15 Evgeny, I never thought about such examples before, because for me like for Tim, a vanishing cycle is more an object of symplectic geometry, we think of them as of little half-dimesnional lagrangean spheres. But in the case of $MG(3,6)$ your approach seems to be completely reasonable. Usually, when you deform a variety, its complex structure and Hodge classes change. But not in the case of $MG(3,6)$ -- the complex structure is unique here. I tried to comment on this varitety in your previous question. – Dmitri Dec 22 2009 at 12:14 | 1,215 | 4,696 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2013-20 | latest | en | 0.914666 |
https://math.stackexchange.com/questions/linked/123524?sort=hot&pagesize=30 | 1,568,798,869,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573264.27/warc/CC-MAIN-20190918085827-20190918111827-00207.warc.gz | 571,469,798 | 26,099 | 16 questions linked to/from Fermat numbers are coprime
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85 views
### Prove that $\gcd(g_a,g_b) = 1$ given that for $n \in Z^{\geq 0}$, define $g_n = 2^{2^n} + 1$ [duplicate]
Prove that $\gcd(g_a,g_b) = 1$ given that for $n \in Z^{\geq 0}$, define $g_n = 2^{2^n} + 1$. I have already proved that $g_0\cdot g_1\cdots g_{n-1} = g_n -2$ if this hint is useful in this proof.
96 views
### $(a^{2^n}+1,a^{2^m}+1)=1 or 2$ [duplicate]
Prove that if $m\not =n,a$ are positive integers then $(a^{2^n}+1,a^{2^m}+1)$ is $1$ if $a$ is even and $2$ if $a$ is odd. I solve the problem in the following way: I assume that $m>n$ then a^{...
65 views
### Problems regarding GCD [duplicate]
If $m > n$ and $a,m,n$ are positive, with $m$ not equal to $n$, find the greatest common divisor of $2^{2^m}+1, 2^{2^n}+1$. Please solve this problem using Euclid's algorithm. I tried to use ...
47 views
### Fermat Prime Numbers Coprime [duplicate]
Fermat numbers are shown by: $F_m = 2^{2^m} + 1$. How can I prove that for any $m ≠ n$, I can have $(F_m, F_n) = 1$, or coprime?
4k views
### Prove that distinct Fermat Numbers are relatively prime [duplicate]
The Fermat numbers are defined by $F_m = 2^{2^m} + 1$. Prove that for $m \ne n$ we have $(F_m, F_n) = 1$. I have to first prove that $F_{m+1} = F_0F_1 \cdots F_m + 2$ by representing $F_{m+1}$ in ...
187 views
### Show that Fermat number $F_n$ and its index $n$ are coprime.
I want to show that $\gcd(F_n,n)=1$, where $F_n=2^{2^n}+1$. How to prove this? I can show that that $\gcd(F_n, F_m)=1$ for any natural $n$ and $m$, and that $F_{n+1}=(F_n)^2-2F_n+2=F_0\dots F_{n-1}+2$...
309 views
### Co prime numbers
How to find a set of numbers which are coprime to each other (all numbers are pairwise co prime) ? the numbers can be assumed to be less than a specific integers. like all numbers in the range [0,x]
171 views
### Show that the only divisors are $1$ and $2$ for $(z^{(2^x)}+1)$ and $(z^{(2^y)}+1)$ where $x,y,z\in\mathbb{N}$
I am trying to show that the only divisors are $1$ and $2$ for both $(z^{(2^x)}+1)$ and $(z^{(2^y)}+1)$ where $x,y,z\in\mathbb{N}$. To start the problem, the logical choice is to use difference of ... | 800 | 2,189 | {"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} | 3.84375 | 4 | CC-MAIN-2019-39 | latest | en | 0.827122 |
https://quilvius.com/solution/what-is-12x19-with-work-shown-800 | 1,701,326,184,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100172.28/warc/CC-MAIN-20231130062948-20231130092948-00349.warc.gz | 565,490,405 | 6,675 | # What is -1/2x+1=9 with work shown.
2 months ago
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## 📚 Related Questions
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Kelly estimates it will take her 35 minutes to make the punch and 45 minutes to set up. Will Kelly finish before the guests arrive if she starts at 11:45 a.m? Explain your answer. (Kelly's guests will arrive at 1:30)
Solution 1
Kelly will have enough time to finish before the guests arrive because 45+35 = 1 hour and 10 mins, which adds that with 11:45, it would be 12:55 when she finishes. Making it possible.
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The mean age of swimmers are all of these is 10 what does a large mad tell you
Solution 1
A large MAD (mean absolute deviation) tells us that there is a lot of variation in the ages of the swimmers. Â It tells us specifically that there are a lot of swimmers that are not close the to the age of 10 (which is the mean).
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steph has 5 1/4 pounds of rice.she wants to place 1/4 pound of rice in each plastic bag.how many bags will she need?
Solution 1
Total amount of rice = 5 1/4
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What is the value of the underlined digit of 2 in 0.812
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You need one more fact either an angle or a side in order to be able to answer. I take it you have neither. But there must be something more you can tell us from the question. List the question exactly as it is asked with a diagram if necessary. I'll follow and use either comments or I'll edit.
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A display that shows how the values in a data set are distributed
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Tyler bought a large bag of peanuts at a baseball game. Is it more reasonable to say that the mass of the peanuts is 1 gram or 1 kilogram?
Solution 1
Would you want only 1 gram of peanuts?
Or one Pound of peanuts? or kilograms
1 Kilo
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Explain why a square is always a rectangle but a rectangle is not always a square.
Solution 1
A rectangle is a plane figure with four straight sides and four right angles. A square is a plane figure with four equal straight sides and four right angles.Â
therefore, because a square has 4 straight sides with 4 right angles, it is a rectangle.Â
and a rectangle can have unequal adjacent sides, so it is not always a square
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Question
Evaluate the profit function at each vertex. P = 0.04x + 0.05y + 0.06(16 – x – y) (8, 1) P = (14, 1) P = (3, 6) P = (5, 10) P =
Solution 1
The given function is:
P = 0.04x + 0.05y + 0.06(16-x-y)
To get the function at each vertex, all you have to do is substitute with the given x and y values in the above equation and get the corresponding value of P as follows:
1- For (8,1):
P = 0.04x + 0.05y + 0.06(16-x-y)
P = 0.04(8) + 0.05(1) + 0.06(16-8-1)
P = 0.79
2- For (14,1):
P = 0.04x + 0.05y + 0.06(16-x-y)
P = 0.04(14) + 0.05(1) + 0.06(16-14-1)
P = 0.67
3- For (3,6):
P = 0.04x + 0.05y + 0.06(16-x-y)
P = 0.04(3) + 0.05(6) + 0.06(16-3-6)
P = 0.84
4- For (5,10):
P = 0.04x + 0.05y + 0.06(16-x-y)
P = 0.04(5) + 0.05(10) + 0.06(16-5-10)
P = 0.76
Hope this helps :)
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3389 | 1,186 | 3,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-50 | latest | en | 0.94154 |
https://www.jiskha.com/display.cgi?id=1354281263 | 1,511,272,751,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806388.64/warc/CC-MAIN-20171121132158-20171121152158-00604.warc.gz | 821,424,715 | 4,251 | # Darin
posted by .
a teacher knows that 80 percent will score no more than 5 points above or below the class average 85 percent on the test which inequality represents the test score of 80 percent of her student.
• Darin -
Darin is not a subject I have ever heard of.
You have not provided the optional answers.
It should look something like
80 < x < 90
Where x is the test score of 85% of the students.
• Darin -
wouldn't it be 5%?
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More Similar Questions | 722 | 2,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-47 | latest | en | 0.946609 |
https://www.get-digital-help.com/2009/12/10/count-unique-values-in-two-lists-combined-in-excel/ | 1,516,480,568,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889733.57/warc/CC-MAIN-20180120201828-20180120221828-00298.warc.gz | 913,369,282 | 24,845 | ### Unique values
Unique values are values existing only once in a list or range. See picture below.
### Count unique values in two lists combined
Array formula in D12:
=SUM(IF(COUNTIF(List1, List1)+COUNTIF(List2, List1)=1, 1))+SUM(IF(COUNTIF(List2, List2)+COUNTIF(List1, List2)=1, 1)) + CTRL + SHIFT + ENTER
Named ranges
List1 (B3:B8)
List2 (D3:D7)
What is named ranges?
How to customize the formula to your excel workbook
Change the named ranges.
Download excel example file
Count-unique-values-in-two-columns.xls
(Excel 97-2003 Workbook *.xls)
Functions used in this blog post:
COUNTIF(range,criteria)
Counts the number of cells within a range that meet the given condition
IF(logical_test;[value_if:true];[value_if_false])
Checks whether a condition is met, and returns one value if TRUE, and another value if FALSE
SUM(number1,[number2],)
Adds all the numbers in a range of cells
Read more related articles | 241 | 923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-05 | longest | en | 0.673078 |
http://list.seqfan.eu/pipermail/seqfan/2018-September/018849.html | 1,582,266,644,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145443.63/warc/CC-MAIN-20200221045555-20200221075555-00288.warc.gz | 87,871,595 | 1,984 | # [seqfan] A closed-form expression
Tomasz Ordowski tomaszordowski at gmail.com
Wed Sep 5 09:29:14 CEST 2018
```Dear SeqFans!
Let a(n) = number of bases b < n such that b^n == b (mod n).
1, 2, 3, 2, 5, 3, 7, 2, 3, 4, 11, 4, 13, 3, 9, 2, 17, 3, 19, 5, 9, ...
For n = 6; 0^6 == 0 (mod 6), 1^6 == 1 (mod 6), and 3^6 == 3 (mod 6).
Note that, for n > 1, a(n) = n if and only if n is a prime or a Carmichael
number.
For n > 1, a(n) > A063994(n) = number of bases b mod n for which b^(n-1) ==
1 (mod n).
Is there any formula for a(n) similar to https://oeis.org/A063994 ?
Best regards,
Thomas
``` | 263 | 600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-10 | latest | en | 0.776062 |
https://mmtom.org/category/millimeters-to-centimeters | 1,653,606,276,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00393.warc.gz | 438,784,403 | 20,144 | Home » Millimeters to Centimeters
Millimeters to Centimeters
In this millimeters to centimeters category you can find our posts explaining the conversion of a particular length in the unit millimeter (mm) to the unit centimeter (cm). We begin each article with the formula, and then give you the result of x mm to cm using different spelling variants. Also included in every article is the equivalence of your length in mm in other common metric and imperial units, including, for example meters, inches and feet. A link to x cm to mm, directions for further infos about the units, and a comment form to ask questions about x millimeter to centimeter are part of every post, too. In addition, every entry in this category explains the use of our search form, something which may also be helpful to locate an entry in these pages here. The FAQs about your specific millimeters to centimeters conversion can be found as well. Along with a mm to cm converter which can swap the units.
1000 mm to cm
Welcome to our post about 1000 mm to cm. Here you can find the answer to how many centimeters in 1000 millimeters? We not only… Read More »1000 mm to cm
999.9 mm to cm
Welcome to our post about 999.9 mm to cm. Here you can find the answer to how many centimeters in 999.9 millimeters? We not only… Read More »999.9 mm to cm
999.8 mm to cm
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999.7 mm to cm
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999.6 mm to cm
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999.5 mm to cm
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999.4 mm to cm
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999.3 mm to cm
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999.2 mm to cm
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999.1 mm to cm
Welcome to our post about 999.1 mm to cm. Here you can find the answer to how many centimeters in 999.1 millimeters? We not only… Read More »999.1 mm to cm | 740 | 2,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-21 | longest | en | 0.892488 |
https://www.solutionsfolks.com/ExpertAnswers/in-matlab-the-following-method-of-computing-pi-is-due-to-archimedes-1-let-a-1-and-pa991 | 1,721,124,333,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514742.26/warc/CC-MAIN-20240716080920-20240716110920-00377.warc.gz | 874,198,552 | 5,756 | Home / Expert Answers / Computer Science / in-matlab-the-following-method-of-computing-pi-is-due-to-archimedes-1-let-a-1-and-pa991
# (Solved): IN MATLAB The following method of computing $$\pi$$ is due to Archimedes: 1. Let $$A=1$$ and $$... IN MATLAB The following method of computing \( \pi$$ is due to Archimedes: 1. Let $$A=1$$ and $$N=6$$ 2. Repeat 10 times, say: Replace $$N$$ by $$2 N$$ Replace $$A$$ by $$\left.\left[2-\sqrt{(} 4-A^{2}\right)\right]^{1 / 2}$$ Let $$L=N A / 2$$ Let $$U=L / \sqrt{1-A^{2} / 2}$$ Let $$P=(U+L) / 2($$ estimate of $$\pi)$$ Let $$E=(U-L) / 2$$ (estimate of error) Print $$N, P, E$$
We have an Answer from Expert | 241 | 650 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.728417 |
https://nbviewer.jupyter.org/gist/justmarkham/6d5c061ca5aee67c4316471f8c2ae976 | 1,550,553,119,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489343.24/warc/CC-MAIN-20190219041222-20190219063222-00590.warc.gz | 647,067,733 | 54,498 | # Logistic Regression with scikit-learn¶
This is an example of logistic regression in Python with the scikit-learn module, performed for an assignment with my General Assembly Data Science class.
## Dataset¶
The dataset I chose is the affairs dataset that comes with Statsmodels. It was derived from a survey of women in 1974 by Redbook magazine, in which married women were asked about their participation in extramarital affairs. More information about the study is available in a 1978 paper from the Journal of Political Economy.
## Description of Variables¶
The dataset contains 6366 observations of 9 variables:
• rate_marriage: woman's rating of her marriage (1 = very poor, 5 = very good)
• age: woman's age
• yrs_married: number of years married
• children: number of children
• religious: woman's rating of how religious she is (1 = not religious, 4 = strongly religious)
• educ: level of education (9 = grade school, 12 = high school, 14 = some college, 16 = college graduate, 17 = some graduate school, 20 = advanced degree)
• occupation: woman's occupation (1 = student, 2 = farming/semi-skilled/unskilled, 3 = "white collar", 4 = teacher/nurse/writer/technician/skilled, 5 = managerial/business, 6 = professional with advanced degree)
• occupation_husb: husband's occupation (same coding as above)
• affairs: time spent in extra-marital affairs
## Problem Statement¶
I decided to treat this as a classification problem by creating a new binary variable affair (did the woman have at least one affair?) and trying to predict the classification for each woman.
Skipper Seabold, one of the primary contributors to Statsmodels, did a similar classification in his Statsmodels demo at a Statistical Programming DC Meetup. However, he used Statsmodels for the classification (whereas I'm using scikit-learn), and he treated the occupation variables as continuous (whereas I'm treating them as categorical).
## Import modules¶
In [1]:
import numpy as np
import pandas as pd
import statsmodels.api as sm
import matplotlib.pyplot as plt
from patsy import dmatrices
from sklearn.linear_model import LogisticRegression
from sklearn.cross_validation import train_test_split
from sklearn import metrics
from sklearn.cross_validation import cross_val_score
## Data Pre-Processing¶
First, let's load the dataset and add a binary affair column.
In [2]:
# load dataset
# add "affair" column: 1 represents having affairs, 0 represents not
dta['affair'] = (dta.affairs > 0).astype(int)
## Data Exploration¶
In [3]:
dta.groupby('affair').mean()
Out[3]:
rate_marriage age yrs_married children religious educ occupation occupation_husb affairs
affair
0 4.329701 28.390679 7.989335 1.238813 2.504521 14.322977 3.405286 3.833758 0.000000
1 3.647345 30.537019 11.152460 1.728933 2.261568 13.972236 3.463712 3.884559 2.187243
We can see that on average, women who have affairs rate their marriages lower, which is to be expected. Let's take another look at the rate_marriage variable.
In [4]:
dta.groupby('rate_marriage').mean()
Out[4]:
age yrs_married children religious educ occupation occupation_husb affairs affair
rate_marriage
1 33.823232 13.914141 2.308081 2.343434 13.848485 3.232323 3.838384 1.201671 0.747475
2 30.471264 10.727011 1.735632 2.330460 13.864943 3.327586 3.764368 1.615745 0.635057
3 30.008056 10.239174 1.638469 2.308157 14.001007 3.402820 3.798590 1.371281 0.550856
4 28.856601 8.816905 1.369536 2.400981 14.144514 3.420161 3.835861 0.674837 0.322926
5 28.574702 8.311662 1.252794 2.506334 14.399776 3.454918 3.892697 0.348174 0.181446
An increase in age, yrs_married, and children appears to correlate with a declining marriage rating.
## Data Visualization¶
In [5]:
# show plots in the notebook
%matplotlib inline
In [6]:
# histogram of education
dta.educ.hist()
plt.title('Histogram of Education')
plt.xlabel('Education Level')
plt.ylabel('Frequency')
Out[6]:
<matplotlib.text.Text at 0x16e48ac8>
In [7]:
# histogram of marriage rating
dta.rate_marriage.hist()
plt.title('Histogram of Marriage Rating')
plt.xlabel('Marriage Rating')
plt.ylabel('Frequency')
Out[7]:
<matplotlib.text.Text at 0x16eac550>
Let's take a look at the distribution of marriage ratings for those having affairs versus those not having affairs.
In [8]:
# barplot of marriage rating grouped by affair (True or False)
pd.crosstab(dta.rate_marriage, dta.affair.astype(bool)).plot(kind='bar')
plt.title('Marriage Rating Distribution by Affair Status')
plt.xlabel('Marriage Rating')
plt.ylabel('Frequency')
Out[8]:
<matplotlib.text.Text at 0x1710c5f8>
Let's use a stacked barplot to look at the percentage of women having affairs by number of years of marriage.
In [9]:
affair_yrs_married = pd.crosstab(dta.yrs_married, dta.affair.astype(bool))
affair_yrs_married.div(affair_yrs_married.sum(1).astype(float), axis=0).plot(kind='bar', stacked=True)
plt.title('Affair Percentage by Years Married')
plt.xlabel('Years Married')
plt.ylabel('Percentage')
Out[9]:
<matplotlib.text.Text at 0x17d83a20>
## Prepare Data for Logistic Regression¶
To prepare the data, I want to add an intercept column as well as dummy variables for occupation and occupation_husb, since I'm treating them as categorial variables. The dmatrices function from the patsy module can do that using formula language.
In [10]:
# create dataframes with an intercept column and dummy variables for
# occupation and occupation_husb
y, X = dmatrices('affair ~ rate_marriage + age + yrs_married + children + \
religious + educ + C(occupation) + C(occupation_husb)',
dta, return_type="dataframe")
print X.columns
Index([u'Intercept', u'C(occupation)[T.2.0]', u'C(occupation)[T.3.0]', u'C(occupation)[T.4.0]', u'C(occupation)[T.5.0]', u'C(occupation)[T.6.0]', u'C(occupation_husb)[T.2.0]', u'C(occupation_husb)[T.3.0]', u'C(occupation_husb)[T.4.0]', u'C(occupation_husb)[T.5.0]', u'C(occupation_husb)[T.6.0]', u'rate_marriage', u'age', u'yrs_married', u'children', u'religious', u'educ'], dtype='object')
The column names for the dummy variables are ugly, so let's rename those.
In [11]:
# fix column names of X
X = X.rename(columns = {'C(occupation)[T.2.0]':'occ_2',
'C(occupation)[T.3.0]':'occ_3',
'C(occupation)[T.4.0]':'occ_4',
'C(occupation)[T.5.0]':'occ_5',
'C(occupation)[T.6.0]':'occ_6',
'C(occupation_husb)[T.2.0]':'occ_husb_2',
'C(occupation_husb)[T.3.0]':'occ_husb_3',
'C(occupation_husb)[T.4.0]':'occ_husb_4',
'C(occupation_husb)[T.5.0]':'occ_husb_5',
'C(occupation_husb)[T.6.0]':'occ_husb_6'})
We also need to flatten y into a 1-D array, so that scikit-learn will properly understand it as the response variable.
In [12]:
# flatten y into a 1-D array
y = np.ravel(y)
## Logistic Regression¶
Let's go ahead and run logistic regression on the entire data set, and see how accurate it is!
In [13]:
# instantiate a logistic regression model, and fit with X and y
model = LogisticRegression()
model = model.fit(X, y)
# check the accuracy on the training set
model.score(X, y)
Out[13]:
0.72588752748978946
73% accuracy seems good, but what's the null error rate?
In [14]:
# what percentage had affairs?
y.mean()
Out[14]:
0.32249450204209867
Only 32% of the women had affairs, which means that you could obtain 68% accuracy by always predicting "no". So we're doing better than the null error rate, but not by much.
Let's examine the coefficients to see what we learn.
In [15]:
# examine the coefficients
pd.DataFrame(zip(X.columns, np.transpose(model.coef_)))
Out[15]:
0 1
0 Intercept [1.48988372957]
1 occ_2 [0.188045598942]
2 occ_3 [0.498926222393]
3 occ_4 [0.25064662018]
4 occ_5 [0.838982982602]
5 occ_6 [0.833921262629]
6 occ_husb_2 [0.190546828287]
7 occ_husb_3 [0.297744578502]
8 occ_husb_4 [0.161319424129]
9 occ_husb_5 [0.187683007035]
10 occ_husb_6 [0.193916860892]
11 rate_marriage [-0.70312052711]
12 age [-0.0584177439191]
13 yrs_married [0.10567679013]
14 children [0.0169195866351]
15 religious [-0.371135218074]
16 educ [0.0040161519299]
Increases in marriage rating and religiousness correspond to a decrease in the likelihood of having an affair. For both the wife's occupation and the husband's occupation, the lowest likelihood of having an affair corresponds to the baseline occupation (student), since all of the dummy coefficients are positive.
## Model Evaluation Using a Validation Set¶
So far, we have trained and tested on the same set. Let's instead split the data into a training set and a testing set.
In [16]:
# evaluate the model by splitting into train and test sets
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3, random_state=0)
model2 = LogisticRegression()
model2.fit(X_train, y_train)
Out[16]:
LogisticRegression(C=1.0, class_weight=None, dual=False, fit_intercept=True,
intercept_scaling=1, penalty='l2', random_state=None, tol=0.0001)
We now need to predict class labels for the test set. We will also generate the class probabilities, just to take a look.
In [17]:
# predict class labels for the test set
predicted = model2.predict(X_test)
print predicted
[ 1. 0. 0. ..., 0. 0. 0.]
In [18]:
# generate class probabilities
probs = model2.predict_proba(X_test)
print probs
[[ 0.3514255 0.6485745 ]
[ 0.90952541 0.09047459]
[ 0.72576645 0.27423355]
...,
[ 0.55736908 0.44263092]
[ 0.81213879 0.18786121]
[ 0.74729574 0.25270426]]
As you can see, the classifier is predicting a 1 (having an affair) any time the probability in the second column is greater than 0.5.
Now let's generate some evaluation metrics.
In [19]:
# generate evaluation metrics
print metrics.accuracy_score(y_test, predicted)
print metrics.roc_auc_score(y_test, probs[:, 1])
0.729842931937
0.74596198609
The accuracy is 73%, which is the same as we experienced when training and predicting on the same data.
We can also see the confusion matrix and a classification report with other metrics.
In [20]:
print metrics.confusion_matrix(y_test, predicted)
print metrics.classification_report(y_test, predicted)
[[1169 134]
[ 382 225]]
precision recall f1-score support
0.0 0.75 0.90 0.82 1303
1.0 0.63 0.37 0.47 607
avg / total 0.71 0.73 0.71 1910
## Model Evaluation Using Cross-Validation¶
Now let's try 10-fold cross-validation, to see if the accuracy holds up more rigorously.
In [21]:
# evaluate the model using 10-fold cross-validation
scores = cross_val_score(LogisticRegression(), X, y, scoring='accuracy', cv=10)
print scores
print scores.mean()
[ 0.72100313 0.70219436 0.73824451 0.70597484 0.70597484 0.72955975
0.7327044 0.70440252 0.75157233 0.75 ]
0.724163068551
Looks good. It's still performing at 73% accuracy.
## Predicting the Probability of an Affair¶
Just for fun, let's predict the probability of an affair for a random woman not present in the dataset. She's a 25-year-old teacher who graduated college, has been married for 3 years, has 1 child, rates herself as strongly religious, rates her marriage as fair, and her husband is a farmer.
In [22]:
model.predict_proba(np.array([1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 3, 25, 3, 1, 4,
16]))
Out[22]:
array([[ 0.77472334, 0.22527666]])
The predicted probability of an affair is 23%.
## Next Steps¶
There are many different steps that could be tried in order to improve the model:
• including interaction terms
• removing features
• regularization techniques
• using a non-linear model | 3,377 | 11,423 | {"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} | 3.296875 | 3 | CC-MAIN-2019-09 | longest | en | 0.943345 |
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# What are we doing? - PowerPoint PPT Presentation
What are we doing? In this activity you and your group will play the role of a 2 parents. You will work in groups of six. 3 students will play the mother 3 students will play the father
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What are we doing?
• In this activity you and your group will play the role of a 2 parents. You will work in groups of six.
• 3 students will play the mother
• 3 students will play the father
• You will use chromosome and gene models to create four offspring and determine their genotypes and phenotypes. Then mathematically, you will determine the probability of having offspring with different traits.
How to Use the Model
You will receive a baggie with pipe cleaners and beads. The pipe cleaners represent chromosomes and the beads represent genes located on the chromosomes. In humans, there are 23 pairs of chromosomes and thousands of genes, but for this exercise, we will only focus on a few.
The traits that we will focus on are:
• Sex of Child
• Hair Color (brown or blue)
• Eye Color (blonde or dark )
• Hemophilia
• Hemophilia is a sex- linked disease
• ****You will create 4 Punnett squares. One for each trait.***
For each Punnett square
the following:
1- What is the Mother’s genotype?
2- What is the Mother’s phenotype?
3- What is the Father’s genotype?
4- What is the Father’s phenotype?
5- What is the offspring's possible genotype?
6- What is the offspring’s possible phenotype?
7- For the hemophilia: Is Mom a carrier?
8- Why doesn’t Dad get an allele for this trait?
The crosses are the following:
• for sex of child: XY x XX
• for eye color of child: Bb x Bb
• for hair color of child: Dd x dd
• for hemophiliac trait : H x Hh
• (H is dominant normal)
• ( h is recessive sick gene)
Hemophilia is carried on the X chromosome. It is called an X linked genetic disorder. A women who is a carrier for hemophilia has the genetic mutation on one of her X chromosomes . She will have another non mutated X chromosome that will usually somewhat compensate for the defect in the other. It is not uncommon for women who carry the hemophilia gene to have low levels of clotting factor and have bleeding problems. A man who has hemophilia has the genetic mutation on his only X chromosome. He does not have another X chromosome to compensate for the defect so he will have hemophilia.
--The “dad” places one set of the homologous pairs (ex: that longer set) behind his back, with a chromosome in each hand. The “mom” picks the hand she wants for the child. Lay this chromosome on the table in front of you and set the other aside.
-- Repeat this procedure for the other homologous pair (ex: shorter set) and for the sex chromosomes. Its should be noted that if the blue chromosome gets chosen from the sex chromosomes, the child in this cross is going to be a boy.
--Now the “mom” places one set of the homologous pairs behind her back and the male chooses.
-- The chromosomes chosen and set on the table in front of you are the genes your first child received.
Arrange the chromosomes into homologous pairs and figure out what phenotypes (appearance or trait) the offspring has.
• What is the sex of the child? _________
• What color eyes does the child have?
• (phenotype) _______ Genotype? ______
• What color hair does the child have? (phenotype) _______ Genotype? ________
• Is the child a hemophiliac? ______
• Is the child a carrier for hemophilia? _____
Data Table
• See the data table, the first group is you and your partner. You are going to have 4 children. Repeat the procedure you used to make you first child to make 3 others.
• Fill out their traits on the table.
• When you are finished, you will post your data on the board. Other groups will also post their children’s data. Fill out the entire chart with all the parents in the class.
Compile Data
• Total number of babies? _______
• Number of girls? _____
• Number of boys? _____
• Number of children with brown eyes _______
• Number of children with blue eyes _______
• Number of children with dark hair ________
• Number of children with blonde hair ______
• Number of girls with hemophilia _______
• Number of boys with hemophilia _______
To get the percents, divide the number you have by the total number and x 100.
Girls ______% Boys ______%
Brown eyes ______%
Blue eyes ______%
Dark hair ______%
Blonde hair ______%
Hemophiliac boys ______%
Hemophilac girls ______% | 1,242 | 5,240 | {"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} | 3 | 3 | CC-MAIN-2018-30 | latest | en | 0.908411 |
http://www.gabormelli.com/RKB/log-odds | 1,597,300,637,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738960.69/warc/CC-MAIN-20200813043927-20200813073927-00121.warc.gz | 141,561,757 | 9,007 | # Log Odds Ratio (Logit) Measure
(Redirected from log-odds)
A Log Odds Ratio (Logit) Measure is a metric based on the logarithm of an odds ratio.
## References
### 2018
• http://en.wikipedia.org/wiki/Odds_ratio#Statistical_inference
• QUOTE: Several approaches to statistical inference for odds ratios have been developed.
One approach to inference uses large sample approximations to the sampling distribution of the log odds ratio (the natural logarithm of the odds ratio). If we use the joint probability notation defined above, the population log odds ratio is :${\log\left(\frac{p_{11}p_{00}}{p_{01}p_{10}}\right) = \log(p_{11}) + \log(p_{00}\big) - \log(p_{10}) - \log(p_{01})}.\,$ ...
... An alternative approach to inference for odds ratios looks at the distribution of the data conditionally on the marginal frequencies of X and Y. An advantage of this approach is that the sampling distribution of the odds ratio can be expressed exactly.
### 2000
• (Bland & Altman, 2000) ⇒ J. Martin Bland, and Douglas G. Altman. (2000). “The Odds Ratio.” In: Bmj, 320(7247).
• QUOTE: The sample odds ratio is limited at the lower end, since it cannot be negative, but not at the upper end, and so has a skew distribution. The log odds ratio,2 however, can take any value and has an approximately Normal distribution. It also has the useful property that if we reverse the order of the categories for one of the variables, we simply reverse the sign of the log odds ratio: log(4.89)=1.59, log(0.204)=−1.59.
We can calculate a standard error for the log odds ratio and hence a confidence interval. The standard error of the log odds ratio is estimated simply by the square root of the sum of the reciprocals of the four frequencies.
### 2000
• (Hosmer & Lemeshow, 2000) ⇒ David W. Hosmer, and Stanley Lemeshow. (2000). “Applied Logistic Regression, 2nd Edition." Wiley. ISBN:0471356328
• QUOTE: ... In summary, the interpretation of the estimated coefficient for a continuous variable is similar to that of nominal scale variables: an estimated log odds ratio. The primary difference is that a meaningful change must be defined for the continuous variable. ...
... In the previous section in this chapter we discussed the interpretation of an estimated logistic regression coefficient in the case when there is a single variable in the fitted model. | 575 | 2,356 | {"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} | 3.953125 | 4 | CC-MAIN-2020-34 | latest | en | 0.816602 |
https://questions.examside.com/past-years/jee/question/plet-s-be-the-set-of-all-solutions-of-the-equation--jee-main-mathematics-trigonometric-functions-and-equations-vj0soiakuxo8en5l | 1,713,147,827,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00431.warc.gz | 445,439,734 | 44,260 | 1
JEE Main 2023 (Online) 1st February Morning Shift
+4
-1
Let $$S$$ be the set of all solutions of the equation $$\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^{2}}\right)=\pi, x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$$. Then $$\sum_\limits{x \in S} 2 \sin ^{-1}\left(x^{2}-1\right)$$ is equal to :
A
$$\pi-2 \sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)$$
B
$$\pi-\sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)$$
C
$$\frac{-2 \pi}{3}$$
D
None
2
JEE Main 2023 (Online) 31st January Evening Shift
+4
-1
Let (a, b) $\subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi)$, holds.
If $\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to :
A
$\frac{\pi}{16}$
B
$\frac{\pi}{48}$
C
$\frac{\pi}{8}$
D
$\frac{\pi}{12}$
3
JEE Main 2023 (Online) 31st January Morning Shift
+4
-1
If $${\sin ^{ - 1}}{\alpha \over {17}} + {\cos ^{ - 1}}{4 \over 5} - {\tan ^{ - 1}}{{77} \over {36}} = 0,0 < \alpha < 13$$, then $${\sin ^{ - 1}}(\sin \alpha ) + {\cos ^{ - 1}}(\cos \alpha )$$ is equal to :
A
16
B
$$\pi$$
C
16 $$-$$ 5$$\pi$$
D
0
4
JEE Main 2023 (Online) 24th January Morning Shift
+4
-1
$${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$$ is equal to :
A
$${\pi \over 2}$$
B
$${\pi \over 3}$$
C
$${\pi \over 6}$$
D
$${\pi \over 4}$$
EXAM MAP
Medical
NEET | 691 | 1,505 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 3.390625 | 3 | CC-MAIN-2024-18 | latest | en | 0.467986 |
https://popflock.com/learn?s=Batting_average_(baseball) | 1,624,516,481,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488551052.94/warc/CC-MAIN-20210624045834-20210624075834-00096.warc.gz | 418,949,567 | 25,926 | Batting Average (baseball)
Get Batting Average Baseball essential facts below. View Videos or join the Batting Average Baseball discussion. Add Batting Average Baseball to your PopFlock.com topic list for future reference or share this resource on social media.
Batting Average Baseball
Reggie Jackson batting at Yankee Stadium in 1979; Jackson batted .297 that season.
In baseball, the batting average (BA) is the number of hits divided by at bats. It is usually rounded to three decimal places and read without the decimal: A player with a batting average of .300 is "batting three-hundred". If necessary to break ties, batting averages could be taken beyond the .001 measurement. In this context, .001 is considered a "point", such that a .235 batter is 5 points higher than a .230 batter.
## History
Henry Chadwick, an English statistician raised on cricket, was an influential figure in the early history of baseball.[1] In the late 19th century he adapted the concept behind the cricket batting average to devise a similar statistic for baseball. Rather than simply copy cricket's formulation of runs scored divided by outs, he realized that hits divided by at bats would provide a better measure of individual batting ability. This is because while in cricket, scoring runs is almost entirely dependent on one's own batting skill, in baseball it is largely dependent on having other good hitters on one's team. Chadwick noted that hits are independent of teammates' skills, so used this as the basis for the baseball batting average. His reason for using at bats rather than outs is less obvious, but it leads to the intuitive idea of the batting average being a percentage reflecting how often a batter gets on base, whereas hits divided by outs is not as simple to interpret in real terms.
## Values
Ted Williams is the most recent MLB player to hit .400 or better in a season (1941).
Ty Cobb has the highest MLB career batting average (.366).
In modern times, a season batting average of .300 or higher is considered to be excellent, and an average higher than .400 a nearly unachievable goal. The last Major League Baseball (MLB) player to do so, with enough plate appearances to qualify for the batting championship, was Ted Williams of the Boston Red Sox, who hit .406 in 1941.[2] Note that batting averages are rounded;[3] entering the final day of the 1941 season, Williams was at 179-for-448, which is .39955 and would have been recorded as .400 via rounding.[4] However, Williams played in both games of a doubleheader, went 6-for-8, and ended the season 185-for-456,[5] which is .40570 and becomes .406 when rounded.[4]
Since 1941, the highest single-season average has been .394 by Tony Gwynn of the San Diego Padres in 1994.[2] Wade Boggs hit .401 over a 162-game span with Boston from June 9, 1985, to June 6, 1986,[6] but never hit above .368 for an MLB season.[7] There have been numerous attempts to explain the disappearance of the .400 hitter, with one of the more rigorous discussions of this question appearing in Stephen Jay Gould's 1996 book Full House.
Ty Cobb holds the record for highest career batting average with .366, eight points higher than Rogers Hornsby who has the second-highest career average at .358.[8] The record for lowest career batting average for a player with more than 2,500 at-bats belongs to Bill Bergen, a catcher who played from 1901 to 1911 and recorded a .170 average in 3,028 career at-bats.[9] Hugh Duffy, who played from 1888 to 1906, is credited with the highest single-season batting average, having hit .440 in 1894.[10] The modern-era (post-1900) record for highest batting average for a season is held by Nap Lajoie, who hit .426 in 1901,[10] the first year of play for the American League. The modern-era record for lowest batting average for a player that qualified for the batting title is held by Chris Davis, who hit .168 in 2018.[11] While finishing six plate appearances short of qualifying for the batting title, Adam Dunn of the Chicago White Sox hit .159 for the 2011 season, nine points lower than the record.[12] The highest batting average for a rookie was .408 in 1911 by Shoeless Joe Jackson.[13]
The league batting average in MLB for the 2018 season was .248, with the highest modern-era MLB average being .296 in 1930, and the lowest being .237 in 1968.[14] For non-pitchers, a batting average below .230 is often considered poor, and one below .200 is usually unacceptable. This latter level is sometimes referred to as "The Mendoza Line", named for Mario Mendoza -- a lifetime .215 hitter but a good defensive shortstop. [15]
Sabermetrics, the study of baseball statistics, considers batting average a weak measure of performance because it does not correlate as well as other measures to runs scored, thereby causing it to have little predictive value. Batting average does not take into account bases on balls (walks) or power, whereas other statistics such as on-base percentage and slugging percentage have been specifically designed to measure such concepts. Adding these statistics together form a player's on-base plus slugging or "OPS". This is commonly seen as a much better, though not perfect, indicator of a player's overall batting ability as it is a measure of hitting for average, hitting for power and drawing walks.
### Anomalies
In 1887, bases on balls were counted as hits by the major leagues in existence at the time. This inflated batting averages, with 11 players batting .400 or better, and the experiment was abandoned the following season. Historical statistics for the season were later revised, such that "Bases on balls shall always be treated as neither a time at bat nor a hit for the batter."[16]
In rare instances, MLB players have concluded their careers with a perfect batting average of 1.000. John Paciorek had three hits in all three of his turns at bat.[17] Esteban Yan went two-for-two, including a home run. Hal Deviney's two hits in his only plate appearances included a triple, while Steve Biras, Mike Hopkins, Chet Kehn, Jason Roach and Fred Schemanske also went two-for-two. A few dozen others have hit safely in their one and only career at-bat.
## Qualifications for the batting title
The MLB batting averages championships (often referred to as "the batting title") are awarded annually to the player in each league who has the highest batting average. Ty Cobb holds the MLB and American League (AL) record for most batting titles, officially winning 11 in his career.[18] The National League (NL) record of eight batting titles is shared by Honus Wagner and Tony Gwynn. Most of Cobb's career and all of Wagner's career took place in what is known as the Dead-Ball Era, which was characterized by higher batting averages by star players (although the overall league batting average was historically at its lowest during that era) and much less power, whereas Gwynn's career took place in the Live-Ball Era.
To determine which players are eligible to win the batting title, the following conditions have been used over the sport's history:[19]
• Pre-1920 - A player generally is required to appear in at least 100 or more games when the schedule was 154 games, and 90 games when the schedule was 140 games. An exception to the rule was made for Ty Cobb in 1914, who appeared in 98 games but had a big lead and was also a favorite of American League President Ban Johnson.
• 1920-1949 - A player had to appear in 100 games to qualify in the NL; the AL used 100 games from 1920 to 1935, and 400 at-bats from 1936 to 1949. The NL was advised to adopt 400 at-bats for the 1945 season, but National League President Ford Frick refused, feeling that 100 games should stand for the benefit of catchers and injured players. (Taffy Wright is often erroneously said to have been cheated out of the 1938 batting title; he batted .350 in exactly 100 games, with 263 ABs. Jimmie Foxx hit .349, in 149 games and 565 AB. But since the AL requirement that year was 400 at-bats, Foxx's batting title is undisputed.)
• 1950-1956 - A player needed 2.6 at-bats per team game originally scheduled. (With the 154-game schedule of the time, that meant a rounded-off 400 at-bats.) From 1951 to 1954, if the player with the highest average in a league failed to meet the minimum at-bat requirement, the remaining at-bats until qualification (e.g., five at-bats, if the player finished the season with 395 at-bats) were hypothetically considered hitless at-bats; if his recalculated batting average still topped the league, he was awarded the title. This standard applied in the AL from 1936 to 1956.
• 1957 to the present - A player has needed 3.1 plate appearances per team game originally scheduled; thus, players were no longer penalized for walking so frequently, nor did they benefit from walking so rarely. (In 1954, for example, Ted Williams batted .345 but had only 386 ABs, while topping the AL with 136 walks. Williams thus lost the batting title to Cleveland's Bobby Ávila, who hit .341 in 555 ABs.) In the 154-game schedule, the required number of plate appearances was 477, and since the era of the 162-game schedule, the requisite number of plate appearances has been 502. Adjustments to this figure have been made during strike-shortened seasons, such as 1972, 1981, 1994, and 1995.
From 1967 to the present, if the player with the highest average in a league fails to meet the minimum plate-appearance requirement, the remaining at-bats until qualification (e.g., five at-bats, if the player finished the season with 497 plate appearances) are hypothetically considered hitless at-bats; if his recalculated batting average still tops the league, he is awarded the title. This is officially Rule 10.22(a), but it is also known as the Tony Gwynn rule because the Padres' player won the batting crown in 1996 with a .353 average on just 498 plate appearances (i.e., he was four shy). Gwynn was awarded the title since he would have led the league even if he'd gone 0-for-4 in those missing plate appearances. His average would have dropped to .349, five points better than second-place Ellis Burks' .344.[20] In 2012, a one-time amendment to the rule was made to disqualify Melky Cabrera from the title. Cabrera requested that he be disqualified after serving a suspension that season for a positive testosterone test. He had batted .346 with 501 plate appearances, and the original rule would have awarded him the title over San Francisco Giants teammate Buster Posey, who won batting .336.[21][22]
### Major League Baseball
Different sources of baseball records present somewhat differing lists of career batting average leaders. There is consensus that Ty Cobb and Rogers Hornsby lead this category, at number one and number two, respectively. Further rankings vary by source, primarily due to differences in minimums needed to qualify (number of games played or plate appearances), or differences in early baseball records. The below table presents the top ten lists as they appear in four well-known sources, with the rankings and degree of precision (decimal places) as provided in the source. The main article linked above is sourced from Baseball-Reference.com, which is also presented here. None of the players listed below are still living; each is an inductee of the Baseball Hall of Fame, with the exception of Lefty O'Doul, Pete Browning, and Shoeless Joe Jackson (who is ineligible due to his alleged role in the Black Sox Scandal of 1919).
Baseball-Reference.com[8] Baseball Almanac[23] ESPN[24] MLB.com[25]
Rank Player Average Rank Player Average Rank Player Average Rank Player Average
1 Ty Cobb .3662 1 Ty Cobb .36636 1 Ty Cobb .366 1 Ty Cobb .367
2 Rogers Hornsby .3585 2 Rogers Hornsby .35850 2 Rogers Hornsby .358 2 Rogers Hornsby .358
3 Shoeless Joe Jackson .3558 3 Shoeless Joe Jackson .35575 3 Shoeless Joe Jackson .356 3 Ed Delahanty .346
4 Lefty O'Doul .3493 4 Ed Delahanty .34590 4 Ed Delahanty .346 4 Tris Speaker .345
5 Ed Delahanty .3458 5 Tris Speaker .34468 5 Tris Speaker .345 5 Ted Williams .344
6 Tris Speaker .3447 6 Ted Williams .34441 6 Billy Hamilton .344 6 Billy Hamilton .344
7 Billy Hamilton .3444 7 Billy Hamilton .34429 Ted Williams .344 7 Dan Brouthers .342
Ted Williams .3444 8 Babe Ruth .34206 8 Dan Brouthers .342 8 Babe Ruth .342
9 Dan Brouthers .3424 9 Harry Heilmann .34159 Harry Heilmann .342 9 Harry Heilmann .342
10 Babe Ruth .3421 10 Pete Browning .34149 Babe Ruth .342 10 Willie Keeler .341
### Minor League Baseball
The highest recorded single-season batting average in Minor League Baseball is .462, accomplished by Gary Redus in 1978, when he played for the Billings Mustangs,[26][27] an affiliate of the Cincinnati Reds in the Rookie Advanced-level Pioneer League. Redus was 117-for-253 in 68 games,[28] as the Pioneer League only plays from June to early September. Redus went on to play in MLB from 1982 through 1994, batting .252 during his MLB career.[29]
### Nippon Professional Baseball
Nori Aoki is the NPB career batting average leader.
In Nippon Professional Baseball (NPB), the leader in career batting average is Nori Aoki, an active player who has hit .326 in his NPB career, as of September 2019.[30] Aoki played in MLB from 2012 to 2017, where he compiled a .285 average.[31] Ichiro Suzuki batted .353 in NPB,[32] but does not have enough NPB career at-bats to qualify for that league's title.
## References
1. ^ Schiff, Andrew (2008). "Henry Chadwick". SABR. Retrieved 2019.
2. ^ a b "MLB Single-Season (Post-1900) Batting Leaders". ESPN. Retrieved 2019.
3. ^ Scott, Paul; Birnbaum, Phil (February 2010). "Do Motivated Players Have Higher Batting Averages?". SABR. Retrieved 2019. recorded batting averages are rounded to three decimal places
4. ^ a b Nowlin, Bill (2013). "The Day Ted Williams Became the Last .400 Hitter in Baseball". SABR. Retrieved 2019.
5. ^ "Ted Williams". Retrosheet. Retrieved 2019.
6. ^ Spaeder, Ryan (May 26, 2016), "Wade Boggs: 26 incredible Red Sox stats for No. 26", Sporting News, retrieved 2019
7. ^ "Wade Boggs". Retrosheet. Retrieved 2019.
8. ^ a b "Career Leaders & Records for Batting Average". Baseball-Reference.com. Retrieved 2019.
9. ^ Dittmar, Joe. "Bill Bergen". SABR. Retrieved 2019.
10. ^ a b "Single Season Leaders for Batting Average". Baseball Almanac. Retrieved 2019.
11. ^ Axisa, Mike (September 29, 2018). "Chris Davis finishes 2018 with the worst batting average in MLB history after Orioles shut him down". CBS Sports. Retrieved 2019.
12. ^ Reiter, Ben (June 4, 2012). "Death, Taxes And Adam Dunn". Sports Illustrated. Retrieved 2019 – via si.com/vault.
13. ^ "Batting Average Records". Baseball Almanac. Retrieved 2019.
14. ^ "Major League Baseball Batting Year-by-Year Averages". Baseball-Reference.com. Retrieved 2019.
15. ^ Landers, Chris (May 22, 2018). "How did Mario Mendoza become a shorthand for batting futility?". MLB.com. Retrieved 2019.
16. ^ Thorn, John (May 4, 2015). "Why Is the National Association Not a Major League ... and Other Records Issues". ourgame.mlblogs.com. Retrieved 2019.
17. ^ Keith, Ted (July 9, 2012). "The Perfect Game". Sports Illustrated. Retrieved 2019 – via si.com/vault.
18. ^ "Year-by-Year League Leaders for Batting Average". Sports Reference, Inc. Archived from the original on 9 February 2007. Retrieved .
19. ^ "Leaderboard Glossary - Baseball". Baseball-Reference.com. Retrieved .
20. ^ Kovacevic, Dejan (August 16, 2012). "Can't crown cheating Cabrera". Pittsburgh Tribune-Review.
21. ^ "Cabrera, Posey are MVPs". The State. Associated Press. 16 November 2012. Archived from the original on 27 June 2013. Retrieved 2018.
22. ^ Baggarly, Andrew. "Melky Cabrera ruled ineligible to win batting crown". CSN Bay Area. Archived from the original on 4 March 2016.
23. ^ "Career Leaders for Batting Average". Baseball Almanac. Retrieved 2019.
24. ^ "MLB Career Batting Leaders". ESPN. Retrieved 2019.
25. ^ "Statistics". MLB.com. Retrieved 2019. All-Time Totals, sorted by AVG
26. ^ Czerwinski, Kevin T. (August 2, 2006). "Redus' .462 in 1978 still Minor League best". MiLB.com. Retrieved 2019.
27. ^ Williams, Doug (April 16, 2013). "In '78, Redus hit .462, a season for the ages". ESPN.com. Retrieved 2013.
28. ^ "Gary Redus Minor Leagues Statistics & History". Baseball-Reference.com. Retrieved 2019.
29. ^ "Gary Redus Stats". Baseball-Reference.com. Retrieved 2019.
30. ^ "Aoki, Norichika". npb.jp. Retrieved 2019.
31. ^ "Nori Aoki". Baseball-Reference.com. Retrieved 2019.
32. ^ Lutz, Eric (March 21, 2019). "Ichiro Suzuki Retires at 45: Inside His Stats, Teams, and Legendary MLB Career". Men's Health. Retrieved 2019. | 4,163 | 16,640 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.958637 |
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### Real Option
Continues for 15 more pages »
# Real Option
By | August 2012
Page 1 of 16
University of Gothenburg
School of Business, Economics and Law
Industrial & Financial Management
Case Study 2
Capital Budgeting and Real Options
Block II: Real Options
Applied Real Options Analysis
PART 1: OPTION VALUATION
PART 2: LABORATORY EXPERIMENTS
PART 3: ESSAY QUESTION
Authors:
EmilAlmefors 860917
Patrik Nilsson 870512
Emil Stribrand 880113
Course responsible: Taylan Mavruk
Innehållsförteckning
Part 1: Option valuation3
Financial option3
1.Call, Put, and Sensitivity Analysis3
Real option6
2.Option to abandon6
3.Option to expand7
4.Option to contract8
5.Option to choose (multiple interacting options)9
6.Sensitivity analysis10
Part 2: Laboratory experiments12
A.Different times to maturity12
B.Uncertainty and option values14
C.The Binomial Model v. the Black-Scholes Model (closed-form)16
D.Dividend/Value leakage and option values18
E.Sequential compound option19
4.Estimating volatility22
Part 3: Essay question25
Reference30
Part 1: Option valuation
Financial option
1. Call, Put, and Sensitivity Analysis
Calculate the price of a six-month European call option on a nondividend-paying stock when the stock price is \$42, the strike price is \$40, the risk-free interest rate is 10% per year, and the volatility is 20% per year.
a) Use a binomial tree with a time interval of one month.
The call option’s value is ~4,82.
b) Use a B-S Model
When we use the B-S model, the value of the option’s value is ~4,76.
c) Application of the Greek Letters.
Calculate Delta, Gamma, Rho, Theta, Vega, and Xi of the call option in question b.
See the table above for values of the Greek letters.
d) What is the price of a put option with the same variables (using...
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